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# combination problem for marble with color and number
suppose I have 3 different color of bags, yellow, blue, red, each contain 2 marbles that are the same color as the bag.
I know the number of combination of picking 2 different color marble out of 3 bags is just 3 pick 2 = 3. But what if the marbles are labeled with number, like this:
Yellow(1), Yellow(2) - bag 1
Blue(1), Blue(2) - bag 2
Red(1), Red(2) - bag 3
How many combination are there for picking 2 different color with the label in mind? That means: Yellow(1) + Blue(1), Yellow(1) + Blue(2) Yellow(2) + Blue(1), Yellow(2) + Blue(2)... There are 12 ways but I am not sure what is the formula I can use.
Suppose you pick any one of the marble then you have $2$ ways to select the different marble
Example: If you pick $\color{red}{\text{Red(1)}}$ then you can pick $\color{yellow}{\text{Yellow(2)}}$ or $\color{blue}{\text{Blue(2)}}$.
And you can pick any of $6$ marbles. Means $6$ marbles each having $2$ choice therefore $$\text{Total number of ways}=6\times2\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=12 \text{ ways}$$
EDIT: (If the marbles are not numbered)
As if only color remains different then if you select any color marble then you have $2$ ways to pick a different marble of different color again you have $6$ marbles then $$\text{Total number of ways}=6\times2\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=12 \text{ ways}$$
(If the marbles are numbered)
If marbles are numbered and only different colored marbles are to be pick. Now, you have $4$ choices with each marble $$\text{Total number of ways}=6\times4\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=24\text{ ways}$$ As the cases such as $\{Yellow(1)+Blue(1)\},\{Blue(1)+Yellow(1)\}$ are counted so we need to divide by $2$ $$\text{Total number of ways}=\frac{24}{2}$$
• @user1701840 Discuss here chat.stackexchange.com/rooms/56810/we-are-friends – Harsh Kumar Apr 18 '17 at 1:08
• what if I have 1 RED, 2 YELLOW, 1 BLACK, 2 BLUE? Then I should think like this: If I pick 1 RED, then I have 3 ways of select different marble. And I can pick any of the 6 marbles, so I am looking at 6 * 3 = 18 ways, but it is not correct. – user1701840 Apr 18 '17 at 1:12
• not sure if I understand that, I can pick Red1 and Black 1, that is the third way – user1701840 Apr 18 '17 at 1:16
• But according to question you can't select the same number. – Harsh Kumar Apr 18 '17 at 1:24
• sorry I edited my question to clarify – user1701840 Apr 18 '17 at 1:27<|endoftext|>
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# How To Solve Physics Problems Gauss' Law problems and solutions
## Gauss' Law
Gauss' law can be tricky. The concepts expressed in mathematical terms often imply considerable mathematical sophistication to work the problems. This is almost always not the case. In general; if you are involved in excessive mathematical manipulations you are doing the problem wrong or do not see an easy way to apply Gauss' law. Remember that Gauss' law represents the density of flux lines through an area. Remember also that most of the difficult looking integrals never occur.
This states that the flux is the sum of the vector (dot) product of electric field over a surface times the area. E represents the electric field, and ds is a vector normal to the surface representing a differential element of surface. The dot product is the component of E parallel to ds, that is, normal to the surface, times ds. If E is a constant (most often the case) then the sum of (integral
of) E.ds over the surface is the component of E normal to the surface area, times the total surface area.
26 1 Calculate the flux through a square plate 1.0m on a side inclined at 120‹ with respect to a constant electric field of 100N/C. You may at this point want to review the definition of dot product in Chapter 1, Vectors, and the concept of the integral in the Introduction, Mathematical Background.
Fig. 26 1
Solution: The surface area ds is represented by a vector normal to the surface. The vector product E.ds is Eds(cos30‹). Integrating this product
This represents the flux through the surface.
<><><><><><><><><><><><>
The construct (concept) gfluxh can be thought of as a collection of lines intercepting this square plate. If the electric field is increased, then there are more gflux linesh per square meter and, if the angle is changed, more or less flux lines are intercepted as the angle is changed.
This integral of a vector product that sounds like a very difficult problem is performed rather easily with an understanding of dot products and integrals.
Electric field lines are a convenient concept or construct to help us visualize the electric field. The electric field lines are envisioned as vectors going from positive charges to negative charges, the direction a unit positive charge would move. The electric field lines between two oppositely charged plates are just lines going from one plate to another. Higher fields would be envisioned as more lines or more lines per cross sectional area.
In the case of a sphere surrounding a unit positive charge the field lines would be pointing radially out, the direction a unit positive charge would move. On spheres with different radii the electric field would be different; the field line density (number per square meter) would be different. However if we could sum all the electric field lines over the different spheres we would find the same number of lines. Their number density would decrease as the radii were increased but the total number over a sphere would remain the same. The formal Gauss' law connects flux to the charge contained again via an integral
The charge q is the net charge enclosed by the integral. The ƒÃo can, for the moment, be thought of as a constant that makes the units come out right.
Place a charge q at the center of a sphere and apply Gauss' law
Fig. 26 2
The little circle on the integral sign serves as a reminder to integrate over an enclosed surface. The sphere we are integrating over in this instance, not a physical sphere, is known as a Gaussian surface a surface where the integral is taken when applying Gauss' law. Now E is a constant over the surface (symmetry) so the integral becomes
and
is just the surface area of the sphere or
and
(26 3) which is just the statement we have from Coulomb's law and the definition of the electric field.
26 2 The electric field at the surface of the earth is approximately 150N/C directed down. Calculate the sign and magnitude of the charge on the earth.
Solution: Gauss' law states that E.ds = q enclosed
Construct a spherical Gaussian surface just outside the physical surface of the earth where E is everywhere normal and of value 150N/C. The integral of E.ds is the sum of this field over the Gaussian surface. E and ds are 180‹ so their dot product is negative.
Fig. 26 3
The field pointing down indicates that the charge is negative as does the dot product being negative. Again notice that this integral over the Gaussian surface is quite easy. These past two problems are very illustrative of the gintegration by inspectionh approach to problems involving Gauss' law.
<><><><><><><><><><><><>
26 3 Calculate the sign and magnitude of the charge contained in a cube 10 cm on a side oriented as shown where the E field is given by
Fig. 26 4
Solution: First calculate the E fields entering and exiting the cube. Watch the algebraic signs carefully.
There is more flux leaving the cube than there is entering it, so the net charge inside the cube must be positive. On the left side E is to the right and the vector representing ds is to the left, so
Remember that the direction of ds is always outward from the enclosed volume. On the right side E is to the right and the vector representing ds is to the right so
The integrals over y and z are zero and the integrals over x add to 1.05 Nm2/C. Now q can be calculated
<><><><><><><><><><><><>
26 4 The total charge (charge density) contained in an electron stream can be calculated with Gauss' law. If the field at the edges of a stream of square cross section 1.0cm on a side is
1.0 x 103 N/C, calculate the charge density in the stream. Assume that the E field in the direction of the stream is constant.
Fig. 26 5
Solution: The field is directed in, so the charge is negative and
reduces to calculating four identical integrals.
The charge in a cube 1.0 cm on a side is now
The charge density is For the electron charge density we need to introduce the charge of the electron.
<><><><><><><><><><><><>
26 5 A planetary probe is traveling radially inward toward the center of the planet. At 600 m above the surface the field is directed up and has value 180 N/C. At 500 m it is still directed up but has value 125 N/C. Find the sign and density of the charge (density) in this region.
Solution: The radius of the planet is sufficiently large so that we can take the region as a cube 100 m on a side. The electric fields are as shown in Fig. 26 6. Gauss' law can be written as
Fig. 26 6
The electric field at the sides is zero. At the top, the electric field and the vector representing ds point in the same direction so the dot product has a plus sign. On the bottom, the electric field vector points up while the vector representing ds points down giving the dot product a minus sign. Putting in the values for the integrals.
the total charge enclosed within the volume is
The charge density then is 4.9 x 10 12 C/m3.
<><><><><><><><><><><><>
26 6 Calculate the electric field between cylinders carrying charges as shown. The +q is on the inner conductor and q on the inside of the outer conductor with another q on the outside of the outer conductor.
Solution: On a surface between the conductors,
applies to the charge contained
and
with 2πrl the surface area of the Gaussian cylinder, corresponding to the point where the field is required, so in this region E points inward and has value E=q/2πƒÃ orl.
Note that the electric field at a point between the cylinders is only due to the charge enclosed by the Gaussian surface. Because of the symmetry of the problem, the charge on the outer cylinder produces no net electric field at any point inside the outer cylinder.
Fig. 26 7
Outside the outer cylinder the E field points inward because the net charge contained is negative, and at any point r (outside the outer cylinder) measured from the center line of the two cylinders E=q/2πƒÃorl.
With this particular arrangement of charge, the field has the same algebraic form between the cylinders and outside the outer cylinder but points outward between the cylinders and inward outside the outer cylinder.<|endoftext|>
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This paper highlights some work of Fluid Dynamics using two very important methods, i.e., Perturbation method and the method of Laplace Transform.
Before getting into the insights of this paper, here are the quick look at the types of continuum mechanics, forces and the governing equations of fluid flow. The importance of the approach of macroscopic study is used in the study of concept of continuum and not the approach of microscopic study.
Fluid Mechanics Solid Mechanics
Ideal fluid Real fluid
Newtonian Non- Newtonian
• Ideal fluid- (i) no viscosity
(iii) zero surface tension
(v) example : water, honey, milk etc.
• Real fluid- (i) existence of viscosity
(iii) existence of surface tension
(iv) example: gases , air etc.
• Newtonian fluid- (i) Linear relationship between stress and strain-rate.
(ii) obeys Newton’s law of viscosity.
(iii) example: Glycerine, air, water etc.
• Non-newtonian fluid- (i) Do not obey Newton’s law of viscosity(Newton’s law of viscosity states that the shear stress that has been applied is linearly proportional with the rate of deformation.)
(ii) example: Ketchup.
GOVERNING EQUATIONS OF FLUID FLOW
Equation of continuity Equation of Momentum Energy Equation
(Volume flow in= Volume flow out) (Rate of momentum change= Pressure energy+ Potential
total forces that are acting on the body) Energy+ Kinetic energy=
• Equation of continuity: The basic principle of this equation is that- conservation of mass is observed. The equation is given by
• Equation of Momentum: Conservation of momentum is the basic principle of this equation.
• Energy Equation: Its basic principle is that energy is conserved.
• Navier-Stokes Equation: This equation is also very important in the study of continuum mechanics. The equation is given by-
Where the 1st term in the L.H.S is the Local Acceleration
the 2nd term in the L.H.S is the Convective acceleration
the 1st , 2nd and 3rd term in the R.H.S is the Pressure Gradient, Body force and Viscous term respectively.
TYPES OF FORCES
Surface Forces Body Forces
• Surface forces: These forces lead to the contact of one fluid with another fluid.
Example- Pressure force, forces due to electric firldet.
• Body Forces: These kind of forces acts throughout the body volume.
Example- Gravity force, centrifugal force etc.
Now, let us have a look at some of the dimensionless numbers:
• Grashof number,
Convective mass transfer
• Sherwood number,
Mass diffusion rate
Viscous diffusion rate
• Prandtl number,
Thermal diffusion rate
Heat transfer (convective)
• Nusselt number,
Heat transfer (conductive)
Advective mass transfer
• Eckert number,
Diffusion rate (viscous)
• Schmidt number,
Diffusion rate (mass)
• Hartmann number,
Here are some important terms that are used in this paper. Let us have a look at them:
• Magnetohydrodynamics: Some fluids such as plasma, electrolytes having the properties of a magnet and behaves as conducting fluids
are known as magnetohydrodynamics.
• Heat transfer: The phenomenon of exchanging heat among those objects that are in physical touch is known as heat transfer. Heat
flows from a body having higher temperature to the body having lower temperature.
• Mass transfer: Some processes such as absorption, evaporation etc., in which there is a total transfer of mass from one phase to
another is termed as mass transfer.
• Skin friction: Skin friction is that friction or that drag force that is observed between the surface of a solid and the fluid through which
it is running.
• Free convective flow: The flow through a surface which is solid in nature and whose temperature can be higher or lower than the
temperature of the surroundings is termed as free convective flow. It can be both laminar and turbulent.
Heat and mass transfer has blowing effects in the industrial use. It varies due to many effects such as the effect in time, effect in velocity, effect in temperature etc. various numbers is also related to it such as the Prandtl number, Grashof number, Schmidt number and so on which is being described in a very brief way above. in the following chapter various methods have been used to study the effects of mass and heat transfer.<|endoftext|>
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# Kseeb Class 6 Mathematics Chapter 1 Knowing Our Numbers Solutions
## Kseeb Class 6 Mathematics Chapter 1 Knowing Our Numbers Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students by providing Solutions for KSEEB Class 6 Mathematics chapter 1 Knowing Our Numbers. Here students can easily find all the solutions for Knowing Our Numbers Exercise 1.1, 1.2 and 1.3. Also here our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 1 solutions. Here all the solutions are based on Karnataka State Board latest syllabus.
Knowing Our Numbers Exercise 1.1 Solution
(1) Fill in the blanks :
(a) 1 lakh = ___ ten thousand.
(b) 1 million = ___ hundred thousand.
(c) 1 crore = ___ ten lakh.
(d) 1 crore = ___ million.
(e) 1 million = ___ lakh.
Solution :
(a) 10 ten thousand.
(b) Ten hundred thousand.
(c) Ten ten lakh.
(d) Ten million.
(e) 10 lakh
(2) Place commas correctly and write the numerals:
(a) Seventy three lakh seventy five thousand three hundred seven.
(b) Nine crore five lakh forty one.
(c) Seven crore fifty two lakh twenty one thousand three hundred two.
(d) Fifty eight million four hundred twenty three thousand two hundred two.
(e) Twenty three lakh thirty thousand ten.
Solution :
(a) 73, 73,307
(b) 9, 05, 00,041
(c) 7, 52, 21,302
(d) 58,423,202
(e) 23, 31,010
(3) Insert commas suitably and write the names according to Indian System of Numeration :
(a) 87595762
(b) 8546283
(c) 99900046
(d) 98432701
Solution :
(a) 8, 75, 95,762
= Eight crore seventy-five lakh ninety-five thousand seven hundred sixty two.
(b) 85, 46,283
= Eighty-five lakh forty-six thousand two hundred eighty-three.
(c) 9, 99, 00,046
= Nine crore ninety-nine lakh forty six.
(d) 9, 84, 32,701
= Nine crore eighty-four lakh, thirty-two thousand seven hundred one.
(4) Insert commas suitably and write the names according to International System of Numeration :
(a) 78921092
(b) 7452283
(c) 99985102
(d) 48049831
Solution :
(a) 78, 921, 092
= Seventy eight million = nine hundred twenty-one thousand ninety-two.
(b) 7, 452, 283
= Seven million four hundred fifty-two thousand two hundred eighty-three.
(c) 99, 985, 102
= Ninety-nine million nine hundred eighty-five thousand, one hundred two.
(d) 48, 049, 831
= Forty-eight million, forty-nine thousand eight hundred thirty-one.
Knowing Our Numbers Exercise 1.2 Solution
(1) A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Solution :
First, second, third and final days were respectively 1094, 1812, 2050 and 2751
∴ Total tickets sold,
= (1094 + 1812 + 2050 + 2751)
= 7707
Therefore, the total number of tickets sold on all the four days is 7707
(2) Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?
Solution :
We will have to find 10000 is how many more than 6980 to find the solution.
Now, 10000-6980 = 3020
Check the answer by adding
6980 + 3020
= 1000 (The answer is right)
Thus, Shekhar need 3020 runs more.
(3) In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Solution :
Clearly 5,77500 is more than 3,48,700. So we will find 577500 is how many more than 3,48,700.
For this 348700 should be subtracted from 5,77,500.
Now, 577500 – 348700 = 228800
Check the answer by adding
348700 + 228800
= 577500 (The answer is right)
Thus, the successful candidate won the election by the margin of 2, 28, 800.
(4) Kirti bookstore sold books worth ₹ 2,85,891 in the first week of June and books worth ₹ 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Solution :
The sale for the two weeks together
= Sale in first week + sale in second week
= 2,85,891 + 4,00,768
= 6,86,659
Now, 285,891+400786 = 686677
Clearly the sale of first week i.e. 285891 is less than the sale of second week i.e. 400786.
We will find the second weeks sale is how many greater than the first week, by subtraction.
Now, 400786 – 285891 = 1,14,895
Thus, The second weeks sale is greater than the first week by 1,14,895
(5) Find the difference between the greatest and the least 5-digit number that can be written using the digits 6, 2, 7, 4, 3 each only once.
Solution :
The greatest and the least 5-digit number that can be written using
The digits 6, 2, 7, 43 each only once, are 76432 and 23,476 respectively.
The difference between the numbers 76432 and 23476 is
= 76432 – 23476
= 52956
∴ 52956 is the answer.
(6) A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Solution :
Since we know that the number of days in the month of January in any year is 31.
The machine manufactures 2825 screws a day.
Therefore, the number of screws that going to manufacture be 31 × 2825 in January 2006.
Now, 2825 × 31
= 87575
Thus correct answer is 87575.
(7) A merchant had ₹ 78,592 with her. She placed an order for purchasing 40 radio sets at ₹ 1200 each. How much money will remain with her after the purchase?
Solution :
The price of each radio set is RS 1200.
The total price of 40 radio is 40×1200
Now, 1200×40 = 48000
Now, she spent RS 48000 on purchasing radio sets.
Thus she will remain with the money is
= 78,582 – 48000
Now, 78592 – 48000 = 30592
∴ She will remain with Rs. 39592
(8) A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer? (Hint: Do you need to do both the multiplications?)
Solution :
After observing the both multiplication we found that student multiplied
65 and 56 by the same number i.e. 7236.
As we know that 65 is greater than 56 by 10.
Thus the answer of multiplication of 7236 × 65 is always be greater
than the multiplication of 7236 × 56 by 7236 × (65-56) I.e. 7236 × 10.
Now, 7236×10 = 72360
Thus the correct answer was greater than the correct answer by 72,360
(9) To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?
Solution :
First of all we will convert the data in CM.
i.e. 40 m = 200 cm + 15 cm = 215 cm
and 40 m = 4000 cm (1m = 100 m)
Each shirt uses 215 cm of cloth
Hence 4000 cm cloth makes 4000 ÷ 215 shirts
Now, 215 ÷ 4000
Quotient = 18
Reminder = 130
Thus, He can stitch 18 shirts and 130 cm cloth will remain i.e. 130 cm = 1m 30 cm.
(10) Medicine is packed in boxes, each weighing 4 kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Solution :
Medicine is packed in boxes, each weighing 4 kg 500g.
Boxes can be loaded in a van which cannot carry beyond 800 kg
= 800 kg / 4 kg 500g
= 177 boxes
Boxes can be loaded in a van which cannot carry beyond 800 kg is 177 boxes.
(11) The distance between the school and a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.
Solution :
The one way distance is 1 km 875 m i.e.
1 × 1000 + 875 = 1875 m
So both ways distance is 1875 × 2 –
Now, 1875×2 = 3750
She walks 3750 m every day.
Thus distance covered in six day is 3750 × 6
Now, 3750 × 6 = 22500
22500 m = 22 km 500 m
∴ Total distance covered by her in six days is 22 km 500 m.
(12) A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Solution :
The glasses capacity is 250 ml.
A vessel has 4 litres and 500 ml of curd if 4000 ml + 500ml = 4500 ml.
Thus the number of glasses to be needed to fill the curd in a vessel is 4500÷25
Now, 25 ÷ 4500
Quotient = 160
Reminder = 0
∴ 160 glasses to be filled with curd.
Knowing Our Numbers Exercise 1.3 Solution
(1) Estimate each of the following using general rule:
(a) 730 + 998
(b) 796 – 314
(c) 12,904 +2,888
(d) 28,292 – 21,496
Make ten more such examples of addition, subtraction and estimation of their outcome.
Solution :
(a) 730+998
= To get a closer estimate, let us try rounding each number to hundreds.
730 rounds off to
+ 998 rounds off to
Estimated addition = 700+1000 = 1700
(b) 796 – 314
= Round off to hundreds
796 is rounds off to
– 314 is rounds off to
Estimated subtraction = 800 – 300 = 500
(c) 12904 + 2888
= Rounds off to thousand
12, 904 is rounds off to
+ 2888 is rounds off to
Estimated sum = 13000 + 3000 = 16000
(d) 28, 292 – 21, 496
= Rounds off to thousand
28, 292 is rounds off to
– 21, 496 is rounds off to
Estimated difference = 28000-21000 = 7000
Ten more such examples of addition, subtraction and estimation of their outcome :
(a) 630 + 752
Round off to hundreds
630 is rounds off to
+ 752 is rounds off to
Estimated sum = 600+800 = 1400
(b) 896 – 225
= 896 is rounds off to
– 225 is rounds off to
Estimated difference = 900 – 200 = 700
(c) 587 + 162
= Round off to hundreds
587 is rounds off to
+ 162 is rounds off to
Estimated sum = 600+200 = 800
(d) 732 – 591
= Rounds off to nearest hundreds
732 is rounds off to
– 591 is rounds off to
Estimated difference = 700 – 600 = 100
(e) 23, 623 + 2932
= Round off to nearest towards
23, 623 is rounds off to
+ 2932 is rounds off to
Estimated addition = 24000 + 3000 = – 27000
(f) 16, 285 – 6923
= Rounds off to nearest thousands
16, 235 is rounds off to
– 6923 is rounds off to
Estimated difference = 16000 – 7000 = 9000
(g) 3,290 + 4,692
Rounds off to nearest thousand
3,290 is rounds off to
+ 4692 is rounds off to
Estimated sum = 3000 + 5000 = 8000
(h) 9999 -6662
Round off to nearest thousand
9999 is rounds off to
– 6662 is rounds off to
Estimated difference = 10000 – 7000 = 3000
(i) 12,302 + 13032
Round of to nearest thousand
12,302 is round off to
+ 13,032 is round off to
Estimated addition = 12000 + 13002 = 15002
(j) 56,023 – 25,011
Round off to nearest thousand
56,023 is round off to
– 25011 is round off to
Estimated difference
= 56000 – 25000
= 31000
(2) Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens) :
(a) 439 + 334 + 4,317
(b) 1,08,734 – 47,599
(c) 8325 – 491
(d) 4,89,348 – 48,365
Make four more such examples.
Solution :
(a) 439 + 334 + 4,317
First we rounding off to closer estimate.
439
+ 334
+ 4,317
——————–
5090 (closer estimate)
5090 rounding off to nearest hundreds is 5000.
(b) 1, 08,734 – 47,599
First we rounding off to closer estimate
1, 08,734
– 47,599
——————–
61135(closer estimate)
61135 rounding off to nearest hundreds is 60000.
(c) 8325 – 491
First we rounding off to closer estimate
8325
– 491
——————–
7834(closer estimate)
7834 rounding off to nearest hundreds is 8000.
(d) 4, 89,348 – 48,365
First we rounding off to closer estimate
4, 89,348
– 48,365
——————–
4, 40,993(closer estimate)
4, 40,993 rounding off to nearest hundreds is 4, 40,980
(3) Estimate the following products using general rule:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Make four more such examples.
Solution :
(a) 578 × 161
= 578 is rounds off to nearest hundred is 600.
161 is rounds off to nearest hundreds is 200.
The estimated product
= 600 × 200
= 12000.
(b) 5281 × 3491
= 5281 is rounded off to 5000 (rounding off to thousands)
3491 is rounded off to 3500 (rounding off to hundreds)
The estimated product
= 5000 × 3500
= 1750000.
(c) 1291 × 592
= 1291 is rounded off to 1300 (rounding off to hundreds)
592 is rounded off to 600 (rounding off to hundreds)
The estimated product
= 1300 × 600
= 780000
(d) 9250 × 29
= 9250 is rounds off to 9300 (rounding off to hundreds)
29 is rounds off to 30 (rounding off to nearest ten)
The estimated product
= 9300 × 30
= 2,79,000.
Four more such examples are :
(a) 978 × 232
= 978 is rounds off to nearest thousand is 1000.
232 is rounds off to 200 (rounding off to nearest hundred)
The estimated product
= 1000 × 200
= 200000.
(b) 9350 × 39
= 9350 is rounds off to 9400 (rounding off to thousands)
39 is rounds off to 40 (rounding off to nearest tens)
The estimated product
= 9400 × 40
= 376000
(c) 4192 × 782
= 4192 is rounds off to nearest thousands i.e. 4000
782 is rounds off to nearest hundred i.e. 800
The estimated product = 4000 × 800 = 3200000
(d) 6383 × 3677
= 6383 is rounds off to 6400 (rounds off to hundreds)
The estimated products,
= 6400 × 3700
= 23680000
Next Chapter Solution :
Updated: July 27, 2023 — 4:39 am<|endoftext|>
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Original content from | Commercial Services | Talent Partnerships
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# How To Teach Long Division
## How To Teach Long Division
Teaching and learning long division could be a breeze. Take it from this video.
Hi, I'm Dr. Shah. I was the National Lecture Competition winner in 1989 and I'm the maths master at Mathscool.
Now, ready for a new way of doing maths? Important technique on division because it's not only used for younger kids but also when you get older, you need to polynomial division and polynomial division uses that same technique. So, we start off with an example, 2558 divided by 12, and I want to do that using long division. So, I start off by writing inside my division 2558 and then outside is the 12.
When I'm looking at this, first thing I want to do is cover up so that I only have two digits here, the same that I have there, and so I'm asking myself, how many times does 12 go into 25. How many times does 12 go into 25? Well, the answer is two times because 2 times 12 is 24. So, I write that underneath there and the next thing I'd do is I then subtract, so 25, subtract 24, is one.
Now, you see you only have one digit here, so I bring down the next digit to join it, so I have two digits here again and I ask myself the same question. How many times does 12 go into 15? So, the answer this time would be one time, so I write 1 on top of there and the reason is 1 times 12 is 12. Do the same thing, subtract it, 15 subtract 12, gives me 3.
Again, I'm left with one digit down here, so I bring down the next digit which is also the last digit, there are no more digits left to bring down to join that. And then I ask myself, how many times does 12 go into 38? And the answer this time is three times because 3 times 12 is 36, and my final subtraction, 38 minus 36 is 2. Now, there are no more numbers to bring down, so this number here is the remainder.
It got left over at the end of our division. And so if somebody asks us, what is 2558 divided by 12, we'd say, well, you'd get 213, that's that number on the top there, but there's this little remainder 2 which got left over and couldn't be divided by 12 so I'll put that 2 still to be divided by 12, so 213 and 2/12 which cancels down, of course, to 1/6. 213 1/6 and in this case, because we got a little number left over that couldn't be divided by, it also tells us that the 12 is not a factor of 2558.
Okay, let's do another example. This time we're going to do 1845 divided by 15. So, same technique as before, start off with our divided-by bracket, put the 1845 inside the bracket and the 15 goes outside the bracket.
So, first thing I have to do is cover up so I'm only looking at two digits because I've only got two digits out here, and I ask myself, how many times does 15 go into 18? The answer is once because one lot of 15 is 15, and then subtract again. 18 subtract 15 gives me 3. Again, I've got one digit here, so I bring down another one from up there to join, and ask myself, how many times does 15 go into 34? And this time, it will go two times and two lots of 15 is 30.
Again, do my subtracting, 34 minus 30 is 4 and bring down the last of these digits, 5, and ask myself the same question. How many times does 15 go into 45? And the answer is 3, three lots of 15 is 45, and so this time, when I do the subtraction, there's nothing left over at all. So, there is no remainder.
And so when somebody asks me, what is 1845 divided by 15, I can say that the answer is 123 and there isn't any remainder to put back on top of the 15. In other words, 15 is a perfect factor of 1845.<|endoftext|>
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Autism is one of the mental, emotional, and behavior disorders that
appears in early childhood. Autism, or autistic disorder, almost always
develops within the first 3 years of a child’s life. Children and adolescents
with autism cannot interact normally with other people. Autism thus affects
many aspects of their development.
Children with and adolescents with autism typically: -have a difficult
time communicating with others -exhibit very repetitious behaviors (like
rocking back and forth, head banging, or touching or twirling objects); -have a
limited range of interests and activities; and -may became upset at a small
change in their environment or daily routine.
Although symptoms of autistic disorder sometimes can be seen in early
infancy, the condition can appear after months of normal development. In most
cases, it is not possible to identify any specific event that triggers autistic
disorder. About 7 in every 10 children and adolescents with autistic disorder
also have mental retardation or other problems with their brain function or
Recent studies estimate that as many as 14 children out of 10,000 may
have autism or a related condition. About 125,000 Americans are affected by
these disorders, and nearly 4,000 families across the country have two or more
children with autism. Three times as many boys as girls have autism.
Researchers are still unsure about what causes autism. Several studies
suggest that autistic disorder might be caused by a combination of biological
factors, including exposure to a virus before birth, a problem with the immune
system, or genetics.
Scientists also have identified chemicals in the brain and the immune
system that may be involved in autistic disorder. As a normal brain develops,
the level of serotonin, a chemical found in the brain, declines. In some
children with autistic disorder, however, the serotonin levels do not decline.
Now researchers are trying to determine whether this happens only to children
with autism and why, and whether other factors are involved.<|endoftext|>
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Most tsunami are caused by large earthquakes on the sea floor when slabs of rock move past each other suddenly, causing the overlying water to move. The resulting waves move away from the source of the earthquake event.
Landslides can happen on the seafloor, just like on land. Areas of the seafloor that are steep and loaded with sediment, such as the edge of the continental slope, are more prone to undersea landslides.
When an undersea landslide occurs (perhaps after a nearby earthquake) a large mass of sand, mud and gravel can move down the slope. This movement will draw the water down and may cause a tsunami that will travel across the ocean.
Tsunami initiated by volcanic eruptions are less common. They occur in several ways:
|Format||Multimedia and Websites|<|endoftext|>
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"The ocean looks blue because red, orange and yellow (long wavelength light) are absorbed more strongly by water than is blue (short wavelength light). So when white light from the sun enters the ocean, it is mostly the blue that gets returned. Same reason the sky is blue."
In other words, the color of the ocean and the color of the sky are related but occur independently of each other: in both cases, the preferential absorption of long-wavelength (reddish) light gives rise to the blue. Note that this effect only works if the water is very pure; if the water is full of mud, algae or other impurities, the light scattered off these impurities will overwhelm the water's natural blueness.
Gross then asks, "So why are sunsets orange?" Several people to wrote in to correct or clarify that comment.
Perhaps the most helpful response came from Michael Kruger of the department of physics at the University of Missouri. He sent the following reaction:
"The answer to why the sky is blue isn't quite correct. The sky is blue not because the atmosphere absorbs the other colors, but because the atmosphere tends to scatter shorter wavelength (blue) light to a greater extent than longer wavelength (red) light. Blue light from the sun is scattered every which way, much more so than the other colors, so when you look up at the daytime sky you see blue no matter where you look. This scattering is called 'Rayleigh scattering'; the amount of scattering goes as the frequency of the light to the 4th power. By the way, this effect is most prevalent when the particles that do the scattering are smaller than the wavelength of light, as is the case for the nitrogen and oxygen molecules in the atmosphere.
"Now we are in a position to figure out why sunsets are reddish! When the sun is setting, the light that reaches you has had to go through lots more atmosphere than when the sun is overhead, hence the only color light that is not scattered away is the long wavelength light, the red.
"We can also answer why clouds, milk, powdered sugar and salt are white. The particles in these materials that are responsible for scattering the light are larger than the wavelength of light. Consequently, all colors of light are scattered by more or less the same amount. Much of the scattering in milk is due to the lipids (fat). If you take out the fat, the milk will not scatter as much light; that is probably why skim milk looks the way it does.<|endoftext|>
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# A child has six blocks, three of which are red and three of which are green. How many patterns can she make by placing
A child has six blocks, three of which are red and three of which are green. How many patterns can she make by placing them all in a line? If she is given three white blocks, how many total patterns can she make by placing all nine blocks in a line?
I was able to figure out the solutions as
a) $$\binom{6}{3\ 3}$$
b) $$\binom{9}{3\ 3\ 3}$$
But I was wondering what is the alternative way to count these situations? I understand this manner in that it is describing the number of ways my total number of positions can be divided into subgroups of the respective sizes. But how would I organize the cases if I say did not know this mulitnomial formula?
Edit: The explanations I have been getting have been based on the multinomial formula. The motivation for my question came from the fashion in which I saw this form of the solution:
I apologize for the link from elsewhere. I was trying to understand how this person deconstructed the cases.
Thanks
• Imagine you had 6 different blocks. You would have 6! combinations. But now you consider 3 of them to be same so you will have to divide your 6! by 3!3!. Jun 3, 2016 at 19:06
Alternate way to think of the second answer:
Apply multiplication principle:
• Choose the locations of the red blocks: $\binom{9}{3}$ options
• From the remaining locations, choose the locations of the green blocks: $\binom{6}{3}$ options
• From the remaining locations, choose the locations of the white blocks: $\binom{3}{3}$ options
There are then $\binom{9}{3}\binom{6}{3}\binom{3}{3}$ total arrangements.
Note that $\binom{9}{3}\binom{6}{3}\binom{3}{3}=\binom{9}{3,3,3}$
For the harder question it is a all possible permutations of 9 blocks divided by permutations of 3 blocks of each color
$\frac{9!}{3!3!3!}$
because all blocks sharing same color are indistinguishable in a line.
An intuition-type argument:
If they were lettered blocks, {ABCDEF} in the first case, there would be $6!$ possible orderings of the blocks. If you take one of those and paint {ABC} blocks solid red so they are indistinguishable, that ordering is now part of a group of $3!$ indistinguishable orderings. Paint the remaining blocks solid green and your have diminished the number of different orderings by another factor of $3!$.
Hence the answer for a) $\dfrac {6!}{3!3!} = \dbinom{6}{3,3}= \dbinom{6}{3}$
And by the same argument, starting with $9$ distinguishable blocks and gradually rendering groups indistinguishable, for b) $\dfrac {9!}{3!3!3!} = \dbinom{9}{3,3,3}$<|endoftext|>
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Asthma Awareness (AFFA.org): #TackleAsthma · Asthma Info · Allergy Types · N.J. Support Groups · Donate
Source: Asthma and Allergy Foundation of America
Asthma affects almost 25 million Americans. It is a chronic disease that causes your airways to become inflamed, making it hard to breathe.
Asthma symptoms can appear when you are exposed to a trigger. A trigger is something you are sensitive to that makes your airways become inflamed. This causes swelling, mucous production and narrowing in your airways. Common asthma triggers are pollen, chemicals, extreme weather changes, smoke, dust mites, stress and exercise. Common symptoms are coughing, shortness of breath, wheezing and chest tightness.
Each day, ten Americans die from asthma. Many of these deaths are avoidable with proper treatment and care.
It is important to know the signs of an asthma attack. Seek medical help immediately for:
– Fast breathing with chest retractions (skin sucks in between or around the chest plate and/or rib bones when inhaling)
– Very pale or blue coloring in the face, lips, fingernails
– Rapid movement of nostrils
– Ribs or stomach moving in and out deeply and rapidly
– Expanded chest that does not deflate when you exhale
– Infants with asthma who fail to respond to or recognize parents
There is no cure for asthma. The best way to manage asthma is to avoid triggers, take medications to prevent symptoms and prepare to treat asthma episodes if they occur.
Allergies are one of the most common chronic diseases. A chronic disease lasts a long time or occurs often. An allergy occurs when the body’s immune system sees a substance as harmful and overreacts to it.
The substances that cause allergic reactions are allergens. When someone has allergies, their immune system makes an antibody called immunoglobulin E (IgE). These antibodies respond to allergens. The symptoms that result are an allergic reaction.<|endoftext|>
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Mathematics » Introducing Decimals » Ratios and Rate
# Finding Unit Rates
## Finding Unit Rates
In the last example, we calculated that Bob was driving at a rate of $$\frac{\text{175 miles}}{\text{3 hours}}.$$ This tells us that every three hours, Bob will travel $$175$$ miles. This is correct, but not very useful. We usually want the rate to reflect the number of miles in one hour. A rate that has a denominator of $$1$$ unit is referred to as a unit rate.
### Definition: Unit Rate
A unit rate is a rate with denominator of $$1$$ unit.
Unit rates are very common in our lives. For example, when we say that we are driving at a speed of $$68$$ miles per hour we mean that we travel $$68$$ miles in $$1$$ hour. We would write this rate as $$68$$ miles/hour (read $$68$$ miles per hour). The common abbreviation for this is $$68$$ mph. Note that when no number is written before a unit, it is assumed to be $$1.$$
So $$68$$ miles/hour really means $$\text{68 miles/1 hour.}$$
Two rates we often use when driving can be written in different forms, as shown:
$$68$$ miles in $$1$$ hour$$\frac{\text{68 miles}}{\text{1 hour}}$$$$68$$ miles/hour$$68$$ mph$$\text{68 miles per hour}$$
$$36$$ miles to $$1$$ gallon$$\frac{\text{36 miles}}{\text{1 gallon}}$$$$36$$ miles/gallon$$36$$ mpg$$\text{36 miles per gallon}$$
Another example of unit rate that you may already know about is hourly pay rate. It is usually expressed as the amount of money earned for one hour of work. For example, if you are paid $$\text{\12.50}$$ for each hour you work, you could write that your hourly (unit) pay rate is $$\text{\12.50/hour}$$ (read $$\text{\12.50}$$ per hour.)
To convert a rate to a unit rate, we divide the numerator by the denominator. This gives us a denominator of $$1.$$
## Example
Anita was paid $$\text{\384}$$ last week for working $$\text{32 hours}.$$ What is Anita’s hourly pay rate?
### Solution
Start with a rate of dollars to hours. Then divide. $$\text{\384 last week for 32 hours}$$ Write as a rate. $$\frac{\384}{\text{32 hours}}$$ Divide the numerator by the denominator. $$\frac{\12}{\text{1 hour}}$$ Rewrite as a rate. $$\12/\text{hour}$$
Anita’s hourly pay rate is $$\text{\12}$$ per hour.
## Example
Sven drives his car $$455$$ miles, using $$14$$ gallons of gasoline. How many miles per gallon does his car get?
### Solution
$$\text{455 miles to 14 gallons of gas}$$ Write as a rate. $$\frac{\text{455 miles}}{\text{14 gallons}}$$ Divide 455 by 14 to get the unit rate. $$\frac{\text{32.5 miles}}{\text{1 gallon}}$$
Sven’s car gets $$32.5$$ miles/gallon, or $$32.5$$ mpg.<|endoftext|>
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Historically, before the advent of commercially produced sugars made from cane and beet, honey was the only natural sugar available and mankind has always loved and craved the sweet taste of golden honey. Collecting honey from wild bee colonies is one of the most ancient of human activities and many indigenous societies in parts of Africa, Asia, Australia, and South America continue this method of harvesting honey to this day.
Some of the earliest evidence depicting humans gathering honey is from rock paintings dating from about 15,000 years ago. However, when honey was gathered from wild bee colonies it usually entailed the destruction of the entire colony. Smoke was used to subdue the bees and the wild hive was broken into and the honeycombs torn out and destroyed; this also resulted in the destruction of the eggs and larvae in the honeycomb. The liquid honey from the destroyed brood nest was saved but the rest of the nest was usually discarded although sometimes the bee larvae were eaten as a source of protein. In settled societies the destruction of a bee colony meant the loss of a valuable resource; there was no continuity of production and no possibility of selective breeding, since each bee colony was destroyed at harvest time, along with its precious all-important queen.
Egyptian art from around 4,500 years ago shows early efforts to domesticate bees – although, as all bees are free to come and go as they please and man merely removes honey and other items of interest to him from their colonies, all bees are wild but they are a managed wild creature. In Egypt simple hives were used and workers used smoke to help remove the honeycombs from the hives. The honey was stored in sealed jars, some of which were found in the tombs of pharaohs such as Tutankhamun. It is said that honey that had been sealed in the jars and then left undisturbed in the tombs was still edible 4,000 years later!
In Europe, before the advent of the modern movable-frame hives, bees were often kept in straw skeps. Because the space available to the bees in these skeps was so limited, they regularly swarmed (which is a natural method of colony reproduction) and this served to replace the colonies which had been killed to obtain the honey. The space for the bees could be enlarged by adding an extra chamber - called an eke (hence the term ‘to eke something out’ which is still used today). Many colonies of honey bees were maintained by religious communities, principally for the wax they yielded. Beautiful cream beeswax candles were produced which burnt without smoke and these were of infinitely superior quality to the candles made from tallow used by the general population. The honey was welcome as a sweetener and, in areas where vines would not grow, fermented honey was used to make the alcoholic drink mead. As a result, in the monasteries, the beekeeper was a person of considerable importance to the monastic community.
During the Middle Ages people were aware that there was a single large bee that was different to all the other bees in the colony. Traditionally, the hierarchy of a bee hive was thought to reflect the structure of Church and State and thus the bee that clearly ruled the colony was believed to be a king bee. It was not until 1586 that it was discovered that the king bee was actually a queen.
In the 18th century European biologists, or natural philosophers, started the scientific study of bee colonies and began to understand the complex social world of a bee colony. Although scientists were aware that queens laid eggs in empty cells, they did not know how a queen was fertilised and it was a Swiss scientist, François Huber (1750 – 1831) who was the first to discover that queens are inseminated by a number of matings with different drones, high in the air at a distance from their hive, during a ‘mating flight’. Huber is universally regarded as "the father of modern bee-science" and in his "Nouvelles Observations sur Les Abeilles” disclosed most of the basic scientific facts on the biology and ecology of honey bees. It was this greater understanding of the colonies and the biology of bees that allowed the construction of the movable-comb hive so that honey could be harvested without destroying the entire colony and the breeding of the bees could be controlled; thus modern bee-keeping was born.
In order to commence beekeeping and be able to sustain a healthy and productive colony of bees, it is important to be aware of the fundamental characteristics and properties of bees. This section of the guide will introduce you to the properties of the honey bees which you will keep, the hierarchical structure within a beehive and how these bees function as a colony.
Apis mellifera is the most widespread species of honey bee in the world. There are many different strains of this honey bee and Apis mellifera mellifera, the European dark bee, was domesticated and bred in modern times. They are predominantly dark brown and black in colour; these bees are docile and can survive throughout the winter in cool temperate zones by storing large amounts of honey, gathered during the summer months. However, much hybridisation has occurred with Italian and other bees making pure bred Apis mellifera mellifera largely a thing of the past and it is no longer the only significant commercial subspecies of the Western honey bee.
The term 'beehive' loosely refers to a structure in which honey bees live and rear their young; examples of natural beehives include tree hollows and rock structures. However, the term 'beehive' in modern parlance usually refers to a manmade structure within which bees are kept with either fixed or movable frames. The bees in these manmade beehives are used to produce honey for food or medical use, to pollinate nearby crops and to preserve the local honey bee population. Within a beehive there exists a complex series of hexagonal cells. These 'honeycombs' are made of beeswax and are used by bees to store not only food, such as honey and pollen, but also to house the brood of eggs, larvae and pupae. In any common honey bee hive there are approximately 5,000 bees during the winter months – the queen bee and worker bees – and a large hive in the summer months can house 50,000 bees – the queen bee, worker bees and drone bees.
A bee hive colony functions according to a strict hierarchical structure. Within this hierarchy there are three types of honey bee: a queen bee, drone bees and worker bees. Each of these types of bee has different tasks to carry out within the hive:
The queen bee lays all of the eggs which populate the colony and is normally the only sexually mature female in the hive; all of the female worker bees and male drones in the hive are her offspring. Royal jelly, a creamy white secretion, produced in glands in the heads of worker bees is responsible for turning an egg into a queen bee rather than a worker bee or drone bee. Royal jelly typically contains about 60% - 70% water, 12% - 15% proteins, 10% - 16% sugars, 3% - 6% fats, and 2% - 3% various vitamins, salts, and amino acids. The queen bee is raised from a normal worker egg, laid in a special elongated ‘queen cell’, but is fed a large amount of royal jelly throughout her development and this results in a radically different growth rate and metamorphosis. All bee larvae are fed royal jelly but after three days the drone and worker larvae are no longer fed in this way and do not experience the rapid growth of the queen.
A queen bee is really just an egg-laying machine; she has a smaller brain than a worker bee and is incapable of even feeding herself. When one buys a new queen bee by post, she arrives in a small box accompanied by her retinue of worker bees without whom she would starve.
A virgin queen, a newly hatched queen, will remain in the hive for 3 – 7 days until making her first orientation flight to mark the position of the hive. During subsequent flights she may mate with a number of male drones on each flight and, during these mating flights, the queen receives and stores sufficient sperm to fertilise hundreds of thousands of eggs. If she does not manage to leave the hive to mate, possibly due to bad weather, she remains infertile and becomes a drone layer, incapable of producing female worker bees. Worker bees sometimes kill a non-performing queen and produce another as, without a properly performing queen, the colony is doomed. After her mating flights the queen will not leave her hive again unless she swarms with her worker bees.
The queen bee is the largest of the bees in a honey bee colony, measuring around 2 cm – which is about twice the length of a worker bee. She walks with a distinctive gait as she proceeds from cell to cell to lay her eggs which can aid in her identification. A queen bee can live for three to five years, during which time she continues to lay eggs, and can lay up to half a million eggs in her lifetime although she will normally lay most eggs in the first season and her laying rate will decrease throughout her life. In April and May a queen bee will lay eggs all day and night, the interior of a hive is dark so day and night do not exist for the queen. This process can result in approximately 2000 eggs being laid each day during her peak laying period. Fertilised eggs become female worker bees and those which remain unfertilised become male bees - drones. In the winter the rate of eggs laid by the queen decreases dramatically as food supplies of pollen and nectar reduce and the colony must survive on the food stores which have been collected by the worker bees during the summer. This reduced food supply results in the reduction of the number of bees in the hive. In winter the number of worker bees drops from 50,000 to about 5,000; there are no drone bees in the hive as they have been evicted from the hive by the worker bees at the end of the queen-raising season when their use has passed and they are a drain on the colony’s resources.
A healthy queen bee will emit pheromones which inform all other bees in the colony that she is present and in good health. As a result of these pheromones, the bees in the colony are aware if an old queen is removed and a new queen introduced by a beekeeper. New queens must be introduced with great care, after the removal of the old queen, particularly if it is a difficult colony who might attempt to kill the interloper. It is the temperament of the queen that dictates the temperament of the whole colony so beekeepers always aim to have a placid, non-swarming strain of queen.
Drone bees are raised in wax cells which are bigger than those constructed by the worker bees for the worker eggs. When the queen lays an egg, she measures the cell and realises that is it larger than a worker cell and lays an unfertilised egg – amazingly she can actually control whether she lays a fertilised egg, which becomes a worker, or an unfertilised egg, which develops into a male bee. Drones are almost twice the size of worker bees and have no sting; they exist within the hive solely in order to mate with the virgin queen. They do not work, do not forage for pollen and cannot collect nectar as they do not have a long enough proboscis to reach the nectar in a flower. They have no other known function than to mate with and fertilise new queens on their mating flights. However, once they have mated with the queen, they will die as a result of this mating. At the end of the summer any remaining drones are evicted from the hive by the worker bees and this eviction is often quite brutal. All through the summer the worker bees have fed and cared for the drones who play no part in the maintenance of the hive; however, in winter they would be too much of a liability on the colony and they will be replaced the following spring. The worker bees stop feeding the drones and the weakened drones are forced from the hive, often having their wings chewed off by the aggressive workers. They are left to die outside the hive while the worker bees continue to prepare the hive for winter.
Worker bees represent nearly all the bees in a hive. They are all female but are not able to reproduce and are all daughters of the single queen bee who rules the colony. Worker bees are smaller than drones but do have a sting. However, having stung once they will die which means that, after defending their hive against intruders, you will often see their bodies lying just outside the hive. They are the general workforce of the hive and carry out all the necessary tasks which enable the hive to function efficiently. They perform different tasks throughout their life, largely dictated by their age.
During the summer months, when activity in the hive is at its peak, worker bees may only live for about 6 weeks. Worker bees start their working life cleaning cells in the hive and caring for and feeding the larvae. It is important for the youngest workers to remain in the hive, as they are still soft bodied and their wings are too delicate for them to fly well. They will then progress to receiving nectar and pollen from incoming foraging bees and then, when they are a little older, they make wax cells – this is a task that is undertaken in teams, as the temperature for wax-building needs to be quite high at 33-35°C. Only when the worker is about three weeks old does she leave the hive to forage for nectar, pollen, propolis and water. The bee colony always has older worker bees, which are designated as dedicated guard bees, on duty. When the hive is under threat, the guard bees release a pheromone scent which alerts the other workers, so that reinforcements can quickly arrive on the scene to protect the hive.
Between winter and early spring worker bees can live for several months as they rarely leave the hive. When external temperatures fall to about 15°C they gather to form a well-defined rugby-ball shaped cluster within the hive; at the heart of the cluster is the queen. The cluster expands and contracts as the weather warms and cools and, as the temperature decreases, the cluster becomes tighter and more compact as the bees cling tightly together on the combs in the hive. The worker bees constantly vibrate their wing muscles in order to generate heat to keep each other and their all-important queen warm and they maintain the temperature at the core of the cluster at about 30°C.
There are several different types of beehive each of which has specific advantages and disadvantages and the choice depends upon your personal beekeeping preferences. Most modern hives are based on a design published by Reverend L. L. Langstroth in America in 1853. The Reverend Langstroth noticed that bees built their wild comb with a specific distance between the combs (the bee space) and he designed a hive with movable frames set apart so that this vital bee space could be maintained. Furthermore, he separated the colony into a brood box, in which the queen laid her eggs, and a honey super in which the colony stored its excess honey; variants of this design are used to this day by most beekeepers.
To help you choose the beehive which best suits your needs, listed below are the different types of hive which you may come across:
The Langstroth hive is the most common beehive used worldwide and by American non-commercial beekeepers. These hives consist of several boxes which are stacked together to form a hive. Because the Langstroth hive is so widely used, it is easy to find supplies and spare parts but it is a large and heavy hive so it can be hard to manipulate the boxes and move the hives.
This eight frame version of the traditional ten frame Langstroth hive is slightly smaller in its construction. Although this means that you will be able to store a smaller amount of honey and the brood box is also smaller, it is easier to assemble and manoeuvre.
If you ask people to draw a beehive it is most likely that they would draw a WBC hive. Named after its creator, William Broughton Carr, it is an attractive design that is popular as a garden ornament as well as being a practical beehive. The double-walled design insulates against extremes of temperature but the brood boxes and super boxes are slightly trickier to manipulate than for other types of hive which is time-consuming for those who have many hives; however, this remains the second most popular hive in Britain.
This is the most popular beehive in Britain and these hives are widely used by amateur and commercial beekeepers alike. The National Hive is similar to the Langstroth hive but is smaller and easier to handle; they take 11 British Standard brood frames and measure 18 1/8″ square externally.
Unlike the Langstroth-style hive which functions vertically, a top-bar hive operates horizontally and the comb hangs from removable bars. These hives are considered by their supporters to encourage a more natural form of beekeeping. Top-bar hives are rectangular in shape and are typically more than twice as wide as the multi-storey framed hives commonly found in English-speaking countries. Top-bar hives allow beekeeping methods that interfere very little with the colony but are usually not readily portable. The brood nest is stored at the back of the nest and the honey is produced at the front of the hive. This small, horizontal structure enables one to keep bees in a small garden or another small storage space. The hives are lightweight if they have to be moved, are easy to maintain, and the bees are able to construct natural cell sizes. However, despite these advantages, top-bar hives can result in poor ventilation for the bees if they are not properly constructed and the bees are also more prone to dying in cold winters. Because these hives are less popular than Langstroth-style hives, it is more difficult to find local support and advice.
A Warré hive consists of several small, square hive bodies and top bars without any frames or foundation. A Warré hive also utilises a quilt and a vented, angled roof. This lack of foundation and the overall size and shape of a Warré hive make it a more natural hive for bees with superior moisture management. However, despite the advantage of minimal inspections by the beekeeper, the frames cannot be moved in a Warré hive as the bees will build and attach comb to the inside of the hive walls. Management of Warré hives calls for the addition of extra boxes to the bottom of the stack, unlike Langstroth hives, causing comb to be regularly harvested and taken out of use. This prevents old comb from being reused and therefore removes any environmental and agricultural chemicals and toxins.
Once you have chosen your preferred type of beehive, you will need to decide where to keep the hive. In order to sustain a healthy and productive hive you will need to select a space where the temperature, prevailing wind and sunlight positively affect your hive. The colony in the hive needs to be able to moderate its own temperature so it is advisable that the hive be sheltered from direct wind or sunlight. The hive should be positioned on level ground but tilted slightly forward so that rain will not run into the hive through its opening. It is recommended that hives should be placed facing the south/south east, where possible, and close to a bush or tree so that they receive morning sunlight and are protected from strong winds. If you intend to maintain multiple hives, ensure that you allocate sufficient room between each hive in order to walk around them and carry out your daily beekeeping tasks – you will need more space than you first think!
Now that you have chosen your beehive and where to place it, you can begin to construct it. Listed below are the various components of a beehive and how they function:
Your hive will require a solid foundation. Many beekeepers choose to use several breeze blocks with wood on top but you can also purchase a hive stand. This structure serves as the bottom part of your hive and may have an angled landing board for your bees.
There are two types of floor which you can use; a solid bottom board and a screened bottom board. The screened bottom board has a screen on the bottom with a removable sticky board which catches Varroa mites and other pests which can harm your colony. A screened bottom board offers all the benefits of a solid bottom board whilst simultaneously aiding in effective hive ventilation.
This appliance is placed between your bottom board and your first brood box to prevent pests and rodents from entering and damaging your hive's structure. These are usually made of wood and many beekeepers only use them in the winter months when the bees are in their cluster and not actively patrolling the entry to their hive. Pictured below you can see the installment of and entrance reducer and one in place on a busy hive.
This appliance is an optional addition to your hive which offers the bees more efficient ventilation and temperature moderation. It gives more room between the entrance of the hive and the brood chamber.
A brood box is a hive box that houses the frames that support the wax foundation from which the bees draw the cells that make up the honeycomb where the queen will lay her eggs and some stores will be kept. The brood box frames are usually made from wood, although sometimes of plastic, and hold a wired wax foundation, stamped with a hexagonal honeycomb pattern to assist the bees in constructing their honeycomb.
A super is a box that holds the frames where the bees will store their honey. Honey supers are much shallower than brood boxes as honey is heavy and, should you be lucky enough to harvest lots of honey, supers that are not too deep make this an easier task. The stamped wax sheets used in the frames in the supers do not need to be wired and, although the bee space is still maintained, may have thicker bars which means that fewer frames are needed in each super.
The queen excluder is a device of either a slotted zinc sheet or wire rods; the width of the slots or between the wires is too small for the queen to pass through but the workers can pass easily. The result is that all the eggs laid by the queen, and hence the brood, are in the box below the excluder. Any boxes with frames placed above the excluder will consist solely of honey and pollen which the beekeeper can remove when required or it can be left as winter supplies for the bees.
The wooden crown board is placed on top of the uppermost super and separates the supers from the roof and stops the bees from sticking the roof down. It will normally have 2 holes and, with additional one-way bee escapes, can be used to remove the bees from the supers when the time has come to harvest the honey.
The hive roof is constructed with the primary aim of keeping the rain off the hive (bees survive cold weather very well, but damp, particularly in winter, often kills a colony). The most common type of roof will fit over the crown board with sides that hang over the hive's upper super.
All of the components of your hive can be purchased from local beekeeping stores or online. Once you have amassed all of the necessary parts then you can begin to construct your beehive from scratch, should you wish. However, most beekeepers limit their DIY work to hammering together the frames and leave the construction to the experts, buying hives that are ready made.
Listed below is a step-by-step guide on how to construct a Langstroth beehive. For information on how to construct other beehives, see the Useful Links section of this guide:
1. Establish a solid foundation for your hive: In order to prevent your hive from tipping over, it is recommended that you create a solid foundation. Breeze blocks are one of the most commonly used foundations; simply lay 4 blocks beside one another on level ground and place another 4 on top on them for a solid base.
2. Lay down the floor of your hive: Place the opening facing towards you.
3. Place one brood box on the floor of your hive: This will serve as the brood chamber wherein your queen will lay her eggs.
4. Install ten frames in the brood box: You will be able to remove these frames at a later date depending on the size of your bee colony and your personal hive preferences.
5. Place a queen excluder on top of the brood box: this will prevent the queen from laying eggs in the honey stores.
6. Place a super on top of the brood box placed and install ten frames within it: This section will serve as the food storage chamber for your hive.
7. Lay the crown board on top of this super and a roof on top of the crown board to complete your beehive: It is also advisable to place a large stone or weight on top of your hive to keep it securely sealed during strong weather conditions.
Over the course of your beekeeping year you might need to add more supers and possibly a further brood box to cope with your growing colony and their honey stores.
Once your beehive has been completed the most exciting moment has arrived and you can add your bees. There are several online stores where you can order bees and have them delivered to your home. Alternatively, you can contact your local Beekeeping Association for information on where to acquire bees in your area; many people start with a swarm, collected by a member of the beekeeping association which is a very cost-effective method of obtaining bees; however, care needs to be takes as the character of the queen is unknown and you could have a bad-tempered hive with a tendency to swarm. If you are in any doubt, the local association is always ready to help new beekeepers.
If you wish to maintain the health of your bees and the productivity of your beehive, you will need to carry out various beekeeping tasks. In order to carry out these tasks you will need to acquire the appropriate clothing and equipment in order to protect yourself and your bees. Your protective clothing should consist of:
This suit should include a protective veil which fully covers your head and face whilst you handle your beehive. As well as providing sufficient protection, this suit and veil should also facilitate clear visibility and easy breathing.
You will need to acquire gloves which offer you protection against bee stings and which can be easily cleaned and thrown away after use. Common examples of beekeeping gloves include disposable latex gloves. This is due to the fact that they do not fester infectious pathogens which could be harmful to you or your colony.
Strong working boots such as wellington boots or boots with protective toe caps are needed in order to protect you from bee stings or from heavy equipment which may fall upon your feet during beekeeping activities.
A clean hive is a healthy hive. Therefore it is pivotal that you regularly clean and disinfect your beekeeping clothing and equipment. There are specialist beekeeping cleaning supplies which you can purchase online or from your local DIY or gardening store.
Bee smokers are designed to produce thick cool smoke that you will puff into the hive. When the bees smell the smoke they assume that their hive is in danger, perhaps near a forest fire, and they then start to eat honey to prepare to leave the hive and this eating calms them down. Smoke also masks communication between bees so that they do not raise the 'alarm' as you open the hive. Smokers have bellows attached and a wire heat shield around the body of the smoker to protect you from burns. Lighting the smoker requires some practice as you are aiming to have smouldering material which will produce the perfect smoke. You can use rotten wood, cardboard, damp leaves, Hessian sack or you can purchase smoker cartridges or pellets.
Hive tools are an absolutely indispensable piece of equipment for a beekeeper. There are various types of hive tool and which type you choose is a question of your preference. Bees make a sort of glue called propolis from the resin of trees. They use the propolis to seal up any tiny cracks and crevices in the hive (larger gaps are filled up with comb). The hive tool really is the Swiss Army knife of beekeeping. It is used as a metal lever which helps you prise open parts of your hive or scrape up any mess that the bees have made. It can also be used to scrape any debris, beeswax and plant resin from your frames and to clean inside your hive surfaces. It is piece of kit that you will find absolutely invaluable – and not just for beekeeping!
In order to produce and sell honey from your hive you will need a honey extractor. You will also need the sufficient equipment to filter your honey of any contaminants or waste before packaging and selling it. All of this equipment can be found online via specific beekeeping manufacturing and suppliers sites.
If your bees do not have sufficient food supplies to survive then you can create your own food supplies to prevent them from starving. You can create a syrup substitute for nectar by mixing white sugar with hot water. Whilst making this nectar substitute, you will need to continue stirring until all of the sugar is dissolved.
The type of food which you give to your bees should alter depending upon the current season. For instance, in the winter months your syrup substitute should have a low water content. This is due to the fact that if the water content is too high then your bees may not have time to dehydrate the food sufficiently in order to prevent fermentation before the cold weather begins. Consequently, if your bees require additional food stores during the winter months then you should feed them baker's fondant which they can eat immediately and which will not ferment.
During the spring and summer months you can provide your bees with food which has a higher water content. An acceptable spring and summer syrup substitute should mix one kilo of sugar with one litre of water, whereas in the autumn months you should mix one kilo of sugar with only half a litre of water.
As well as moderating the water content of your syrup substitute, you should also carefully select the type of sugar you use. You should only ever use refined white sugar because unrefined or brown sugar can cause dysentery in bees. Moreover, if you use pure sugar syrup then it will have no odour. As a result, you bees may not even notice your syrup substitute in the hive. To rectify this issue, add a little honey to your mix and apply it into their brood chamber so that they have a food trail to follow.
You can supply your food substitute to your bees via containers known as 'feeders' which can be placed above your brood chamber. These feeders facilitate restricted access to prevent your bees from falling into the feeder and drowning. To prevent your bees from drowning, you should never place an open container of syrup into your hive. Furthermore, you should place your feeders onto your hive in the evening. This will prevent the overexcitement of your bees as well as reducing the chances of external pests depleting your hive's food stores.
Throughout the year there are several beekeeping tasks which you can carry out to facilitate the growth of your hive and the health of your bee colony. These tasks can be categorised by season as follows:
Spring: The springtime is the best season to start a new hive and feed your pre-existing bees if necessary. During the winter they will have relied upon their summer food stores which may be depleted by spring. Alternatively, if your hive has any excess honeycomb stores left over the winter, now is the time to harvest it. If you require to add any new chambers to your hive or carry out any repairs then springtime is also the ideal season during which to carry out these tasks.
Summer: During the summer your bees will be working at their fastest rate; building honeycomb, bringing nectar and pollen back to the hive and caring for their brood. As a result, there are not many tasks which you will need to carry out during the summer months. However, it is worthwhile checking on your bees regularly in order to prevent any future problems from occurring. For instance, if you are using a top-bar hive or if your hive does not have a foundation then you should check that all of your combs are hanging straight. Summer is also the time to harvest any excess honey stores which your bees are producing.
Autumn: During the autumn months you will need to begin preparing your hive for the cold winter months. This will require providing your bees with sufficient food stores to survive through the winter and into the spring, as well as reducing the hive entrance, facilitating effective ventilation channels and applying guards against pests and rodents. Moreover, the autumn is the prime time for collecting the honey which your bees have produced during the summer. However, it is important to leave enough honey stores to enable your bees to survive and feed themselves throughout the winter.
Winter: The winter is the season during which your bees are highly susceptible to the cold and severe weather conditions. On a regular basis your bees will need to maintain temperatures between 22 and 25 degrees Celsius for the general colony, 34 to 35 degrees Celsius for the brood chamber, and up to 40 degrees Celsius for honey maturation. As a result, it is vital that you cover your hive with a protective jacket. You can purchase custom jackets online or construct your own using a thick plastic bag which has been lined with a strip of R19 insulation. You should fasten this jacket to the bottom of your hive with tape or staples to ensure it does not fall off during high winds, rain or snow.
You should also order your new bees in the winter months for their arrival in the following April. It is possible to order bees from specific online stores or to check with your local Beekeeping Association for suppliers near you or for any local swarms. If you are purchasing bees from a local or online supplier, then it is advisable to order 5 or 6 frame nucleus of bees. Nucleus, or 'nucs' as they are commonly known, will consist of a frame of honey bees along with a brood, a queen bee and food stores.
It is advisable that you purchase bees which have been bred locally within the UK in order to avoid a foreign hybrid species. As well as checking online or with your local Beekeeping Association, you can locate a reputable Bee Breeder via beekeeping magazines and journals through which to purchase complete colonies or nucleus.
The average hive produces 25lbs of honey; which in a good season can increase up to 60lbs of honey. It is important that you extract these honey supplies efficiently in order to reap the maximum benefits from storing and selling your honey. To help you extract your honey successfully, listed below is a step-by-step guide to extracting honey from your beehive:
Each cell of honey will be 'capped' or sealed by a mould of wax. Ensure that you extract each frame individually in order to prevent any damage to your hive's structure.
You can purchase an uncapping tank from a local beekeeping supplier or online site. These appliances facilitate the uncapping of honey cells for the extraction of honey. Hold each frame vertically over your uncapping tank and tip it forward gently in order to help the capping fall away from the comb as you slice them.
You can do so by using an electric uncapping knife or a serrated kitchen knife. When slicing your wax cappings you should commence a quarter of the way from the bottom of the comb and slice upward to expose the honey cells. After doing so slice the knife downwards to expose the honey cells on the lower section of the frame. If any honey cells remain you can use an uncapping fork, or 'cappings scratcher' as it is commonly known, to expose any outstanding cells. The cranked needles on an uncapping fork enable you to extract honey with minimal wastage.
Flip your frame over and extensively remove any wax cappings in order to expose the honey cells and harvest as much honey as possible. After uncapping each frame, place it vertically into your extractor device. An extractor device can be purchased online or from a local beekeeping supplier and is used to spin the honey from the uncapped cells. The spun honey is dispensed into a holding tank.
Once you have uncapped enough frames and your extractor is full, you can place the extractor lid on, seal it and begin to turn the crank. You should begin spinning your extractor slowly and then accelerate gradually in order to extract your honey gently from the wax combs. After approximately five minutes of spinning, you should turn all of your frames over in order to expose the opposite sides to the spinning process. After another five minutes your combs will be empty and you can return your frames to your Honey Super.
You should open the valve at the bottom of your extractor which will allow your honey to filter through a honey strainer and into your bottling container. If you purchase a specific bottling container or 'bottling bucket' then you can use its integrated valve to fill your honey jars.
It is important to purchase appropriately sized jars within which to store your honey. Each shallow Honey Super yields a honey harvest of approximately 30 pounds so it is important to purchase jars according to your hive size. Fortunately, standard honey jars are available in one, two and five pound sizes to accommodate your individual preferences.
As your hive begins to grow and expand you may find yourself with a surplus supply of honey. Many beekeepers choose to sell this surplus honey and make a modest additional income to bolster their regular wage. Given that you do not need a license to package and sell honey, it is an extremely promising venture for both small scale and large scale beekeepers. If you intend to bottle, package and sell your honey for commercial use then there are several factors which you will need to consider. For instance, you will need to include certain information on the labels for your jars which certifies that your honey is legitimate and safe to consume. The information which you will need to state on your honey jar labels includes:
Under EU law, certain information must be included on every jar of honey that you sell.
All food products must contain a 'best before date' which indicates when they will expire. Honey lasts approximately two years from the point of bottling so you will need to mark the date, month and year when it is expected to expire.
If you have produced your honey within the UK then your honey labels should bear the words 'Product of the UK' and include the producer's address. You should check the regulations of your local area to ensure your honey fulfils the necessary regional requirements. Otherwise, your honey must bear the words 'Product of more than one Country'.
Your lot number identifies that you have produced each jar of honey and that each jar can be traced back to your colony and the date of processing. You should also include the producer's name and address on your honey jar labels.
Your label must display the weight of honey in the jar. The weight contained in the jar must be equal or greater than the weight displayed on the label.
There are specific reserved descriptions for different types of honey. For instance; blossom or nectar honey (which has been obtained from the nectar of plants), honeydew honey (which has been obtained from the excretions of insects or plants), comb honey (which has been stored in the cells of freshly built brood-less combs), chunk or cut comb in honey (which contains pieces of comb honey), drained honey (which has been sourced by draining de-capped brood-less combs), extracted honey (which has been sourced by centrifuging de-capped brood-less combs), pressed honey (which has been sourced by pressing brood-less combs with or without the addition of heat), filtered honey (which has been sourced by removing foreign inorganic or organic matters in order to extract pollen) and baker's honey (which is suitable for use as an ingredient or in other processed foodstuffs, which has foreign tastes or odours, which has been overheated or which has begun to ferment). It is important to correctly categorise your honey if you intend to sell it commercially. If you are unsure of your honey's type, then you can contact your local Beekeeping Association or Food Hygiene Administration for more information.
As well as the national Food Hygiene regulations which you will need to consider, your honey jar labels should also promote your honey to potential customers. As a result, you should consider how you wish to brand your honey. If you can invent a unique and eye-catching logo or slogan which will pique customer interest, then you can begin to successfully sell your honey. If you intend to sell your honey on a larger scale, then it is worthwhile investigating patents and trademarks in order to register your particular brand of honey and its logo. Alternatively, you can acquire standard packaging labels from your local beekeeping supplier.
Once you have bottled and labelled your honey then you can begin to sell them. If you intend to turn your beekeeping hobby into a commercial business venture, then there are several avenues which you should investigate in order to solidify a strong customer base. One of the most lucrative resources from which you can benefit is your local community. Health food stores and local farmers markets are always looking for a source of fresh, local honey. If you visit these stores and markets as well as other local gift stores, gardening centres and fairs then you can find a sustainable outlet through which to sell your honey. You can even contact your local Beekeeping Association for advice. By conversing with local beekeepers you can gain excusive insights on how to successfully bottle, package and sell honey in your local area.
As well as selling your honey locally you can also capitalise upon the enormous resource that is the Internet. By creating a simple website you can sell your honey internationally to a substantially larger audience. You can even create business cards which include your website address on them. By handing out these cards in your local area you can encourage people to purchase your honey and recommend it to others. Alternatively, if you do not have a personal website then you can create an account on auction sites such as eBay through which you can sell your honey. However, if you intend to sell your honey via the Internet, then you will need to carefully package your honey so that it does not break during transit.
There are several pests and diseases which have been known to harm bee colonies in the UK in recent years. This section of the guide will introduce you to some of the most common honey bee pests and diseases which could adversely affect your hive, as well as illustrating effective treatment methods in order to protect your colony.
These are the most common pests in the UK that are known to infect honey bee colonies. These external parasitic mites attach themselves to the body of honey bees. Once attached, the varroa mite sucks hemolymph out of the honey bee. Hemolyph is the blood of honey bees and consequently varroa mites will weaken your bees and spread viral ' varroosis' diseases throughout your hive; which in turn can result in the death of your bee colony. Varroa mites reproduce on a 10 day cycle wherein they enter the brood cell of a hive and lay eggs on the honey bee larvae. Once the honey bees develop, the mites attach themselves to the bees. As these bees spread throughout the hive the mites can then begin to feed off of other bees and larvae.
Subsequently, it is important to regularly check your hive for varroa mites. It is worthwhile scrutinising your hive's drone bees as varroa mites are known to primarily target drone bee cells. Once you have identified a varroa mite infestation there are several treatment methods which you can utilise. There are several 'miticide' chemicals on the market which you can buy from your local beekeeping supplier in order to treat your colony. Always read the instructions on the miticide label carefully. Moreover, varroa mites can develop a resistance to miticides so it is advisable to alternate the products which you use each season. Common synthetic miticides include;
As well as synthetic miticides, there are several naturally occurring chemicals which you can use to treat a varroa mite infestation. Some of the most effective naturally occurring chemicals include;
As well as natural and synthetic chemicals you can also use a drone comb to capture and remove varroa mites from your bee hive. You can purchase a drone comb online or from your local beekeeping supplier. This special drone comb foundation frame has large hexagons imprinted in the sheet. As a result, your bees will only construct drone comb on these specific sheets. Given that varroa mites prefer to attach themselves to drone bees, they will be attracted to these sheets. Once the drone cells are caped you can remove the foundation frame and place it in the freezer overnight to kill off the varroa mites from the cells. You can then return this sheet to your hive and repeat the process until all varroa mites are eradicated.
Small Hive Beetles or Aethina tumida are extremely destructive pests for honey bee colonies. They are reddish or dark brown in colour and can cause damage to the comb, stored honey and pollen within the hive. This due to the fact that small hive beetle larvae will travel throughout honey combs; feeding, defecating and causing the honey to ferment and discolour as they travel throughout the combs. Moreover, if your hive has been contaminated by a heavy small hive beetle infestation it may cause your bees to abandon their hive.
If you regularly check your hive for small hive beetles and larvae you can actively prevent a large scale infestation from occurring. Once you have detected small hive beetles within your hive, there are several measures which you can put into place. As well as using the miticides mentioned above for treating varroa mites, there are a variety of traps which you can purchase online or from your local beekeeping supplier in order to remove small hive beetles from your hive. Amongst some of the most popular beetle traps include;
European Foulbrood or Melissococcus plutonius is a bacterium which affects the gut of honey bees who have eaten contaminated food. Symptoms include dead or dying larvae which appear curled upwards and which have turned a brown or yellowish colour. Your bees may also appear melted or deflated with dried out, rubbery or apparent tracheal tubes. European Foulbrood is considered as a stress disease which affects unhealthy bee colonies. Therefore if you ensure your bee colony is strong and healthy then they are far less susceptible to contract European Foulbrood. If you suspect your bees are suffering from European Foulbrood then you should contact your local Beekeeping Association immediately for treatment advice. Your colony can be treated by antibiotic medications such as oxytetracycline or via a Shook Swarm husbandry method. To perform this method you will need;
Before carrying out this method you will need to ensure that all of your equipment is sterilised and that you are wearing protective clothing. Listed below is a step-by-step guide on how to carry out the Shook Swarm husbandry method;
1. Shift your hive so that it can be shaken to one side, 2. Place the clean floor with entrance block on the original stand, 3. Place your spare queen excluder on the floor, 4. Place your fresh brood chamber with frames and foundation on top of the queen excluder, 5. Remove a few of the frames from the centre of the brood chamber to create a small, dark space into which you can insert your bees, 6. Dismantle your original hive and place your queen bee in between the frames and foundation, 7. Remove each brood frame and shake the frame diagonally across your new brood chamber, 8. Place the old, empty frames into the container. If these frames are infected with European Foulbrood then they must be burned to remove all traces of the disease, 9. Repeat this process with all of your remaining frames, 10. Insert the spare frames into the central space of the new brood chamber, 11. Insert your queen into the new brood chamber.
Colony Collapse Disorder is a phenomenon in which worker bees from a European honey bee colony disappear for no apparent reason. Although the causes for colony collapse disorder remain unclear, many researchers have cited possible causes which include; varroa mites, the use of pesticides, harmful pathogens, genetic factors, loss of habitat, malnutrition and changing beekeeping practices. Initial symptoms of colony collapse disorder include;
Once colony collapse disorder has taken place you may notice the following signs;
Although the primary cause of colony collapse disorder is as yet unknown, there are several measures you can put into place to prevent it adversely affecting your hive. As well as these measures, you can contact your local Beekeeping Association for more information and advice;
Do not use synthetic chemicals or pesticides in your garden which may be harmful for your colony.
If you wish to keep bees in the UK then you need to abide by the local and national rules and regulations for beekeeping. The principle UK legislation regarding beekeeping is the 1980 Bees Act. This legislation was introduced to control, inspect and treat the various pests and diseases which affect honeybee colonies throughout the UK. The 1980 Bees Act enables the Ministers or the Secretary of State to create Orders which will control pests and diseases. The Act also enables these Ministers or the Secretary of State to grant authorised persons the power to remove and/or destroy bee colonies which are infested with harmful pests or diseases. Under the act, the FERA (Food and Environment Research Agency) National Bee Unit is deemed responsible for apiary surveillance as well as pests and disease control across England and Wales.
A series of statutory orders have been established since the introduction of the 1980 Bees Act. The most notable of these orders is the Bee Diseases and Pests Control Order of 2006. The Order was introduced in order to replace the Bee Diseases Control Order of 1982 and revoke the Importation of Bees Order of 1997. The Order lists in detail the various pests and diseases for which statutory action must be taken to control them. Amongst these pests and diseases are included American foul brood (AFB), European foul brood (EFB), and the Small Hive Beetle (Aethina tumida) and Tropilaelaps mites – the latter two not yet being a threat to British bees. Within the Order are both revised and new measures on how to treat these diseases should an infestation occur. As well as national and local rules and regulations, there are also a series of beekeeping associations, local groups and training courses which you can use to broaden your beekeeping experience and expertise. Organisations such as BeeBase, the Bee Farmers Association, the National Farmers Union and the British Beekeepers' Association offer advice and ongoing support to both small- and large-scale beekeepers. For instance, the British Beekeeper's Association is a registered charity which was founded in 1874. To this day, this association represents the interests of 24,000 amateur beekeepers and the 3 billion honey bees for which they care. Furthermore, if you have a particular issue which you need resolved, then you can contact the Bee Diseases Insurance Ltd (BDI). As well as funding research into various bee diseases and offering training courses on how to treat diseases, the BDI provides insurance for the replacement of beekeeping equipment if it has been destroyed as a direct result of a virulent beekeeping disease such as European Foulbrood or American Foulbrood. If you wish to consolidate your knowledge of beekeeping, there are several training courses on offer. You can study for and receive a National Diploma in Beekeeping (NDB) as well as attend various other short term and advanced courses. Moreover, if you wish to remain informed of the latest beekeeping news and phenomena, you can subscribe to various beekeeping magazines and online journals. Some of the most popular beekeeping resources include:<|endoftext|>
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# Dividing unit fractions and whole numbers
### Dividing unit fractions by whole numbers (practice) Khan
• Dividing a unit fraction by a whole number Our mission is to provide a free, world-class education to anyone, anywhere. Khan Academy is a 501(c)(3) nonprofit organization
• Dividing unit fractions by whole numbers Our mission is to provide a free, world-class education to anyone, anywhere. Khan Academy is a 501(c)(3) nonprofit organization
• ator always works! This is because, for any numbers a and b with b nonzero, we have. a divided by (1/b) = a times (b/1) = (a/1) times (b/1) = ab/1 = ab. Have a blessed, wonderful day
• drawings to divide a whole number by a unit fraction. Revisit the Essential Question. To promote understanding of the relationships among the parts of a division problem involving fractions, discuss how the area model relates to the numbers in the division problem. The 2 larger regions represent the whole number, 2
A. Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) & 4, and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that (1/3) ÷ 4 = 1/12 because (1/12) × 4 = 1/3 Dividing Fractions By Whole Numbers Multiply the bottom number of the fraction by the whole number. To Divide a Fraction by a Whole Number: Step 1 Welcome to Dividing Whole Numbers by Unit Fractions with Mr. J! Need help with how to divide whole numbers by unit fractions? You're in the right place!Whet.. Unit fractions are fractions with 1 as the numerator. Below are our grade 5 math worksheet with word problems involving the division of whole numbers by unit fractions, or vice versa. Some problems include superfluous data to encourage students to think about the problems carefully, rather than trying to apply a pattern to the solutions To help, you can use the Dividing Fractions & Whole Numbers Pre-Assessment Class Checklist on page A12.6 if you like. By compiling results for your entire class, you can get a sense of the areas in which the class as a whole may need extra support. Activity 1 Dividing Fractions & Whole Numbers Pre-Assessment (cont.
Start studying Dividing Unit Fractions with Whole Numbers. Learn vocabulary, terms, and more with flashcards, games, and other study tools This lesson shows how to divide unit fractions by whole numbers using models. For more videos and instructional resources, visit TenMarks.com. TenMarks is a. Welcome to Dividing Unit Fractions by Whole Numbers with Mr. J! Need help with how to divide unit fractions by whole numbers? You're in the right place!Whet.. Fifth Grade Common Core Math Jeopardy Game - 5 NF.7 Divide Unit Fractions by Whole Numbers 5.NF.7 Practice provides two ways for students to practice and show mastery of their ability to divide unit fractions by whole numbers and whole numbers by unit fractions. 5.NF.B.7 5.NF.B.7a 5.NF.B.7b 5.NF
### Dividing a unit fraction by a whole number (video) Khan
1. Dividing a whole number by a unit fraction Our mission is to provide a free, world-class education to anyone, anywhere. Khan Academy is a 501(c)(3) nonprofit organization
2. Example dividing a whole by a unit fraction. Example dividing a whole by a unit fraction
3. Math worksheets: Dividing proper fractions by whole numbers. Below are six versions of our grade 5 fractions worksheet on dividing proper fractions by whole numbers. These math worksheets are pdf files. Open PDF. Worksheet #1 Worksheet #2 Worksheet #3 Worksheet #4 Worksheet #5 Worksheet #6
Dividing unit fractions by whole numbers visually Our mission is to provide a free, world-class education to anyone, anywhere. Khan Academy is a 501(c)(3) nonprofit organization Use a colored pencil to shade 1 column. (This shows the one part indicated by the numerator of the unit fraction.) Use horizontal lines to divide the box into the number of equal parts shown by the divisor (the whole number), and then use a darker colored pencil to shade 1 part or piece of the shaded column
### Dividing a whole number by a unit fraction (video) Khan
Example diving a unit fraction by a whole number. Example diving a unit fraction by a whole number Improve your math knowledge with free questions in Divide unit fractions and whole numbers: word problems and thousands of other math skills Improve your math knowledge with free questions in Divide whole numbers and unit fractions and thousands of other math skills So, a fraction written as 4/5 simply indicates to 4 pieces out of 5 equal sections. View PDF. Use calculator to divide the fraction's numerator by the denominator. Step 2. 5.92 / Divide Unit Fractions by Whole Numbers Up to 20 I. Divide whole numbers by fractions Our mission is to provide a free, world-class education to anyone, anywhere
Dividing Unit Fractions by Whole Numbers. Let's find the quotient of: 1. 5. ÷. 2. First of all, make a model of fraction. 1. 5 Divide Unit Fractions and Whole Numbers Divide a Unit Fraction by a Whole Number Anchor Chart - I use this Anchor Chart as a visual to show my students how to divide fractions using a number line and a model. Common Core Standard: 5.NF.B. Dividing Fractions By Whole Numbers Worksheet - Do you know how to maximize the use of numbers worksheets to aid increase your child's reading and writing skills? I will reveal how by supplying you with a preliminary understanding of some great benefits of numbers worksheets. When you understand the advantages of employing a printable pack format, you will be able to make use of this load.
After getting comfortable with dividing unit fractions by a whole number, children should be able to flip things over and learn how to divide a whole number by a unit fraction. Cool Fact. Dividing a fraction by a whole number is actually multiplying in disguise. For e.g.: 1/3 divided by 5 is actually 1/3 x 1/5 To divide fractions by a whole number, the first step is to find the reciprocal of the second fraction, then, multiply the two numerators, and the two denominators, separately. Lastly, simplify the fractions if needed Topic Definition. Division is a process of equally distributing or equally sharing.Division may be splitting one group of things into a specified number of groups or into groups with a specified number of items per group.Division may also be splitting one whole item into equal parts. Dividing is sharing in equal groups or parts. A fraction is a number in the form a/b
unit fraction by a whole number Multiplying Fractions by Whole Numbers Explained! 3rd Grade Math 8.6, Relate Fractions and Whole Numbers Dividing Whole Numbers by Fractions Using Models | Math with Mr. A unit fraction is a fraction whose numerator is 1. It can also be considered as a special type of proper fraction. Arithmetic way of dividing a unit fraction by a whole number. To divide a unit fraction by a whole number, first, we take the reciprocal of the whole number and simultaneously change the division operation to the multiplication Interpret division of a whole number by a unit fraction, and compute such quotients. For example, create a story context for 4 ÷ (1/5), and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that 4 ÷ (1/5) = 20 because 20 × (1/5) = 4. 5.NF.B.7c To divide any fraction by a whole number, divide the unit fraction by the whole number and multiply the result by the numerator of the fraction (i.e., to divide 2 3 by 4, consider 1 3 divided into 4 parts so each part is 1 12; thus, 2 3 divided by 4 would be 2 times 1 12 or 2 12
### Divide Unit Fractions by Whole Numbers (examples
• understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions. a. Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for 1 ··3 4 4, and use a visual fraction model to show the quotient. Use the relationshi
• Try out this fifth grade level math lesson for dividing whole numbers and unit fractions practice with your class today! Dividing Whole Numbers and Unit Fractions. Login. Time. Progress. Score. 0. end %. %. Times up! Times up! You may finish answering the current question and then proceed to your score!.
• Fraction Bars Activity 2 Problems for division of unit fractions by whole numbers 1. Two cyclists are on a trip and one has 1/5 gallon of water. If this water is shared equally between the two people, what fraction of a gallon will each person have? a. Sketch a 1/5 bar with one part out of 5 shaded to represent 1/5 gallon of water
• Khan Academy: Dividing Unit Fractions and Whole Numbers. Lesson Frame: We will divide AND represent division of a unit fraction by a whole number, and the division of a whole number by a unit fraction. I will divide AND use models to represent division of a unit fraction by a whole number and a whole number by a unit fraction
• ator that is the same in two or more fractions. Two or more fractions that are equal. A fraction that can not be simplified any further. Any positive number that can be written as a quotient where the deno
• Steps. Write the problem. The first step to dividing a fraction by a whole number is to simply write out the fraction followed by the division sign and the whole number you need to divide it by. Let's say we're working with the following problem: 2/3 ÷ 4. Change the whole number into a fraction
### Dividing Fractions By Whole Numbers - mathsisfun
• Here you can enter a fraction and a whole number (integer). We will show you step-by-step how to divide the fraction by the whole number. Please enter your math problem below so we can show you the solution with explanation: ÷ Here are some examples of math problems that our Fraction Divided By Whole Number Calculator can explain and solve:.
• Divide Whole Numbers by Unit Fraction. - Year 5 Maths. Practise Now. Divide a whole number by a unit fraction. Before using a written method for dividing a whole number by a unit fraction, students must interpret the meaning of this division by visual models. The Divide Whole by a Unit Fraction Worksheet includes problems based on intermediate.
• Whole Number Divided By Fraction Calculator. Welcome to our Whole Number Divided By Fraction Calculator. Here you can enter a whole number (integer) and a fraction. We will show you step-by-step how to divide the whole number by the fraction. Please enter your math problem below so we can show you the solution with explanation
• 5.NF.7 - Divide Unit Fractions By Whole Numbers & Whole Numbers By Unit Fractions. Kindergarten. Overview. Counting and Cardinality. Operations and Algebraic Thinking. Numbers and Operations in Base Ten. Measurement and Data. Geometry. Grade 1
• Dividing Unit Fractions by Whole Numbers Objective : 5.NF.A.1.a Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) ÷ 4, and use a visual fraction model to show the quotient
• I will reveal how by supplying you with a preli
divide a unit fraction by a whole number greater than 1. Prerequisites. Students should already be familiar with. the role of numerators and denominators, specifically with reference to unit fractions, representing unit fractions using bar models/tape diagrams, number lines, and shape-based models Things to Know. Unit fraction: a fraction that has a numerator of 1 Algorithm: sequence of steps to follow when completing a math operation The reciprocal of any fraction is when the numerator and the denominator are flipped. For example, the reciprocal of 2/1 is 1/2. To divide a unit fraction by a whole number, you keep the first fraction, switch the division sign to a multiplication sign.
Use visual models for division of whole numbers by unit fractions. You'll gain access to interventions, extensions, task implementation guides, and more for this instructional video. In this lesson you will learn how to divide whole numbers by unit fractions by using visual models Dividing Unit Fractions By Non Zero Whole Numbers - Displaying top 8 worksheets found for this concept. Some of the worksheets for this concept are Division division of unit fractions by whole numbers, Unit fraction division, Georgia standards of excellence curriculum frameworks, Divide a unit fraction by a whole number, Division word problems. Printable Fraction Worksheets Unit Fractions Of Numbers 5 Gif 780 1009 Unit Fractions Fractions Worksheets Fractions Unit fraction worksheets - To observe the image more plainly in this article, you could click on the preferred image to look at the picture in its original dimension or in full
### Dividing Whole Numbers by Unit Fractions Math with Mr
• e each quotient using a diagram. Show and explain your work. A) 1 2 ÷ 4 B) 3 ÷ 5 C) 1 3 ÷ 4 D) 2 ÷ 5 3. Use the relationship between multiplication and division to check your work. Example: A jug of water is full
• 5.NF.B.8 Missouri Grade 5 Math Dividing Unit Fractions By Whole Numbers - Extend the concept of division to divide unit fractions and whole numbers by using visual fraction models and equations. (a) Calculate and interpret the quotient of a unit fraction by a non-zero whole number. (b) Calculate and interpret the quotient of a whole number by a unit fraction
• e the size of each area. Q2: David devoted 1 4 of his day to study, exercise, play a video game, and surf the web. Given that he spent the same amount of time on each of these.
• To divide (a/b) by a whole number c, students can consider dividing the unit fraction (1/b) into c parts, then multiplying that number by a. What to look for Asking students how the representation of the unit square is connected to the process of dividing by a whole number can lead to a productive discussion about their interpretation both of.
• Name Date DIVIDING WHOLE NUMBERS BY UNIT FRACTIONS 4 Work out the missing fractions and numbers. 1) ___ ÷ 1 4 = 8 16) ___ ÷ 1 6 = 6 2) 5 ÷ 1 3 = ____ 17) 4 ÷ = 32 3) ___ ÷ 1 2 = 14 18) 4 ÷ 1 7 = ____ 4) 7 ÷
• above bars for a unit fraction to represent the same amount (½ of a pizza, or 1/3 of a bag of popcorn, etc). Finally, ask them to make up a word problem about the number of times the amount for their unit fraction bar fits into the amount for 3 whole bars and to write a division equation to express 3 divided by the unit fraction
• Divide and Grab is a fun jumping game that let students practice dividing fractions with whole numbers in a different way. In this game, they need to divide fractions with whole numbers. For that, they just need to solve the given expression using division operation and from the given options, simply click on the correct response to jump on it
Divide a Whole Number by a Unit Fraction. By (date), when given a non-zero whole number, (name) will divide by a unit fraction (e.g.; using a visual model and/or...equation) to calculate the quotient with (80%) accuracy (i.e, 4 out of 5 correct) for (2 out of 3) sets of fractions. You are not authorized to perform this action Divide Whole Numbers by Unit Fractions (5nf7b) Videos, examples, solutions, and lessons to help Grade 5 students learn to apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions. A. Interpret division of a unit fraction by a non-zero whole number, and compute such quotients
### Division with unit fractions word problems K5 Learnin
1. Mixed Numbers Calculator (also referred to as Mixed Fractions): This online calculator handles simple operations on whole numbers, integers, mixed numbers, fractions and improper fractions by adding, subtracting, dividing or multiplying. The answer is provided in a reduced fraction and a mixed number if it exists
2. Interpret division of a whole number by a unit fraction, and compute such quotients. For example, create a story context for 4 ÷ (1/5), and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to justify conclusions (e.g., 4 ÷ (1/5) = 20 because 20 x (1/5) = 4)
3. Use this helpful guide to learn and practice multiplying fractions. Students are going to take a deeper dive into fractions in this unit! Learners will apply previous understanding of finding equivalent fractions, and converting between fractions and mixed numbers to work with fractions in more complex ways. Students will continue to use visual.
### Dividing Unit Fractions with Whole Numbers Flashcards
fractions make sense. (Note: this is limited to the case of dividing unit fractions by whole numbers and whole numbers by unit fractions.) Critical Area #2 Extending division to 2‐digit divisors, integrating decimal fractions into the plac Grade 5 » Number & Operations—Fractions » Apply and extend previous understandings of multiplication and division. » 7 Print this page. Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions. USA » Common Core State Standards » Math » Grade 5 » Number and Operations—Fractions » Apply and extend previous understandings of multiplication and division. » (5.NF.B.7) Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.1 » (5.NF.B.7.B) Interpret. Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among.
What will you always have when you divide a fraction by a whole number. a fraction. 500. What is the mixed number for 7/2. What is 3 1/2? 500. Devon hiked 4 miles. Each day he hiked 1/3 mile. How many days did he hike? Katie said one-third divided by four is the equation to solve the problem. Kevin said four divided by one-third is the equation. 5th Grade's Dividing With Whole Numbers & Unit Fractions. Mixed Operations with Fractions. Dividing Whole Numbers by Fractions. Dividing Fractions by Whole Numbers. Word Problems. Mystery Questions. 100. 2/5 x 1/4 = Write your answer in its simplest form. What is 1/10 Dividing Mixed Numbers N-Gen Math 6.Unit 2.Lesson 9.More Work Dividing FractionsDividing fractions and whole number word problems Finding the quotient of two mixed numbers Fractions-Multiplying and Dividing Fractions Dividing Fractions On A Number Line Dividing Fractions with Fraction Models Dividing Fractions
Unit 2 - Fraction Division Course Documents Unit 1 - Whole Number and Decimal Operations Unit 2 - Fraction Division Unit 3 - Understanding and representing rational numbers Unit 4 - Graphing Rational Numbers on the Coordinate Grid Unit 5 - Algebraic Expressions Unit 6 - Understanding, Writing and Solving Equation 5th grade Multiplying and Dividing Fractions Printable Worksheets. Learn how to convert mixed fractions to improper fractions in this worksheet. Click the checkbox for the options to print and add to Assignments and Collections. Use this helpful guide to learn and practice multiplying fractions These fractions worksheets are great for working with dividing fractions and Whole Numbers. The fractions worksheets may be selected for two different degrees of difficulty. The answer worksheets will show the progression on how to solve the problems. These worksheets will generate 10 fraction division problems per worksheet Using this Dividing Unit Fractions By Whole Numbers Activity: This resource supports student understanding of dividing unit fractions by whole numbers. These 10 questions are perfect as a quiz, review, or homework activity! This resource some with two pages (one sheet) for questions, and two pages for answers, making this activity super easy. Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) ÷ 4, and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that (1/3) ÷ 4 = 1/12 because (1/12) × 4 = 1/3
Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) ÷ 4, and use a visual fraction model to show the quotient Save today on Math U See. Multi-Sensory homeschool math curriculum. Low price The Dividing unit fractions by whole numbers exercise shows up in the 5th grade (U.S.) Math Mission, Arithmetic essentials Math Mission, Pre-algebra Math Mission and Mathematics I Math Mission.This exercise practices dividing a fraction by a whole number. Types of Problems. There is one type of problem in this exercise: Divide the numbers: This problem provides a division problem with a. Lesson 13: Divide Whole Numbers by Unit Fractions. Standards: NC.5.NF.3 Use fractions to model and solve division problems. Interpret a fraction as an equal sharing context, where a quantity is divided into equal parts. Model and interpret a fraction as the division of the numerator by the denominator. Solve one-step word problems involving the.
### Dividing Unit Fractions by Whole Numbers: 5
The Dividing whole numbers by unit fractions exercise appears under the 5th grade (U.S.) Math Mission, 6th grade (U.S.) Math Mission, Arithmetic essentials Math Mission, Pre-algebra Math Mission and Mathematics I Math Mission. This exercise practices dividing whole numbers by fractions. There is one type of problem in this exercise: Divide the numbers: This problem provides a division problem. Dividing whole numbers and unit fractions 6.NS.A.1 - Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions. To link to this Dividing whole numbers and unit fractions page, copy the following code to. First, let's review dividing whole numbers by whole numbers. (We will soon see that dividing whole numbers by unit fractions has the same meaning.) $$6 \div 3$$ means: How many groups of 3 are in 6. As you can see, there are 2 groups of 3 in 6 The division involving a whole number and a fraction is done as follows. Rules of division. The whole number, at first, is written as a fraction. The division then becomes division of two fractions. Dividing by a number is same as multiplying with its reciprocal
Dividing Whole Number By Unit Fractions. Dividing Whole Number By Unit Fractions - Displaying top 8 worksheets found for this concept.. Some of the worksheets for this concept are Dividing fractions t1s1, Grade 5 fractions work, Fraction word problems unit fraction, Dividing a fraction by a whole number, Division division of unit fractions by whole numbers, Eureka lesson for 6th grade unit one. Welcome to the Multiply and Divide Fractions section at Tutorialspoint.com.On this page, you will find worksheets on multiplying and dividing of fractions, product of a unit fraction with a whole number, product of fractions and whole numbers, fraction multiplication, reciprocals of numbers and fractions, modeling of multiplication and division of fractions, words problems, fact families for. Dividing Unit Fractions by whole numbers will result in a Unit Fraction. When a Unit fraction say 1/3 is divided by a whole number 4 then the whole number will get multiplied with the Denominator so the numerator remains the same. 1/3 ÷ 4 = 1/16 which is still a Unit fraction Big Idea: We can draw a picture to divide whole numbers by unit fractions. This lesson introduces students to the concept of dividing a whole number by a unit fraction. This task gives students a situation where they must divide several wholes by fractional pieces. The suggested method of drawing a picture will help them understand that instead of just thinking of it as dividing by a fraction.
When a whole number is divided by a unit fraction the result is each unit being divided into smaller parts. Dividing each unit of 6 into four equal parts results in 24 equal parts of one- fourths, so the qotient will be a whole number greater than the dividend. of the whole is divided into 6 equal parts Divide Whole Numbers by Unit Fraction. - Year 6 Maths. Practise Now. Divide a whole number by a unit fraction. Before using a written method for dividing a whole number by a unit fraction, students must interpret the meaning of this division by visual models. The Divide Whole by a Unit Fraction Worksheet includes problems based on intermediate. Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions. 1 a. Interpret division of a unit fraction by a non-zero whole number , and compute such quotients. For example, create a story context for (1/3) ÷ 4, and use a visual fraction mode Dividing Whole Numbers by Unit Fractions 15 PPT / Google Slides 5F27. On PowerPoint or Google Slides, kids match card images showing fractions, fraction circle models, division expressions, and verbal expressions. By matching cards on 15 digital screens, kids will learn to make sense of math Dividing Whole Numbers by Unit Fractions How can you divide a whole number by a fraction? Reteaching 2. 3. Think: How can I divide two into one-thirds? 2 Two is the sum of one plus one. Each one is the sum of three one-thirds. Count the number of one-thirds. 3 3 6 Check To divide a whole number by a fraction, multiply the whole number by th
### Dividing Unit Fractions by Whole Numbers Math with Mr
Make sense of problems and persevere in solving them. Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions. Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. Complete an exit slip to help you determine instructional next steps Dividing fractions by whole numbers quiz for online practice. Children in 4th, 5th, 6th and 7th grade will find this a useful resource. This could also serve as a math game, math quiz or math test as per user. Teachers and parents can use this to supplement their fraction lessons. Improve mental math skills with out math trivia questions The Divide Whole Numbers By Unit Fractions learning objective — based on CCSS and state standards — delivers improved student engagement and academic performance in your classroom, as demonstrated by research. This learning objective directly references 5.NF.B.7.b as written in the common core national math standards Dividing Whole Numbers and Unit Fractions RC 2 TEKS 5(3)(L) Dividing Whole Numbers and Unit Fractions Content Objective I can solve problems involving division with whole numbers and unit fractions. Language Objective I can represent and explain division with whole numbers and unit fractions. Key Questions 1 Fraction word problems division of whole numbers grade 5 word problems worksheets read and answer each question. These grade 5 word problem worksheets have problems involving the division of whole numbers with fractional results such as sharing 3 pizzas between 10 people. Dividing fractions word problems
5 (Common Core 5.NF.B.7.B: Interpret division of a whole number by a unit fraction, and compute such quotients) 6 (Common Core 6.NS.A.1: Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions, e.g., by using visual fraction models and equations to represent the problem 4 ÷ 1 = 4. 2 ÷ 1 / 2 = 4. This means that one half goes into two 4 times or that 4 halves make 2. Turn the division into a multiplication and flip the fraction you are dividing by. Multiply the whole number by the top of the fraction and then divide by the bottom of the fraction. We keep the whole number 6 the same The Divide Unit Fractions By Whole Numbers learning objective — based on CCSS and state standards — delivers improved student engagement and academic performance in your classroom, as demonstrated by research. This learning objective directly references 5.NF.B.7.a as written in the common core national math standards WHOLE NUMBERS 1. Adding 2. Subtracting 3. Multiplying 4. Dividing 5. Order of Operations FRACTIONS 6. Mixed Numbers 7. Prime Factorization 8. Least Common Multiple 9. Simplifying Fractions 10. Add & Subtract Fractions 11. Add & Subtract Mixed Numbers 12. Multiply Fractions 13. Divide Fractions DECIMALS 14. Decimals to Fractions 15. Add.
MGSE5.NF.7 Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.1 a. Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) ÷ 4, and use a visual fraction model to show the quotient Divide whole numbers by unit fractions. 5th Grade, Math, Common Core: 5.NF.7b Students will learn how to divide whole numbers by fractions using a number line. Related Content. Find the area of a rectangle with fractional. By: Learn Zillion. Find the product of a fraction and a mixed number For example, the approach used for a unit fraction should be the same as a multiple fraction and for a mixed whole number or fraction. The examples below illustrate a uniform teaching approach across both multiplication and division, as well as situations within each operation. Multiplication of fractions Dividing Fractions by Whole Number - Online. This activity requires students to solve division of fractions problems. Show all questions. 1 / 12. 9 ÷ 1/3 =. 14 ÷ 2/3. Check. 3 ÷ 1/2 =. Check Multiply and Divide Unit Fractions and Whole Numbers. Share Share by Jmjones2. G5 Math. Like. Edit Content. Embed. More. Log in required. Theme. Log in required. Options. Leaderboard. Show more Show less . This leaderboard is currently private. Click Share to make it public
Thorough and structured, our printable worksheets on dividing fractions help children stop foraging for resources to practice and learn how to divide fractions. Access our pdf practice resources to give the learners in grade 5, grade 6, and grade 7 a newfound drive to divide fractions with other fractions, whole numbers, and mixed numbers 3. Multiply Whole Numbers 4. Divide Whole Numbers 8 Week(s) Numbers and Operations- Fractions 1. Compare and Order Fractions 2. Add and Subtract Fractions 3. Multiply Fractions 4. Compare Decimals and Fractions 7 Week(s) Geometry 1. Classify Shapes 2. Measurement of Angles 3. Lines of Symmetry 4 Week(s) Measurement and Data 1 Important facts about dividing fractions for grade 6. As a matter of fact, the divide whole numbers by unit fractions exercise is modelled in a very clear and captivating way, essential to enhance a deeper understanding of fractions to our 6 th graders.. Super dividing fraction skills useful in daily life and advanced math concept Dec 5, 2019 - Divide Unit Fractions and Whole Numbers BundleThis bundle includes everything that you will need to teach:•Real world math problems•Dividing unit fractions using models•Dividing a whole number by a unit fraction•Dividing a unit fraction by a whole number•Dividing unit fractions using the standard al.. This article has been viewed 183,990 times. To divide a whole number with a fraction, make the whole number into a fraction by putting it over a denominator of 1. Next, reverse the numerator and denominator of the fraction you're dividing the whole number with. Multiply this new fraction and the whole number
5.NF.2.3 Interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem MCC5.NF.7: Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions. a. Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) ÷ 4, and use a visual fraction model to show the quotient Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions. Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) ÷ 4, and use a visual fraction model to show the quotient Lesson 10: Dividing Whole Numbers by Unit Fractions Main Idea: Students divide whole numbers by fractions. They learn that one way to find the quotient of mixed numbers is to change the calculation to an equivalent one involving multiplication of improper fractions. Standard (s): 5.NF.B.7b Interpret division of a whole number by a unit fraction. Unit Goals. Students extend multiplication and division of whole numbers to multiply fractions by fractions and divide a whole number and a unit fraction. Section A Goals. Recognize that $\frac{a}{b} \times \frac{c}{d}=\frac{a \ \times \ c}{b \ \times \ d}$ and use this generalization to multiply fractions numerically<|endoftext|>
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# Estimation of Sums of Mixed Numbers and Fractions
## Use benchmarks of 0, 1/2 and 1 whole to estimate sums of fractions and mixed numbers.
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Estimation of Sums of Mixed Numbers and Fractions
Remember the blueberries from the Addition of Fractions Concept? Well, there is another blueberry dilemma here.
Teri and Ren finished baking pies and muffins and have moved on to scones.
Teri tells Ren that they need $4 \frac{3}{4}$ cups of blueberries for the scones. Plus she will need an additional $\frac{3}{4}$ of a cup for decorating the topping.
About how many blueberries are need in all?
You can figure out an estimate of this sum. Pay attention to this Concept and you will learn exactly how to do this.
### Guidance
Estimation is a method for finding an approximate solution to a problem.
The sum of 22 and 51 is exactly 73. We can estimate the sum by rounding to the tens place and adding $20 + 50$ . Then we can say the sum of 22 and 51 is “about 70.”
Previously we worked with three benchmarks (0, $\frac{1}{2}$ and 1) to get a sense of the approximate value of different fractions. We can use this same technique to estimate sums of two or more fractions and mixed numbers.
First, we approximate the value of each fraction or mixed number using the benchmarks 0, $\frac{1}{2}$ and 1.
Next, we find the sum of the approximate values.
When you are approximating the value of mixed numbers, first figure out the approximate value of the fraction and then add it to the whole number. For example, the approximate value of $2 \frac{3}{4}$ is 3 because the approximate value of $\frac{3}{4}$ is $1 (2 +1 = 3)$ .
Even when you are asked to find an exact answer, estimation is a useful way to get an idea of a reasonable solution to a problem. Once you have finished solving for an exact answer to a problem, you can check your answer against the estimate. Refer back to the following steps if necessary.
Estimating sums of fractions and mixed numbers:
1. Approximate the value of each of the fractions or mixed numbers by using the benchmarks 0, $\frac{1}{2}$ and 1
2. Add these approximate values to get an estimated sum
Write these steps in your notebook and then continue with the Concept.
Estimate the following sum, $\frac{5}{9} + \frac{1}{77}$
First, we approximate the value of the individual fractions. $\frac{5}{9}$ is approximately $\frac{1}{2}$ and $\frac{1}{77}$ is approximately 0. Now, we rewrite the problem substituting the approximate values: $\frac{1}{2} + \frac{0.5}{9} + \frac{1}{77}$ is about $\frac{1}{2}$ .
Estimate the following sum, $3 \frac{6}{7} + 1 \frac{4}{9}$
First, we approximate the value of each of the mixed numbers. The approximate value of $3 \frac{6}{7}$ is 4 because the approximate value of the fraction $\frac{6}{7}$ is $1 (3 + 1 = 4)$ . The approximate value of $1 \frac{4}{9}$ is $1 \frac{1}{2}$ because the approximate value of $\frac{4}{9}$ is $\frac{1}{2}$ . We rewrite the problem with the approximate values of the mixed numbers and it looks like this: $4 + 1 \frac{1}{2}$ . We estimate that the sum of $3 \frac{6}{7}$ and $1 \frac{4}{9}$ is about $5 \frac{1}{2}$ .
Try a few on your own. Estimate the following sums.
#### Example A
$\frac{6}{7}+\frac{1}{2}$
Solution: $1 + \frac{1}{2} = 1 \frac{1}{2}$
#### Example B
$\frac{29}{30}+7 \frac{8}{10}$
Solution: $8 + 1 = 9$
#### Example C
$1 \frac{1}{2}+ 3 \frac{5}{6}$
Solution: $1 \frac{1}{2} + 1 = 2 \frac{1}{2}$
Now back to the blueberries.
Teri and Ren finished baking pies and muffins and have moved on t0 scones.
Teri tells Ren that they need $4 \frac{3}{4}$ cups of blueberries for the scones. Plus she will need an additional $\frac{3}{4}$ of a cup for decorating the topping.
About how many blueberries are need in all?
To figure out this estimate, we must first use benchmarks to estimate each fraction or mixed number.
$4 \frac{3}{4} = 5$
$\frac{3}{4} = 1$
$5 + 1 = 6$
Teri and Ren will need about 6 cups of blueberries. This is our estimate.
### Guided Practice
Here is one for you to try on your own.
$\frac{18}{20} + 5 \frac{9}{10}$
First, we have to estimate each fraction or mixed number by using the common benchmarks.
$\frac{18}{20} = 1$
$5 \frac{9}{10} = 6$
Here is our solution.
$1 + 6 = 7$
This is our estimate of the sum.
### Explore More
Directions: Estimate the sums.
1. $\frac{1}{29} + \frac{4}{5}$
2. $\frac{9}{11} + \frac{4}{10}$
3. $\frac{2}{5} + \frac{12}{13}$
4. $\frac{2}{71} + \frac{1}{29}$
5. $\frac{1}{29} + \frac{4}{5}$
6. $\frac{3}{20} + \frac{14}{15}$
7. $\frac{6}{7} + \frac{1}{5}$
8. $\frac{9}{18} + \frac{5}{6}$
9. $\frac{12}{13} + \frac{1}{25}$
10. $\frac{7}{9} + \frac{1}{30}$
Directions: Estimate the sums.
11. $3 \frac{6}{7}+ 2 \frac{10}{11}$
12. $8 \frac{1}{12} + 6 \frac{3}{7}$
13. $2 \frac{9}{10} + 3 \frac{1}{17}$
14. $1 \frac{2}{12} + \frac{44}{46}$
15. $8 \frac{1}{29} + 10 \frac{4}{5}$
### Vocabulary Language: English
Mixed Number
Mixed Number
A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$.<|endoftext|>
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Many parents are pulling their kids out of athletic programs because they worry about how the increased level of competition in today’s sports makes them more prone to injuries, particularly concussions.
Concussions are mild traumatic brain injuries that affect proper brain function. It’s a blow to the head that typically causes the trauma. However, violent shaking of the head and upper body can also cause concussions. In fact, the word concussion comes from the Latin word concutere which means “to shake violently.”
However, according to Medscape, only 40% of youth concussions are caused by contact sports. The reality is that as kids are so active, they can get a concussion from everyday play or accidents. This doesn’t change the fact that many concussions are from youth sports because of the physicality involved.
The important thing is to educate both athletes and parents about recognizing the signs of a concussion so that they can be reported and treated sooner.
Some people who suffer a concussion may lose consciousness; however, the majority do not. The best person to recognize if they have a concussion is the one who suffered the blow to the head which is why it is vital to educate young athletes to know the signs. Left untreated, those who suffer from a concussion can experience one or a combination of symptoms.
Here are the most common signs and symptoms of a concussion. Some will be noticeable immediately after the blow to the head; however, some symptoms can last for days if left untreated:
- Temporary loss of consciousness
- Seeing “stars”
- Persistent neck pain
- Ringing in the ears
- Changes in behavior and personality
- Slurred speech
- Difficulty remembering
- Problems concentrating
- Blurred vision
- Difficulty sleeping
- Delayed response to questions
- Sensitivity to light and noise
There are also three different grades for concussions:
Grade 1 – Mild: With grade 1 concussions, there is no loss of consciousness, and the symptoms typically last for less than 15 minutes.
Grade 2 – Moderate: With grade 2 concussions, there is no loss of consciousness yet the symptoms last longer than 15 minutes.
Grade 3 – Severe: In grade 3 concussions, the person loses consciousness for anywhere between a few seconds to just under a minute.
Some head injuries are more severe than others. Concussions are typically not life-threatening; however, leaving them unchecked can lead to something more serious. If you suspect that your child has suffered a concussion, continually monitor them for any signs. Explain what the symptoms are so they can help monitor themselves. Observe if there are changes in their behavior and if there is repeated vomiting or if their headache worsens.
Most importantly, never return an athlete who may have suffered a traumatic blow to the head back on the field or court. Before they return to play, they should be medically evaluated by a health professional and cleared.<|endoftext|>
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Learning to play musical instruments not only helps children increase their ability to sense their emotions, but also helps develop learning and social skills. Not only that, instrumental classes also help children become familiar with Physics.
For example, strumming a guitar string or violin also makes it easier for children to visualize the phenomenon of resonance and stop when learning Physics. In addition, wireless instruments – such as drums or flutes – also help children explore the scientific principles mentioned above.
Learning to play musical instruments not only helps your child both ambidextrous, perform difficult coordination movements. Especially when participating in physical activities, children will acquire coordination skills for body parts, estimation precise time and good reflexes in performing movements, such as dancing, dancing, passing balls, …
Some music schools offer group music lessons, teachers will teach how to play in groups and assign a exercise to be done together. At the same time, each student in each group is assigned a different task to contribute to that exercise.
So whatever you do, children need to pay attention to contributing to the common goal. It is a microcosm for real interactions in society. We always need to practice team interaction skills and problem solving.
Play musical instruments to teach children how to control their emotions. Private lessons and practice time at home require a high concentration of children every 10 minutes. For example, the violin, before pulling the strings, the child must learn to hold the strings, hold the strings, and how to stand properly.
Besides, when defining a specific goal such as remembering how to play a complete piece of music or performing with an orchestra on stage, I will patiently practice many hours, months, even years. It is this that helps children get perseverance and discipline with themselves.
Moreover, if you are a member of the orchestra, children will learn to wait patiently for their turn and how to coordinate with everyone. In the meantime it is time for me to listen to the play of my classmates. Since then, children have been able to practice respect for others, how to keep quiet and focused for a certain period of time.<|endoftext|>
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# Turunan Fungsi Eksponen
x
= e sin 2 x 3
1. y
dy
2. y
dy
dx
e x sin 2 x 3 d
dx
sin 2 x 3.e x
x
= e sin 2 x 3 2 cos 2 x 3
= e 5 x 2 2 x 1
= e 5 x 2 x 1 dy
5x 2 2 x 1
dx
dx
2
= (10x + 2) e 5 x 2 x 1
2
## Turunan Fungsi Trigonometri
= sin (2x2+3) + cos 3x
3. y
dy/dx = cos (2x2+3) d/dx (2x2+3) + (sin 3x) d/dx (3x)
= 4x cos (2x2+3) 3sin 3x
2
4. Y = sec 5x
misalkan u = sec 5x Y = u2
dy/du = 2 u ; v = 5x
du/dx = sec v tg v dv/dx = 5
dy/dx = dy/du . du/dv . dv/dx
= 10 sec25x tg 5x
5. y
dy
= log 5 sin 2x
## dx = log 5e . 2 sin x. cos x
sin 2x
= log 5e . 2 cos x
sin x
= 2 ctg x log 5e
= In 3x3 + In 2 x
1
1
1
dy
9x 2
.
3
2x
x
dx = 3x
= 4/x
6. y
7. y
dy
dx
x
= 2log 5
1
= 5
x2 2
d x2 2
5
In 2 dx
5 . In 5. 2x
In 2
5x
x2 2
= 2x In 5
In 2
Invers Fungsi
8. Diketahui f ( x) x 1
x2
a. Periksa apakah f mempunyai invers
Jawab :
a.
3
1.( x 2) 1.( x 1)
0, x Df
2
( x 2) 2
(
x
2
)
Karena f selalu naik(monoton murni) maka f mempunyai invers
f ' ( x)
b. Misal, y
x 1
x 2
2y 1
xy 2 y
xx
xy
12 y 1 x
y 1
f
( y)
## Turunan Fungsi Invers
9. Diketahui,
Jawab :
2y 1
y 1
f ( x ) x 5 2 x 1 , tentukan : ( f
( x)
2x 1
x 1
)' (4)
## ,y=4 jika hanya jika x=1
f ' ( x) 5 x 4 2
1
1
( f 1 )' (4)
f ' (1) 7
Fungsi Logaritma Asli
## 10. Diberikan f ( x ) ln(sin(4 x 2))
Maka
Jika
f ' ( x)
1
Dx (sin(4 x 2)) 4 cos(4 x 2)
sin( 4 x 2)
y ln | x | , x 0
ln x , x 0
ln( x ) , x 0
y ln x y '
y ln( x) y '
d
1
(ln | x |)
, x 0.
dx
x
Dari Sini Diperoleh :
1
dx ln | x | C
x
11. 4 x 2
0 x 3 2dx
u x 3 2 du 3 x 2 dx
x2
x 2 du 1 1
1
dx
x3 2
u 3x 2 3 u du 3 ln | u | c
4
4
x2
1
3
1
1
0 x 3 2dx 31 ln
ln || xx 3 22 || 0c (ln 66 ln 2) ln 33.
3
3
3
1
x
1
1
x
x
2 x 1
2sin 2 x
12. f ( x ) 3
## f ' ( x ) 2.32 x 1 ln 3 2.2sin 2 x cos 2 x ln 2
13.
x2
.xdx
u x 2 du 2 xdx dx 21x du
2
du 1 4 u
4x
4 .xdx 4 2 2 ln 4 C 2 ln 4 C
Fungsi Logaritma Umum
x2
3
2
14. f ( x ) log( x 1)
f ' ( x)
15.
ln( x 2 1)
ln 3
2x
1
x 1 ln 3
2
ln( xx 11 )
x 1
)
x 1
ln 4
1 1
x 1
f ' ( x)
Dx
(
)
ln 4 xx 11
x 1
f ( x) 4 log(
1 x 1 x 1 ( x 1)
ln 4 x 1 ( x 1) 2
1
2
ln 4 ( x 1)( x 1)
### Dapatkan aplikasi gratis kami
Hak cipta © 2021 Scribd Inc.<|endoftext|>
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# Video: AQA GCSE Mathematics Higher Tier Pack 4 • Paper 3 • Question 13
A group of 36 students timed how long they can hold their breath while swimming underwater. The histogram shows information about the times achieved. Work out an estimate of the interquartile range. Show all your working.
06:22
### Video Transcript
A group of 36 students timed how long they can hold their breath while swimming underwater. The histogram shows information about the times achieved. Work out an estimate of the interquartile range. Show all your working.
The difference between a histogram and a frequency diagram or bar chart is it on the 𝑦-axis we have the frequency density instead of the frequency. This means that we can calculate the frequency or number of students in this case in each bar by working out its area. The area of any rectangle is calculated by multiplying its length by its width or its base by its height.
This means that the area of the first bar can be calculated by multiplying 25 by 0.2. This is equal to five. Therefore, there are five students in the first bar. The area of the second bar can be calculated by multiplying 10 by one. This is equal to 10. Therefore, there are 10 students in the second bar. The area of the third bar is equal to 10 multiplied by 1.5. This is equal to 15. Therefore, there are 15 students represented by the third bar. Finally, the area of the fourth bar is 15 multiplied by 0.4. This is equal to six. Therefore, there are six students in the fourth bar.
As these four numbers represent the number of students, they must all be integers. They must also have a sum of 36 as there were 36 students altogether. Five plus 10 plus 15 plus six is equal to 36. We could display this information in a grouped frequency table as shown. There were five students who held their breath for between zero and 25 seconds, 10 students between 25 and 35 seconds, 15 between 35 and 45 seconds, and six students between 45 and 60 seconds.
We were asked to estimate the interquartile range. This is equal to the upper quartile minus the lower quartile. The upper quartile is equal to the 75th percentile and the lower quartile is equal to the 25th percentile. This is because 25 percent equals one-quarter and 75 percent is equal to three-quarters. One-quarter of 36 is equal to nine. This means that when the times are listed from smallest to largest, the ninth person will be the lower quartile. Three-quarters of 36 is equal to 27. This means that the upper quartile will be the 27th person.
There were five students who held their breath for less than 25 seconds. This means there were 15 students that held their breath for less than 35 seconds as five plus 10 is equal to 15. The lower quartile or ninth person must therefore be in the group between 25 and 35 seconds. We can go one stage further and say that they would be the fourth person in this group as five plus four is equal to nine.
To calculate the lower quartile, we need to be four tenths of the way into this group of 10 people. As the lower bound of this group was 25, the lower quartile can be calculated by adding four tenths of 10 to 25. Four tenths of 10 is equal to four. Therefore, the lower quartile is 25 plus four. This is equal to 29. We could also have worked this out from the graph. As there are 10 squares between 25 and 35, we need to be four squares along from 25. This gives us a lower quartile value of 29.
The upper quartile was the 27th person. As there were 30 students who took less than 45 seconds, we know that the upper quartile is in the group 35 to 45. Once again, we can calculate the position within this group. The upper quartile will be the 12th person in this group as five plus 10 plus 12 is equal to 27. We need to be twelve fifteenths into this group. We need to calculate twelve fifteenths of 10 as the class width was 10 between 35 and 45 seconds.
The lower limit of this group was 35. Therefore, we need to add twelve fifteenths of 10 to 35. Twelve fifteenths can be simplified to four-fifths. And four fifths of 10 is equal to eight. Adding eight to 35 gives us 43. This means that our upper quartile is 43 seconds. Once again, we could have used the histogram to work this out. As four-fifths of 10 was equal to eight, we need to be eight squares along from 35. This confirms that the upper quartile is 43.
The interquartile range is, therefore, 43 minus 29. This is equal to 14. Therefore, the interquartile range of times is 14 seconds.<|endoftext|>
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As neighboring countries, the United States and Mexico have a long history of bilateral agreements. In this installment of “Time & Place,” historian Alicia Barber looks at one cooperative program introduced during World War II to benefit residents on both sides of the border.
By the summer of 1942, the United States was fully engaged in the Second World War, and millions of Americans were involved in some form of military service. With so many people and resources dedicated to the war effort, the home front was beginning to feel the impact, and certain industries were being hit especially hard.
During the war, Armando Martini, known as “Barney,” was a teenager living on his family’s ranch at Vista, just outside of Sparks, Nevada. Interviewed in 2005, he explained how the massive deployment of men overseas had begun to affect the region’s agricultural community.
“A lot of the farm boys that were working the ranches were in the service,” Martini said. “It was during the war, and we were raising potatoes and onions and grain, and you needed help to harvest all of that.”
The labor shortage was serious enough that the federal government decided to step in. At first, they tried to recruit workers from cities to help out the farms, but that didn’t pan out. After intensive negotiations with Mexico, the U.S. government introduced the Mexican Farm Labor Agreement as an emergency measure to bring in temporary contract workers from south of the border.
It was informally known as the Bracero Program, using a Spanish term for a manual laborer. The workers’ contracts lasted for a number of months and were renewable. Five hundred workers were sent from Mexico to the Central Valley of California in September 1942, and the following summer, several hundred arrived in western Nevada. The effort was coordinated on the state level by the University of Nevada’s agricultural extension service.
“The Farm Bureau used to bring in quite a few of the Mexican workers, and they’d assign so many to each ranch that needed help, and we’d have to cook and supply living quarters for them,” Martini said. “We always had four or five men.”
In addition to providing housing, the host ranchers were also responsible for paying their guest workers the region’s prevailing wage for the work they did. That made the contracts especially appealing to the Mexican citizens who signed up, and to Mexican government officials, who hoped that the money the workers earned could help strengthen their own country’s economy when they returned.
It was for many on both sides, their first time interacting with someone of each other’s nationality. But as Martini explained, they all made an effort to overcome the language barrier:
“We gradually learned. They learned some of ours [language], we learned theirs [Spanish], and we made out. But it was a little bit confusing at first. Sometimes we’d have to call the Farm Bureau office to send an interpreter to find out what was going on.”
In addition to farm labor, the program also brought 2,000 Mexican citizens to work on Nevada railroads. When the war ended in 1945, so did the domestic labor shortage, and the wartime contracts were not renewed. But the Bracero program continued in some form until 1964, employing approximately 4.6 million Mexican citizens and filling a critical need for some major American industries.
Historian Alicia Barber is the editor of the smart phone app and website Reno Historical.org. Oral history clips for this segment were provided by the Special Collections Department of the University of Nevada, Reno Libraries. The full transcript for Armando Martini’s interview can be found at the Unviersity of Nevada, Reno Oral History Program.<|endoftext|>
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Congratulations - you have completed . You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%%
Your answers are highlighted below.
Which woman challenged the authority of the Puritan clergy and was a key figure in the development of religious freedom in the American colonies?
Question 1 Explanation:
Born in England in 1591, Anne Hutchinson emigrated to Boston in 1634. Her adherence to the Free Grace theology was at odds with the established Puritan clergy in the Boston area, and she was tried for slander. She was excommunicated and settled in New Hampshire. She, and most of her family, died during an American Indian attack in 1643.
Who was the first female writer in the American colonies to be published?
Question 2 Explanation:
Anne Bradstreet was born in England in 1612 and emigrated to Massachusetts in 1630. In 1650, she published a book of poetry titled The Tenth Muse Lately Sprung up in America.
Which wife of a Founding Father wrote a letter to her husband in 1776, urging him to “...remember the ladies, and be more generous and favorable to them than your ancestors”?
Question 3 Explanation:
Abigail Smith was born in Massachusetts in 1744 and married her third cousin John Adams in 1764. She advocated for better educational opportunities for women and demanded more property rights for married women. She served as first lady between 1797 and 1801.
Who is credited with making the first American flag in 1776?
Question 4 Explanation:
Betsy Ross was born in Philadelphia in 1752, and her husband supported the Patriot cause. According to family lore, Ross created the first American flag in 1776. However, there is no archival evidence supporting the claim.
Which woman is credited with servicing a Patriot cannon during the Battle of Monmouth in 1778 during the American Revolution?
Question 5 Explanation:
Molly Pitcher is the nickname given to Mary Ludwig Hays resulting from her participation in the American Revolution, when she carried pitchers of water to help American soldiers. She appeared on a postage stamp in 1928 in honor of the 150th anniversary of the Battle of Monmouth.
Which woman was an advocate for reform in prisons and mental institutions in the 1800s?
Question 6 Explanation:
Born in Maine in 1802, Dorothea Dix investigated abuse in mental institutions and prisons. She published reports that led to reforms in New Jersey and Massachusetts, and during the Civil War she supervised the nursing service for the Union army.
Which Quaker woman published Letters on the Equality of the Sexes and the Condition of Women in 1837?
Question 7 Explanation:
Born in South Carolina in 1792, Sarah Moore Grimké was an abolitionist and a member of the women's suffrage movement. She became a Quaker in 1821. She influenced many other activists, including Lucretia Mott and Elizabeth Cady Stanton, before she died in 1873.
Who organized the 1848 women’s rights convention in Seneca Falls, New York?
Lucretia Mott & Elizabeth Cady Stanton
Elizabeth Cady Stanton
Question 8 Explanation:
Mott and Stanton were both abolitionists who met at an anti-slavery convention in London in 1840. They organized the 1848 Seneca Falls Convention, the first women's rights convention.
Who wrote Uncle Tom’s Cabin?
Harriet Beecher Stowe
Question 9 Explanation:
Born in Connecticut in 1811, Harriet Beecher Stowe later moved to Cincinnati. She married an abolitionist and published Uncle Tom’s Cabin in serial form in 1851. In its first year, it sold more than 300,000 copies. The book energized anti-slavery forces in the North, while provoking widespread anger in the South.
Clara Barton served as a nurse during what conflict?
World War I
World War II
The Civil War
Question 10 Explanation:
Born in Massachusetts in 1821, Barton became a teacher and nurse. During the Civil War, she tended to wounded soldiers during several important battles, including Antietam and Fredericksburg. She is best known as the founder and first president of the American Red Cross.
Which woman went on trial for attempting to vote illegally in the US presidential election of 1872?
Susan B. Anthony
Question 11 Explanation:
Born in Massachusetts in 1820, Susan B. Anthony was a social reformer and feminist activist who played a pivotal role in the women's suffrage movement. She was arrested when she tried to vote in the 1872 presidential election and was convicted. She was ordered to pay a $100 fine, but refused to do so. She died in 1906. In 1979, the United States Mint began issuing the Susan B. Anthony dollar coin.
Which woman was a key figure in the American temperance movement?
Question 12 Explanation:
The temperance movement was a movement to curb the consumption of alcohol in late 18th and early 19th century America. Frances Willard was born in New York in 1839, and in 1874 she helped found the Women’s Christian Temperance Union. She argued that alcohol destroyed families and that liquor should be banned.
Which woman was an anarchist who also advocated for free love and atheism in the United States during the early 1900s?
Question 13 Explanation:
Born in Lithuania in 1869, Emma Goldman emigrated to the United States in 1885. She became an anarchist after the 1886 Haymarket Riot. During World War I she was arrested for opposing the draft and was eventually deported to Russia.
Which industrial disaster occurred in the early 1900s and killed more than 100 young women?
Port Chicago disaster
Three Mile Island accident
Great Molasses Flood
Triangle Shirtwaist fire
Question 14 Explanation:
The Triangle Shirtwaist Factory fire occurred on the 8th, 9th, and 10th floors of a factory in Greenwich Village. Most of the workers were young immigrant women, and 123 women died during the inferno. Many doors were locked to prevent theft by employees, and the disaster led to new regulations and better working conditions.
In 1916, which woman opened the first birth control clinic in the United States?
Question 15 Explanation:
Margaret Sanger was born in New York in 1879. She was a birth control activist, sex educator, writer, and nurse. She opened the United States’ first birth control clinic in Brooklyn. Birth control was illegal, and nine days after the clinic opened she was arrested. Publicity surrounding her arrest and conviction sparked birth control activism across the country. In 1946, Sanger helped found the International Committee on Planned Parenthood.
When did American women gain the right to vote?
1791 with the Bill of Rights
1789 with the US Constitution
1919 with the 18th Amendment
1920 with the 19th Amendment
Question 16 Explanation:
A constitutional amendment that granted women’s suffrage was proposed in the US House of Representatives in January 1918. President Woodrow Wilson supported the amendment, and it finally passed the US Senate in June 1919. Tennessee was the 36th state to ratify the amendment, which provided the final ratification necessary to add the 19th Amendment to the Constitution in August 1920.
“Flappers” are most closely associated with what time period?
Question 17 Explanation:
“Flappers” flaunted social norms during the 1920s. They wore short skirts, bobbed their hair, listened to jazz, smoked cigarettes, drank alcohol, and drove automobiles. The flapper style disappeared after the Stock Market Crash of 1929.
Who was the first woman to win a Pulitzer Prize?
Question 18 Explanation:
Edith Wharton was born in New York in 1862, and attempted her first novel at age 11. She was a prolific writer who authored 15 novels, 7 novellas, and 85 short stories. She won the Pulitzer Prize for fiction in 1921 with her novel Age of Innocence.
Which American woman was a prominent social and political activist who won the Nobel Peace Prize in 1931?
Question 19 Explanation:
Addams is known as the "mother" of social work. She was a settlement activist, social worker, public philosopher, sociologist, author, and a leading advocate of women's suffrage and world peace. In 1889, Jane Addams co-founded Chicago's Hull House, a settlement house that provided educational and recreational facilities for European immigrant women and children.
Who was the first woman appointed to a presidential cabinet?
Ruth Bader Ginsburg
Question 20 Explanation:
Born in Massachusetts in 1880, Frances Perkins was a sociologist and workers-rights advocate. Franklin D. Roosevelt appointed her Secretary of Labor in 1933, and she served until 1945.
Rosie the Riveter played a symbolic role for women during which event?
World War II
The Great Depression
World War I
Question 21 Explanation:
Rosie the Riveter became an icon that represented women’s contributions to the home front during World War II. Three million women entered the workforce and many of them worked in factories that produced war material. Rose Will Monroe, a riveter at a B-24 factory in Ypsilanti, Michigan, is thought to be the inspiration for Rosie the Riveter.
Which woman helped draft the Universal Declaration of Human Rights in 1948?
Olympe de Gouges
Question 22 Explanation:
Born in New York in 1884, Eleanor married her fifth cousin Franklin Roosevelt in 1905. She maintained a highly visible role as First Lady between 1933 and 1945, and became the first chairperson of the UN Commission on Human Rights, influencing the adoption of the Universal Declaration of Human Rights.
Who wrote The Second Sex?
Olympe de Gouges
Simone de Beauvoir
Question 23 Explanation:
Simone de Beauvoir was a French writer. In 1949 she published The Second Sex, which examines the treatment of women throughout history. The book was translated into English and played an important role in the development of feminism in America.
Who wrote Silent Spring?
Question 24 Explanation:
Born in Pennsylvania in 1907, Rachel Carson was a marine biologist, a writer, and an environmentalist. In 1962, she published Silent Spring, which examined the effects of DDT on the environment. Her book led to a ban on the use of DDT for agricultural purposes in 1972.
Who wrote The Feminine Mystique?
Question 25 Explanation:
Betty Friedan was born in Illinois in 1921. She studied psychology at Smith College and Berkeley, and later worked as a journalist. She published The Feminine Mystique in 1963. The book examined why many housewives felt dissatisfied with their lives. It is widely credited with sparking second-wave feminism in the United States. She was also the co-founder and first president of the National Organization for Women (NOW).
What landmark piece of legislation prohibited employment discrimination against women?
The Equal Pay Act of 1963
The Civil Rights Act of 1964
The 19th Amendment
The Equal Rights Amendment
Question 26 Explanation:
Originally, the Civil Rights Act of 1964 only outlawed discrimination based on race, color, religion, or national origin, but Virginia Democratic Congressman Howard Smith offered an amendment that added sex as a protected category. His amendment passed with the votes of Republicans and Southern Democrats.
Who was the first African American woman elected to Congress?
Question 27 Explanation:
Born in New York in 1924, Shirley Chisholm became a teacher and later served in the New York State Assembly. She was elected to Congress in 1968 and served until 1983. In 1972, she made an unsuccessful bid for the Democratic presidential nomination.
Which Supreme Court decision lifted restrictions on access to abortion services?
Roe v. Wade
Plessy v. Ferguson
Loving v. Virginia
Dred Scott v. Sandford
Question 28 Explanation:
Norma McCorvey, under the alias of Jane Roe, filed suit in the US District Court to overturn a Texas law that restricted access to abortion. The case went to the US Supreme Court in 1971, and the justices issued a 7–2 ruling that the Texas law violated an implied right to privacy in the 14th Amendment’s due process clause.
Which of the following banned gender discrimination in education?
Equal Rights Amendment
Civil Rights Act of 1964
Question 29 Explanation:
Title IX was a portion of the 1972 Education Amendments, which banned discrimination against women in education. It greatly expanded women’s participation in sports at the high school and college level.
The “Battle of the Sexes” between athletes Billie Jean King and Bobby Riggs in 1973 occurred in what sport?
Question 30 Explanation:
Bobby Riggs had been a tennis star during the 1940s, and he came out of retirement in 1973 to play against Billie Jean King. The match was held at the Houston Astrodome, and had a worldwide audience of 90 million viewers. Riggs initially took a lead, but King eventually won.
Who was the first woman to serve on the US Supreme Court?
Sandra Day O’Connor
Ruth Bader Ginsburg
Question 31 Explanation:
Sandra Day O’Connor was born in El Paso in 1930 and graduated from Stanford Law School. She served as the Attorney General, a legislator, and a judge in Arizona before President Ronald Reagan appointed her to the Supreme Court in 1981. She retired in 2005.
Who was the first American woman to go into space?
Question 32 Explanation:
Sally Ride was born in Los Angeles in 1951 and earned a Ph.D. in physics from Stanford. She joined NASA in 1978 and became the first American woman to go into space as a crew member on the space shuttle Challenger in 1983. Ride remains the youngest American astronaut to have traveled to space, having done so at the age of 32.
Who was the first woman to run as the vice-presidential nominee for a major political party?
Question 33 Explanation:
Geraldine Ferraro was born in New York in 1935 and worked as a public school teacher before training as a lawyer. She served in the US House of Representatives from 1979 to 1985. In 1984, Walter Mondale, the Democratic nominee for president, chose Ferraro as his running mate. They lost the election to Ronald Reagan. She ran unsuccessfully for the US Senate in 1992, but later served as Ambassador to the UN Commission on Human Rights.
Who was the first woman to serve as Speaker of the House in Congress?
Question 34 Explanation:
Born in Maryland in 1940, Nancy Pelosi later moved to San Francisco and ran for Congress in 1987. Before being elected Speaker of the House, she served as the House minority whip. She served as Speaker from 2007 to 2011.
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 34 questions to complete.<|endoftext|>
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# Find a particular solution satisfying the given condition $(1+x^2)\large\frac{dy}{dx}$$+2xy=\large\frac{1}{1+x^2}$$;y=0\;when\;x=1$
$\begin{array}{1 1}(A)\;y(1+x^2) = \tan^{-1} x -\large\frac{ \pi}{2} \\(B)\;y(1+x^2) = \tan^{-1} x -\large\frac{ \pi}{4} \\(C)\;y(1+x^2) = \tan^{-1} x -\large\frac{ 2\pi}{3} \\ (D)\;y(1+x^2) = \tan^{-1} x +\large\frac{ \pi}{4} \end{array}$
Toolbox:
• To solve the first order linear differential equation of the form $\large\frac{dy}{dx}$$+ Py = Q • (i) Write the given equation in the form of \large\frac{dy}{dx}$$ + Py = Q$
• (ii) Find the integrating factor (I.F) = $e^{\int Pdx}$.
• (iii) Write the solution as y(I.F) = $\int\:Q(I.F) dx + C$
Step 1:
Using the information in the tool box, let us rewrite the equation
dividing throughout by $(1+x^2)$ we get,
$\large\frac{dy}{dx } + \frac{2xy}{(1+x^2)}$$= 1 Here P =\large\frac{ 2x}{(1+x^2)} and Q = \large\frac{1}{(1+x^2)^2} Let us find the integrating factor(I.F) \int Pdx = \int\large\frac{ 2x}{(1+x^2)}$$= \log(1+x^2)$
Hence $I.F = e^{\log(1+x^2)} = (1+x^2)$
Step 2:
Hence the required solution is $y.(1+x^2) =\int\big[\large\frac{1}{(1+x^2)^2}$$. (1+x^2)\big] \:dx+ C \int\large\frac{ 1}{(1+x^2)}$$dx = \tan^{-1}x + C$
Hence the required solution is $y.(1+x^2) =\tan^{-1}x + C$
Step 3:
To evaluate the value of C, let us substitute the given values of $x = 1$ and $y =0$
$0 = \tan^{-1}(1) + C$
$\tan^{-1}(1)=\large\frac{\pi}{4}$
Therefore $C = \large\frac{- \pi}{4}$
substituting this we get,
$y(1+x^2) = \tan^{-1} x -\large\frac{ \pi}{4}$
This is the required solution.
edited Feb 11, 2014<|endoftext|>
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WASHINGTON, D.C., March 28, 2017 -- When tilling soil, the blade of the tillage tool cuts through the dirt, loosening it up in preparation for seeding. The dirt granules are pushed aside in a way that certainly looks random -- but might not be.
Now, from studying soil tilling, researchers have found a way to distinguish whether such a process is truly random, or only appears so and is actually predictable and deterministic -- which can lead to a deeper understanding and the ability to precisely control the process.
This mathematical method, the researchers say, could be useful in a range of applications beyond soil, from drying and sorting grains to seismology. They describe the analysis this week in the journal Chaos, from AIP Publishing.
During tilling, soil particles stick to and rub against one another. These interactions subject the granules to forces that oscillate in strength. Under some conditions, these fluctuations may be deterministic, which means calculations can predict how the resultant forces will behave at a later time.
For example, these fluctuations can be similar to the oscillatory behavior of a box sitting on a conveyor belt while being tethered to a spring. As the conveyor belt moves, the box pulls away, stretching the spring. When the spring force becomes greater than the friction force holding the box in place, the box slides back, only to be pulled again by the conveyor belt. This back-and-forth motion -- called stick-slip dynamics -- also appears in the interactions between grains and other particles, the strike-slip faults that trigger earthquakes, and even in the friction of a gecko's feet.
Stick-slip behavior is deterministic, but it also can be chaotic in certain instances. A slight change in the initial conditions of the system can lead to wildly different outcomes.
But such chaotic behavior often appears random, and knowing whether it is truly random is essential for understanding the system.
"If the system was deterministic chaos, we can expect to start to develop a method to predict and control the behavior of the system with higher accuracy," said Kenshi Sakai of the Tokyo University of Agriculture and Technology in Japan. "However, if the system was stochastic, the only thing we can do is to estimate the probability of the occurrence of the behavior."
There are mathematical tools that distinguish deterministic chaos, like in the stick-slip model, from pure randomness. But, Sakai said, those tools don't work as well when statistical noise contaminates the data -- such as the noise inherent to the messy reality of soil tilling.
So the researchers sought a new technique, which they tested on a series of soil-tilling experiments. In the experiment, they used a device developed at the University of California, Davis that measures forces as it cuts through the soil. They made these measurements on Yolo loam soil under four conditions: tilled and dry, tilled and wet, untilled and dry, and untilled and wet.
By analyzing the data, the researchers calculated a parameter called the normalized deterministic nonlinear prediction (NDNP). It turns out that when this number is less than one, the system is random. But if it's greater than one, it's deterministic. Only the case in which the soil was tilled and dry was deterministic. The tilled and wet case was uncertain, with a NDNP value of one.
These results show that NDNP is a relatively simple way to determine whether a system is deterministic, Sakai said. One important use would be for analyzing seismic data and, perhaps, to better anticipate future earthquakes.
The article, "Chaos emerging in soil failure patterns observed during tillage: Normalized deterministic nonlinear prediction (NDNP) and its application," is authored by Kenshi Sakai, Shrinivasa K. Upadhyaya, Pedro Andrade-Sanchez and Nina Sviridova. The article will appear in the journal Chaos March 28, 2017 (DOI: 10.1063/1.4978027). After that date, it can be accessed at http://aip.
ABOUT THE JOURNAL
Chaos is devoted to increasing the understanding of nonlinear phenomena in all disciplines and describing their manifestations in a manner comprehensible to researchers from a broad spectrum of disciplines. See http://chaos.<|endoftext|>
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This New Zealand-based citizen science project aims to collect data about the types and numbers of common garden birds in our own backyard. This is done once annually during a particular window of time (29 June to 7 July 2019) and the results contribute to New Zealand’s knowledge and monitoring of garden bird species and the health of the environment we live in.
Participation in the project requires you to be able to identify types of birds, with plenty of resources to support this. Opportunities for discussions around critiquing evidence (How many birds have we seen?, How do we know that we didn’t count one bird twice?) are huge, as is the potential for practising interpreting data from previous years’ surveys.
Reach: Regional, national
Nature of science focus: Online citizen science (OCS) projects can be used to develop any of the Nature of Science (NoS) substrands. What is important is to identify aspects of NoS that your students need to be better at or understand more fully. Then frame your unit to be very clear about these things when you do them.
Science capability focus: Gather and interpret data, Critique evidence, Interpret representations
Science focus: Ecology - bird identification and species distribution
Some suggested science concepts:
- Birds can be identified by their external features.
- Birds have adaptive features – structural (beaks, feet) or behavioural (calls, migration) – that enable them to survive.
- Each bird species can be classified as being native, endemic, introduced, and/or endangered.
- Bird species are adapted to live in their particular habitats.
- Different birds live in different habitats. For example, introduced birds are more common in our towns than native species – their adaptations better enable them to live in close contact with people and mammalian predators.
- Changes in habitat can affect the survival of living organisms in an area, and the relationships between them.
- Populations are living organisms of the same species living in the same area at the same time.
Many concepts could be learned – focusing on a few can often be more powerful. Develop your learning outcomes and success criteria from these concepts as well as the Nature of Science strand and the science capabilities.
Some examples of learning outcomes:
- collect bird observations independently
- discuss the strengths and weaknesses of the data collection method
- identify local birds by their external features
- compare adaptive features and explain their role in a bird’s survival.
About Garden Bird Survey
The NZ Garden Bird Survey is a national project helping our scientists monitor bird populations in gardens, parks and schools over time. Students can learn about birds in their area and participate in a bird count during the national week of data collection. Resources are provided online to assist with this.
Ko te manu i kai ana i te miro nōna te ngahere,Māori proverb
ko te manu i kai ana i te mātauranga nōna te ao.
The bird that consumes the miro berry owns the forest,
the bird that consumes knowledge owns the world.
The New Zealand Garden Bird Survey has been running since 2007, providing extensive data to look back through and consider. While you can’t access your own data or compare data between schools (locally or nationally), this is a direction Manaaki Whenua – Landcare Research would like to take the website in the future. In the meantime, there are fabulously displayed results available for rich class discussions and developing the science capability ‘Interpret representations’.
Several resources are provided to support teachers throughout the survey, such as a series of short videos including a 2-minute clip likening gardens to layer cakes and showing that different birds can be found in different layers depending on what they like to eat. Discussions around scale and the comparative size and weight of birds is supported by another short video introducing the Chocolate Fish Index. A LEARNZ Virtual Field Trip and DOC’s ‘experiencing birds in green spaces’ resource about the New Zealand Garden Bird Survey are also available. There are also online quizzes for testing your bird identification skills and knowledge including Māori bird names.
Through the website, it is possible to connect with a real scientist, and there are opportunities to share your work, findings and questions through social media.
Nature of science
Using this OCS gives opportunities to discuss how New Zealand scientists might use the collective data from sightings to monitor species over time, including population numbers and distributions. Students can also consider the challenges for scientists in collecting large datasets themselves and appreciate how involving citizen scientists makes the scientists’ findings more valid.
The Hub has an extensive range of resources featuring birds including from Native bird adaptations, Birds’ roles in ecosystems and Predation of native birds and articles on conservation of our native species and bird classification. You can also find out more about our native birds such as the kiwi, takahē, New Zealand ducks, penguins, godwits and kererū. For all of our articles and activities, browse through our birds topic. For more on populations, see Population biology. Protecting native birds references the important role of citizen scientists.
Participate in eBird to log bird sighting data year round and compare data from around the world. The iNaturalist online citizen science project uses Seek, a species identification app. You can also participate in The Great Kererū Count, an annual 10-day event in September.
Here are some planning tips for when you intend to use a citizen science project with your students.
LEARNZ field trip on the Garden Bird Survey.
The Ministry of Education’s Building Science Concepts series includes Book 3: Birds: Structure, Function, and Adaptation.<|endoftext|>
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# Video: Finding the Unknown That Makes an Exponential Function Increase on Its Domain
What condition must there be on π§ for π(π₯) = (π§/7)^π₯ to be an increasing function?
02:31
### Video Transcript
What condition must there be on π§ for π of π₯ equals π§ over seven to the power of π₯ to be an increasing function?
Whatβs the definition of an increasing function? A function π of π₯ is said to be an increasing function if π of π is less than or equal to π of π whenever π is less than π. This is the formal definition of an increasing function, but it turns out, like many things, that itβs easier to understand what it means for a function to be an increasing function by looking at its graph.
Letβs consider for a moment the related function π of π₯ equals π to the power of π₯. We can see that π of π₯ itβs just a function with π replaced by π§ over seven. Letβs see if we can graph this function for different values of π. When π is equal to one, π of π₯ is equal to one to the power of π₯ and one to the power of any number is just one. So we have the constant function π of π₯ equals one, whose graph is shown.
When π is greater than one, we have an exponential growth function; for example, π of π₯ could be two to the power of π₯ or π to the power of π₯. This function is increasing; as π₯ increases, the value of π¦ also increases. And looking at our formal definition, if we pick any two values of π and π, where π is less than π, then we see that π of π is less than or equal to π of π. This shows that our formal definition is satisfied, but really, itβs best to look at the graph and see if π¦ increases as π₯ increases.
And finally, when π is less than one, we have exponential decay. So this is a decreasing function: as π₯ increases, the value of π¦ decreases. To summarize, when π is greater than one, the function is increasing, when π is equal to one, the function is constant, and when π is less than one, the function is decreasing.
We are interested in the condition that makes our function π of π₯ an increasing function. We therefore need our value of π to be greater than one. And looking at the definition of our function π of π₯, we can see that our value of π is π§ over seven, so the condition we require is π§ over seven is greater than one. And we can simplify this condition by multiplying both sides by seven to get the condition π§ is greater than seven.<|endoftext|>
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Ukrainian War of Independence
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|Ukrainian War of Independence|
|Part of the Eastern Front of World War I|
and the Russian Civil War
A pro-Ukrainian People's Republic demonstration in Kiev's Sofia Square, 1917.
Part of a series on the
|History of Ukraine|
The Ukrainian War of Independence, a period of sustained warlike conflict, lasted from 1917 to 1921 and resulted in the establishment and development of a Ukrainian republic – later a part of the Soviet Union as the Ukrainian Soviet Socialist Republic of 1922–1991. The war consisted of a series of military conflicts between different governmental, political and military forces. Belligerents included Ukrainian nationalists, anarchists, Bolsheviks, the forces of Germany and Austria-Hungary, the White Russian Volunteer Army, and Second Polish Republic forces. They struggled for control of Ukraine after the February Revolution (March 1917) in the Russian Empire. The Allied forces of Romania and France also became involved. The struggle lasted from February 1917 to November 1921 and resulted in the division of Ukraine between the Bolshevik Ukrainian SSR, Poland, Romania, and Czechoslovakia. The conflict is frequently viewed[by whom?] within the framework of the Russian Civil War of 1917–1922, as well as the closing stage of the Eastern Front of the First World War of 1914–1918.
During the First World War Ukraine was in the front lines of the main combatants: the Entente-allied Russian Empire and Romania, and the Central Powers of the German Empire and Austria-Hungary. By the start of 1917 – after the Brusilov Offensive – the Imperial Russian Army held a front line which partially reclaimed Volhynia and eastern Galicia.
The February Revolution of 1917 encouraged many ethnic groups in the Russian Empire to demand greater autonomy and various degrees of self-determination. A month later, the Ukrainian People's Republic was declared in Kiev as an autonomous entity with close ties to the Russian Provisional Government, and governed by a socialist-dominated Tsentralna Rada ("Central Council"). The weak and ineffective Provisional Government in Petrograd continued its loyalty to the Entente and the increasingly unpopular war, launching the Kerensky Offensive in the summer of 1917. This offensive was a complete disaster for the Imperial Russian Army. The German counter-attack caused Russia to lose all their gains of 1916, as well as destroy the morale of its army, which caused the near-complete disintegration of the armed forces and the governing apparatus all over the vast Empire. Many deserting soldiers and officers – particularly ethnic Ukrainians – had lost faith in the future of the Empire, and found the increasingly self-determinant Central Rada a much more favorable alternative. Nestor Makhno began his anarchist activity in the south of Ukraine by disarming deserting Russian soldiers and officers who crossed the Haychur River next to Huliaipole, while in the east in the industrial Donets Basin there were frequent strikes by Bolshevik-infiltrated trade unions.
Ukraine after the Russian revolution
All this led to the October Revolution in Petrograd, which quickly spread all over the empire. The Kiev Uprising in November 1917 led to the defeat of Russian imperial forces in the capital. Soon after, the Central Rada took power in Kiev, while in late December 1917 the Bolsheviks set up a rival Ukrainian republic in the eastern city of Kharkov (Ukrainian: Kharkiv) – initially also called the "Ukrainian People's Republic". Hostilities against the Central Rada government in Kiev began immediately. Under these circumstances, the Rada declared Ukrainian independence on January 22, 1918 and broke ties with Russia.
The Rada had a limited armed force at its disposal (the Ukrainian People's Army) and was hard-pressed by the Kharkov government which received men and resources from the Russian Soviet Republic. As a result, the Bolsheviks quickly overran Poltava, Aleksandrovsk (now Zaporizhia), and Yekaterinoslav (now Dnipro) by January 1918. Across Ukraine, local Bolsheviks also formed the Odessa and Donetsk-Krivoy Rog Soviet Republics; and in the south Nestor Makhno formed the Free Territory – an anarchist region – then allied his forces with the Bolsheviks. Aided by the earlier Kiev Arsenal Uprising, the Red Guards entered the capital on February 9, 1918. This forced the Central Rada to evacuate to Zhytomyr. In the meantime, the Romanians took over Bessarabia. Most remaining Russian Imperial Army units either allied with the Bolsheviks or joined the Ukrainian People's Army. A notable exception was Colonel Mikhail Drozdovsky, who marched his White Volunteer Army unit across the whole of Novorossiya to the River Don, defeating Makhno's forces in the process.
German intervention and Hetmanate, 1918
Faced with imminent defeat, the Rada turned to its still hostile opponents – the Central Powers – for a truce and alliance, which was accepted by Germany in the first Treaty of Brest-Litovsk (signed on February 9, 1918) in return for desperately needed food supplies which Ukraine would provide to the Germans. The Imperial German and Austro-Hungarian armies then drove the Bolsheviks out of Ukraine, taking Kiev on March 1. Two days later, the Bolsheviks signed the Treaty of Brest-Litovsk, which formally ended hostilities on the Eastern Front of World War I and left Ukraine in a German sphere of influence. Ukrainian troops took control of the Donets Basin in April 1918. Also in April 1918 Crimea was cleared of Bolshevik forces by Ukrainian troops and the Imperial German Army. On March 13, 1918 Ukrainian troops and the Austro-Hungarian Army secured Odessa. On April 5, 1918 the German army took control of Yekaterinoslav, and 3 days later Kharkov. By April 1918 all Bolshevik gains in Ukraine were lost; this was due to the apathy of the locals and the then-inferior fighting skills of the Red Army compared to their Austro-Hungarian and German counterparts.
Yet disturbances continued throughout Eastern Ukraine, where local Bolsheviks, peasant self-defense groups known as "green armies", and the anarchist Revolutionary Insurrectionary Army of Ukraine refused to subordinate to Germany. Former Imperial Russian Army General Pavlo Skoropadsky led a successful German-backed coup against the Rada on April 29. He proclaimed the conservative Ukrainian State (also known as the "Hetmanate") with himself as monarch, and reversed many of the socialist policies of the former government. The new government had close ties to Berlin, but Skoropadsky never declared war on any of the Triple Entente powers; Skoropadsky also placed Ukraine in a position that made it a safe haven for many upper- and middle-class people fleeing Bolshevik Russia, and was keen on recruiting many former Russian Army soldiers and officers.
Despite sporadic harassment from Makhno, the territory of the Hetmanate enjoyed relative peace until November 1918; when the Central Powers were defeated on the Western Front, Germany completely withdrew from Ukraine. Skoropadsky left Kiev with the Germans, and the Hetmanate was in turn overthrown by the socialist Directorate.
Resumed hostilities, 1919
Almost immediately after the defeat of Germany, Lenin's government annulled their Brest-Litovsk Treaty – which Leon Trotsky described as "no war no peace" – and invaded Ukraine and other countries of Eastern Europe that were formed under German protection. Simultaneously, the collapse of the Central Powers affected the former Austrian province of Galicia, which was populated by Ukrainians and Poles. The Ukrainians proclaimed a Western Ukrainian People's Republic (WUNR) in Eastern Galicia, which wished to unite with the Ukrainian People's Republic (UNR); while the Poles of Eastern Galicia – who were mainly concentrated in Lwów (Ukrainian: Lviv) – gave their allegiance to the newly formed Second Polish Republic. Both sides became increasingly hostile with each other. On January 22, 1919, the Western Ukrainian People's Republic and the Ukrainian People's Republic signed an Act of Union in Kiev. By October 1919, the Ukrainian Galician Army of the WUNR was defeated by Polish forces in the Polish–Ukrainian War and Eastern Galicia was annexed to Poland; the Paris Peace Conference of 1919 granted Eastern Galicia to Poland for 25 years.
The defeat of Germany had also opened the Black Sea to the Allies, and in mid-December 1918 some mixed forces under French command were landed at Odessa and Sevastopol, and months later at Kherson and Nikolayev (Ukrainian: Mykolaiv). The cause and purpose of French intervention was not entirely clear; French military leaders quickly became disillusioned by internal quarrels within the anti-Bolshevik forces that prevented effective collaboration against Bolshevik pressures, and they particularly criticized the White Russian Volunteer Army for its arrogance towards the local population. Strong anti-foreigner feelings among Ukrainians convinced French officers that intervention in this climate of hostility was doomed without massive support. When the French government failed to supply enough equipment and manpower for extensive military operations, the French army faced defeat at the hands of pro-Bolshevik forces and French officers counseled Paris to withdraw the expedition from Odessa and Crimea.
A new, swift Bolshevik offensive overran most of Eastern and central Ukraine in early 1919. Kiev – under the control of Symon Petliura's Directorate – fell to the Red Army again on February 5, and the exiled Soviet Ukrainian government was re-instated as the Ukrainian Soviet Socialist Republic, moving to Kiev on March 15. The Ukrainian People's Republic (UNR) faced imminent defeat against the Bolsheviks – it was reduced to a strip of land along the Polish border with its capital moving from Vinnytsia to Proskurov (now Khmelnytskyi), then to Kamianets-Podilskyi, and finally to Rivne. But the UNR was saved when the Bolshevik armies had to regroup against a renewed White Russian offensive in South Russia and the Urals, which threatened the very existence of Bolshevism – and so required more urgent attention. During the spring and summer of 1919, Anton Denikin's Volunteer Army and Don Army overran all of central and Eastern Ukraine and made significant gains on other fronts. Yet by winter the tide of war reversed decisively, and by 1920 all of Eastern and central Ukraine except Crimea was again in Bolshevik hands. The Bolsheviks also betrayed and defeated Nestor Makhno, their former ally against Denikin.
Polish involvement, 1920
Again facing imminent defeat, the UNR turned to its former adversary, Poland; and in April 1920, Józef Piłsudski and Symon Petliura signed a military agreement in Warsaw to fight the Bolsheviks. Just like the former alliance with Germany, this move partially sacrificed Ukrainian sovereignty: Petliura recognised the Polish annexation of Galicia and agreed to Ukraine's role in Piłsudski's dream of a Polish-led federation in Eastern Europe.
Immediately after the alliance was signed, Polish forces joined the Ukrainian army in the Kiev Offensive to capture central and southern Ukraine from Bolshevik control. Initially successful, the offensive reached Kiev on May 7, 1920. However, the Polish-Ukrainian campaign was a pyrrhic victory: in late May, the Red Army led by Mikhail Tukhachevsky staged a large counter-offensive south of Zhytomyr which pushed the Polish army almost completely out of Ukraine, except for Lviv in Galicia. In yet another reversal, in August 1920 the Red Army was defeated near Warsaw and forced to retreat. The White forces, now under General Wrangel, took advantage of the situation and started a new offensive in southern Ukraine. Under the combined circumstances of their military defeat in Poland, the renewed White offensive, and disastrous economic conditions throughout the Russian SFSR – these together forced the Bolsheviks to seek a truce with Poland.
End of hostilities, 1921
Soon after the Battle of Warsaw the Bolsheviks sued for peace with the Poles. The Poles, exhausted and constantly pressured by the Western governments and the League of Nations, and with its army controlling the majority of the disputed territories, were willing to negotiate. The Soviets made two offers: one on 21 September and the other on 28 September. The Polish delegation made a counteroffer on 2 October. On the 5th, the Soviets offered amendments to the Polish offer, which Poland accepted. The Preliminary Treaty of Peace and Armistice Conditions between Poland on one side and Soviet Ukraine and Soviet Russia on the other was signed on 12 October, and the armistice went into effect on 18 October. Ratifications were exchanged at Liepāja on 2 November 1920. Long negotiations of the final peace treaty ensued.
Meanwhile, Petliura's Ukrainian forces, which now numbered 23,000 soldiers and controlled territories immediately to the east of Poland, planned an offensive in Ukraine for 11 November but were attacked by the Bolsheviks on 10 November. By 21 November, after several battles, they were driven into Polish-controlled territory.
On March 18, 1921, Poland signed a peace treaty in Riga, Latvia with Soviet Russia and Soviet Ukraine. This effectively ended Poland's alliance obligations with Petliura's Ukrainian People's Republic. According to this treaty, the Bolsheviks recognized Polish control over Galicia (Ukrainian: Halychyna) and western Volhynia – the western part of Ukraine – while Poland recognized the larger central parts of Ukrainian territory, as well as eastern and southern areas, as part of Soviet Ukraine.
Having secured peace on the Western front, the Bolsheviks immediately moved to crush the remnants of the White Movement. After a final offensive on the Isthmus of Perekop, the Red Army overran Crimea. Wrangel evacuated the Volunteer Army to Constantinople in November 1920. After its military and political defeat, the Directorate continued to maintain control over some of its military forces; in October 1921, it launched a series of guerrilla raids into central Ukraine that reached as far east as the modern Kiev Oblast ("Kiev province"). On November 4, the Directorate's guerrillas captured Korosten and seized a cache of military supplies. But on November 17, 1921, this force was surrounded by Bolshevik cavalry and destroyed.
In the current Chyhyryn Raion of Cherkasy Oblast (then in the Kiev Governorate), a local man named Vasyl Chuchupak led the "Kholodny Republic" which strived for Ukrainian independence. It lasted from 1919 to 1922, making it the last territory held by armed supporters of an independent Ukrainian state before the incorporation of Ukraine into the Soviet Union as the Ukrainian SSR.
In 1922, the Russian Civil War was coming to an end in the Far East, and the Communists proclaimed the Union of Soviet Socialist Republics (USSR) as a federation of Russia, Ukraine, Belarus and Transcaucasia. The Ukrainian Soviet government was nearly powerless in the face of a centralized monolith Communist Party apparatus based in Moscow. In the new state, Ukrainians initially enjoyed a titular nation position during the nativization and Ukrainization periods. However, by 1928 Joseph Stalin had consolidated power in the Soviet Union. Thus a campaign of cultural repression started, cresting in the 1930s when a massive famine – the Holodomor – struck the republic, claiming several millions of lives. The Polish-controlled part of Ukraine had a different fate – there was very little autonomy, both politically and culturally – but it was not affected by famine. In the late 1930s the internal borders of the Ukrainian SSR were redrawn, yet no significant changes were made.
The political status of Ukraine remained unchanged until the Molotov–Ribbentrop Pact between the USSR and Nazi Germany in August 1939, in which the Red Army allied with Nazi Germany to invade Poland and incorporate Volhynia and Galicia into the Ukrainian SSR. In June 1941, Germany and its allies invaded the Soviet Union and conquered Ukraine completely within the first year of the conflict. Following the Soviet victory on the Eastern Front of World War II, to which Ukrainians greatly contributed, the region of Carpathian Ruthenia – formerly a part of Hungary before 1919, of Czechoslovakia from 1919 to 1939, of Hungary between 1939 and 1944, and again of Czechoslovakia from 1944 to 1945 – was incorporated into the Ukrainian SSR, as were parts of interwar Poland. The final expansion of Ukraine took place in 1954, when the Crimea was transferred to Ukraine from Russia with the approval of Soviet leader Nikita Khrushchev.
Many folk songs were written from 1918 to 1922 that were inspired by people and events of this conflict. "Oi u luzi chervona kalyna" and "Oi vydno selo" were inspired by the Ukrainian Sich Riflemen unit of the Austro-Hungarian Army, which became the core battalion of the West Ukrainian People's Republic's Ukrainian Galician Army. "Pisnya pro Tiutiunnyk" was inspired by events surrounding Ukrainian People's Army brigade commander Yuriy Tiutiunnyk. Another song written at this time was "Za Ukrayinu". These "war songs" started to be sung publicly again in the western part of the Ukrainian SSR after the introduction of glasnost by Soviet leader Mikhail Gorbachev, and regained popularity throughout Ukraine after independence – especially during the current Russian military intervention.
Another musical legacy of this period was the Ukrainian Republic Capella (later the Ukrainian National Chorus), set up in early 1919 by the Directorate government of Symon Petliura. Under the direction of Oleksandr Koshetz, the Capella/Chorus toured Europe and North America from 1919 to 1921 and while in exile from 1922 to 1927; popularising the songs "Shchedryk" and "Oi khodyt son, kolo vikon" – which influenced the composition of the popular English language songs "Carol of the Bells" and "Summertime", respectively.
The coat of arms of Skoropadsky's Ukrainian State ("Hetmanate"), 1918.
- Ukrainian (Soviet) People's Republic at WMS Archived 2006-06-23 at the Wayback Machine (in Russian)
- J. Kim Munholland. "Ukraine". Encyclopædia Britannica. Retrieved 2007-11-08.
- Reid, Anna (2000). Borderland : A Journey Through the History of Ukraine. Westview Press. p. 33. ISBN 0-8133-3792-5.
- (in Ukrainian) 100 years ago Bakhmut and the rest of Donbass liberated, Ukrayinska Pravda (April 18, 2018)
- Tynchenko, Yaros (March 23, 2018), "The Ukrainian Navy and the Crimean Issue in 1917-18", The Ukrainian Week, retrieved October 14, 2018
- Germany Takes Control of Crimea, New York Herald (May 18, 1918)
- War Without Fronts: Atamans and Commissars in Ukraine, 1917-1919 by Mikhail Akulov, Harvard University, August 2013 (page 102 and 103)
- Arkadii Zhukovsky. "Struggle for Independence (1917–1920)". Encyclopedia of Ukraine. Retrieved 2007-11-08.
- (in Polish) Wojna polsko-bolszewicka. Entry at Internetowa encyklopedia PWN. Retrieved 27 October 2006.
- Text in League of Nations Treaty Series, vol. 4, pp. 8–45.
- Ukraine: A Concise Encyclopedia, Volume I (1963). Edited by Volodymyr Kubiyovych. Toronto: University of Toronto Press. pp. 831–833 and pp.872–874
- (in Ukrainian) In honor of the chieftain blessed knives, Gazeta.ua (21 April 2010)
- (in Ukrainian) Gerashchenko offers a National Park "Cold Yar", Ukrinform (19 October 2016)
- "Russian-Ukrainian war never stopped", Gazeta.ua (12 December 2014)
- "Russian Civil War 1918-1820". OnWar.com. Retrieved 2007-11-08.
- "31 серпня 1919 року. Як галичани з денікінцями Київ звільняли (August 31, 1919. How Galicians and Denikians liberated Kiev)" (in Ukrainian). Ukrayinska Pravda.
- J. Kim Munholland. "The French army and intervention in Southern Russia, 1918–1919". Cahiers du monde russe (in French). Editions Ehess. Archived from the original on 2007-06-10. Retrieved 2007-11-08.
- "The beginning of the Civil War and war intervention". RKKA (in Russian). Retrieved 2007-11-08.
- Der Vormarsch der Flieger Abteilung 27 in der Ukraine (in German) (The advances of Flight Squadron 27 in the Ukraine). This portfolio, comprising 263 photographs mounted on 48 pages, is a photo-documentary of the German occupation and their military advances through southern Ukraine in the spring and summer of 1918.<|endoftext|>
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Today’s lesson will build upon the work that we did yesterday. We will continue to try to find roots of polynomial functions, but we will be focusing on functions that have imaginary roots. To get started, show students the second slide of the PowerPoint (Notes - Day 2 of Finding Roots) and ask students how they could quickly figure out how many roots each function has. They will likely say to check the graph, so have them use their graphing calculator to look for the number of x-intercepts. Ask them if they see a relationship between the equation of the function and the number of roots. They will likely have past knowledge from other math classes and may oversimplify to say that the exponent corresponds to the number of roots. Don’t correct them – this will make them think even deeper when we move on to the next slide! It is also fine if a student mentions the imaginary roots – it will lead nicely to the next slide.
The problem on the third slide of the (Slides 3 to 5) can be tricky if students oversimplify and say that all roots are x-intercepts (and I bet many of them will). Now this is true if the roots are all real, but now we have a function (y = x3 – 4x2 + 9x – 36) that has real and imaginary roots. Give students about 5 minutes to work on this problem in their groups. They will likely begin by graphing (Slide 3 Graph); some students may be surprised to see that there is only one x-intercept when they feel there should be three. If they get stuck, push them to still use the procedure that we learned yesterday – divide out the root we know and see if they can do anything with what is left. After they work in groups, come back together as a class and talk about the problem. If a student found the two imaginary roots, have them show their work and see if a different student can explain the work.
Next give students the problem from the fourth slide (y = x4 + 2x3 + 5x2 + 8x + 4). Start by asking students how many roots the function should have. Then ask them to graph to see if they can find any. They will know that there are four roots and they will see one from the graph (Slide 4 Graph), so they may think that three of the roots will be imaginary. Again, give students about five minutes to work on this problem with their group. Some students may remember from a previous lesson that a root is repeated an even number of times if it “bounces off” the x-axis, but it may be something you will revisit during the class discussion. Again, choose a student to show their work and have another student try to explain the work.
As a summary of the work we did today, show the last slide. This gets them thinking about the past two days and how today’s lesson felt very different than yesterday’s because of the imaginary roots. Revisit the rule about how the exponent of the function relates to the number of roots and see if students can be more specific if they didn’t get it right before the lesson.
Imaginary numbers probably seem very unnatural for students. Surprisingly, mathematicians did not arbitrarily decide that the square root of –1 is equal to i; the imaginary number i came out of necessity. Since this is our first time talking about i in this class, I want students to read an article (The Roots of Complex Numbers - Katz) that gives them some history about it. The article is taken from Math Horizons, published by the Mathematics Association of America in November 1995. I remember studying non-Euclidean geometry in college and how the history became an integral part of the curriculum and helped me to conceptualize many of the topics; I want my students to feel the same way. Give them time to read the article themselves. Although the math is somewhat complicated, they can still appreciate the information without making sense of all of the calculations. Having a quick class discussion after they read would be a good way to end the class.
Finally, an assignment is given to synthesize the last two days of class.<|endoftext|>
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(Grades 7-8) Kaye Stacey and Susie Groves
Learning to solve problems is now considered to be one of the main reasons for doing mathematics. Many teachers now have considerable experience of using a wide range of problems with their classes and it is widely acknowledged that a balanced approach to teaching mathematics will include problems that make students think, as well as exercises to help them polish algorithmic skills. The authors believe that successful problem solving involves the following strategies and good habits:
In Strategies for Problem Solving some of these strategies and good habits are explored systematically by building on the experience of tackling intriguing questions.
The version 2 Mathomat will be available in June 2018, and is a more powerful product - being the same size as the current Mathomat but with many new functions and features built into its design. These include:
The new large regular octagon, and the resized large regular pentagon combine to form a group of regular polygons with 15mm sides to compliment the very popular existing regular polygons with 10mm sides. Students now have more creative freedom in 2-D designs to suit their project style.
The Mathomat V2 manual is a rich source of creative drawing ideas; for classroom or individual student use.
Sometimes students want to fill the whole page with their drawing. At other times its best to leave space for the remainder of a presentation. The Mathomat V2 offers both creative options.
The solution to a senior school trigonometry problem. Finding the hours of the day at which it is safe to cross the harbour bar of a fishing village by boat
The larger sine curve makes a striking improvement to hand drawn sketches. The integrated unit circle allows students to find angle values as distance travelled around its circumfrence before locating them on the x-axis of the graph of that function.Lesson 11 in Maths with Mathomat in the free resources section of this website provides a comprehensive plan for using Mathomat with the unit circle method of teaching trigonometry.
A new 1:2 scale ruler for detailed drawings, especially useful in engineering drawings
Ask students to redraw this section at a scale of 1:2 using their Mathomat V2 template
The many circles on Mathomat have a revised set of graduations to create interesting fractions such sevenths, and ninths, to challenge students to think flexibly about numbers
A new kite, non isosceles trapezium and concave quadrilateral (arrow head) for drawing
The polygons in the Mathomat V2 template flow in a clockwise direction in hierarchical order according to shape property. Students are encouraged to consider shape property continuously while working with the new V2 template<|endoftext|>
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# How to Convert Room Size into Square Feet
Save
Whether your planning to put in new flooring, tile the walls or paint the ceiling, you need to know the area of the relevant surface so you can buy enough materials to cover it. Although the mathematics for computing area depends on the geometrical shape of the room, you can reduce most measurements to approximate rectangles, and the mathematics for calculating the area of a rectangle is simple.
## Rectangular Rooms
Most rooms are rectangular, and to find the area of a rectangle, you need to measure its length and width and multiply those together. For example, to find how much hardwood you need to cover a rectangular floor, measure the length and width of the floor in inches and multiply those measurements to get the area in square inches. You need to know the square footage, though, so divide that number by the conversion factor to go from square feet to square inches: 144. If the floor is 123 inches wide (10 feet, 3 inches) and 149 inches (12 feet, 5 inches) long, the area is 18,327 square inches, or 127.3 square feet.
## Useful Mathematical Formulas
When a room has features that make the walls, floor and ceiling shapes other than rectangular, you need to know two important mathematical formulas:
• The area of a triangle is given by the expression A = 1/2 b•h, where "b" is the base of the triangle and "h" is the height, or distance from the base to the apex.
• The area of a circle is given by the expression A = pi•r2, where "r" is the radius of the circle and pi is a constant equal to approximately 3.14.
## Computing the Area
The first step in making use of these mathematical tools is to map out a rectangle that covers the bulk of the floor and compute its area. You'll be left with triangles -- if the room has more than four walls or the ceiling or floor have pointed features -- or semicircles -- if the walls are rounded or the ceiling is arched.
### Triangular Sections
If you're left with a triangle after mapping out a rectangle on a floor or ceiling:
• Measure the distance between the two points from which the triangle begins to slope away from the rectangle and the distance from the line demarcating the rectangle to the apex of the triangle -- both in inches.
• Plug the numbers into the formula to compute the area of the triangle.
• Add this area to that of the rectangle before you make the conversion to square feet.
• Repeat the procedure for every triangular section, and and convert to square feet after adding all the results.
### Curved Sections
If you're trying to find area in a room with curved walls or an arched ceiling:
• Measure the distance between the points from which the curve slopes away from the main rectangle.
• Divide that distance by 2.
• Substitute it in the formula for the area of a circle.
• Divide the result of that computation by 2 because the curve is a semicircle, not a full circle.
• Add that number to the area of the rectangle.
• Repeat the operation for each curve on the surface you're measuring, add all the results to the area of the rectangle and convert from square inches to square feet.
The result is approximate, but it's close enough to allow you to estimate materials.
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Is DIY in your DNA? Become part of our maker community.<|endoftext|>
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# How Many Degrees in an Octagon: Understanding the Mathematics and Design
## Introduction
When it comes to understanding geometry and design, one shape that often comes up is the octagon. Whether you’re interested in architecture, landscaping, or simply want to explore geometric shapes, understanding how many degrees are in an octagon is important. This article will explore the mathematics behind this shape, its use in design, and ways to visualize and observe it in the natural world.
## Starting with the Basics: What is an Octagon?
Before we dive into the math and design aspects of the octagon, it’s important to understand what this shape is. An octagon is a two-dimensional shape with eight sides and eight angles. It is a polygon, which means it is a closed shape with straight sides. The octagon is unique in that it has eight sides of equal length, making it a regular octagon.
There are different types of octagons, including concave and irregular octagons. The degrees in these types of octagons may vary, but for the purpose of this article, we will focus on regular octagons. Understanding the degrees in a regular octagon can help in a variety of fields, including design and construction.
## Mathematical Properties of an Octagon: Understanding the Degrees
To understand the degrees in an octagon, we first need to calculate the total number of degrees that make up this shape. To do this, we can use a simple formula: (n-2) x 180, where “n” is the number of sides in the shape. For an octagon, this formula would be (8-2) x 180 = 1080 degrees.
Now that we know the total degrees in an octagon, we can find the degrees in individual angles within the shape. To do this, we can use another formula: (n-2) x 180 / n, where “n” is the number of sides in the shape. For an octagon, this would be (8-2) x 180 / 8 = 135 degrees. This means that each angle in an octagon measures 135 degrees.
If you want to practice finding the degrees in individual angles within an octagon, try these sample problems:
1. What is the measure of one angle in a regular octagon?
2. What is the sum of all the angles in a regular octagon?
3. What is the measure of each exterior angle in a regular octagon?
## Designing an Octagon: Factors to Consider
Octagons are commonly used in design and construction, particularly in architecture and landscaping. Understanding the degrees in an octagon can impact the design process, as it can ensure that the shape is harmonious and balanced. It can also ensure that the angles are accurate, which is important for stability and safety.
When incorporating octagons into designs, there are a few factors to consider. For example, the size and proportion of the octagon will impact how it fits in with surrounding shapes and structures. The materials used to create the octagon, such as wood or stone, can also impact its appearance and function.
## Visualizing an Octagon: Using Geometry to Enhance Learning
When learning about geometry, it can be helpful to use visualization techniques to enhance understanding. This is no different when it comes to understanding the degrees in an octagon. There are a variety of visualization techniques you can use, such as creating 3D models or using interactive geometry software. These tools can help you see the relationships between shapes and better understand how the degrees in an octagon relate to one another.
Some benefits of using visualization tools for learning geometry include improved spatial reasoning skills, increased engagement with the material, and the ability to see problems from multiple perspectives. If you’re interested in using visualization tools to learn more about the degrees in an octagon, there are many resources available online.
## Octagons in Nature: Finding Symmetry and Angles
While octagons are often associated with man-made structures, they can also be found in nature. For example, crystal formations may take on octagonal shapes. Observing these shapes in nature can enhance your understanding of the degrees in an octagon and how they relate to other shapes and structures.
If you’re interested in exploring your local environment and observing octagonal shapes in nature, there are many resources available online to help you get started. Some suggestions include visiting local rock formations or observing the shapes of leaves and flowers.
## From Octagon to Circle: Understanding the Relationship of Angles
Understanding the degrees in an octagon can also help you understand the degrees in other shapes, such as a circle. While circles and octagons may seem very different, there is actually a relationship between the degrees in these shapes.
The formula for finding the degrees in a circle is 360 degrees divided by the number of sides. Since a circle has an infinite number of sides, we can use the formula to find the degrees in other shapes that are similar to circles. For example, a regular octagon inscribed in a circle will have angles that measure 45 degrees.
## Conclusion
Understanding the degrees in an octagon is an important aspect of learning about geometry and design. By knowing the total number of degrees in the shape and the degrees in individual angles, you can better understand how octagons fit into designs and structures. A variety of visualization techniques can enhance your learning experience, as can exploring natural structures that take on octagonal shapes. Ultimately, understanding the degrees in an octagon can help you see the relationships between shapes and angles and give you a greater appreciation for the underlying mathematics of our world.
To continue learning about geometry and related topics, consider exploring resources online or consulting with a math or design professional.
Proudly powered by WordPress | Theme: Courier Blog by Crimson Themes.<|endoftext|>
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As a political movement, progressivism arose at the local and state levels in the 1890s. Urban reformers attacked political machines run by corrupt bosses and monopolies in municipal services such as electricity or gas. To address these problems, they promoted professional city managers and advocated public ownership of utilities.
The social settlement movement, which originated in cities in the 1890s, also became a force for progressive reform at the local level. Settlement houses offered social services to the urban poor, especially immigrants. Pioneering settlement houses, such as Hull House, founded by Jane Addams and Ellen Gates Starr in 1889, provided nurseries, adult education classes, and recreational opportunities for children and adults. Settlements spread rapidly. There were 100 settlement houses in 1900, 200 in 1905, and 400 in 1910. Settlement leaders joined the battle against political machines and endorsed many other progressive reforms.
At the state level, progressives campaigned for electoral reforms to allow the people to play a more direct role in the political process. Some Western states adopted practices that expanded voter rights, including the initiative, the referendum, and the recall. Under the initiative, citizens could sign petitions to force legislatures to vote on particular bills. With the referendum, a proposal could be placed on the ballot to be decided by a vote at election time. Using the recall, voters could petition to oust officials from their jobs. Progressives also supported the 17th Amendment, ratified in 1913, which provides for election of U.S. senators directly by vote of the people, rather than indirectly by state legislatures.
Progressive reformers used the states as laboratories of reform. For instance, Wisconsin governor Robert La Follette, who held office from 1901 to 1906, introduced progressive changes such as establishing a commission to supervise railroad practices and raising state taxes on corporations.
Following Wisconsin’s example, one state after another passed laws to regulate railroads and businesses. Progressives also focused on labor reform at the state level. They sought to eliminate (or at least regulate) child labor, to cut workers’ hours, and to establish a minimum wage. By 1907 progressive efforts had led 30 states to abolish child labor. In Muller v. Oregon (1908), the Supreme Court upheld a state law that limited women factory workers to a ten-hour day, and many states began to regulate women’s working hours. Progressives also endorsed workmen’s compensation (an insurance plan to aid workers injured on the job) and an end to homework (piecework done in tenements). In New York’s Triangle Fire of 1911, many women leapt to their deaths from a burning shirtwaist factory. The tragedy reminded people of the need for higher safety standards in factories and the need to protect workers from unscrupulous employers. Some progressive reformers supported causes that had a coercive or repressive dimension, such as Prohibition, a movement to prevent the manufacture, sale, or use of alcohol. The Woman’s Christian Temperance Union (WCTU), founded in 1874, had long campaigned against alcohol. In 1895 the Anti-Saloon League of America joined the crusade. Together they worked to gain support for the 18th Amendment, which provided for Prohibition. The amendment was ratified in 1919 and remained law until 1933, when the 21st Amendment repealed it. Progressive moral fervor also emerged in campaigns to combat prostitution and to censor films.
Finally, some progressives endorsed other restrictive causes, now seen as ungenerous or inhumane, such as a campaign against immigration or support for eugenics, a movement to control reproduction in order to improve the human race.
Progressive causes won support from a broad section of the middle class—editors, teachers, professionals, and business leaders—who shared common values. Progressive supporters appreciated order, efficiency, and expertise; they championed investigation, experimentation, and cooperation. Many, including some progressive employers, sought regulations to make business practices more fair and break up monopolies. To regulate business, however, progressives had to wield influence on the national level. "USA" © Emmanuel BUCHOT, Encarta, Wikipedia.
Photos of European countries to visit
Photos of Asian countries to visit
Photos of America<|endoftext|>
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Representation of rational numbers on number line
Draw a straight line. Take a point O on it. Call it 0 (zero). Set off equal distances on its right as well as on the left of 0. Such a distance is known as a units length. Clearly, the points A, B, C, D, E represent the integers 1, 2, 3, 4, 5 respectively and the points A', B', C', D', E' represent the integers –1, –2, –3, –4, –5 respectively.
Thus, we may represent any integer by a point on the number line. Clearly, every positive integer lies to the right of O and every negative integer lies to the left of O.
Illustration 2
Solution:
Draw a line, take a point O on it, let it represent 0.
From O, set off unit distances OA, AB and BC to the right of O. Clearly, the points A, B and C represent the integers 1, 2 and 3 respectively. Now, take 2 units OA and AB, and divide the third unit BC into 5 equal parts. Take 3 parts out of these 5 parts to reach at a point P. Then the point P represents the rational number
Again, from O, set off unit distances to the left. Let these segments be OA’, A’ B’, B’C’, etc. Then, clearly the points A’, B’ and C’ represent the integers –1,–2,–3 respectively.
Take 2 full unit lengths to the left of O. Divide the third unit B’C’ into 5 equal parts. Take 3 parts out of these 5 parts to reach a point P’.
Then, the point P’ represents the rational number
Thus, we can represent every rational number by a point on the number line.<|endoftext|>
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The shallow, scalloped depression in the center of this picture from NASA's Galileo spacecraft is a caldera-like feature 5 to 20 kilometers(3 to 12 miles) wide on Jupiter's largest moon, Ganymede.
Calderas are surface depressions formed by collapse above a subsurface concentration of molten material. Some shallow depressions in bright, smooth areas of Ganymede have some overall similarities to calderas on Earth and on Jupiter's moon Io.
On Ganymede, caldera-like depressions may serve as sources of bright, volcanic flows of liquid water and slush, an idea supported by a Ganymede photo obtained by Galileo during its seventh orbit and available at PIA01614. In the more recent image here, from Galileo's 28th orbit, a tall scarp marks the western boundary of a caldera-like feature. The western scarp is aligned similarly to older tectonic grooves visible in the image, suggesting the feature has collapsed along older lines of weakness. The interior is mottled in appearance, yet smooth compared to most of Ganymede's bright terrain seen at high resolution. The eastern boundary of the caldera-like feature is cut by younger, grooved terrain. Small impact craters pepper the scene, but the lack of a raised rim argues against an impact origin for the caldera-like feature itself. Instead, water-rich icy lava may have once flowed out of it toward the east. If so, later tectonism could have erased any telltale evidence of volcanic flow fronts. Direct evidence for icy volcanism on Ganymede continues to be elusive.
North is to the top of the picture and the Sun illuminates the surface from the left. The image, centered at -24 degrees latitude and 318degrees longitude, covers an area approximately 162 by 119 kilometers(101 by 74 miles). The resolution is 43 meters (141 feet) per picture element.
This image and other images and data received from Galileo are posted on the Galileo mission home page at http://www.jpl.nasa.gov/galileo. Background information and educational context for the images can be found at http://www.jpl.nasa.gov/galileo/sepo.
The Jet Propulsion Laboratory, a division of the California Institute of Technology in Pasadena, manages the Galileo mission for NASA's Office of Space Science, Washington, D.C.
This image was produced by Brown University, Providence, R.I.,http://www.planetary.brown.edu/.<|endoftext|>
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As part of your skills development in GCSE geography, you will have learnt about map scales, working out distances and directions from Ordnance Survey and other maps. Without a scale, a map is of little use. It can show you how the features in an area are related to one-another but it gives no idea of distance. The earliest maps were all hand drawn and scales were not particularly accurate which made using them quite difficult.
A map represents a sector of landscape. Large scale maps show a small area in great detail whilst a small scale map shows a much larger area but in less detail. This can be a little confusing at first as the scale number of a large scale map is smaller. A 1:100,000 scale map is a small scale map but 1:5,000 is a large scale map.
The scale number on an OS map indicates how many centimetres on the ground are represented by a centimetre on the map. On a 1:100,000 scale map, one centimetre on the map represents 100,000 cm on the ground, in other words, one centimetre on the map represents one kilometre in reality. A scale of 1:5,000 therefore means that a centimetre on the map represents a distance in real life of 5,000 centimetres (50 metres). This method of representing the scale of a map is called the fractional method, but you will also see graphical representations or written representations like 2 cm = 1 km.
Ordnance Survey maps are printed with north at the top. East is therefore to the right, south to the bottom and west to the left. The four points north, east, south and west are called the cardinal points of the compass. There are four other directions exactly half-way between the cardinal points - north-east, south-east, south-west and north-west. These are often abbreviated to NE, SE, SW and NW. Between these and the cardinal points, there are eight others, each beginning with the closest cardinal point e.g. the direction between north and north-east is called north-north-east. Using these descriptions for a direction is usually sufficient unless you are trying to navigate in difficult terrain. Then you need to use exact compass bearings. North is a bearing of zero degrees, east is ninety degrees, south is one hundred and eighty degrees and west is two hundred and seventy degrees. If you were required to follow a bearing of thirty degrees, you would be travelling in a direction somewhere between NNE and NE.<|endoftext|>
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# Question 9.17: (a) Find the inverse of the matrix D = ( 1 0 -2 2 2 3 1 3 2)......
(a) Find the inverse of the matrix
D = $\left(\begin{matrix} 1 & 0 & -2 \\ 2 & 2 & 3 \\ 1 & 3 & 2 \end{matrix} \right)$
by Gauss–Jordan elimination.
(b) Show that $DD^{–1} = I$.
Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
(a) Write out the augmented matrix consisting of the matrix D and the unit matrix of the same dimension:
$\left(\begin{array}{ccc|ccc} 1 & 0 & -2&1 & 0 & 0 \\ 2 & 2 & 3& 0 & 1 & 0 \\ 1 & 3 & 2& 0 & 0 & 1 \end{array} \right)$
Carry out Gauss–Jordan elimination on the matrix D by the method given in Worked Example 9.9.
The original matrix, D, is now reduced to the unit matrix. The inverse of D is given by the transformed unit matrix: the 4th, 5th and 6th columns of the augmented matrix.
$D^{-1} = \left(\begin{matrix} \frac {5}{13} & \frac {6}{13} & -\frac {4}{13} \\ \frac {1}{13} & -\frac {4}{13} & \frac {7}{13} \\ – \frac {4}{13} & \frac {3}{13} & -\frac {2}{13}\end{matrix} \right) = \frac {1}{13} \left(\begin{matrix} 5 & 6 & -4 \\ 1 & -4 & 7 \\ -4 & 3 & -2 \end{matrix} \right)$
Note: If division by zero arises at any stage, then D does not have an inverse.
(b) Every element in $D^{−1}$ is multiplied by 1/13. Before multiplying $D^{−1}$ by D, factor out this scalar to simplify the arithmetic involved
$DD^{−1} = \left(\begin{matrix} 1 & 0 & -2 \\ 2 & 2 & 3 \\ 1 & 3 & 2 \end{matrix} \right) \frac {1}{13} \left(\begin{matrix} 5 & 6 & -4 \\ 1 & -4 & 7 \\ -4 & 3 & -2 \end{matrix} \right) = \frac {1}{13} \left(\begin{matrix} 13 & 0 & 0 \\ 0 & 13 & 0 \\ 0 & 0 & 13 \end{matrix} \right) = \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)$
Action Augmented matrix Calculations $\left(\begin{array}{ccc|ccc} 1 & 0 & -2& 1 & 0 & 0 \\ 2 & 2 & 3 &0 & 1 & 0\\ 1 & 3 & 2&0 & 0 & 1 \end{array} \right)\begin{matrix} (1) \\ (2) \\ (3) \end{matrix}$ row 2 + (–2 × row 1) row 3 + (–1 × row 1) $\left(\begin{array}{ccc|ccc} 1 & 0 & -2& 1 & 0 & 0 \\ 0 & 2 & 7 &-2 & 1 & 0 \\ 0 & 3 & 4&-1 & 0 & 1 \end{array} \right)\begin{matrix} (1) \\ (2)¹ \\ (3)¹ \end{matrix}$ Calculate (–2 × row 1) →(–2 0 4 –2 0 0) Calculate (–1 × row 1) →(–1 0 2 –1 0 0) row 2¹ × $\frac {1}{2}$ $\left(\begin{array}{ccc|ccc} 1 & 0 & -2&1 & 0 & 0 \\ 0 & 1 & \frac {7}{2}&-1 & \frac {1}{2} & 0 \\ 0 & 3 & 4& 0 & 0 & 1 \end{array}\right)\begin{matrix} (1) \\ (2)² \\ (3)¹ \end{matrix}$ row 3¹ + (–1 × row 2²) $\left(\begin{array}{ccc|ccc} 1 & 0 & -2&1 & 0 & 0 \\ 0 & 1 & \frac {7}{2}&-1 & \frac {1}{2} & 0 \\ 0 & 0 & -\frac {13}{2}&2 & -\frac {3}{2} & 1 \end{array} \right)\begin{matrix} (1) \\ (2)² \\ (3)² \end{matrix}$ Calculate (–3 × row 2²) →(0 -3 –$\frac {21}{2}$ 3 –$\frac {3}{2}$ 0) row 3² × $\frac {2}{13}$ $\left(\begin{array}{ccc|ccc} 1 & 0 & -2& 1 & 0 & 0 \\ 0 & 1 & \frac {7}{2}&\ -1 & \frac {1}{2} & 0 \\ 0 & 0 & 1 &-\frac {4}{13} & \frac {3}{13} & -\frac {2}{13}\end{array}\right)\begin{matrix} (1) \\ (2)² \\ (3)³ \end{matrix}$ row 1 + (2 × row 3³) row 2² + (−$\frac {7}{2}$ × row 3³) $\left(\begin{array}{ccc|ccc} 1 & 0 & 0&\frac {5}{13} & \frac {6}{13} & -\frac {4}{13} \\ 0 & 1 & 0 &\frac {1}{13} & -\frac {4}{13} & \frac {7}{13} \\ 0 & 0 & 1&- \frac {4}{13} & \frac {3}{13} & -\frac {2}{13} \end{array} \right)\begin{matrix} (1)¹ \\ (2)² \\ (3)³ \end{matrix}$ Calculate (–2 × row 3³) →(0 0 2 $-\frac {8}{13}$ $\frac {6}{13}$ $-\frac {4}{13}$) Calculate ($-\frac {7}{2}$ × row 3³) →(0 0 $-\frac {7}{2}$ $\frac {14}{13}$ $\frac {10.5}{13}$ $\frac {7}{13}$)
Question: 9.7
## Solve the following equations by Gaussian elimination: x + y − z = 3 (1) 2x + y − z = 4 (2) 2x + 2y + z = 12 (3) ...
All three equations must be written in the same fo...
Question: 9.20
## Given the input/output table for the three-sector economy: ...
Step 1: Use the underlying assumption total input=...
Question: 9.18
## (a) Find the inverse of the matrix D = (1 0 -2 2 2 3 1 3 2) (b) Show that DD^0−1 = I. ...
(a) Use the definition of the inverse given in equ...
Question: 9.12
## Given the supply and demand functions for two related goods, A and B,Good A :{Qda = 30 − 8Pa + 2Pb {Qsa = −15 + 7Pa Good B :{Qdb = 28 + 4Pa − 6Pb {Qsb = 12 + 2Pb (a) Write down the equilibrium condition for each good. Hence, deduce two equations in Pa and Pb . (b) Use Cramer’s rule to find the ...
(a) The equilibrium condition for each good is tha...
Question: 9.9
## Solve the following equations by Gauss–Jordan elimination: 2x + y + z = 12 6x + 5y − 3z = 6 4x − y + 3z = 5 ...
Rearrange the equations to have variables on the L...
Question: 9.8
## Solve the following equations by Gaussian elimination: 2x + y + z = 12 6x + 5y − 3z = 6 4x − y + 3z = 5 ...
The equations are already written in the required ...
Question: 9.2
## A company manufactures two types of wrought iron gates. The number of labour-hours required to produce each type of gate, along with the maximum number of hours available, are given in Table 9.3. ...
(a) For x type I gates and y type II gates, the nu...
Question: 9.21
## (a) Find the inverse of A = (2 1 1 6 5 -3 4 -1 3) by the elimination method, hence solve the equations 2x + y + z = 12 6x + 5y − 3z = 6 4x − y + 3z = 5 (b) Solve the equations in (a) by Gauss–Jordan elimination. ...
(a) Set up the augmented matrix, consisting of the...
Question: 9.19<|endoftext|>
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More than 60 percent of infectious diseases that affect humans are caused by pathogens that we share with wild and domestic animals. Such diseases, known as “zoonoses,” are transmitted from other vertebrate animals to humans in a process known as “spillover,” and pose a significant threat to human health. Spillover is more common in developing countries, which experience rapid environmental change as their human populations grow. Some zoonoses, such as rabies and anthrax, occur only when there is transmission directly from animals to humans. However, there is the potential for a shift from animal-to-human to human-to-human disease transmission. The result may be a localised outbreak of disease, such as the periodic emergences of Ebola virus in Zaire/ Democratic Republic of the Congo, or a global spread of epidemic proportions such as the recent outbreak of Severe Acute Respiratory Syndrome (SARS).
In order to better anticipate, contain, and even prevent such outbreaks, it is first necessary to improve our understanding of the basic epidemiology of zoonoses, including the factors that cause new diseases to emerge, and those that cause re-emergence of known diseases. The first step in this process is considering the anthropogenic activities that have been linked with the dramatic increase in incidence and frequency of zoonoses over the past 30 years.
Changes in human population size and density
Pathogens are more likely to occur in populations above a certain density; below this “threshold density,” pathogens cannot survive. Before the Second World War, most human settlements in tropical developing countries were scattered, with few large cities. This pattern has changed over the past decades: areas that previously consisted of scattered settlements are now occupied by large mega-cities surrounded by semi-urban settings, with only small areas of undisturbed forest area remaining near croplands and degraded lands. In 2006, there were 18 mega-cities in the world, cumulatively containing more than 10 million inhabitants. By 2025, Asia alone is predicted to have 10 mega-cities; by 2030, these areas may be home to as many as 2 billion people. Both the size and high density of human populations and their associated domestic animals increases the risk of new disease emergence. These conditions also allow rapidly reproducing pathogens more chances to undergo mutation and develop new traits that will promote survival and infectiousness. Even if there are no immediate outbreaks or emergences of new diseases, pathogens will adapt to conditions within the infected host. Later, under favorable circumstances such as a weakened host immune system, the resident microbes can cause an infection.
Land use and environmental changes
Another consequence of human population growth is an increase in deforestation, the conversion of forested land to non-forested areas such as cropland, plantation and urban habitats. This allows humans to settle in formerly isolated areas rich in previously unknown pathogens. This is probably what led to the spread of human immunodeficiency virus (HIV), which originated as simian immunodeficiency virus (SIV) in chimpanzees.
Deforestation is frequently associated with forest fragmentation, which reduces the number of predators that can occupy the habitat. Without predators to control their population, rodents and biting insects thrive; because these animals can act as reservoirs for human disease, their increase allows zoonoses to flourish. As the movement of humans and wildlife species between remnant forest and human habitation increases, pathogenhost interactions also increase and result in the spread of infections.
Global biosphere and climate change
Anthropogenic changes to the global biosphere include shifts in land and water use, biodiversity loss, and introduction of new chemicals. These processes alter environmental factors such as temperature, humidity levels, and water availability. Disease vectors such as ticks and mosquitoes, which carry disease from one organism to another, are sensitive to such changes in the environment. The resultant fluctuations in vector population can facilitate the emergence of vector-borne zoonoses such as West Nile virus.
Climate change, a phenomenon associated with global biosphere change, is likely to increase populations of vector-borne pathogens in cold regions, which are expected to become warmer. At the same time, it may decrease the transmission of infections in areas that become so warm that vectors are no longer able to survive in high numbers. Climate change is also predicted to change patterns of human activity. For example, continued warming of the Earth’s surface temperature could reduce the quality of pasture lands, causing a decline in livestock numbers and human-animal interactions.
This changing context presents many uncertainties. For example, droughts may decrease mosquito populations and, therefore, the incidence of mosquito-borne diseases; at the same time, the accumulation of dead vegetation might act as a reservoir for other pathogens. The effect of climate change on the spread of zoonotic disease is likely to vary according to geographic location and local habitat.
Human consumption of animals
Recent advances in the way food is produced, processed, and preserved can also play a role in the emergence of zoo-notic disease. One especially influential practice is the use of antimicrobial feeds and drugs. The latter are particularly to blame for an increase in the number of antimicrobial-resistant pathogens that can be transmitted from animals to humans through the food chain. In human Salmonella isolates collected in the United States, for example, resistance to antimicrobial agents rose from 0.4% to 1.0% between 1996 and 2001.
Among animals reared for agricultural purposes, repeated breeding of genetically similar stocks leads to weakened immune systems that leave animals less resistant to infections that may then be transmitted to humans. Further, these animals are often housed and transported in crowded conditions that increase the chances of exposure to, and transmission of, infections.
Another risk factor for cross-species spread of infection is subsistence hunting. The bush-meat trade–the tracking, capture and butchering of animals in the wild, and transportation of meat— brings humans dangerously close to potential vectors and the microbes they carry. Risk of infection is especially high during interactions with non-human primates such as chimpanzees; these close relatives of humans may transmit pathogens such as Ebola virus. The amount of annual wild meat consumption has been calculated to be around one billion kilograms for central African countries alone. In Cameroon, Central, and West Africa, bush-meat hunters and other persons who handle vertebrate pets are at a higher risk of zoonotic transmissions as a result of bites, cuts and exposure to the bodily fluids and tissues of infected animals. Bush-meat hunters are commonly infected by simian foamy virus; luckily, human-to-human transmission of this pathogen has not yet been established.
Zoonotic diseases in developing countries
Zoonoses are most likely to emerge where humans come into close contact with animals. Such encounters are most likely in tropical regions, which are characterised by increasing human population densities and rapid urbanisation. Because of the high number of low-income individuals in these countries, a significant proportion of the population is faced with poor sanitation, substandard housing, inadequate disease control and management, and rapid urbanisation. Globally, over 600 million people are dependent on livestock for their income, and up to 70% of these people live in marginalised developing regions. Two-thirds of the workforce in sub-Saharan Africa and South Asia is involved in agriculture; livestock are vital as a source of both income and food. Unfortunately, they may also be a source of pathogens and, to compound the problem, there are few incentives for farm-level management and control of infections.
Once infected, people in this socioeconomic group often lack access to proper medical care. Health centers are few and far between, and potential patients lack the time and money needed to visit them often. As a result, most zoonotic infections are chronically under-diagnosed.The burden of looking after a seriously ill family member may push the household further into poverty and illness.
Another major challenge associated with disease management in developing countries is the lack sufficient information to make decisions, both at an individual and a national level. In India, for example, around 68% of the national workforce depends on farming, yet most people are unable to tell whether they are working with infected or healthy animals. Because of ineffective data collection and poor disease management practices, there are no data on death rates associated with zoonotic infections across the country. Further, there seems to be a reluctance to recognise that the study and management of zoonoses requires the combined efforts of medical and veterinary professionals, who have, historically, worked in isolation from each other. The transmission of swine flu across the country can be used as an example: Even though many control measures were taken to prevent infections among humans, few or no measures were taken at the veterinary level to prevent infections among livestock. This lack of coordination and information is only exacerbated by widespread illiteracy, poverty and unsafe living conditions.
Learning from the ‘Nipah outbreak’ in Bangladesh
The emergence of new infections and the reemergence of known infectious diseases are both major global concerns, particularly in developing countries that lack adequate medical surveillance, disease management practices, and financial resources. Successful management of zoonoses requires collaborations between experts at the local, national, and global levels, as currently being demonstrated in Bangladesh in association with the emergence of the Nipah virus.
Nipah infections in Bangladesh are very different to the first emergence of this zoonosis in pig farms in Malaysia in 1998, when pigs were exposed to flying fox urine, faeces or saliva carrying Nipah virus. The disease quickly spread to humans throughout Malaysia and Singapore as infected pigs were transported to slaughterhouses. Since 2001, periodic outbreaks of Nipah virus, with approximately 200 human fatalities, have been reported from Bangladesh and northern regions of India; however, these occurrences of zoontoc disease resulted from increased habitat loss which placed humans in closer contact with flying foxes. In Bangladesh, the infections coincide with the date palm sap harvesting season from November to March because the most common transmission pathway for Nipah virus is human consumption of sap contaminated with flying fox saliva and urine.
Date palm sap is collected through a tap or funnel that drains into a clay pot, often left in place overnight and frequently visited by flying foxes that enjoy the sugar-rich sap.The harvester, their families and friends drink the raw sap the next morning. Shields to keep fruit bats away from the sap collection pots are known to local harvesters, but rarely used until scientists from United States Centers for Disease Control and Prevention, and International Centre for Diarrhoeal Disease Research, Bangladesh, determined that homemade skirts covering the sap-producing surface and mouth of the collecting pot prevented most flying fox and bats from contaminating the sap. Community intervention trials are underway to determine if changes in sap harvesting practices can reduce local spillover of Nipah virus from flying foxes to humans, minimising the human-to-human transmission which may occur during the care for an infected patient.
Prediction and prevention of the future zoonotic disease
As human populations grow, and our interactions with the environment change, there is a greater likelihood of emerging disease at the human-animal interface. However, greater awareness and collaboration among doctors, veterinarians, wildlife carers, biologists and communities can identify potential diseases at the source, before there is the potential for diseases to be carried to other regions and have a greater impact on human health and economies.
As shown by the case of Nipah virus in Bangladesh, minimising the risk of zoonoses is best accomplished with the involvement of local people. Global co-operation to support monitoring programs in tropical developing countries, hotspots for disease emergence, and new molecular methods of identification will help to quickly isolate potential new pathogens. Advanced communications technology allows outbreaks to be reported quickly so susceptible communities can be informed of transmission pathways and prevention strategies. Our best weapons against future zoonoses are understanding the origin and dynamics of pathogens in wildlife, reacting quickly to spillover events so that the disease has limited time to be transmitted among the human populations.
Khan SU, ES Gurley, MJ Hossain, N Nahar, MAY Sharker & SP Luby. 2012. A Randomized Controlled Trial of Interventions to Impede Date Palm Sap Contamination by Bats to Prevent Nipah Virus Transmission in Bangladesh. PLoS ONE 7: e42689. doi:10.1371/journal.pone.0042689.
Aswathy Vijayakumar is a recent graduate (BSc) and current Teaching Assistant at Asian University for Women, Bangladesh, and hopes to enter a graduate program in molecular biology in the near future, [email protected].
Andrea D Phillott, an Associate Professor in Biological Sciences at Asian University for Women, studies zoonoses and disease transmission at the human-animal interface in Australia and Bangladesh, [email protected].
Illustrations: Kalyani Ganapathy<|endoftext|>
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Table of Contents
The main difference between Demand and Supply is that Demand refers to how much buyers and Supply (quantity of a product or service) represents how much the market can offer.
Demand vs. Supply
The balance between the price and the quantity demanded of a product or the commodity at a certain period is called demand. To the contrary, the balance between the price of the product or goods and the quantity that supplied at a given period called as supply. While the paying aptitude and the devotion of the buyer at a specific price is demand, while the quantity that is said by the producers of those products to its customers or consumers at a particular price supplied demand has an opposite or indirect relationship with the price that is if the price of the goods increases the demand decreases. Likewise, if the price of the product decreases then the demand increases, however, on the flip side, the price has a direct association with supply, that is if price decreases then the supply will also decrease and if the price increases supply also increases. Demand do represent the consumer or the customer’s preferences and taste for a product or the commodity that is demanded by him, on the other hand, Supply does represent the firms, which is how much of the good or the commodity is offered by those producers in that huge market.
What is Demand?
Demand is a commercial or economic principle referring to a consumer’s desire and willingness to pay the price for definite product or service. Holding all other elements constant, an increase in the price of a good or service will decrease demand, and vice versa. Think of demand as your readiness to go out and buy a definite product. For example, market demand is the summing what everyone in the market wants. Businesses often spend a substantial amount of money to determine the amount of demand the public has for their goods and services. Incorrect estimations moreover result in money left on the table if demand is disrespect or losses if demand overestimated. There are five determinants of demand. The most important is the price of the good or service itself. The second is the price of related products, whether they are substitutes or complementary. Circumstances drive the next three determinants. These are consumers’ incomes, their tastes, and their expectations.
Basis of Classification
- Individual Demand and Market Demand
- Total Market Demand and Market Segment Demand
- Derived Demand and Direct Demand
- Industry Demand and Company Demand
- Short-Run Demand and Long-Run Demand
- Price Demand
- Income Demand
- Cross Demand
What is Supply?
Supply is an essential economic impression that describes the total amount of a specific product or service that is available to consumers. Supply can link to the amount available at a specific price or the amount available cover a range of prices displayed on a graph. This connects closely to the demand for a product or service at a particular price; all else being equal, the supply presented by producers will increase if the price rises because all firms look to maximize profits. The concept of supply in economics is complicated with many mathematical formulas, practical applications, and contributing factors. While supply can relate to anything in demand that sold in a competitive marketplace, supply is most used to relate to goods, services, or labor. One of the major important factor that influence supply is the good’s price. Generally, if a good’s price rise so will the supply.
- Market Supply
- Short-term Supply
- Long-term Supply
- Joint Supply
- Composite Supply
- The stability between the quantity demanded and the price of a product at a given time is known as demand. On the other hand, the stability between the quantity supplied and the price of a commodity at a given time known as supply.
- Demand is the readiness and gainful capacity of a buyer at a specific price while Supply is the quantity provided by the producers to its customers at a specific price.
- Demand has an oblique relationship with the price, i.e., if price increases the demand decreases, as well as if the price decreases the demand increases, despite that, the price has a straight or direct relationship with supply, i.e., if cost or price increases the supply will also increase, in addition, if the price decreases supply also decreases.
- While demand curve slopes downward, the supply curve is upward sloping.
- Demand illustrates the customer’s inclination and preferences for a particular product demanded by him, though supply renders the firms.
The market flooded with several substitutes in each product category, and quick rise or fall in the prices will have an influence on these goods, and their demand and supply may increase or decrease. Demand and supply refer to the relationship price have with the quantity consumers demand and the quantity supplied by producers. As for price increases, quantity demanded decreases and quantity supplied increases.<|endoftext|>
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# What is Pythagoras Theorem?
Geometry is fun!
Right?
Okay, while you ponder that over, we think geometry is one of those subjects that make maths time so much more fun and interesting! What with their head-scratching diagrams and theorems, you are always left trying to figure out the most logical and diagrammatical way to reach that conclusion.
And the Pythagoras theorem is just one of those famous gripping head-scratchers!
With a rich history that dates back to 2500 BCE, it is believed that Pythagoras, the Greek philosopher whose name forms the first half of this theorem, studied all about it when he was a student in Egypt.
Should we make it more interesting? Well, the fun fact is that the Egyptians are believed to have known all about this fascinating theorem a thousand years before Pythagoras made it famous! Kudos to the Egyptians for being one of the most scientific and advanced civilizations of their time.
Anyway, cut to the chase, let us learn all about this fascinating theorem, shall we? Also known as Pythagoras law or Pythagorean theorem, there is a lot of logic and calculations in its framework. So let us dive straight in!
Also Read: What Is Tessellation In Geometry
## Pythagoras theorem: Decoding what it is all about
Okay, so to establish what exactly happens in a Pythagoras theorem, let us get the basics straight.
• You have a right angled triangle in Pythagoras law.
• The Pythagoras triangle is made up of three sides, vis a vis, the base (no explanation needed there), the altitude (which is the right angle side and is also known as the height or the perpendicular) and the hypotenuse (which is the longest side of the triangle).
• Also, remember that a right-angled triangle is one with a 90-degree angle. Never forget that. That angle is always going to be straight as an arrow.
• Now for the labels because these will form the foundation of the equation of this law. So you label the right angle side of the triangle as “a”, the base as “b” and the hypotenuse as “c”.
So far, so clear? You are with us? Okay, good. Now let us get to the formula for Pythagoras theorem.
## Pythagoras theorem equation: Figuring it all out
So this theorem, in the simplest of terms, is all about figuring out one simple thing: the relationship between all the sides of this right-angled triangle. That is it. That is the core. The crux of it.
But the question is how? How do we crack the formula for Pythagoras theorem?
For that, we have to look at its core statement, which goes something like this:
The square of the hypotenuse = square of the base + square of the altitude.
To spell it out:
c2 = a2 + b2
Now remember, this holds true for any right angled triangle, okay? Burn that in your memory.
The Pythagoras theorem equation will always be c2 = a2 + b2. And it will always conclusively prove that the square of the hypotenuse equals the sum of the square of the base and the square of the altitude.
## Pythagoras theorem proof: A brief how-to
Now let us come to the meat of the matter. How do we prove this theorem in geometry? Is there a way?
Yes and yes!
So there is a method called “similar triangles”, which you can use to establish the very truth (tried and tested) of the Pythagoras theorem and it goes something like this:
NOTE: This way works on a sort of relational analysis. So say you have two triangles with corresponding angles that measure the same and have corresponding sides of the same ratio, you can call them “similar”. Thus, you can sine law to prove c2 = a2 + b2.
• Step 1: Let us take a right-angled triangle. Let us call it ABC. The right angle in this triangle is B. Clear?
• Step 2: Cool. Now let us draw a perpendicular from B and call it BD. This perpendicular will meet AC at D, ergo BD.
• Step 3: Now you have two triangles △ABD and △ACB. What is common between them? For starters, ∠A is common to both. Also, ∠ADB = ∠ABC because both of them are right angles.
• Step 4: So you can effectively conclude that △ABD ∼ △ACB (because of the AA similarity).
• Step 5: Now you can also prove △BCD ∼ △ACB. How? Because we have already proven △ABD ∼ △ACB, you can assume that AD/AB = AB/AC. By that logic, you can also assume AD × AC = AB2.
• Step 6: Making sense so far? Good. Now similarly, you know that △BCD ∼ △ACB. Now going by that logic, you can safely assume that CD/BC = BC/AC and CD × AC = BC2.
• Step 7: Now what do you do? Any guesses? Now you add the two equations you have. Which is AB2 + BC2 = (AD × AC) + (CD × AC).
• Step 8: And now we are finally at the conclusion because by the above logic, AB2 + BC2 = AC (AD + CD). So AB2 + BC2 = AC2. Proven! You did it! You now know the Pythagoras triangle well!
Also Read: What Is A 30-60-90 Triangle?
## Pythagoras law: Real world applications
Phew! That was complicated, right? We totally get it if you are wondering whether the Pythagoras theorem will ever come in handy! But the truth is, it will. It is used in a lot of different fields. Let us check out some of them where the Pythagoras theorem equation finds its application:
1. In the field of construction and engineering
2. Did you know that architects actually frequently use this theorem? It helps them find what are known as unknown dimensions. It actually makes their jobs a lot simpler as they can measure unknown lengths or breadths pretty easily with Pythagoras theorem. Even engineers use it very commonly.
3. In face recognition systems of security cams
4. So here is the mind-blower! Each time you look into a security camera at someone’s door, the camera uses the Pythagoras theorem to calculate the distance between the camera and you. And then it processes that data to the lens to capture security footage. Fascinating!
5. In interior designing and woodworking
6. Woodworkers and interior designers often use this theorem to find the natural balance of the different elements at play in designing objects.<|endoftext|>
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1. ## proportions
Hello, I have a hard time figuring out where to start for this problem:
If $\displaystyle \frac{a}{b} = \frac{c}{d}$ prove that
$\displaystyle \frac{ a^2c + ac^2}{b^2d + bd^2} = \frac{(a+c)^3}{(b+d)^3}$
2. Originally Posted by Percent09
Hello, I have a hard time figuring out where to start for this problem:
If $\displaystyle \frac{a}{b} = \frac{c}{d}$ prove that
$\displaystyle \frac{ a^2c + ac^2}{b^2d + bd^2} = \frac{(a+c)^3}{(b+d)^3}$
From given, we know $\displaystyle b\ne 0,\;d\ne 0$. Now we consider 2 cases for this proof:
(1) If $\displaystyle a=c=0$, the prove is trivial.
(2) Now let's consider the case $\displaystyle a\ne 0$ and $\displaystyle c\ne 0$. By a simply manipulation, we found that showing $\displaystyle \frac{ a^2c + ac^2}{b^2d + bd^2} = \frac{(a+c)^3}{(b+d)^3}$ is equivalent to showing $\displaystyle \frac{ac}{bd} = \frac{(a+c)^2}{(b+d)^2}$. Now let's start from the only given identity and see what we can get.
$\displaystyle \frac{a}{b} = \frac{c}{d}\Rightarrow \frac{a}{c} = \frac{b}{d}\Rightarrow \frac{a+c}{c}=\frac{b+d}{d}\Rightarrow \frac{a+c}{b+d}=\frac{c}{d}\;\;\dagger$
Also realize that $\displaystyle \frac{a}{c} = \frac{b}{d}\Rightarrow\frac{c}{d}=\frac{a}{b}\;\;\ ddagger$
Combine $\displaystyle \dagger$ and $\displaystyle \ddagger$, we have
$\displaystyle \frac{a+c}{b+d}=\frac{c}{d}$ and $\displaystyle \frac{a+c}{b+d}=\frac{a}{b}$, multiply these two identities gives us $\displaystyle \frac{ac}{bd} = \frac{(a+c)^2}{(b+d)^2}$.
Roy<|endoftext|>
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Independence Day in Mongolia comes on 29 December to remember the date when Mongolia declared its independence from China in 1911.
|2019||29 Dec||Sun||Independence Day|
|2020||29 Dec||Tue||Independence Day|
The history of Mongolia’s struggle for independence goes back for centuries. After the great Mongolian Khanate empires finally weakened and crumbled, in the early 1600’s, Mongolia was threatened by the Manchus. Therefore, they went to the Chinese Ming Dynasty for help in defending themselves. However, the Manchus ended up taking control of China under the Qing Dynasty and also taking over Mongolia.
Finally, the Qing Dynasty fell in 1911, and Mongolia declared its independence that year on 29 December. Not yet succeeding in gaining their freedom on their own, in 1915, Mongolia got Russia to intervene in their behalf during the Khyagta Talks. The end result was that “Inner Mongolia”, still a province of China today, was separated from “Outer Mongolia”, which is modern day Mongolia. Both were given limited autonomy within the Chinese Empire.
In 1919, however, with Russia busy with a civil war, China cancelled the autonomy and invaded Mongolia. But the Soviet Union invaded Mongolia and defeated them, so that Outer Mongolia has remained a separate country ever since.<|endoftext|>
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Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### MEAP Preparation - Grade 7 Mathematics1.123 Integers - Positive and Negative Integers
Natural Numbers: Natural Numbers are 1,2,3,4,5,................ Whole Numbers: Whole Numbers are 0,1,2,3,4,5,............... Positive Numbers: Positive numbers are, 1,2 ,3 ,4 ,5................. Negative Numbers: Negative numbers are, ............-3, -2, -1. They are read as negative three, negative two, negative one. Integers: A set containing the positive numbers, 1, 2, 3, 4, ...., and negative numbers,............-3, -2, -1, together with zero is called a set of integers. In other words, Integers are defined as set of whole numbers and their opposites. Z = {..., -3, -2, -1, 0, 1, 2, 3, ...} {1, 2, 3, . . .} is the set of positive integers { . . . -3, -2, -1} is the set of negative integers Zero is not positive nor negative, but is both. A positive integer has a (+) or addition sign in front of the number and looks like ( 5 or (+ 5). Also note that a positive integer doesn't need a sign. A negative integer has a (-) or negative sign in front of the number and looks like (- 7). A negative integer can be thought of as owing someone something. Think of it as (-8) means owing someone 8 items or things. Zero in neither positive or negative and is always larger than a negative integer. Positive integers are always larger than zero and any negative number. The larger a positive integer is, the bigger it is. Therefore a +7 is greater than a +5, written another way +7 > +5. The larger a negative integer the smaller is its value. Therefore, (-6) < (-5). Directions: Answer the following questions. Write your own examples of positive and negative integer problems.
Q 1: whole numberA positive integer.A member of the set of positive integers and zero.An integer.all the others Q 2: Natural numbers are positive integers.truefalse Q 3: Zero is neither positive nor negative.falsetrue Q 4: If an integer is less than zero, we say that its sign is negative.falsetrue Q 5: Positive integers are all the whole numbers greater than zero: 1, 2, 3, 4, 5, ... .falsetrue Q 6: _________ numbers are used for counting.NaturalNegativeReal Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!<|endoftext|>
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Salmonella bacteria flip an electric switch as they hitch a ride inside immune cells, causing the cells to migrate out of the gut toward other parts of the body, according to a new study publishing on April 9 in the open-access journal PLOS Biology by Yaohui Sun and Alex Mogilner of New York University and colleagues. The discovery reveals a new mechanism underlying the toxicity of this common food-borne pathogen.
Salmonella are among the commonest, and deadliest, causes of food poisoning, causing over 400,000 deaths every year. Many of those deaths result when the bacteria escape the gut inside immune cells called macrophages. Macrophages are drawn to bacteria in the gut by a variety of signals, most prominently chemicals released from the site of infection. Once there, they engulf the bacteria as a regular part of their infection-fighting job. However, rather than remaining there, bacteria-laden macrophages often leave the site and enter the bloodstream, disseminating the bacteria and greatly increasing the gravity of the infection.
Tissues such as the gut often generate small electrical fields across their outer surfaces, and these electrical fields have been known to drive migration of cells, including macrophages. In the new study, the authors first showed that the lining of the mouse cecum (the equivalent of the human appendix) maintains a cross-membrane electrical field, and that Salmonella infection altered this field and contributed to the attraction of macrophages. Measurements of the polarity of the local charge indicated that the macrophages were attracted to the anode, or positively charged pole within the field. Once they engulfed bacteria, however, they became attracted to the cathode and reversed their migratory direction, moving away from the gut lining, toward vessels of the circulatory system. This switch was driven by a in the composition of certain charged surface proteins on the macrophages; the mechanism by which bacterial engulfment triggers this change is still under investigation.
"Dissemination, rather than localized infection, is the greatest cause of mortality from Salmonella (and other food-borne bacteria), and so understanding more about this polarity switch is likely to help develop new treatments to reduce deaths from food-borne bacterial infections," said Mogilner.
In your coverage please use this URL to provide access to the freely available article in PLOS Biology: http://journals.
Citation: Sun Y, Reid B, Ferreira F, Luxardi G, Ma L, Lokken KL, et al. (2019) Infection-generated electric field in gut epithelium drives bidirectional migration of macrophages. PLoS Biol 17(4): e3000044. https:/
Funding: This work was supported by US Army Research Office grant W911NF-17-1-0417 to A. M., by AFOSR FA9550-16-1-0052 to M.Z. (Program PI: Wolfgang Losert), by inter-department seed grant S-MPIDRGR from UC Davis School of Medicine to M. Z., Y. S. and R.M.Tsolis. F. F. was supported by Fundaça~o para a Ciência e a Tecnologia. (SFRH/BD/87256/2012). E.M. was supported by an early career award from the Burroughs Wellcome Fund and by NIH 1DP2OD008752. The funders had no role in study design, data collection and analysis, decision to publish, or preparation of the manuscript.
Competing Interests: The authors have declared that no competing interests exist.<|endoftext|>
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Paul's Online Notes
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### Section 2.6 : Infinite Limits
8. Find all the vertical asymptotes of $$\displaystyle g\left( x \right) = \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}}$$.
Hint : Remember how vertical asymptotes are defined and use the examples above to help determine where they are liable to be for the given function. Once you have the locations for the possible vertical asymptotes verify that they are in fact vertical asymptotes.
Show Solution
Recall that vertical asymptotes will occur at $$x = a$$ if any of the limits (one-sided or overall limit) at $$x = a$$ are plus or minus infinity.
From previous examples we can see that for rational expressions vertical asymptotes will occur where there is division by zero. Therefore, it looks like we will have possible vertical asymptote at $$x = - 5$$ and $$x = 9$$.
Now let’s verify that these are in fact vertical asymptotes by evaluating the two one-sided limits for each of them.
Let’s start with $$x = - 5$$. We’ll need to evaluate,
$\mathop {\lim }\limits_{x \to \, - {5^{\, - }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}}\hspace{0.25in}\,\,\,\,{\rm{and}}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{x \to \, - {5^{\, + }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}}$
In either case as $$x \to - 5$$ (from both left and right) the numerator is a constant -8.
For the one-sided limits we have the following information,
\begin{align*}x \to - {5^ - }\hspace{0.25in} \Rightarrow \hspace{0.25in}x < - 5\hspace{0.25in}\, \Rightarrow \hspace{0.25in}x + 5 < 0\\ x \to - {5^ + }\hspace{0.25in} \Rightarrow \hspace{0.25in}x > - 5\hspace{0.25in}\, \Rightarrow \hspace{0.25in}x + 5 > 0\end{align*}
Also, note that for $$x$$’s close enough to -5 (which because we’re looking at $$x \to - 5$$ is safe enough to assume), we will have $$x - 9 < 0$$.
So, in the left-hand limit, the numerator is a fixed negative number and the denominator is positive (a product of two negative numbers) and increasingly small. Likewise, for the right-hand limit, the denominator is negative (a product of a positive and negative number) and increasingly small. Therefore, we will have,
$\mathop {\lim }\limits_{x \to \, - {5^{\, - }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}} = - \infty \hspace{0.25in}\,\,\,\,{\rm{and}}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{x \to \, - {5^{\, + }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}} = \infty$
Now for $$x = 9$$. Again, the numerator is a constant -8. We also have,
\begin{align*}x \to {9^ - }\hspace{0.25in} \Rightarrow \hspace{0.25in}x < 9\hspace{0.25in}\, \Rightarrow \hspace{0.25in}x - 9 < 0\\ x \to {9^ + }\hspace{0.25in} \Rightarrow \hspace{0.25in}x > 9\hspace{0.25in}\, \Rightarrow \hspace{0.25in}x - 9 > 0\end{align*}
Finally, for $$x$$’s close enough to 9 (which because we’re looking at $$x \to 9$$ is safe enough to assume), we will have $$x + 5 > 0$$.
So, in the left-hand limit, the numerator is a fixed negative number and the denominator is negative (a product of a positive and negative number) and increasingly small. Likewise, for the right-hand limit, the denominator is positive (a product of two positive numbers) and increasingly small. Therefore, we will have,
$\mathop {\lim }\limits_{x \to \,{9^{\, - }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}} = \infty \hspace{0.25in}\,\,\,\,{\rm{and}}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{x \to \,{9^{\, + }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}} = - \infty$
So, as all of these limits show we do in fact have vertical asymptotes at $$x = - 5$$ and$$x = 9$$.<|endoftext|>
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## How To Teach A Child Meters Litre Kilograms
#### Math Mammoth Metric Measuring Worktext
A kilogram is a unit of mass in the Metric System. The symbol for kilogram is kg. The base unit for a kilogram is gram and the prefix is kilo. The prefix kilo is derived from the Greek chilioi meaning thousand and is symbolized as k in the Metric System.
#### Lesson Unit Conversions Centimeters & Meters BetterLesson
Children use the bathroom scales to weigh objects in kilograms. Other lessons teach grams and estimating weight. The goal is to let students become familiar with kilograms, and have an idea of how many kilograms some common things weigh. Volume The lessons include many hands-on activities so that the student gets first-hand experience in measuring, and has a basic knowledge of how “big
#### Kids Math Learning About Centimeters and Metric Measurement
Inspiring Your Child to Learn and Love Math Resource Guide / Introduction 2 Inspiring Your Child to Learn and Love Math Introduction This Parent Tool Kit was created speci?cally for parents of children in the elementary grades (junior kindergarten to grade 8) in Ontario. The goal of this resource is to provide parents with the most essential, research-based information to help them be the
#### Measurement Grams and Kilograms Worksheet Education.com
A father is trying to teach his child to ice skate. As the child stands still, the father pushes him forward with an acceleration of 2.0 meters per second2 north. If the child's mass is 20 kilograms, what is the force with which the father is pushing. (Since they are on ice, you can ignore friction.)
How to teach a child meters litre kilograms
#### Math Mammoth Metric Measuring Worktext
For instance, the conversion factor for kilograms and grams is 1,000 because 1 kilogram = 1,000 grams. To convert three kilograms to grams, multiply three by 1,000 to get 3,000 grams (3 x 1,000 = 3,000).
#### Kilograms (kg) Weight / Mass Conversions - CheckYourMath
To convert meter to centimeter, multiply the number by 100. To perform the reverse conversion, divide the number by 100. To perform the reverse conversion, divide the number by 100. Type 1:
#### Why is it important to weigh ourselves in kg? Metric Views
how to convert between kilograms and grams Students complete the task by following these steps. Measure and record (in grams) the mass of each item in their lunch box.
#### Converting between grams and kilograms worksheet by
A father is trying to teach his child to ice skate. As the child stands still, the father pushes him forward with an acceleration of 2.0 meters per second2 north. If the child's mass is 20 kilograms, what is the force with which the father is pushing. (Since they are on ice, you can ignore friction.)
#### Math Quiz for children measurement quiz kilograms
Teacher holds up an empty two-liter bottle and says, “So if I had a two-liter container of soda, how many milliliters of soda would I have?” [2,000 milliliters] Teacher writes 2 liters = 2,000 milliliters on the board and says, “That’s great, so we know that 2 liters is equal to 2,000 milliliters.
#### Lesson Unit Conversions Liters & Milliliters BetterLesson
Inspiring Your Child to Learn and Love Math Resource Guide / Introduction 2 Inspiring Your Child to Learn and Love Math Introduction This Parent Tool Kit was created speci?cally for parents of children in the elementary grades (junior kindergarten to grade 8) in Ontario. The goal of this resource is to provide parents with the most essential, research-based information to help them be the
#### Why is it important to weigh ourselves in kg? Metric Views
A key life skill children learn within this maths topic is how to measure accurately, using standard units (metres, centimetres, grams, kilograms, millilitres and litres). When introducing standard units it’s a good idea to mention some of the problems (noted above) of using non-standard measures to give meaning to the use of rulers, scales and measuring cups.
#### Meters Liters and Grams teachertube.com
Teacher writes centimeter ? 100 = meter and meter x 100 = centimeter on the board and says, “Can someone guess why I put a 100 in both formulas?” [Because there are 100 centimeters in a meter] This is a very important point to remember when we are converting Metric units of measurement. The Metric System is based on multiples of 10, just like place value. Therefore, multiplying and
#### Measurement Grams and Kilograms Worksheet Education.com
Length/distance: measured in meters, 1km is one kilometer = 1000 meters ~ approximately 1093 yd Volume: measured in liters, 1l is one liter = 1000 milliliter or 1000 ml ~ approximately 2.1134 US pt lqd.
### How to teach a child meters litre kilograms - Topic B Measuring Weight and Liquid Volume in Metric Units
#### how to teach mapping and practical skils
This unit describes the performance outcomes, skills and knowledge required to teach junior players fundamental Rugby League skills for modified games. It requires the ability to plan, conduct and evaluate drills, activities and games which focus on junior player development in the fundamental skills …
#### how to use an observor
I am using the TICC2640 for this purpose. I am trying to use the simple_observer example from the SDK to listen to beacons. I am able to scan the devices and print their MAC addresses.
#### how to search starly gts
No, only by using a cheat device, hacking or trading on GTS or with friends. If you want, you could ask a friend to breed a baby Chespin at the day care in exchange for a baby Froakie or Fennekin.
#### how to use track my golf
Using the arrows on the touch screen, you need to tell the app how you hold the club so it can more accurately track your swing. This is a nice feature, but as Robby pointed out, it's a little
#### how to see time.spent good in realm grinder
Since you will be dealing with huge amounts of data, it is a good thing to know such things as memory consumption, time spent in executing the cells, especially the cells that are doing most of
#### monitor riser how to use
Product Overview Fellowes Smart Suites Black/Clear Monitor Riser 8020101 The Fellowes Black Smart Suites Monitor Riser is a height adjustable monitor base that helps to position your monitor at a customisable height to reduce aches and strains associated with extensive computer use.
### You can find us here:
Australian Capital Territory: Goomburra ACT, Tuggeranong ACT, Moncrieff ACT, Denman Prospect ACT, Brookfield ACT, ACT Australia 2682
New South Wales: Kulnura NSW, Mayfield West NSW, Long Beach NSW, Murrumbateman NSW, Abercrombie NSW, NSW Australia 2075
Northern Territory: Alyangula NT, Nhulunbuy NT, Mataranka NT, Alyangula NT, Wurrumiyanga NT, Ngukurr NT, NT Australia 0861
Queensland: Bajool QLD, Dysart QLD, East Brisbane QLD, Wallumbilla QLD, QLD Australia 4037
South Australia: Stanley Flat SA, Loxton SA, Magill SA, East Melbourne SA, Thebarton SA, Willunga South SA, SA Australia 5089
Tasmania: Bell Bay TAS, Perth TAS, Orielton TAS, TAS Australia 7043
Victoria: Tinamba VIC, Glenorchy VIC, Curyo VIC, Newry VIC, Codrington VIC, VIC Australia 3005
Western Australia: Robinson WA, Loongana WA, Mooterdine WA, WA Australia 6054
British Columbia: Williams Lake BC, Cumberland BC, Cumberland BC, White Rock BC, Creston BC, BC Canada, V8W 6W4
Yukon: Frances Lake YT, Eagle Plains YT, Whitefish Station YT, Little Teslin Lake YT, Grand Forks YT, YT Canada, Y1A 4C5
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Nova Scotia: Westville NS, Oxford NS, Bridgetown NS, NS Canada, B3J 4S8
Prince Edward Island: Clyde River PE, Crapaud PE, Georgetown PE, PE Canada, C1A 9N8
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England: Batley ENG, Slough ENG, Bebington ENG, Dewsbury ENG, Maidenhead ENG, ENG United Kingdom W1U 8A7
Northern Ireland: Derry (Londonderry) NIR, Bangor NIR, Craigavon (incl. Lurgan, Portadown) NIR, Newtownabbey NIR, Craigavon (incl. Lurgan, Portadown) NIR, NIR United Kingdom BT2 2H7
Scotland: Aberdeen SCO, Cumbernauld SCO, Glasgow SCO, Paisley SCO, Aberdeen SCO, SCO United Kingdom EH10 4B8
Wales: Wrexham WAL, Neath WAL, Swansea WAL, Cardiff WAL, Cardiff WAL, WAL United Kingdom CF24 4D7<|endoftext|>
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Disinfection is defined as the reduction in the number of viable microorganisms on a surface of a medical device, to a level previously specified.
Medical devices are disinfected in order to render them safe to use. Thermal disinfection is often used in washer disinfectors to render surgical instruments safe to handle after washing. Chemical disinfection is used to process semi-critical medical devices. Semi critical medical devices are devices that come into contact with intact mucous membranes or non-intact skin. Examples of semi-critical medical devices include flexible endoscopes, rigid endoscopes, and ultrasound probes. Surgical instruments that come into contact with a sterile cavity should rather be sterilized and not disinfected. Chemical disinfection is achieved by soaking a device in a disinfectant. There are various products available on the market that can be purchased to disinfect semi-critical medical devices. Disinfectant formulations may and sometimes do sound similar but they can vary in their antimicrobial activity (efficacy), surface compatibility and safety aspects.
Always follow the manufacturers’ instruction for use carefully when using chemical disinfectants. The manufacture will stipulate how long the device should soak for and how many times the device should be rinsed with sterile water afterwards. Always wash, rinse and dry instruments thoroughly before immersing them in the disinfectant. If this is not done the disinfectant can easily be diluted, and may no longer be effective.
Some disinfectants remain active for long periods of time whilst other disinfectants need to be mixed on a daily basis. There are advantages and disadvantages associated with this. Disinfectants that remain active for long periods of time may be become soiled with patient debris, may become diluted and are often ‘topped up’ which is not good clinical practice. Disinfectants that remain active for one day must be mixed on a daily basis.
It is safe to use high level disinfection if the manufactures instructions are followed precisely, and as long as only semi-critical items are being disinfected.<|endoftext|>
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Tigrinya is a Semitic language that is widely spoken in Ethiopia and Eritrea. It is also spoken in many parts of the world where there are Eritrean and Ethiopian communities.
There are two major variants: Eritrean Tigrinya and Ethiopian Tigrinya.
Tigrinya is written with an adapted version of the Ge’ez script. Ge’ez is the ancient classical language of Ethiopia, now a liturgical language of the Ethiopian Orthodox Church.
When translating from Tigrinya into English, the word count tends to increase.
The church is a central feature of Tigray communities and their daily life. Christianity was introduced to the Tigrays hundreds of years before most of Europe. Churches were cut into cliffs or into single blocks of stone. Many Tigray holidays are therefore associated with the religious calendar. Christmas is celebrated on 7 January, Epiphany on 19 January, and Easter in the month of April.<|endoftext|>
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Magnesium and calcium are equally important in preventing Osteopenia and Osteoporosis
Osteoporosis is a worldwide health concern. International Osteoporosis Federation (IOF) estimates that it affects 75 million people in the United States (US), Europe and Japan. As Europe’s population ages and becomes more sedentary, the number of people affected by osteoporosis is expected to increase significantly – hip fractures alone are expected to double in the next 50 years.
What are Osteopenia and Osteoporosis?
Osteopenia is a condition where bone mineral density is lower than normal. It is considered a precursor to osteoporosis. People with osteoporosis suffer from reduction in their bone mass and bone quality—their bones become fragile, leading to an increased risk of fractures.
In medical terms, Osteopenia is a situation where bone density is lower than the norm and T score (which defines the standard deviation from a healthy bone) is between -1.0 and -2.5. In Osteoporosis the T-score value is lower than -2.5 and consequences are more dangerous.
Wrongly often thought of as an “old woman’s disease”, osteoporosis affects not only one in three Postmenopausal women, but also one in five men over the age of 50,younger women and even children. Because there are often no symptoms until a fracture occurs, Osteoporosis is known as the “silent disease”. Once a fracture has occurred, the risk of a future fracture is at least doubled within one year.
Risk factors for developing osteopenia and osteoporosis include:
- Age – Decreasing oestrogen levels in older women and decreasing testosterone levels in older men can lead to deterioration in bone density. This is why the more senior members of society have to be particularly careful of their bones.
- Genetics – Those with a family history of osteoporosis, osteopenia and bone fractures are more likely to inherit these traits. Europeans and Asians are two ethnic groups that have been noted for their predisposition to develop osteoporosis.
- Gender – Women are at an elevated risk of developing osteopenia and osteoporosis. Women start with lower bone densities and have a quicker loss of bone mass.
- Pre-existing fractures – People that already have bone fracture or fractures are two times more likely to get another fracture. This is especially true for the elderly
Magnesium and Bone Health
Magnesium is the fourth most abundant mineral in the body. Approximately 50 per cent of it is present in our bones. Magnesium helps prevent osteopenia and osteoporosis in the following ways:
- Magnesium helps control hundreds of enzymatic reactions in cells that influence bone density
- Magnesium is required for the formation of proteins that help form bone
- Magnesium serves as a calcium regulator. Magnesium is important in calcium metabolism because it is required for secretion of Parathyroid Hormone (PTH). PTH increases the production of the active form of vitamin D, and plays a role in the absorption of calcium and phosphorus (another important bone mineral). It teams up with vitamin B6 to regulate the absorption of calcium into the bone. Magnesium also helps transport calcium out of the gastrointestinal tract to form bone tissue
- Magnesium deficiency alters calcium metabolism and the hormones that regulate calcium, so individuals with chronically low blood levels of calcium may actually be deficient in magnesium
- Researchers at Yale University School of Medicine, United States found that adolescent girls who were given magnesium supplements had significant increases in body mineral content in the hip bones and spine. Another study carried out in Turkey found that Magnesium supplementation increased bone mineral density when used in the treatment of osteoporosis.
Magnesium Supplements for Osteoporosis
If you have been diagnosed with osteopenia and osteoporosis or are at risk to develop it, you should increase your intake of Magnesium. Magnesium is a balancing mineral for calcium and it is recommended that you take both in the ratio of at least 1:1.
You can either try magnesium only supplements like Mag365 or opt for a supplement that offers both calcium and magnesium. Mag365 is a magnesium chloride-based supplement with high bioavailability of magnesium. Mag365 plus Calcium is a superior Magnesium-Calcium formulation with magnesium and calcium in ratio of 3:2. Both of these can give much needed boost to the health of your bones.
In addition to taking magnesium supplements, you must engage in regular physical activity that includes weight bearing exercises, such as low-impact aerobics, jogging, and walking. Moderate weight training will put some stress on the bones; encouraging them to become stronger over time. Regular check-ups to monitor bone loss are recommended in people over 50.
- Osteoporosis in Europe: Indicators of progress. http://www.iofbonehealth.org/download/osteofound/filemanager/publications/pdf/eu-report-2005.pdf
- Wikipedia.org, “Osteopenia”, accessed 2011-10-08. URL: http://en.wikipedia.org/wiki/Osteopenia
- Wikipedia.org, “Osteoporosis”, accessed 2011-10-08. URL: http://en.wikipedia.org/wiki/Osteoporosis
- PubMed.gov, “Epidemiology Worldwide”, accessed 2011-10-08. URL: http://www.ncbi.nlm.nih.gov/pubmed/12699289
- EveryDayHealth.com, “Why Osteoporosis is More Common in Women”, accessed 2011-10-08. URL: http://www.everydayhealth.com/osteoporosis/osteoporosis-and-gender.aspx
- PubMedCentral.gov, “History of fractures as predictor of subsequent hip and nonhip fractures among older Mexican Americans”, accessed 2011-10-08. URL: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2569658/?tool=pmcentrez
- Calcium and Bone Disorders in Children and Adolescents. Physiology of Calcium, Phosphate and Magnesium. Allgrove, J., Shaw, N.J. (eds): Calcium and Bone Disorders in Children and Adolescents. Endocr Dev. Basel, Karger, 2009, vol. 16, pp. 8-31
- Short-term oral magnesium supplementation suppresses bone turnover in postmenopausal osteoporotic women www.ncbi.nlm.nih.gov/pubmed/19488681?ordinalpos=15&itool=EntrezSystem2.PEntrez.Pubmed.Pubmed_ResultsPanel.Pubmed_DefaultReportPanel.Pubmed_RVDocSum<|endoftext|>
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# From set with n elements how many ways to select two disjoint subsets of size k and r?
I am new to combinatorics and and struggling with the following questions.
If you have a set of n elements and you need to select two disjoint subsets containing k and r elements respectively. In how many ways can you select these subsets?
{the following text was added on 20 February 2017 after a request was received for more details and context}
Herewith an example of such a problem using n = 5 and k = r = 2: Consider a set S with 5 elements:
N = {1,2,3,4,5}
Calculating all the ways in which a subset of k (for example 2) elements can be selected from this set can be calculated using the standard formula for $${n \choose k}$$ and results in 10 ways.
If we for some reason need to count the number of ways in which two disjoint subsets (example 2 and 2) can be selected from set N I might use the following method. One way to think about this problem is as follows (Thank you drhab for your guidance on this):
I can first do a standard n choose k formula. 5 choose 2 = 10. Then I can re-do the formula for the second set using (5 - 2) choose 2 which gives 3. Finally I multiply the two results to get an answer of 10 * 3 = 30. This gives 30 ways to select 2 disjoint subsets of size 2 from a set of 5 elements.
Application of this logic to another problem as an example: How many ways can you choose two disjoint subsets from a set of 10 elements where the subsets have exactly 4 and 3 elements each? The answer (if the above logic is correct) is $${10 \choose 4} * {6 \choose 3}$$ which gives 210 * 20 = 4200 ways.
This feels correct to me ,however, I am not 100% sure. Please confirm whether it is.
• Start by selecting $k$ elements of $n$ at the first round. Can you figure out how many possibilities there are? After that select $r$ elements of the $n-k$ elements that were not selected at the first round. – drhab Feb 19 '17 at 14:34
• @drhab If $r=k$ then you double count. – ajotatxe Feb 19 '17 at 14:50
• @ajotatxe You are correct (I overlooked that possibility). – drhab Feb 19 '17 at 14:55
First, select a subset with $r+k$ elements. There are $\binom n{r+k}$ ways to do this.
Then, if $r\neq k$, select a subset with $r$ elements from the previously chosen ones. There are $\binom{r+k}r$ ways to do this.
But if $r=k>0$ then we must divide by $2$ because choosing first $r$ elements is the same as choosing the other $r$ elements.
To sum up, the solution is $$\begin{cases} \frac12\binom n{2r}\binom{2r}r\text{ if }k=r>0 \\\binom{n}{r+k}\binom{r+k}r\text{ otherwise}\end{cases}$$<|endoftext|>
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Where Mosquitoes Come From
The three main types of mosquitoes illustrate where they originate:
- Floodwater mosquitoes lay their eggs in soil that became moist from irrigation, rain, or soil close to main water sources such as ponds, lakes, streams, or rivers. The eggs need to dry out with the soil and then become wet again to trigger the cue to hatch. In some cases this will happen in the same season, but eggs laid in areas with infrequent rains and eggs laid at the end of the season will settle into protective cracks and crevices in the soil and lie dormant until the next year. Some species of floodwater mosquitoes can survive multiple years until the right conditions occur for the eggs to hatch.
- Permanent water mosquitoes need a constant source of water for the life cycle to be completed. Their eggs typically cannot withstand a drying out period. Eggs are laid directly on the water, in reeds, or on rafts and typically hatch after 24 hours.
- Container mosquitoes lay their eggs in anything large enough to hold water. In natural scenarios, this includes tree holes, bromeliad leaf axils, bamboo trunks, or other water filled cavities. Artificial containers are vast and include items such as old tires, kiddy pools, bird baths, wheelbarrows, toys, or even gutters with blockages providing small areas of water containment.
Rove’s Mosquito Services
Prevention – A mosquito technician will identify the type of mosquitoes affecting your property and help determine the adjustments that may be made to minimize the reproductive opportunities. This source reduction may be any combination of adjustments to habits, landscaping, or structures.
Larva control – In cases where landscape, structural, and other alterations are not feasible, or where alterations need supplemental treatment, there are treatments that may be done to prevent larva from completing their life cycle. The method of this treatment will vary based off of the biology of the mosquitoes of concern.
Adult control – In situations where sources are outside of the control of the property owner, it may be necessary to focus on the adult mosquitoes. The method of treating the adults will depend on the goal of mosquito control as well as the biology of the specific mosquitoes in question. Some mosquitoes may travel miles from their breeding site while others will remain very close. Different options are available for those that need to prepare for specific outdoor events than those that are near a continuous source of mosquitoes. A Rove Pest Control technician will work closely with you to determine goals and thresholds to be achieved and will customize a mosquito control system to fit your individual needs.
Out of nearly 3,000 species of mosquitoes throughout the world, 176 species are recognized in the US. Only a few of those are considered vectors of malaria, yellow fever, dengue, filariasis, St. Louis encephalitis (SLE), Western Equine encephalitis (WEE), LaCrosse encephalitis (LAC), Eastern Equine encephalitis (EEE) and West Nile virus (WNV), and dog heartworm.
In addition to taking efforts to control mosquitoes around the workplace and home, it is important to take other cautions when going to areas where mosquitoes may be of concern. Wearing clothing that provides greater coverage and supplementing with ambient or personal repellents will minimize the chance of bites. A simple rule is to simply go somewhere else if mosquito activity is high.
Read Our Mosquito Blog Posts:
- Why do mosquitoes bite some people and not others
- Is your landscaping hiding pests?
- Does a long winter reduce mosquito numbers
- How to use moisture management to reduce pest activity
- Are snow mosquitoes real?
- Why do some bug bites hurt while others just itch?
- How long can mosquitoes survive?
- What do gutters have to do with pest control?<|endoftext|>
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How many equal angles does an equilateral triangle have
An equilateral triangle is a type of triangle where all three sides are of equal length. This means that all three angles of an equilateral triangle are also of equal measure. So, how many equal angles does an equilateral triangle have? The answer is simple – an equilateral triangle has three equal angles, each measuring 60 degrees.
To understand why an equilateral triangle has three equal angles, let’s delve into its properties. Firstly, because all three sides are of equal length, all three angles must also have the same measure. This is true for any triangle – the sum of all angles must be 180 degrees. Since an equilateral triangle has three equal angles, each must measure 60 degrees to satisfy this property.
Furthermore, the concept of an equilateral triangle can be visualized by imagining an equilateral polygon with infinitely many sides. As the number of sides increases, the shape approaches that of a circle. A circle has 360 degrees in total, so dividing this by the number of sides would yield 60 degrees for each angle of the equilateral triangle.
Equilateral triangles have a variety of fascinating properties, including their ability to tessellate a plane and the fact that they are the only type of triangle that can be inscribed in a circle with all three vertices on the circle. These unique characteristics make equilateral triangles an intriguing subject of study in geometry.
What is an equilateral triangle?
An equilateral triangle is a special type of triangle where all three sides are equal in length. It is classified as an acute triangle, meaning that all angles inside the triangle are less than 90 degrees.
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In an equilateral triangle, all three angles are also equal to each other. This means that each angle measures 60 degrees. The sum of the angles in an equilateral triangle is always 180 degrees, just like any other triangle.
Because all sides and angles are equal in an equilateral triangle, it exhibits a high degree of symmetry. The equilateral triangle is often used in architecture, engineering, and design due to its aesthetically pleasing and balanced proportions.
Definition and properties
An equilateral triangle is a special type of triangle where all three sides are congruent, meaning they have the same length.
Some key properties of an equilateral triangle:
1. Shape: An equilateral triangle has three straight sides and three vertices. Its shape resembles the letter “E” rotated.
2. Angles: Since all three sides are congruent, all three angles in an equilateral triangle are equal.
3. Angle measure: Each angle in an equilateral triangle measures 60 degrees.
4. Perpendicular bisectors: The perpendicular bisectors of the sides in an equilateral triangle all intersect at a single point called the circumcenter.
5. Height and base: The height of an equilateral triangle is a line segment perpendicular to the base and passing through the vertex opposite the base. The height divides the equilateral triangle into two congruent right triangles.
6. Area: The area of an equilateral triangle can be calculated using the formula:
Area = (side length2 × √3) / 4.
7. Perimeter: The perimeter of an equilateral triangle is simply three times the length of one side.
Equilateral triangles have a number of unique properties that make them important in various math and geometry applications, including tessellations, symmetry, and trigonometry. They are also commonly found in nature and design due to their pleasing symmetry and balanced appearance.
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How many angles does an equilateral triangle have?
An equilateral triangle is a special type of triangle where all three sides have the same length and all three angles are equal. This makes an equilateral triangle unique as it is the only triangle with all equal angles. Therefore, an equilateral triangle has three equal angles.
Each angle in an equilateral triangle measures 60 degrees. This means that the sum of all the angles in an equilateral triangle is 180 degrees. You can calculate the measure of each angle by dividing the total sum of the angles by the number of angles, which in this case is 3.
It is important to note that the property of having equal angles in an equilateral triangle is a result of its symmetry. Since all three sides are equal, the triangle forms a balanced shape where each angle contributes equally to its overall structure.
The equal angles in an equilateral triangle make it a fundamental shape in geometry, and it is often used as a building block for more complex shapes and constructions. The equilateral triangle also has a number of interesting properties and applications in various fields including art, architecture, and mathematics.
In conclusion, an equilateral triangle has three equal angles, with each angle measuring 60 degrees. This property of having equal angles is a defining characteristic of an equilateral triangle and makes it a unique and important shape in geometry.
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Understanding equal angles
Angles are a fundamental concept in geometry that describe the measure of rotation between two lines or between a line and a plane. In mathematics, an angle is typically measured in degrees or radians.
An equilateral triangle is a special type of triangle with three equal sides and three equal angles. Each angle in an equilateral triangle measures 60 degrees.
When we say that the angles of an equilateral triangle are equal, we mean that all three angles are the same size. The equal angles are formed at each vertex of the triangle where two sides meet.
The fact that an equilateral triangle has equal angles is a consequence of its symmetry. Because all three sides are congruent and all three angles are congruent, the triangle is symmetric about its center. This symmetry ensures that all the angles are the same size.
The measure of each equal angle in an equilateral triangle is always 60 degrees. This means that regardless of the size of the triangle, the three angles will always be equal.
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For example, imagine scaling up an equilateral triangle to be really big or scaling it down to be really small. Even though the size of the triangle changes, the angles within the triangle remain the same- all measuring 60 degrees.
Understanding the concept of equal angles is key to understanding and working with equilateral triangles. By knowing that equilateral triangles have three equal angles, mathematicians can make conclusions and solve problems related to these unique triangles.
Harrison Clayton
Meet Harrison Clayton, a distinguished author and home remodeling enthusiast whose expertise in the realm of renovation is second to none. With a passion for transforming houses into inviting homes, Harrison's writing at https://thehuts-eastbourne.co.uk/ brings a breath of fresh inspiration to the world of home improvement. Whether you're looking to revamp a small corner of your abode or embark on a complete home transformation, Harrison's articles provide the essential expertise and creative flair to turn your visions into reality. So, dive into the captivating world of home remodeling with Harrison Clayton and unlock the full potential of your living space with every word he writes.<|endoftext|>
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Mathwizurd.com is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.
## Nov 2 Efficient Way to Calculate Determinants
Before we do this, we have to define what a determinant is.
# Determinant
MathJax TeX Test Page The determinant of a $2\times2$ matrix $\begin{bmatrix}a & b\\c & d\end{bmatrix}$ is $ad - bc$. We can define the determinant of an $n \times n$ as the cofactor expression across any row or down any column. Row i: $$\det A = a_{i1}(-1)^{i + 1}\det(A_{i1}) + ... + a_{in}(-1)^{i + n}\det(A_{in})$$ Column j $$\det A = a_{1j}(-1)^{1 + j}\det(A_{1j}) + ... + a_{nj}(-1)^{n + j}\det(A_{nj})$$
# Important Facts
MathJax TeX Test Page $$\det(A^T) = \det(A)$$ $$\text{Row reduction maintains the determinant, meaning } \begin{vmatrix}a & b\\c & d\end{vmatrix} = \begin{vmatrix}a & b\\ka + c & kb + d\end{vmatrix}$$ $$\text{Row swapping negates the determinant, meaning } \begin{vmatrix}a & b\\c & d\end{vmatrix} = -\begin{vmatrix}c & d\\a & b\end{vmatrix}$$ $$\text{Row scaling scales the determinant, meaning } k\cdot\begin{vmatrix}a & b\\c & d\end{vmatrix} = \begin{vmatrix}a & b\\kc & kd\end{vmatrix}$$ $$\det(AB) = \det(A)\cdot\det(B)$$
# Example One
MathJax TeX Test Page $$\det\left(\begin{vmatrix}4 & 3 & 9\\0 & 2 & 7\\-5 & 10 & 10\end{vmatrix}\right)$$ $$=-5 \cdot \begin{vmatrix}4 & 3 & 9\\0 & 2 & 7\\1 & -2 & -2\end{vmatrix}$$ $$\text{ Remember that if you scale inside, you must scale outside}$$ $$-5 \cdot \begin{vmatrix}0 & 11 & 17\\0 & 2 & 7\\1 & -2 & -2\end{vmatrix} \text{r_1 = r_1 + -4r_2}$$ $$-5 \cdot \begin{vmatrix}0 & 7 & 3\\0 & 2 & 7\\1 & -2 & -2\end{vmatrix} \text{r_1 = r_1 - 2r_2}$$ $$\text{Note- I didn't eliminate the first row, I just reduced the size of the numbers, making the calculation easier}$$ $$-5 \cdot \begin{vmatrix}7 & 3\\2 & 7\end{vmatrix} = -5 \cdot (49-6) = -215$$
# Example Two
MathJax TeX Test Page $$\det\left(\begin{vmatrix}5 & -1 & 8\\2 & -2 & -8\\-7 & -2 & -2\end{vmatrix}\right)$$ Remember, the determinant of the transpose is the same as the determinant of the regular matrix, so we can do column operations!!! (basically, you transpose, then do a row operation, the transpose back) $$\det\left(\begin{vmatrix}0 & -1 & 8\\-8 & -2 & -8\\-17 & -2 & -2\end{vmatrix}\right) \text{C_1 = C_1 + 5C_2}$$ $$\det\left(\begin{vmatrix}0 & -1 & 0\\-8 & -2 & -24\\-17 & -2 & -18\end{vmatrix}\right) \text{C_3 = C_3 + 8C_2}$$ $$-6\det\left(\begin{vmatrix}0 & -1 & 0\\-8 & -2 & 4\\-17 & -2 & 3\end{vmatrix}\right) \text{C_3 = C_3/6}$$ $$-6\det\left(\begin{vmatrix}0 & -1 & 0\\0 & -2 & 4\\-11 & -2 & 3\end{vmatrix}\right) \text{C_1 = C_1 + 2C_3}$$ $$-6\det\left(\begin{vmatrix}0 & 4\\-11 & 3\end{vmatrix}\right) = -6 \cdot 44 = \boxed{-264}$$<|endoftext|>
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Augmented reality is computer simulation triggered by a real world context. The big question for educational institutions is: how does this foster deep learning. For the 8th Innovation Room CEL invited four (groups of) experts in the field to present their findings and experiences.
Eric Klopfer: Teaching and learning by playing and creating augmented reality games
Keynote speaker is MIT professor Eric Klopfer, author of the book Augmented Learning. His research focuses on serious gaming and simulations in learning. The question is: ‘How do we use the real world in learning and how can we improve and enrich this with virtual objects? Before giving answers he shares with the audience the components he thinks are important to successful use of games in education.
Edutainment (‘Where play is the reward for learning’) is useless in learning; a game without a relation with learning material only serves as a reward in itself and if you skip the reward (game) the learner’s motivation disappears. This principle is called ‘chocolate covered broccoli’: A child will eat the broccoli but will not be motivated by the chocolate to do this more often. Klopfer’s example is the game called ‘Math Blaster’: it is fun to do but has nothing to do with mathematical problems, whereas the goal is to give the learner a learning experience by playing the game. Hereto you should look for a ‘pleasantly frustrating experience’; gaming is extremely suitable for this, as long as it is based on good design principles, Klopfer emphasizes.
The way and extent in which the game is structured has a huge effect on the enjoyment of the learner. Eric gives five design principles:
- The learner should be able to make interesting and relevant choices in the game
- These choices should have consequences in the game
- The game has to have clear goals
- Visual feedback is an important stimulator
- The system/model of the game should be clear and coherent.
To keep a game interesting you should consider the ‘four freedoms of play‘ when designing a game: players should have the opportunity to fail, experiment, adopt several identities and control the extend of energy they put in the game.
The use of augmented reality in educational games provides chances to combine social and sociological aspects with science. Making the connection between the real and the virtual world is the main challenge. How do you reinforce the learning experience with virtual elements and how do you prevent these elements to distract the learner from the learning materials? To design AR games MIT developed the Taleblazer platform. Eric shows an example of an AR game in an open air museum where players go through the game at their own pace but interact with each other. By using a positioning tool virtual elements appear at the right moment. When GPS cannot be used as positioning tool, iBeacons is used. At the end of his presentation Klopfer pleas for learners to make their own AR games, a nice suggestion!
What motivates the learner to play the game in the first place?, a member of the audience would like to know. ‘Two elements are key’, according to Klopfer: ‘1. a narrative (e.g. about empowerment, rebellion, depending on the age of the player) and 2: a series of interesting problems to keep them going.’
Another question was: How can we use prior knowledge in games? This turns out to be a real challenge. Klopfer: ‘This would be an important next step.’
Beerend Hierck and Thomas Hurkxkens: Learning in mixed reality with the Microsoft Hololens
Beerend Hierck (teaches anatomy at Leiden University Medical Centre) en Thomas Hurkxkens (New media lab Leiden University) present their project ‘With holographic projection to customized education’ which was one of the winners of the Surf Innovation Challenge 2016/2017.
Inspired by Holonatomy app of Case Western University in Cleveland and the case of Marco van Basten, the famous Dutch soccer player who suffered from severe ankle problems, Hierck en Hurkxkens developed the idea of developing a virtual model of the ankle and pair this to movements of the ankle of students.
The next step is to combine the AR learning tool with story telling techniques, using the story of soccer player Marco van Basten: ‘How can we help Marco van Basten to play soccer again?’ The big question for Hiercks is: will these educational innovations turn students into better doctors?
Programming a good virtual ankle was not the biggest problem, but the use of a real ankle to control the virtual model appears to be very complicated. This is not possible with the Hololens and Kinect turns out to give a huge delay, so some challenges still have to be defeated. When the app is finished, it will be available for anybody. The Erasmus Medical Centre is currently developing a similar app to use in plastic hand surgery. At Leiden University, archaeologists are also involved in using the Hololens in education. LDE CEL is looking for opportunities to pair these initiatives to develop generic learning scenarios and feedback tools.
Nelson Jorge, Sofia Dopper and Johanetta Gordijn: Augmented and virtual reality for today’s learners
After a short break Nelson Jorge, Sofia Dopper en Johanetta Gordijn of the Delft Extension School put the audience to work. The message is that AR and VR are in the pockets of us all; we all have a smartphone with recording devices. Which lightweight applications can the audience invent in order to be able to apply AR/VR in education at a low cost base? Each table gets an ‘AR & VR Learning Experience Canvas’ to elaborate the ideas.
The audience came up with various ideas: rebuild the Rotterdam city centre using augmented reality on a smartphone, use the Hololens in vocational education about arthritis of the hand, use AR in a public transport setting.
Pasi Vahimaa and Juha Eskelinen : Sm4rtlab: internet of things meets augmented reality in laboratories
‘Using one another’s labs will be the future of all labs’ Pasi Vahimaa en Joha Eskelinen (both University of East Finland) predicted, when presenting their project ‘Sm4rtlab’. The idea of Sm4rtlab is to maximise the use of expensive labs and infrastructures by making it possible to control the devices and cooperate on distance via an internet-of-things-cloud-solution.
They virtualised Pashi’s optometric lab in Finland with Unity, so experiments in the lab can be done at a distance. In the virtual environment equipment can be controlled via the Internet of Things. Webcams show what happens in the lab. With the Hololens one can project and control your own environment in the virtualised lab.
If the virtualised lab is not connected to the real lab it is possible to do more experiments at the same time. Virtualised components can also be duplicated: you buy one lens and one laser and you are able to create a setup with 100 lenses.
We would like to thank Pieter van Rooij of the Ixperium centre of expertise for his contribution to this report.<|endoftext|>
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Reading, Writing, Arithmetic – and ‘Oracy’ and Critical Thinking
While the 3 R’s – reading, writing and arithmetic – are undoubtedly important, they’re not the only skills that students need to succeed. Some K-12 schools are modifying the core curriculum to reflect that. For example, at School 21, a public school in London, eloquence is a core value, and the school teaches “oracy” or oral communication in every grade. Others are adding critical thinking.
What does oracy entail? Students learn voice projection, tonal variation, fluency and pace of speech, and rhetorical technique. But beyond the physical and linguistic aspects, these pupils also learn the cognitive, social, and emotional facets of oral communication. What accounts for changes in the core curriculum? Several factors. No less an authority than Warren Buffet recently highlighted the importance of public speaking skills. And beyond that, a report by MindEdge reveals that almost half of Millennials get an “F” in critical thinking.
As part of these oracy classes, students learn how to listen and respond, take turns, gauge the audience, and manage interactions. They are also taught how to structure their ideas, build on the opinions of others, ask questions for clarity, and provide reasons to support their ideas. In other words, they learn the foundation for critical thinking.
Bringing more skills to U.S. students
Closer to home, GEMS World Academy Chicago, part of the international GEMS Education network, is a pre-K through 12 school that teaches emotional and social skills, in addition to problem-solving and critical thinking techniques. GEMS believes that children should be taught these strategies at an early age: Learning how to analyze data and make decisions based on facts instead of emotion lays a foundation for success in every area of life.
“In a global world, children encounter diverse thoughts and opinions,” explains Denise Gallucci, CEO of GEMS Education and The Education Partners. “As educators, it is our job to ensure that students are learning how to develop their own thoughts, communicate effectively with their peers, and accept differing opinions,” Gallucci tells GoodCall®.
She believes it’s never too early to start teaching these skills. “Kids need to learn how to develop arguments rooted in fact as early as kindergarten,” Gallucci says. “Five- and six-year-olds are still developing cognitively, and this is the time where students are learning how to work together as a team while also figuring out where they fit in the world.”
While many adults struggle to receive negative feedback, these youngsters are learning the benefits of criticism in honing their views and arguments. “At GEMS, students learn to provide and receive peer feedback, equipping them with the skills to defend their work and grow from constructive criticism.”
Collaborative, diverse experiences are also a part of the GEMS experience, and students are taught how to have meaningful interactions with fellow peers from a variety of backgrounds. “GEMS World Academy Chicago works with seven other world academies: our students spend four days in the school and one day in the field, and students have collaborative working groups with students in Dubai, France, and Singapore, providing a truly global perspective.”
Never too young or too soon
Some people might think that students are not able to firmly grasp oracy and other concepts at such a young age. However, Gallucci explains, “Critical thinking skills should be developed at an early age and can be challenged and strengthened over time.” The school uses various types of projects that include technology, global studies, and language to stimulate imagination and inquisitiveness. “As students transition in grade level, their studies become deeper and more comprehensive – by building on previous skills, we encourage complex thinking.”
There is strong support for stimulating analytical and critical thinking skills among young students. For example, Jeff Gray, Ph.D., a professor in the department of computer science at the University of Alabama, is an advocate for teaching kids to code and tells GoodCall® that coding helps to develop problem-solving and computational skills at an early age.
At GEMS, field studies are also used to foster an analytical mindset. “Recently, first- and fifth-grade GEMS students collaborated on releasing trout into Lake Michigan,” Gallucci says. “The field study focused on helping students examine why fresh water is an important resource and why they have a responsibility to preserve it.”
Gallucci believes that critical thinking should be taught to students of every age, and notes that it can help them become future innovators. In fact, MIT has a program that helps high school students become entrepreneurs. However, Gallucci explains, “This unique approach to education requires a greater integration between schools and new technologies of learning.” And she believes that organizations such as GEMS and School 21 are creating schools of the future. “We are developing education ecosystems that act as a strong catalyst preparing the next generation to engage in an innovative economy.”
Collision course for critical thinking
Ironically, GEMS and School 21 are teaching critical thinking at a time when students and young adults are less likely to gain exposure from other sources. Dr. Jennifer L. Schneider, the Eugene H. Fram Chair in Applied Critical Thinking at Rochester Institute of Technology, tells GoodCall®, “Fundamentally, it takes work to think critically, and very often, time; in our fast paced, just-in-time world, this type of consideration is not rewarded.”
That doesn’t mean these skills are not needed, but due to a variety of factors, they’re just less likely to be cultivated. “Our flattening management structures with less hierarchies and formal mentoring, gig economy, ever changing technological impacts and disruptions are contributing to a society that builds less of a supportive environment to practice critical thinking, just when we collectively need it the most,” Schneider explains.”
Critical thinking is important on an individual level, but its significance extends beyond personal benefits. “Critical thinking skills are the basis of being a well-functioning member of society as it allows you to be truly informed and engaged in the world around you, and make quality decisions for yourself and your world . . . in other words, if you cannot critically think, you cannot contribute to your own wellbeing and success and then, certainly, you cannot contribute meaningfully to our larger world,” Schneider concludes.<|endoftext|>
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## Reflection: High Expectations Absolute Value Equations and Inequalities - Section 1: Workshop
I wanted to try throwing in a new skill and a new set of ideas even though I did not give students much time to learn this.
In thinking about the Common Core, I have been thinking about these key mathematical concepts:
Expressions, Equations and Functions
I am trying to understand the best way to put these together for students. One idea is to start with expressions, which I guess are like phrases: "two more than x", and are the most basic in some ways. The expressions are the building blocks for both equations and functions.
Beyond this, to me, equations are born from functions. Equations can be created by looking at specific cases of a function--a function is a relationship between two variables, and an equation comes from fixing one of those variables with a certain value, and then figuring out what the other variable must be to make the relationship hold.
I was trying to use this idea to help my students make sense of equations, and I think this was a good attempt, though I would like to investigate further. Basically, my thinking was that if they could understand a relationship like this:
"y is equal to 3 times the distance between x and -2."
I thought that they could extend this understanding to a relationship like this:
"12 is 3 times the distance between x and -2."
Or even:
"12 is less than 3 times the distance between x and -2."
I want to spend more time within each unit having students apply their understanding of functions to equations and inequalities. This lesson was a start, and it was pretty fluid and straight-forward for students. Hopefully by building this into more units, students will be able to understand the relationship between functions and equations.
Extending Understanding
High Expectations: Extending Understanding
# Absolute Value Equations and Inequalities
Unit 3: Absolute Value Functions and More Piecewise Functions
Lesson 8 of 9
## Big Idea: Put it all together -- students use a lot of strategies and make a lot of connections to hopefully demonstrate their mastery of the skills and ideas of this unit.
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Standards:
Subject(s):
Math, Graphing (Algebra), Algebra, absolute value functions, review lesson
35 minutes
### Hilary Yamtich
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###### Where are the Functions Farthest Apart? - Day 1 of 2
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Big Idea: Function combinations and maximization problems collide to create a challenging and mathematically rich task.
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###### Graphing Linear Functions Using Given Information
Algebra I » Graphing Linear Functions
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|Books & Reports|
Do you have a question?
Improving ESL Learners' Listening Skills: At the Workplace and Beyond
Carol Van Duzer
Listening is a critical element in the competent language performance of
adult second language learners, whether they are communicating at school,
at work, or in the community. Through the normal course of a day, listening
is used nearly twice as much as speaking and four to five times as much
as reading and writing (Rivers, 1981). In a recent study of Fortune 500
Corporations, Wolvin and Coakley (1991) found that listening was perceived
to be crucial for communication at work with regards to entry-level employment,
job success, general career competence, managerial competency, and effectiveness
of relationships between supervisors and subordinates. Yet listening remains
one of the least understood processes in language learning despite the recognition
of the critical role it plays both in communication and in language acquisition
(Morley, 1991). As language teaching has moved toward comprehension-based
approaches, listening to learn has become an important element in the adult
English as a second language (ESL) classroom (Lund, 1990).
What other processes are at work?
At the same time, two types of cognitive processing are also occurring: bottom-up and top-down processing.
Top-down processing refers to utilizing schemata (background knowledge and global understanding) to derive meaning from and interpret the message. For example, in preparing for training on the operation of a new floor polisher, top-down processing is activated as the learner engages in an activity that reviews what the learner already knows about using the old floor polisher. This might entail discussing the steps in the polishing process; reviewing vocabulary such as switch, on, off, etc.; or generating a list of questions that the learner would like answered in the training.
Bottom-up processing refers to deriving the meaning of the message based on the incoming language data, from sounds, to words, to grammatical relationships, to meaning. Stress, rhythm, and intonation also play a role in bottom-up processing. Bottom-up processing would be activated as the learner is signaled to verify comprehension by the trainer/teacher asking a question using the declarative form with rising intonation ("You see that switch there?"). Practice in recognizing statements and questions that differ only in intonation help the learner develop bottom-up processing skills.
Learners need to be aware that both of these processes affect their listening comprehension, and they need to be given opportunities to practice employing each of them.
How can listening help the adult learner acquire English?
Current research and theory point to the benefit of providing a silent or pre-speaking period for the beginning-level learner (Dunkel, 1991). Delaying production gives learners the opportunity to store information in their memories. It also spares them the trauma of task overload and speaking before they are ready. The silent period may be long or short. It could comprise several class periods of listening activities that foster vocabulary and build comprehension such as in the Total Physical Response (TPR) approach. In this approach, the teacher gives a series of commands while demonstrating each one. Learners then show their comprehension by acting out the commands as repeated by the teacher. Learners themselves begin to give the commands as they feel comfortable speaking. Or, the silent period may consist of learners listening to a tape-recorded conversation two or three times before answering questions about the content. A listening period consistent with the demands of the following productive task works to enhance rather than inhibit language acquisition and helps the more advanced-level learner as well as the beginner.
What should be considered when selecting listening techniques and activities?
What is known about the listening process and the factors that affect listening can be a guide when incorporating listening skill development into adult ESL classes. The following guidelines have been adapted from a variety of sources including Brod (1996), Brown (1994), Dunkel (1991), Mendelsohn (1994), Morley (1991), Peterson (1991), Richards (1983), and Rost (1991).
Listening should be relevant.
Because learners listen with a purpose and listen to things that interest them, accounting for the goals and experiences of the learners will keep motivation and attention high. For example, if learners at a worksite need to be able to understand new policies and procedures introduced at staff meetings, in class they should be helped to develop the abilities to identify main ideas and supporting details, to identify cause and effect, to indicate comprehension or lack of comprehension, and to ask for clarification.
Material should be authentic.
Authenticity should be evident both in language and in task. The language should reflect real discourse, including hesitations, rephrasing, and a variety of accents. Although the language needs to be comprehensible, it does not need to be constantly modified or simplified to make it easier for the level of the listener. Level of difficulty can be controlled by the selection of the task. For example, in a unit on following instructions, at the beginning level, the learner might hear a command ("May I borrow your hammer?") and respond by choosing the correct item. At an intermediate level, the learner might hear a series of instructions ("Go to the broom closet, get the floor polisher, take it to the hall in front of the cafeteria, polish the floor there, then go to the . . .") and respond appropriately by tracing the route on a floor plan of the worksite. An advanced-level learner might listen to an audio tape of an actual work meeting and write a summary of the instructions the supervisor gave the team. Use of authentic material, such as workplace training videos, audio tapes of actual workplace exchanges, and TV and radio broadcasts, increases transferability to listening outside of the ESL classroom context--to work and to community.
Opportunities to develop both top-down and bottom-up processing skills should be offered.
As mentioned above, top-down oriented activities encourage the learners to discuss what they already know about a topic, and bottom-up practice activities give confidence in accurate hearing and comprehension of the components of the language (sounds, words, intonation, grammatical structures).
The development of listening strategies should be encouraged.
Predicting, asking for clarification, and using non-verbal cues are examples of strategies that increase chances for successful listening. For example, using video can help learners develop cognitive strategies. As they view a segment with the sound off, learners can be asked to make predictions about what is happening by answering questions about setting, action, and interaction; viewing the segment again with the sound on allows them to confirm or modify their hypothesis (Rubin, 1995).
Activities should teach, not test.
Teachers should avoid using activities that tend to focus on memory rather than on the process of listening or that simply give practice rather than help learners develop listening ability. For example, simply having the learners listen to a passage followed by true/false questions might indicate how much the learners remembered rather than helping them to develop the skill of determining main idea and details. Pre- and post-listening task activities would help the learners to focus attention on what to listen for, to assess how accurately they succeeded, and to transfer the listening skill to the world beyond the classroom.
What are the steps in a listening lesson? The teacher can facilitate the development of listening ability by creating listening lessons that guide the learner through three stages: pre-listening, the listening task, and post-listening.
Engage the learners in a pre-listening activity.
This activity should establish the purpose of the listening activity and activate the schemata by encouraging the learners to think about and discuss what they already know about the content of the listening text. This activity can also provide the background needed for them to understand the text, and it can focus attention on what to listen for.
Do the listening task itself.
The task should involve the listener in getting information and in immediately doing something with it.
Engage in a post-listening activity.
This activity should help the listener to evaluate success in carrying out the task and to integrate listening with the other language skills. The teacher should encourage practice outside of the classroom whenever possible.
For example, at a worksite where schedule changes are announced at weekly team meetings, learners may need practice recognizing details such as their names, times, and dates within a longer stream of speech. A tape of such announcements may be used along with any pertinent forms or a weekly calendar. The lesson stages might proceed as follows:
What kinds of listening tasks are appropriate?
There are numerous activities to choose from for developing listening skills. Lund (1990) has categorized them according to nine responses that can be observed as comprehension checks:
Assisting learners in the development of listening comprehension is a challenge. It is a challenge that demands both the teacher's and the learner's attention because of the critical role that listening plays, not only in communication, but also in the acquisition of language. Knowledge of the listening process and factors that affect listening enable teachers to select or create listening texts and activities that meet the needs of the their adult ESL learners. Teachers, then, must weave these listening activities into the curriculum to create a balance that mirrors the real-world integration of listening with speaking, reading, and writing.
Brod, S. (1996). Teaching listening in the workplace English language training program at the Spring Institute. Unpublished manuscript.
Brown, G., & Yule, G. (1983). Teaching the spoken language. Cambridge: Cambridge University Press.
Brown, H.D. (1994). Teaching by principles: An interactive approach to language pedagogy. Englewood Cliffs, NJ: Prentice Hall Regents.
Dunkel, P. (1986). Developing listening fluency in L2: Theoretical principles and pedagogical considerations. The Modern Language Journal, 70(2), 99-106.
Dunkel, P. (1991). Listening in the native and second/foreign language: Toward an integration of research and practice. TESOL Quarterly, 25(3), 431- 457.
Lund, R.J. (1990). A taxonomy for teaching second language listening. Foreign Language Annals, 23, 105-115.
Mendelsohn, D.J. (1994). Learning to listen: A strategy-based approach for the second-language learner. San Diego: Dominie Press.
Morley, J. (1991). Listening comprehension in second/foreign language instruction. In M. Celce-Murcia (Ed.), Teaching english as a second or foreign language (2nd ed.) (pp. 81-106). Boston: Heinle and Heinle.
Nunan, D., & Miller, L. (Eds.). (1995). New Ways in Teaching Listening. Alexandria, VA: Teachers of English to Speakers of Other Languages. (ERIC Document Reproduction Service No. ED 388 054)
Peterson, P.W. (1991). A synthesis of methods for interactive listening. In M. Celce-Murcia (Ed.),Teaching English as a second/foreign language (2nd ed.) (pp.106-122). Boston: Heinle and Heinle.
Richards, J. (1983). Listening comprehension: Approach, design, procedure. TESOL Quarterly, 17(2), 219-240.
Rivers, W.M. (1981). Teaching foreign language skills (2nd ed.). Chicago: University of Chicago Press.
Rost, M. (1991). Listening in action: Activities for developing listening in language teaching. New York: Prentice Hall.
Rubin, J. (1994). A review of second language listening comprehension research. The Modern Language Journal. 78(2),199-221.
Rubin, J. (1995). The contribution of video to the development of competence in listening. In D. Mendelsohn & J. Rubin (Eds.), A guide for the teaching of second language listening (pp. 151-165). San Diego: Dominie Press.
Wolvin, A., & Coakley, C. (1991). A survey of the status of listening training in some Fortune 500 Corporations. Communication Education, 40, 152-164.
This document was produced by the Project in Adult Immigrant Education, funded by the Andrew W. Mellon Foundation through a grant to the Center for Applied Linguistics (4646 40th Street, NW, Washington, DC 20016 202-362-0700). Additional funding was from the U.S. Department of Education (ED), Office of Educational Research and Improvement, under contract no. RR 93002010, The opinions expressed in this report do not necessarily reflect the positions or policies of ED or the Andrew W. Mellon Foundation. This document is in the public domain and may be reproduced without permission.<|endoftext|>
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- Around 466 million people worldwide have disabling hearing loss (1), and 34 million of these are children.
- It is estimated that by 2050 over 900 million people will have disabling hearing loss.
- Hearing loss may result from genetic causes, complications at birth, certain infectious diseases, chronic ear infections, the use of particular drugs, exposure to excessive noise, and ageing.
- 60% of childhood hearing loss is due to preventable causes.
- 1.1 billion young people (aged between 12–35 years) are at risk of hearing loss due to exposure to noise in recreational settings.
- Unaddressed hearing loss poses an annual global cost of US$ 750 billion. Interventions to prevent, identify and address hearing loss are cost-effective and can bring great benefit to individuals.
- People with hearing loss benefit from early identification; use of hearing aids, cochlear implants and other assistive devices; captioning and sign language; and other forms of educational and social support.<|endoftext|>
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What Is A Septic Tank?
The septic tank was patented in London around 1900. Websters Dictionary defines the septic tank as "a tank in which waste
matter is decomposed through bacterial action." The modern septic tank is a watertight box usually made of precast concrete, concrete blocks, or reinforced fiberglass.
The septic system is a small, on-site treatment and disposal system buried
in the ground. the septic system has two essential parts: (1) the septic tank and (2) the soil absorption area. When household waste enters the septic tank several things occur:
- Organic solid material floats to the surface and forms a layer of what
is commonly called "scum." Bacteria in the septic tank biologically convert this material to liquid.
- Inorganic or inert solid materials and the by-products of bacterial
digestions sink to the bottom of the tank and form a layer commonly known as "sludge."
- Only clear water should exist between the scum and sludge layers. It
is this clear water - and only this clear water - that should overflow into the soil absorption area.
Solid material overflowing into the soil absorption area should be avoided at
all costs. It is this solids overflow that clogs soil pores and causes system to fail. Two main factors cause solid material to build up enough to
overflow: (1) bacterial deficency, and (2) lack of sludge removal.
Bacteria MUST BE PRESENT in the septic tank to break down and digest the
organic solids. Normal household waste probides enough bacteria to digest the solids UNLESS any harm is done to the bacteria. Bacteria are very
sensitive to environmental changes. Check teh lables of products you normally use in home. Products carrying harsh warnings such as "HARMFUL OR FATAL IF SWALLOWED" will harm bacteria.
- Cleaning compunds
- Toilet cleaners
- Caustic drain openers
People rarely think of the effect of these products on the septic tank system when the products go down the drain. What kind
of effect to you think anti-septics have on your septic tank?
Bacteria must be present to digest the scum. If not digested, the scum will accumulate untill it overflows, clogging the soil
The sludge in the septic tank - inorganic and inert material - is not biodegradable and will not decompose. If not removed, the
sludge will accumulate until it eventually overflows, again clogging the soil absorption area.
How does it work?
A typical septic system has four main components: a pipe from the home, a
septic tank, a drainfield/leachfield, and the soil. Microbes in the soil digest or remove most contaminants from wastewater before it eventually reaches groundwater.
Pipe from the house:
All of your house hold wastewater exits your home through a pipe to
the septic tank.
The septic tank is buried, watertigh container typically made of
concrete, fiberglass, or polyethylene. It holds the wastewater long enough to allow solids to settle out, forming sludge, and oil and grease to float to the surfance as scum. It also allows partial
decompositions of the solid materials. Compartments and a T-shaped outlet in the septic tank prevent the sludge and scum from leaving the tank and traveling into the leachfield/drainfield area.
The wastewater exits the septic tank and is discharged into the leachfield/drainfield for futher treatment by the soil. The partially
treated wastewater is pushed along into the leachfield/drainfield for further treatment everytime new wastewater enters the tank.
The most common leachfield/drainfield consists of a series of trenches
containing perforated pipe surrounded by septic rock or gravel, and covered with mesh and dirt. The effluent entering the leachfield/drainfield is partially absorbed into the soil and partially
evaporated. The leachfield/drainfield shoud not be driven on or covered by a driveway or patio.
If the leachfield/drainfield is overloaded with too much liquid it will
flood, causing sewage to flow to the ground surface or create backups in plumbing fixtures and prevent treatment of all wastewater.
Septic tank wastewater flows to the leachfield/drainfield, where it
percolates into the soil, which provides final teatment by removing harmful bacteria, viruses, and nutrients. Suitable soil is necessary for successful wastewater treatment.
Plumbing and waste system
If you are like most people, you may be unfamiliar with your septic tank system. This is undertandable. In urban and suburban
areas there are sewers to carry household waste to manicipal wastewater treatment facilities. In more rual areas, however,
septic tank systems provide the functions of both sewers and treatment facillites
All household waste is disposed of through the septic system. The proper operation of the septic system is essential to public
and private health, to propery values, and to the environment. To see if you know enough about your septic system, answer
the following questions. IF you cannot answer all the questions, your septic system could become a huge aggravation, public nuisance, health hazard and finacial burden.
- Do you know what a septic tank is and how it works?
- Do you know what kind of soil absorption area you have and how it works?
- Do you know what causes septic systems to fail?
- Do you know what it costs to replace a failed septic system?
- Do you know how to keep a septic system from failing?
- Do you know that failed and failing septic systems contribute to pollution?
These are very serious questions. The health of your family and the value of your propery rely heavily upon the answers to
"An ounce of prevention is worth a pound of cure" was never more true then it is with septic tank care. A small commitment to
the care of your septic system will protect you indefinitely from the nightmare created by a failing system.<|endoftext|>
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Capacitors (Latin term: condensus = compressor) are capacitive, i.e. they store electric charge. The physical unit of measure for capacity [C] (Latin term: capacitas = capacity) is Farad [F] (in honour of the English physicist and chemist Michael Faraday). Capacitors consist of two electrodes (surfaces conducting electricity) which are arranged close to each other, and a dielectric (insulating layer) in between.
Capacitors (abbr. cap) are frequency-dependent resistors. This is an important property for audio applications because capacitors can filter out low frequencies (i.e. low tones) from music signals. As the filter effect decreases with increasing frequency, the reverse conclusion is: The lower the capacity, the higher the filter effect (i.e. the higher the separating frequency).<|endoftext|>
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Adolf Hitler, the chancellor of Germany, led the Germans to take over other countries and brutally murder millions of Jews.In January of 1933, Adolf Hitlerwas named chancellor in Germany. Only about three months later he opened the first concentration camp, with many to follow. That is when the Jewish people started to lose their rights. By April of the same year, all Jews working had to retire. In 1934, Hitler became the president and commander-in-chief of the German army. Then in 1935 the Jews were denied citizenship. By the end of 1936, Hitler signed a document to form the Rome-Berlin axis. Four years later Japan will get into the alliance, too.
- He promised to solve all of Germany’s economic problems.
- He said that he would build up the army to what it had once been and better, in order to be able to defend Germany but also to attack other weaker countries.
- He told the people that he would ignore the treaty of Versailles, which was the very reason that Germany was in such a bad state economically: the treaty had essentially ruined Germany.
- He assured the people that he could provide them with strong leadership skills, something they had been missing for a long time.
- Finally, he swore that if voted into power, Hitler would make Germany great once again. This was one promise that all Germans felt good about: they wanted to be great again as much as Hitler did.
After he built up the army, he invaded Poland that is when World War 2 began. Then France and Britain declaredwar on Germany. In 1940, the German army attacked Denmark, which was defeated in one day, and Norway, it took almost two months to defeat them. While they were attacking Norway, they attacked Luxembourg (one day), Holland (five days),and Belgium (it took eighteen days to surrender). In June of the same year, France had surrendered to Germany. When Germans ally, Japan, attacked Pearl Harbor, the United States declared war on Japan. Germany thought the United States were crazy so both Germany and Italy declared was on the U.S.
A month later the Germans came upwith a “final solution”, which means that they were going to kill the rest ofthe Jews in Europe. Then started some resistance in the camps, both concentration and death. The first place of the resistance was the Warsawghetto and the Baily ghetto. By April of 1943 the Warsaw ghetto had revolted. Laterthat year, the Danish helped some 7,000 Jews escape to Sweden. However, in 1944, 400,000 Hungarian Jews were killed in the time span of less than onemonth at the Auschwitz death camp. Then the next year, Hitler committed suicide.<|endoftext|>
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The white-tailed deer population in Missouri has seen significant change over the past 100 years, and so have management priorities.
Presettlement, white-tailed deer were found throughout the state. However, like many wildlife species in the latter half of the 19th century, deer numbers declined with European settlement. The decline occurred at a time when humans were affecting the Missouri landscape on a scale never before experienced. Throughout much of Missouri, forests were cut, most accessible land was grazed or farmed, and people were scattered on small parcels across the rural landscape. Deer numbers declined to a low of approximately 400 deer in 1925.
In the early 20th century, attitudes toward wildlife shifted from a utilitarian to a more conservation-oriented emphasis. As a result, we entered the modern era of wildlife management and made species recovery a priority.
Although there had been small increases in deer numbers since the low in 1925, the creation of the Department of Conservation and the Conservation Commission initiated the first significant and successful efforts to protect and restore deer and many other wildlife species. In 1938, the Commission put into place several programs that stimulated rapid growth of the deer population, and by 1944 there were an estimated 15,000 deer in Missouri. As a result, the Conservation Commission established the first modern-day firearms hunting season in 1944.
Deer management at the time was relatively simple because the primary objective was to increase deer populations, which could be accomplished in large part by protecting does from harvest.
By the late 1980s, deer populations across much of the state were growing rapidly, leading to increased crop damage, deer/vehicle collisions, and the emergence of urban deer issues. This era of rapid population growth was met with liberalization of regulations and expanding hunting opportunities by lengthening seasons, establishing new portions to seasons, increasing bag limits and permit availability, and implementing restrictions on buck harvest in the form of the antler point restriction. All of these changes were intended to slow population growth by increasing harvest pressure on does.
Concerns over hunter recruitment and retention began to emerge as many states began to see declines in hunter numbers. Hunters are the primary tool of deer management and fewer hunters means a reduced capacity to manage a growing deer population. Therefore, numerous efforts to recruit new hunters were implemented.
By 2010, changes in regulations (longer seasons, new portions, antler point restriction, increased antlerless permits) were affecting deer populations. Significant losses due to hemorrhagic disease in 2007, 2012, and 2013 resulted in stable or reduced deer populations in many parts of central, northern, and western Missouri. Across much of southern Missouri, deer populations continue to grow slowly, though they largely remain below biological and social carrying capacity.
Traditionally, deer management focused on increasing deer numbers on a large geographic scale, which was relatively simple to accomplish through limited harvest quotas. Now, the focus has shifted to achieving localized population goals, which is more complicated.
The goal of the deer program is to use science-based wildlife management to maintain biologically and socially balanced deer populations throughout the state that provide recreational opportunities and minimize human-deer conflicts and potential negative effects on ecosystem health. Reaching deer management goals is more challenging today due to the complexity of interrelated factors such as land use, ownership, hunter density, and human population levels. Therefore, the Department of Conservation’s regulatory process incorporates both science-based information and citizen feedback. The Department uses hunter surveys, production landowner surveys, bowhunter observation surveys, deer population simulations, biological data, harvest summaries, and public comments when determining deer management goals for a particular county. If goals aren’t being met, regulation changes are proposed.
In many areas where deer populations are low, we will propose a reduction in firearms antlerless permits for the 2014-2015 deer season to allow populations to stabilize or increase. However, in response to growing deer populations in parts of southern Missouri we will propose increases in antlerless permits in select counties.
In response to the evolving challenges to deer management in the 21st century, we have drafted a deer management plan to outline the current priorities of Missouri’s deer management program and direct deer management over the next 10 years. The plan has four primary goals for deer management in Missouri: 1) Deer Population Management, 2) Hunting and Recreation, 3) Deer Health and Disease, and 4) Education, Communication, and Public Engagement.
In next month’s issue of the Conservationist we will discuss the draft deer management plan, proposed approaches to deer management in the future, and detail our plans to gather public input.
Editor In Chief - vacant
Managing Editor - Nichole LeClair Terrill
Art Director - Cliff White
Staff Writer/Editor - Brett Dufur
Staff Writer - Jim Low
Photographer - Noppadol Paothong
Photographer - David Stonner
Designer - Stephanie Thurber
Circulation - Laura Scheuler<|endoftext|>
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# Vectors - Span RR^2?
## Any help here is much appreciated
May 3, 2018
A requirement for any two vectors to span ${\mathbb{R}}^{2}$ is that the vectors are linearly independent .
For convenience we normally use a natural basis for vectors based on a standard cartesian coordinate system. Thus we normally use standard vectors $\boldsymbol{\underline{\hat{i}}}$ and $\boldsymbol{\underline{\hat{j}}}$, or in column format the basis:
$\boldsymbol{{B}_{1}} = \left\{\boldsymbol{\underline{\hat{i}}} , \boldsymbol{\underline{\hat{j}}}\right\} = \left\{\begin{matrix}\begin{matrix}1 \\ 0\end{matrix} \\ \begin{matrix}0 \\ 1\end{matrix}\end{matrix}\right\}$
Using this basis $\boldsymbol{{B}_{1}}$ we can can represent any vector by just using its standard cartesian coordinates in the column vector, so that the coordinate $\left(3 , 2\right)$ is trivially represented as:
$\left(\begin{matrix}3 \\ 2\end{matrix}\right) = 3 \boldsymbol{\underline{\hat{i}}} + 2 \boldsymbol{\underline{\hat{j}}} = 3 \left(\begin{matrix}1 \\ 0\end{matrix}\right) + 2 \left(\begin{matrix}0 \\ 1\end{matrix}\right)$
However we can readily show that the vectors:
$\boldsymbol{\underline{u}} = \left(\begin{matrix}1 \\ 1\end{matrix}\right) \setminus \setminus$ and $\boldsymbol{\underline{v}} = \left(\begin{matrix}- 3 \\ 2\end{matrix}\right)$
are also linearly independent, and as such can also be used as a basis:
$\boldsymbol{{B}_{2}} = \left\{\boldsymbol{\underline{u}} , \boldsymbol{\underline{v}}\right\} = \left\{\begin{matrix}\begin{matrix}1 \\ 1\end{matrix} \\ \begin{matrix}- 3 \\ 2\end{matrix}\end{matrix}\right\}$
And now to represent the coordinate $\left(3 , 2\right)$, we seek $l a m \mathrm{da} , \mu \in \mathbb{R}$ st:
$\left(\begin{matrix}3 \\ 2\end{matrix}\right) = l a m \mathrm{da} \boldsymbol{\underline{u}} + \mu \boldsymbol{\underline{v}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = l a m \mathrm{da} \left(\begin{matrix}1 \\ 1\end{matrix}\right) + \mu \left(\begin{matrix}- 3 \\ 2\end{matrix}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}1 & - 3 \\ 1 & 2\end{matrix}\right) \left(\begin{matrix}l a m \mathrm{da} \\ \mu\end{matrix}\right)$
And solving this system, yields the solution:
$l a m \mathrm{da} = \frac{12}{5} \setminus \setminus$, and $\mu = - \frac{1}{5}$
Thus we can write the coordinate using this spanning basis $\boldsymbol{{B}_{2}}$ as:
$\left(\begin{matrix}3 \\ 2\end{matrix}\right) = \frac{12}{5} \left(\begin{matrix}1 \\ 1\end{matrix}\right) - \frac{1}{5} \left(\begin{matrix}- 3 \\ 2\end{matrix}\right)$<|endoftext|>
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This week in science is a review of the most interesting scientific news of the week.
This week two scientific collaborations exploring the outer space have released their collected data, necessary software, and tutorials to the public, so they can get all the possible help in hunting for new planets. This was mainly due to the huge amount of data collected, that requires more work to be analyzed than scientists can do by themselves.
The first collaboration is the Backyard Worlds: Planet 9, between NASA, UC Berkeley, the American Museum of Natural History in New York, Arizona State University, the Space Telescope Science Institute in Baltimore, and Zooniverse, a collaboration of scientists, software developers, and educators. They are sharing videos collected by NASA's Wide-field Infrared Survey Explorer (WISE) mission between 2010 and 2011, that highlight any object that has gradually moved across the sky, currently the most comprehensive survey at mid-infrared wavelengths.
The data provided by the Backyard Worlds: Planet 9 project can be easily analyzed by anyone using only their eyes. That is because it is easy for us to recognize the important moving objects on the images while ignoring the artifacts. As stated by Aaron Meisner, a postdoctoral researcher at the University of California, Berkeley:
"Backyard Worlds: Planet 9 has the potential to unlock once-in-a-century discoveries, and it's exciting to think they could be spotted first by a citizen scientist."
This data can be used to search for any unknown object in and beyond our own solar system, but the current main goal is to find the so-called Planet 9, or Planet X. Last year, astronomers at Caltech in Pasadena, California, have demonstrated that an as-yet-undetected planet is gravitationally interfering on the orbits of several distant solar system objects.
The second collaboration is headed by the Carnegie Institution for Science and includes the Massachusetts Institute of Technology (MIT). They have released the, at present, largest collection of observations made with a technique called radial velocity. This technique allows the detection of the small movements a star makes that, if made in a regular pattern, can indicate the presence of an exoplanet orbiting it, whose detection is the main goal of the project.
The data set was acquired over two decades by the High-Resolution Echelle Spectrometer (HIRES), at the W.M. Keck Observatory in Hawaii and is being provided with the open-source software necessary to process it along with an online tutorial so the public can learn how to help the hunting. And you can be sure there is work to be done: the enormous data set contain almost 61,000 measurements of more than 1,600 nearby stars. As stated by Jennifer Burt, a Torres Postdoctoral Fellow in MIT's Kavli Institute for Astrophysics and Space Research:
"This is an amazing catalog, and we realized there just aren't enough of us on the team to be doing as much science as could come out of this dataset."
Among the data already available are over 100 highlighted stars that are likely to host exoplanets but require closer inspection. By further analyzing this data, for example, the public can greatly help the scientists accelerate their work. Also, HIRES will continue to record new data that the collaboration promises to make available in the future to expand the current data set.
Scientists are not only on the hunt for exoplanets but also for more elusive components of our universe, like theoretical dark matter. First deduced through calculations by Swiss astronomer Fritz Zwicky in 1933, researchers around the world are now in a race to directly detect it.
A collaboration comprising 38 institutions from around the globe and about 220 participating scientists and engineers are now working on the LUX-ZEPLIN (LZ) experiment. The construction of the experiment is expected to be finished by April 2020 at the Sanford Underground Research Facility (SURF) in Lead, South Dakota, and will try to detect WIMPs (weakly interacting massive particles), theoretical candidates for dark matter particles.
This is not the first time a WIMP detector was built, though. The Large Underground Xenon (LUX) experiment was running at the same place at SURF where LZ, which is at least 50 times more sensitive, is going to be built. Also, Italy plans to upgrade its XENON1T experiment and China wants to advance the work on its PandaX-II experiment, both with the same goal and a similar schedule as the LZ experiment. As stated by Carter Hall, the spokesperson for the LZ collaboration and an associate professor of physics at the University of Maryland:
"The science is highly compelling, so it's being pursued by physicists all over the world. It's a friendly and healthy competition, with a major discovery possibly at stake."
Another development in the search for dark matter came from the Haloscope At Yale Sensitive To Axion Cold Dark Matter (HAYSTAC) project. Instead of searching for WIMPs, this project wants to detect axions, which are another theoretical candidates for dark matter particles.
This time, though, the team led by Yale physicist Steve Lamoreaux has already published their first results in the renowned scientific journal Physical Review Letters. According to Lamoreaux:
"Our major breakthrough was making the detector colder and quieter than ever before, by adapting amplifiers developed for quantum computing research whose noise performance approaches the fundamental limits imposed by the laws of quantum mechanics. With the first data from our detector, we have set limits on the interactions of dark matter axions and opened a new portion of the allowed axion mass range to experimental investigation."
The team has developed a new instrument built at Yale's Wright Lab for this project. By using it, the scientists could determine the sensitivity an instrument must have to detect axions up to 10 times heavier than those currently being targeted.<|endoftext|>
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What Is The Gcf Of 28 And 12 // cadrugdetoxcenters.com
# What is the greatest common factor of 12 and 28? - Answers.
Greatest common factor GCF of 12 and 28 is 4. GCF12,28 = 4. We will now calculate the prime factors of 12 and 28, than find the greatest common factor greatest common divisor gcd of the numbers by matching the biggest common factor of 12 and 28. To get the Greates Common Factor GCF of 28 and 12 we need to factor each value first and then we choose all the copies of factors and multiply them: The Greates Common Factor GCF is: 2 x 2 = 4.
The GCF of 12 and 28 is 4. One way to determine the greatest common factor is to find all the factors of the numbers and compare them. The factors of 12 are 1, 2, 3, 4, 6, and 12. The GCF is the largest common positive integer that divides all the numbers 12, 28 without a remainder. The GCF is also known as: Greatest common divisor gcd. Find the prime factorization of 12 12 = 2 × 2 × 3 Find the prime factorization of 28 28 = 2 × 2 × 7 To find the gcf, multiply all the prime factors common to both numbers: Therefore, GCF = 2 × 2. To find the greatest common factor of two numbers just type them in and get the solution. Greatest Common Factor GCF of and SOLVE To get the Greates Common Factor GCF of 12 and 28 we need to factor each value first and then we choose all the copies of factors and multiply them. Find the prime factorization of 28 28 = 2 × 2 × 7 Find the prime factorization of 12 12 = 2 × 2 × 3 To find the gcf, multiply all the prime factors common to both numbers: Therefore, GCF = 2 × 2.
To calculate the GCF you first have to split the numbers into their prime factors: 28 = 2x2x7 12 = 2x2x3 20 = 2x2x5 The next step is to identify any common factors. In this case, all three numbers. GCF44,12,28 Greatest Common Factor is: 4 Calculate Greatest Common Factor for: 44, 12 and 28. Factorize of the above numbers: 44 = 2 2 • 11 12 = 2 2 • 3 28 = 2 2 • 7 Build a prime factors table. The greatest common factor GCF or GCD or HCF of a set of whole numbers is the largest positive integer that divides evenly into all numbers with zero remainder. For example, for the set of numbers 18, 30 and 42 the GCF = 6. GCF12,28,24 Greatest Common Factor is: 4 Calculate Greatest Common Factor for: 12, 28 and 24. Factorize of the above numbers: 12 = 2 2 • 3 28 = 2 2 • 7 24 = 2 3 • 3 Build a prime factors table.
Apr 23, 2016 · The greatest common factor of 12, 28, and 40 is 4. To find the greatest common factor of numbers, you first need to split these numbers into their prime factors: 12 = 2x2x3. 28 = 2x2x7. 40 =. GCF12,28,40 Greatest Common Factor is: 4 Calculate Greatest Common Factor for: 12, 28 and 40. Factorize of the above numbers: 12 = 2 2 • 3 28 = 2 2 • 7 40 = 2 3 • 5 Build a prime factors table.<|endoftext|>
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What is the Meaning of Interpersonal Communication?
Interpersonal communication is communication between several people. Inter- is a Latin prefix that means ‘between’, whilst personal means pertaining to people. Communication comes from the Latin verb communicare, which means to share.
So, putting all of these etymological connections together we can say that interpersonal communication means several people sharing something. What is shared during interpersonal communication can be many things, including words, thoughts, information and feelings.
Interpersonal communication can take many forms. For example, it can be:
- Verbal: words spoken face to face or on the phone.
- Written: emails, letters, text messages and quick scribbled notes are all examples of written communication between people.
- Gestural: our body language can often say a lot about us.
- Based on silence and listening: sometimes, silence speaks volumes, and much more that words do.
- Facial: facial expressions are another, very subtle, form of interpersonal communication.
In addition, we can divide up the category of interpersonal communication into formal and informal communication. Here, the context in which the communication takes place is important. The workplace is usually the place for formal interpersonal communication, whilst parties and other social gatherings are usually places where informal interpersonal communication is more appropriate.
Interpersonal communication is so important in very many aspects of our lives. Below, you will find 7 key reasons why this is. As you will see from the below, interpersonal communication has many uses in society. These range from the legal to the social, and from the romantic to the information. How many of these uses of interpersonal communication will you have guessed already, one might wonder?
Importance of Interpersonal Communication.
1. Promulgating laws.
The laws of the society will have no effect unless citizens are able to read, learn about and understand them. This is one example of how interpersonal communication is important for maintaining the fabric of society. Almost all societies have some form of written law that – if everything is to be done fairly – ought to be accessible to all citizens so that they are fully aware of what the laws of the land are.
2. Calling for help
It is said that a problem shared is a problem halved. Sometimes, the mere act of sharing a problem can help that problem to decrease or even to go away entirely. Calling a helpline or the doctor, or simply pouring our heart out to our friends, are all examples of the ways in which interpersonal communication can help to lighten the load on our minds. In addition, if we want, we can receive advice in return. Once we have told another person what is wrong, anything is possible: a solution will be in reach!
3. Disseminating the news.
Media outlets use various kinds of interpersonal communication – including printed pages and social media outlets – to let everybody know what the news is. Then, it is common for people to discuss the news via various types of interpersonal communication, giving their opinions about it and hearing the opinions of others.
4. Making friends.
Sharing stories and experiences, and comparing our opinions on different topics, is a very common way of making friends. Friendships are founded on communication, and are friends are very often the people that we talk to first about new developments in our lives. Friendships may be said to be not only founded but sustained by keeping the channels of communication open.
5. In marriages and romantic relationships.
Good interpersonal communication skills are absolutely essential to maintaining healthy relationships. That way, the couple can resolve any disagreements right away. If this does not happen, arguments can just grow and grow until they are extremely difficult to solve. It is also important for a couple to be open about their thoughts and feelings: that is the best way for them to be able to support each other. After all, how can someone celebrate their spouse’s achievements if they do not know what those achievements are? And how can they support them throughout their troubles if they do not know what those troubles are?
6. Getting jobs done at work.
Employees will only work in an efficient and effective way if they know exactly what it is they are supposed to be working on. That is why clear and concise communication is crucial in the workplace. Managers need to be able to state what tasks they want their teams to perform, when the deadlines are and so on. In addition, employees will need to feed back on how their work is going, and bosses and CEOS need to be able to listen fairly to their employee’s queries and complaints and to give advice and constructive criticism whenever it is needed. From work emails to chats in the communal coffee area, and from presentations given at training events to official directives from the very top of the chain, interpersonal communication is happening all the time at work.
7. Helping others out.
We can use our interpersonal communication skills to help other people out in numerous other ways. For example, if a stranger stops us in the street to ask us for directions, we can communicate with them in order to help them to find the right way to their destination. Or, if someone tells us their problems, we can listen carefully and sympathetically to show our support – after all, interpersonal communication is as much about active listening as it is about talking. If we have good interpersonal communication skills, we can be ready to lend a helping hand to anyone who needs it.
Interpersonal communication is both a useful tool and a joy in itself. Communication with another person and sharing thoughts and feelings with them can give us plenty of positive feelings and can also help to relieve our troubles and worries. In addition, interpersonal communication is also very useful for holding society together – for example, by disseminating news items that are of interest to citizens or by ensuring that everyone in a society knows what the law is. Are there any ways in which you might be able to improve your interpersonal communication skills? You never know when you might need them in order to help out one of your fellow human beings.
Also read: Types of Interpersonal Communication.<|endoftext|>
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###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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# Exponential Functions - Problem 4
Norm Prokup
###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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One of the problems you are going to see a lot of when you are studying exponential functions is, find the exponential function that matches the situation. Well here, the situation is a graph. We want to find an exponential function f of x equals 8 times b to the x that matches this graph.
So the first thing you want to do is you have to know the formula of an exponential function. You have to know this form. And if you know that the graph passes through these points, you can use that to figure out what the coefficients a and b are.
So for example, using the point (1,54) f of x is going to be 54, and x is going to be 1. So 54 equals 8 times b to the 1. That’s one equation. And using a second point, f of x is going to be 24 and x is going to be 3. So you have two equations with 2 unknowns. Easy to solve, all I need to do is take one equation, and divide both sides by the other equation. This is kind of like the method of elimination.
And you see that a lot of cancellation that’s going to happen here. There a’s will cancel, some of the b’s will cancel, and I can also reduce this fraction; 24 over 54. 24 is 2 times 12, this is 2 times 27. So I have 12 over 27, but that reduces even any further. Both of these have a factor of 3. 3 times 4 over 3 times 9, cancel the 3’s. So b² is 4 over 9. That means b is plus or minus 2/3. Now we know that the base of an exponential function has to be positive. So b has to be 2/3. All I have to do is find a.
Let’s use this equation because I'm going to have to raise b to the first power. So it will be a little easier to find a. 54 equals a times b. Equals a times 2/3. Now I just multiply both sides by 3/2. I get 3 times 27, 81 equals a, a equals 81. And so my function is f of x equals 81 times 2/3 to the x.
That’s it. All you have to do when you are finding the equation of an exponential function is, use the points that you are given, (1,54) and (3,24), to set up equations that involve the variables that are the coefficients of your exponential function and solve for them.<|endoftext|>
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# Square of The Difference of Two Binomials
How to find the square of the difference of two binomials?
(a - b) (a - b) = a(a - b) - b(a - b)
= a2 - ab - ba + b2
= a2 - 2ab + b2
= a2 + b2 - 2ab
Therefore, (a - b)2 = a2 + b2 - 2ab
Square of the difference of two terms = square of 1st term + square of 2nd term - 2 × fist term × second term
This is called the binomial square.
It is stated as: the square of the difference of two binomials (two unlike terms) is the square of the first term plus the second term minus twice the product of the first and the second term.
Worked-out examples on square of the difference of two binomials:
1. Expand (4x - 7y)2 using the identity.
Solution:
Square of 1st term + square of 2nd term - 2 × fist term × second term
Here, a = 4x and y = 7y
= (4x)2 + (7y)2 - 2 (4x) (7y)
= 16x2 + 49y2 - 56xy
Therefore, (4x + 7y)2 = 16x2 + 49y2 - 56xy.
2. Expand (3m - 5/6 n)2 using the formula of (a - b)2.
Solution:
We know (a - b)2 = a2 + b2 - 2ab
Here, a = 3m and b = 5/6 n
= (3m)2 + (5/6 n)2 - 2 (3m) (5/6 n)
= 9 m2 + 25/36 n2 - 30/6 mn
= 9 m2 + 25/36 n2 - 5 mn
Therefore, (3m - 5/6 n)2 = 9 m2 + 25/36 n2 - 5 mn.
3. Evaluate (999)2 using the identity.
Solution:
(999)2 = (1000 – 1)2
We know, (a – b)2 = a2 + b2 – 2ab
Here, a = 1000 and b = 1
(1000 – 1)2
= (1000)2 + (1)2 – 2 (1000) (1)
= 1000000 + 1 – 2000
= 998001
Therefore, (999)2 = 998001
4. Use the formula of square of the difference of two terms to find the product of (0.1 m – 0.2 n) (0.1 m – 0.2 n).
Solution:
(0.1 m – 0.2 n) (0.1 m – 0.2 n) = (0.1 m – 0.2 n)2
We know (a – b)2 = a2 + b2 – 2ab
Here, a = 0.1 m and b = 0.2 n
= (0.1 m)2 + (0.2 n) 2 - 2 (0.1 m) (0.2 n)
= 0.01 m2 + 0.04 n2 - 0.04 mn
Therefore, (0.1 m – 0.2 n) (0.1 m – 0.2 n) = 0.01 m2 + 0.04 n2 - 0.04 mn
From the above solved problems we come to know square of a number means multiplying a number with itself, similarly, square of the difference of two binomial means multiplying the binomial by itself.
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### Fun with Uniform Random Numbers
Q: You have two uniformly random numbers x and y (meaning they can take any value between 0 and 1 with equal probability). What distribution does the sum of these two random numbers follow? What is the probability that their product is less than 0.5.
The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists
A: Let z = x + y be the random variable whose distribution we want. Clearly z runs from 0 to 2. Let 'f' denote the uniform random distribution between [0,1]. An important point to understand is that f has a fixed value of 1 when x runs from 0 to 1 and its 0 otherwise.
So the probability density for z, call it P(z) at any point is the product of f(y) and f(z-y), where y runs from 0 to 1.
However in that range f(y) is equal to 1. So the above equation becomes
From here on, it gets a bit tricky. Notice that the integral is a function of z. Let us take a look at how else we can simply the above integral.
It is easy to see that f(z-y) = 1 when (z-y) is between [0,1]. This is the same as saying
Likewise, f(z-y) = 1 when y is lesser than z and greater than 0. ie
Combining the two cases above results in a discontinuous function as
which is a triangular function.
Now that we done with the sum, what about the product xy? A quick way to go about it is to visualize a 2 dimensional plane. All the points (x,y) within the square [0,1]x[0,1] fall in the candidate space. The case when xy = 0.5 makes a curve
The area under the curve would represent the cases for which xy <= 0.5 (shown shaded below). Since the area for the square is 1, that area is the sought probability.
The curve intersects the square at [0.5,1] and [1,0.5]. The area under the curve would the be sum of the 2 quadrants (1/4 each) along with the integral of y = 0.5/x under the range 0.5 to 1 yielding
\begin{align*} P(xy\lt 0.5) &= \frac{1}{2} + \int_{0.5}^{1}\frac{0.5}{x}dx \\ &= \frac{1}{2} + \frac{1}{2}ln2 \approx 0.85 \end{align*}
If you are looking to buy some books in probability here are some of the best books to learn the art of Probability
Here are a few
Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)
This book is a great compilation that covers quite a bit of puzzles. What I like about these puzzles are that they are all tractable and don't require too much advanced mathematics to solve.
Introduction to Algorithms
This is a book on algorithms, some of them are probabilistic. But the book is a must have for students, job candidates even full time engineers & data scientists
Introduction to Probability Theory
An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition
The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (and Everyone Else!)
Introduction to Probability, 2nd Edition
The Mathematics of Poker
Good read. Overall Poker/Blackjack type card games are a good way to get introduced to probability theory
Let There Be Range!: Crushing SSNL/MSNL No-Limit Hold'em Games
Easily the most expensive book out there. So if the item above piques your interest and you want to go pro, go for it.
Quantum Poker
Well written and easy to read mathematics. For the Poker beginner.
Bundle of Algorithms in Java, Third Edition, Parts 1-5: Fundamentals, Data Structures, Sorting, Searching, and Graph Algorithms (3rd Edition) (Pts. 1-5)
An excellent resource (students/engineers/entrepreneurs) if you are looking for some code that you can take and implement directly on the job.
Understanding Probability: Chance Rules in Everyday Life A bit pricy when compared to the first one, but I like the look and feel of the text used. It is simple to read and understand which is vital especially if you are trying to get into the subject
Data Mining: Practical Machine Learning Tools and Techniques, Third Edition (The Morgan Kaufmann Series in Data Management Systems) This one is a must have if you want to learn machine learning. The book is beautifully written and ideal for the engineer/student who doesn't want to get too much into the details of a machine learned approach but wants a working knowledge of it. There are some great examples and test data in the text book too.
Discovering Statistics Using R
This is a good book if you are new to statistics & probability while simultaneously getting started with a programming language. The book supports R and is written in a casual humorous way making it an easy read. Great for beginners. Some of the data on the companion website could be missing.<|endoftext|>
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3. Force diagrams or generalized coordinates?
Typically you are taught in high school that in order to solve problems with interacting bodies you need to draw force diagrams, and write down the force balance equations (based on Newton II law) for x and y components (for three-dimensional problems, also the z-component). However, for problems which are more difficult than the elementary ones, this is typically far from being the simplest approach. Meanwhile, there is a very powerful method based on generalized coordinates, which provides in most cases the easiest route to the solution. The basic idea of the method is as follows.
Suppose the state of a system can be described by a single parameter , which we call the generalized coordinate (the method can be also applied with two or more parameters, but this will complicate things, and in most cases, one parameter is perfectly enough). Then, what you need to do is to express the potential energy of the system in terms of , , and the kinetic energy in terms of , the time-derivative of : . Then, if there is no dissipation and external forces, the net energy is conserved: . Upon taking time-derivative of this equality, we obtain , from where we can express the acceleration of the generalized coordinate:
Note that most often, is constant, because the kinetic energy is proportional to , and plays the role of an effective mass . In some cases, it may happen that depends also on and/or depends also on ; then, the above formula will not work, but the technique itself remains still applicable (cf. the example of rotating spring below).
In order to illustrate this method, let us start with a simple wedge problem. Consider a system where a ball of mass lays on a wedge of mass , and is attached with a weightless rope and pully to a wall as depicted in Figure; you are asked to find the acceleration of the wedge, assuming that all the surfaces are frictionless, and there is a homogeneous gravity field .
When using the force diagram method, it would be a good idea to use the (non-inertial) reference frame associated with the wedge (introducing thereby the inertial forces and ), because otherwise, it would be difficult to write down equation describing the fact that the ball will remain on the inclined surface of the wedge. Here, however, we leave this for the reader as an exercise, and describe the state of the system via the displacement of the wedge. Then, the velocity of the wedge is ; the velocity of the ball with respect to the wedge is also , implying that the vertical component of the ball's velocity is , and the horizontal component is . Hence, we find that
Upon taking time derivative of this equation and cancelling out , we obtain an expression for the wedge acceleration:
As another example, let us consider an old IPhO problem (5th IPhO in Sofia, 1971, Problem No 1). The set-up is quite similar to the previous problem, but there is no wall, there are two bricks instead of one ball, and the wedge has two inclined surfaces (see Figure); we ask again, what is the acceleration of the wedge.
You might think that the method does not work here, because there are two degrees of freedom: the wedge can slide on the table, and the bricks can slide with respect to the wedge. However, if we make use of the conservation of the centre of mass (there are no external horizontal forces), we can express the displacement of the bricks (with respect to the wedge) via the displacement of the wedge :
What is left to do, is to write
substitute by , take time derivative of the full energy, and express . Well, there is some math do be done, but that is actually just an algebra. If you do it correctly, you obtain
A really simple example is provided by water level oscillations in U-tube. Let the water occupy length of the U-tube, and let us use the water level height (with respect to the equilibrium level) as the generalized coordinate. For a state with , a water column of height from one arm has been lifted by an height difference and moved into the other arm of the U-tube, which corresponds to the potential energy ; meanwhile, . So, upon applying our technique we obtain , which describes an harmonic oscillator of circular frequency .
Actually, when in hurry and oscillation frequency is needed, two steps of the scheme (taking time derivative and writing the equation of motion) can be skipped. Indeed, for an harmonic oscillator, both and need to be quadratic in and , respectively, ie. should have form and , where and are constants; then, .
Next, the technique can be used to analyse oscillations in simple rotating systems, such as, for instance, a system of two balls of mass , connected with a spring of length and stiffness , rotating with angular momentum (which is perpendicular to the spring). Here, again, an additional (to the energy) conservation law (of angular momentum) reduces the effective number of degrees of freedom down to one. Let us use the deformation of the spring as the generalized coordinate. Then,
This case is different in that the kinetic energy depends not only on , but also on ; in effect, the second term of the kinetic energy behaves as a potential one, and can be combined into an effective potential energy in the expression for the full energy. Following our technique,
This equation of motion can be linearised around the state of equilibrium (such that for , the right-hand-side turns to zero), by introducing . Linearisation means approximating a non-linear function with a linear one, and is typically done by neglecting in the Taylor expansion quadratic and higher terms, ie. by substituting with ; this is legitimate if the argument varies in a narrow range, in this case for . As a result, we obtain
which gives us immedieately the circular frequency of small oscillations, .
What we did here can be also called a linear stability analysis (which is a very popular technique in physics). Indeed, it is easy to see that regardless of the parameter values, the circular frequency is always a real number, ie. the circular trajectories of the balls are always stable (meanwhile, imaginary circular frequency would mean that the solution includes a component which grows exponentially in time, ie. the regular motion along the circular trajectory would be unstable).
Note that almost exactly the same analysis which was done here for the rotating spring, was used in the "official" solution of the Problem 1 (subquestion 3) of IPhO-2011. However, it appears that for the mentioned problem, this technique cannot be applied as easily: there is one mistake in the solution, and another one among the assumptions of the problem; for more details, see the mosaic tile "Are trojans stable?".
Up til now we have dealt with problems where the task was to find an acceleration. What to do, if you are asked to find a force? For instance, a sphere and a wedge are placed on two facing ramps as shown in Figure; all the surfaces are frictionless. Find the normal force between the wedge and the sphere.
Well, it would be very easy to find the acceleration of the ball (or that of the wedge) using the method of generalized coordinates (ball displacement can be used as the coordinate). But once we know the acceleration, it is also easy to find the normal force between the wedge and the ball from the Newton II law! (The answer is .)
The method of generalized coordinates is designed to work for dissipation-less systems.. However, in some cases it is also possible to take into account the friction. To illustrate this, let us modify the previous problem so that the right ramp remains frictionless, but the left ramp has high friction, so that the ball will rotate along it, and the friction between the wedge and the ball is described by kinetic friction coefficient .
The idea here is to "fix" the energy conservation law by adding the work performed by the friction force. Initally, such an equation will involve the normal force as a parameter, but it can be determined later: we express the normal force in the same ways as for the previous problem, and this will be the equation for finding . So, and ; the contact point leaves "traces" both on the wedge (of length ) and on the ball (of length ), corresponding to the net work of . So, the energy conservation law is written as
Now, assuming that we have heavy wedge, and the system moves leftwards, the Newton II law for the wedge can be written as
As a final example illustrating this method, let us consider a somewhat more difficult problem, posed by W.H. Besant in 1859, and solved by Lord Rayleigh in 1917: in an infinite space filled with an incompressible liquid of density at pressure , there is a spherical "bubble" of radius , which has vacuum inside. Due to the pressure (far away, it is kept equal to ), the "bubble" starts collapsing; find the collaps time of the "bubble". Here we use the radius of the "bubble" as the generalized coordinate; there is no potential energy, but there is work done by the pressure, . What is left to do, is to express the kinetic energy of the fluid in terms of . Due to the incompressibility of the fluid, the volume flux of liquid through any spherical surface of radius around the centre of the "bubble" is independent of : . So, the kinetic energy can be found as
So, the energy balance can be written as
This equation could be used to find the acceleration ; however, we need to know the collapse time; so we put , and express in terms of and :
Thus, we were able to obtain an answer, which contains a dimensionless integral: substituting allowed us to get rid of the dimensional quantities under the integral (if possible, always use this technique to convert integrals into dimensionless numbers). This result could be left as is, since finding an integral is a task for mathematicians. The mathematicians, however, have been up to the task: where denotes the gamma function. So, we can write
Finally, to close the topic of the generalized coordinates, it should be mentioned that this technique can be developed into generic theories – Lagrangian and Hamiltonian formalisms, which are typically taught as a main component of the course of theoretical mechanics. In particular, the Hamiltonian formalism makes it possible to prove the conservation of adiabatic invariant, as well as the KAM (Kolmogorov-Arnold-Mozer) theorem, as well as to derive conservation laws from the symmetry properties of the Hamiltonian (or Lagrangian) using the Noether's theorem. The Hamiltonian approach differs from what is described here by using the generalized momentum , instead of the generalized velocity . For the most typical cases when the kinetic energy is proportional to the square of the generalized velocity, one can just use the effective mass (defined above): . Then, the expression for the full energy is considered as a function of and , , and is called the Hamiltonian; the equation of motion is written in the form of a system of equations, , . However, for the practical application of problem solving, the simplified approach to the generalized coordinates provided above is just enough!
Jaan Kalda, Academic Committee of IPhO-2012<|endoftext|>
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## GeoGebra Basic Construction 5– Rectangle
In this GeoGebra tutorial, we use the Perpendicular Line and Parallel Line tools to construct a rectangle. The idea is to construct segment AB, construct two lines perpendicular to it, one passing through segment A and the other through segment B. Next, we will construct point on the line passing through B, then construct a line parallel to AB passing through C. The fourth intersection will be our point D. If you want to follow this construction step-by-step, click here to open GeoGebra on your browser.
Step-by-Step Construction
1. Open GeoGebra and select Geometry from the Perspective menu on the side bar. 2. To automatically show the labels of points and not the other objects, click the Options menu, click Labeling, then click New Points Only. 3. To construct a rectangle, select the Segment between Two Points tool and click two distinct places on the Graphics view to construct segment AB. 4. We construct two lines that are both perpendicular to AB, one passing through A and the other through B. To do this, click the Perpendicular Line tool, click point A and click the segment. To construct another line, click point B then click the segment. After these steps, your drawing should look like the one shown in Figure 1. Figure 1 5. Next, we construct a point on the line passing through point B. To do this, click the New Point tool, and then click on the line passing through B. 6. Now, to create a line parallel to AB, passing through point C, click the Parallel line tool, click on point C and then click on the segment. Figure 2 Your drawing should look like the one shown in Figure 2 after this step. 7. Using the intersect tool, we construct the intersection of the line passing C (parallel to AB) and the line passing through A. To do this, click the Intersect Two Objects tool and click the two lines mentioned. 8. Now, we hide the three lines. To do this, click the Show/Hide Object tool and click the three lines. Notice that the lines are highlighted. Click Move tool to hide the lines. 9. Use the Segment between Two Points tool to connect the points to construct the rectangle. 10. Drag the points of the vertices of the rectangle. What do you observe?<|endoftext|>
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The blue morpho butterfly is principally native to Mexico, Central America and South America including Venezuela, Brazil and Costa Rica. This species of butterfly inhabits the forest canopy layer and rarely roams in the understory layers or near the forest floor. This species of butterfly has characteristic brilliant blue wings, which are visible at a distance. The blue morpho butterfly exhibits numerous adaptations, which allow it to survive and continue to reproduce.
The blue morpho butterfly flies in a highly specific manner. The flight of this butterfly is designed to show the blue color of the wing as little as possible, so as not to draw attention to the insect. The upper surface of this butterfly’s wing consists of reflective scales. The blue morpho flashes its blue reflective wings to blind predators, such as young jaguars. The underside of the morpho’s wings has cryptic colors that blend into the vegetation. When resting, the blue morpho folds its wings in such a manner that only the underside shows. The blue morpho butterfly also folds its wings in this manner when it sleeps at night and is therefore less noticeable to nocturnal predators.
The blue morpho butterfly has a varied diet. This feeding preference is a survival strategy, as the butterfly will still be able to grow and thrive even if one of the food items has become scarce or has disappeared all together. This butterfly species, unlike most others, feeds on fermented fruits, fungi, tree sap, mud and the body fluids of dead forest creatures. The blue morpho butterfly develops a proboscis or long protruding mouth part after changing from the caterpillar stage into its adult form. This straw-like mouth adaptation allows the butterfly to feed on the numerous food items that it requires to survive.
Sciencing Video Vault
The blue morpho butterfly enhances its survival as a species by laying large numbers of eggs. The principle behind this reproductive strategy is that even if many eggs are predated upon or if a number of lava or pupa die for whatever reason, enough will survive to continue the reproduction cycle.
The blue morpho butterfly has bronze colored eye spots on the brown underside of its wings. Eye spots frighten off predators, who believe that the butterfly is considerably larger than it is. The eye spots will also confuse predators as to the identity of the creature. The blue morpho butterfly will escape while the predator is confused enough to delay its attack.<|endoftext|>
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Think back to lunch. Did you finish everything on your plate? If you did, bravo! If you didn’t, did your food end up in the trash can? It’s okay if you answered yes – we’re all guilty of it. But according to a new study from the Potsdam Institute for Climate Impact Research, our wasteful habits are having a pretty staggering effect on the climate.
According to the study, up to 14 percent of greenhouse gas emissions from agriculture in 2050 could be avoided if we more efficiently managed food use and distribution. “Agriculture is a major driver of climate change, accounting for more than 20 percent of overall global greenhouse gas emissions in 2010,” said co-author Prajal Pradhan, according to the Guardian. “Avoiding food loss and waste would, therefore, avoid unnecessary greenhouse gas emissions and help mitigate climate change.”
The report states, “Avoiding food loss and waste may counteract the increasing food demand and reduce greenhouse gas (GHG) emissions from the agricultural sector. This is crucial because of limited options available to increase food production. In the year 2010, food availability was 20 percent higher than was required on a global scale.”
In other words, we’re wasting energy producing food that never even makes it to our plates. In fact, it’s estimated that one-third of all food produced worldwide goes to waste. That’s a whole lot of resources used just to put something in the garbage can, isn’t it? Not to mention it’s hurting our wallets, too – $1 trillion worth of waste is generated every year. And if you aren’t convinced food waste is a MAJOR problem yet, consider this: in the U.S., organic food waste is the second highest component of landfills, which are the largest source of methane emissions. Methane has a global warming potential 20 times higher than carbon dioxide.
You’re convinced you’ve gotta stop wasting food! What can you do?
We might not all have the power to influence policy, but we can all combat food waste with our daily actions. Just check out these six ways you can cut down on food waste, ten ideas to adopt a zero-waste lifestyle, and these simple ways to minimize waste in cooking. If you can’t stop waste at the source, you can always try composting! Check out this guide to composting 101 to get started. Together we can keep more food out of landfills – a win for people, animals, and the planet all around!
Image Source: Patrick Feller/Flickr<|endoftext|>
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The 1-Sample Z conducts a hypothesis test of the mean based on sampled data when the population standard deviation, s is known. It also calculates a confidence interval of the mean. This procedure is based on normal distribution. For larger samples (>30), the sample standard deviation may be used in the absence of the population standard deviation. For smaller samples, make sure the data is drawn from a normal distribution or the t-test may be more appropriate.
|Test||Null Hypothesis, H0||Alternate Hypothesis, H1|
|One-tailed||μ = μ0||μ < μ0 or μ > μ0|
|Two-tailed||μ = μ0||μ ≠ μ0|
where μ is the population mean and μ0 is the hypothesised population mean<|endoftext|>
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### Chapter 1: Introduction to Statistics
```COURSE: JUST 3900
TIPS FOR APLIA
Chapter 5:
z-Scores
Developed By:
John Lohman
Michael Mattocks
Aubrey Urwick
Key Terms and Formulas: Don’t
Forget Notecards
Describing z-Scores
Question 1: Identify the z-score value corresponding to
each of the following locations in a distribution.
Below the mean by 3 standard deviations.
Above the mean by 1/4 standard deviations.
Below the mean by 2.50 standard deviations.
Question 2: Describe the location in the distribution for
each of the following z-scores.
z = - 1.50
z = 0.25
z = - 3.50
z = 1.75
Describing z-Scores
z = -3.00
z = 0.25
z = -2.50
Below the mean by 1.50 standard deviations.
Above the mean by ¼ standard deviations.
Below the mean by 3.50 standard deviations.
Above the mean by 1.75 standard deviations.
Describing z-Scores
−
The numerator in our z-score formula ( =
)
describes the difference between X and µ. Therefore, if
a question asks you to calculate the z-score for a score
that is above the mean by 4 points and has a standard
deviation of σ = 2, you cannot calculate (X - µ) - - i.e.,
you cannot find the original values for X or µ and
calculate the difference between the two. In this case,
it has already been provided for you because the
question tells you the distance between X and µ
(X - µ) = 4. Thus, your formula should read z = 4/2,
which comes out to be z = 2.00.
Transforming X-Values into
z-Scores
Question 3: For a distribution of µ = 40 and σ = 12, find
the z-score for each of the following scores.
a)
b)
c)
X = 36
X = 46
X = 56
Question 4: For a population with µ = 30 and σ = 8, find
the z-score for each of the following scores.
a)
b)
c)
X = 32
X = 26
X = 42
Transforming X-Values into
z-Scores
Using z-Scores to Compare
Different Populations
Question 5: A distribution of English exam scores has
µ = 70 and σ = 4. A distribution of history exam scores
has µ = 60 and σ = 20. For which exam would a score of
Using z-Scores to Compare
Different Populations
−
−
=
78−70
4
78−60
20
=
=
For the English exam, X = 78 corresponds to z = 2.00, which is a
higher standing than z = 0.90 for the history exam.
=
=
8
= 2.00
4
18
= 0.90
20
=
Remember that 95% of all scores fall between ± 2.00. Thus, a score
of +2.00 means that over 95% of the class scored below 78 on the
English exam.
Using z-Scores to Compare
Different Populations
Question 6: A distribution of English exam scores has
µ = 50 and σ = 12. A distribution of history exam scores
has µ = 58 and σ = 4. For which exam would a score of
Using z-Scores to Compare
Different Populations
−
−
=
62−50
12
62−58
4
=
=
The score X = 62 corresponds to z = 1.00 in both distributions.
The score has exactly the same standing for both exams.
=
=
12
= 1.00
12
4
= 1.00
4
=
z-Scores and Standardized
Scores
Question 7: A population of scores has µ = 73 and σ = 8.
If the distribution is standardized to create a new
distribution with µ = 100 and σ = 20, what are the new
values for each of the following scores from the original
distribution?
a)
b)
c)
d)
X = 65
X = 71
X = 81
X = 83
z-Scores and Standardized
Scores
z-Scores and Standardized
Scores
Question 8: A population with a mean of µ = 44 and a
standard deviation of σ = 6 is standardized to create a
new distribution with µ = 50 and σ = 10.
a)
b)
What is the new standardized value for a score of X = 47 from
the original distribution?
One individual has a new standardized score of X = 65. What
z-Scores and Standardized
Scores
X = 47
z = 0.50
X = 55
Old Distribution
32
38
44
50
56
z-Score Distribution
New Standardized Distribution
z-Scores and Standardized
Scores
X = 53
z = 1.50
X = 65
32
38
44
Old Distribution
50
56
z-Score Distribution
New Standardized Distribution
Measure of Relative Location
and Detecting Outliers
Question 9: A sample has a mean of M = 30 and a
standard deviation of s = 8.
a)
b)
Would a score of X = 36 be considered a central score or an
extreme score in the sample?
If the standard deviation were s = 2, would X = 36 be central or
extreme?
Measure of Relative Location
and Detecting Outliers
a)
=
1.
b)
=
1.
−
=
36−30
8
=
6
8
= 0.75
X = 36 is not an extreme score because it is within two standard
deviations of the mean.
−
=
36−30
2
=
6
2
= 3.00
In this case, X = 36 is an extreme score because it is more than
two standard deviations above the mean.
WARNING!!!
The book defines an extreme score as being more than
TWO standard deviations away from the mean.
However, Aplia defines extreme scores as being more
than THREE standard deviations from the mean.
When using Aplia, use the THREE definition of standard
deviation for extreme scores.
On in class exercises and on the test, use the TWO
definition of standard deviation for extreme scores.
FAQs
How do I find the z-scores from a raw set of scores?
1)
X = 11, 0, 2, 9, 9, 5
Find the mean:
1)
2)
=
=
11+0+2+9+9+5
6
=
36
6
=6
Find SS:
1)
=
−
2
X
X-µ
(X - µ)2
11
11 – 6 = 5
(5)2 = 25
0
0 – 6 = -6
(-6)2 = 36
2
2 – 6 = -4
(-4)2 = 16
2
9–6=3
(3)2 = 9
9
9–6=3
(3)2 = 9
5
5 – 6 = -1
(-1)2 = 1
=
−
2
= 96
FAQs
3)
Find σ2:
1)
4)
=
96
6
= 16
Find σ:
1)
5)
2 =
=
=
96
6
=
16 = 4
Find z-score for each X:
1)
=
2)
=
3)
=
4)
=
5)
=
6)
=
−
−
−
−
−
−
=
=
=
=
=
=
11−6
5
=
= 1.25
4
4
0−6
−6
=
= −1.50
4
4
2−6
−4
=
= −1.00
4
4
9−6
3
=
= 0.75
4
4
9−6
3
=
= 0.75
4
4
5−6
−1
=
= −0.25
4
4
```<|endoftext|>
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# Area under a graph Car travelling at 70 mph Area = This is the distance travelled, 140 miles 2 70 = 140 v mph t hours 0 2 70.
## Presentation on theme: "Area under a graph Car travelling at 70 mph Area = This is the distance travelled, 140 miles 2 70 = 140 v mph t hours 0 2 70."— Presentation transcript:
Area under a graph Car travelling at 70 mph Area = This is the distance travelled, 140 miles 2 70 = 140 v mph t hours 0 2 70
Car accelerating steadily from 0 to 45 mph in 10 seconds Area = Distance travelled = 110 yards = 110 v mph t seconds 0 10 45 45 mph = = 22 yards per second 22
Car travelling between 2 sets of traffic lights 0 Area of A = t (s) v (ms -1 ) 0 0 2 5 4 8 8 8 10 5 12 0 6 9 v ms -1 t seconds 12 624810 A A CC B B = 70 Area of C == 17 Area of B = = 13 Total Area = 5 Distance travelled = 70 metres
y x 0 a b Integration As x 0 Integration is the inverse of differentiation. lim x 0 A = As A = y x A = xx y y x AA The gradient of the graph of A against x.
Rules of Integration = x y = 5x = 5 y = c = 0 Function Derivative Reverse to give integral = 5x y = = x 2 y = = x 3 y = + c
13 Example Find the area under y = 2x 3 - 9x 2 + 12x + 6 between x = 1 and x = 3 y x 0 y = 2x 3 - 9x 2 + 12x + 6 A = + 6x - 3x 3 + 6x 2 + 6 3 -3 3 3 + 6 3 2 + 6 1 -3 1 3 + 6 1 2 = 31.5- 9.5 = 22 square units
Distance = The distance travelled is 140 miles = 140 Car travelling at 70 mph v mph t hours 0 2 70 Velocity-time graphs v = 70
Car accelerating steadily from 0 to 22 yd/s (45 mph) in 10s v yd/s t seconds 0 10 22 gradient = v = 2.2t intercept = 0 = 2.2 Distance = The distance travelled is 110 yards = 110
Car travelling between 2 sets of traffic lights t024810126 v0588509 v ms -1 t seconds 0 12 Distance travelled = 72 metres v = 0.25t(12 - t) Distance = = 72
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Since interest is compounded quarterly, the principal amount will change at the end of the first 3 months(first quarter). Other than the first year, the interest compounded annually is always greater than that in case of simple interest. Compound interest is the interest calculated on the principal and the interest accumulated over the previous period. Let us solve various examples to understand the concepts in a better manner. 10000, Rate = 10%, and Time = 2 years, From the table shown above it is easy to calculate the amount and interest for the second year, which is given by-, Amount($$A_{2}$$) = $$P\left (1+\frac{R}{100} \right )^{2}$$, $$A_{2}$$= $$= 10000 \left ( 1 + \frac{10}{100} \right )^{2} = 10000 \left ( \frac{11}{10} \right )\left ( \frac{11}{10} \right )= Rs.12100$$, Compound Interest (for 2nd year) = $$A_{2} – P$$ = 12100 – 10000 = Rs. https://byjus.com/cbse-sample-papers-for-class-8-maths/ We can also reduce the formula of compound interest of yearly compounded for quarterly as given below: $$CI =P(1+\frac{\frac{R}{4}}{100})^{4T}-P$$. 6050\), Interest (Second Year) = A – P = 6050 – 5000 = Rs.1050. Example of Compound Interest Formula Suppose an account with an original balance of $1000 is earning 12% per year and is compounded monthly. 2100. Compound interest finds its usage in most of the transactions in the banking and finance sectors and also in other areas as well. The price of a radio is Rs 1400 and it depreciates by 8% per month. for the first year: Amount after first year = $$P~+~SI_1$$ = $$P ~+~ \frac{P~×~R~×~T}{100}$$ = $$P \left(1+ \frac{R}{100}\right)$$ = $$P_2$$, Amount after second year = $$P_2~+~SI_2$$ = $$P_2 ~+~ \frac{P_2~×~R~×~T}{100}$$ = $$P_2\left(1~+~\frac{R}{100}\right)$$ = $$P\left(1~+~\frac{R}{100}\right) \left(1~+~\frac{R}{100}\right)$$ Some of its applications are: To understand the compound interest we need to do its Mathematical calculation. Illustration 3: Calculate the compound interest to be paid on a loan of Rs.2000 for 3/2 years at 10% per annum compounded half-yearly? What will be its total population in 2005? For the depreciation, we have the formula A = P(1 – R/100)n. Thus, the price of the radio after 3 months = 1400(1 – 8/100)3, = 1400(1 – 0.08)3 = 1400(0.92)3 = Rs 1090 (Approx.). For the total accumulated wealth (or amount), the formula is given as: Compound interest is when a bank pays interest on both the principal (the original amount of money)and the interest an account has already earned. Find the value of the investment after the two years if the investment earns the return of 2 % compounded quarterly. Due to being compounded monthly, the number of periods for one year would be 12 and the rate would be 1% (per month). To calculate compound interest use the formula below. To calculate compound interest we need to know the amount and principal. From the data it is clear that the interest rate for the first year in compound interest is the same as that in case of simple interest, ie. For the decrease, we have the formula A = P(1 – R/100)n, Therefore, the population at the end of 5 years = 10000(1 – 10/100)5, = 10000(1 – 0.1)5 = 10000 x 0.95 = 5904 (Approx.). Its population declines at a rate of 10% per annum. P is the principal; that's the amount you start with. Any link to worksheets/assignments/practice tests? Example, 6% interest with " monthly compounding " does not mean 6% per month, it means 0.5% per month (6% divided by 12 months), and is worked out like this: FV = PV × (1+r/n)n =$1,000 × (1 + 6%/12)12 = \$1,000 × (1 + 0.5%)12 Solution: Principal, $$P$$ = $$Rs.2000$$, Time, $$T’$$ = $$2~×~\frac{3}{2}$$ years = 3 years, Rate, $$R’$$ = $$\frac{10%}{2}$$ = $$5%$$, amount, $$A$$ can be given as: $$A = P ~\left(1~+~\frac{R}{100}\right)^n$$, $$A = 2000~×~\left(1~+~\frac{5}{100}\right)^3$$, = $$2000~×~\left(\frac{21}{20}\right)^3 = Rs.2315.25$$, $$CI = A – P = Rs.2315.25~ –~ Rs.2000$$ = $$Rs.315.25$$. Simple Interest (S.I.) For detailed discussion on compound interest, download BYJU’S -The learning app. It is the difference between amount and principal. $$\frac{PR}{100}$$. Principal (P) = Rs.5000 , Time (T)= 2 year, Rate (R) = 10 %, We have, Amount, $$A = P \left ( 1 + \frac{R}{100} \right )^{T}$$, $$A = 5000 \left ( 1 + \frac{10}{100} \right )^{2} = 5000 \left ( \frac{11}{10} \right )\left ( \frac{11}{10} \right ) = 50 \times 121 = Rs. A town had 10,000 residents in 2000. Compound Interest Formula Example #3 Case of Compounded Quarterly Fin International Ltd makes an initial investment of 10,000 for a period of 2 years. The population of the town decreases by 10% every year. This interest varies with each year for the same principal amount. We can see that interest increases for successive years. Students can also use a compound interest calculator, to solve compound interest problems in an easier way. First, we will look at the simplest case where we are using the compound interest formula to calculate the value of an investment after some set amount of time. nice questions , but some hard questions must be added. Interest (I1) = \(P\times \frac{R}{100} = 5000 \times \frac{10}{100} =500$$, Interest (I2) = $$P\times \frac{R}{100}\left (1 + \frac{R}{100} \right ) = 5000 \times \frac{10}{100}\left ( 1 + \frac{10}{100} \right ) = 550$$, Total Interest = I1+ I2 = 500 + 550 = Rs. Compound Interest Formula The formula for the Compound Interest is, This is the total compound interest which is just the interest generated minus the principal amount. = $$P \left(1~+~\frac{R}{100}\right)^2$$. Thus, it has a new population every year. Putting these variables into the compound interest formula would show CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, https://byjus.com/maths/important-questions-class-8-maths/, https://byjus.com/cbse-sample-papers-for-class-8-maths/, CBSE Previous Year Question Papers Class 12 Maths, CBSE Previous Year Question Papers Class 10 Maths, ICSE Previous Year Question Papers Class 10, ISC Previous Year Question Papers Class 12 Maths.
Hence, we can conclude that the interest charged by the bank is not simple interest, this interest is known as compound interest. This data will be helpful in determining the interest and amount in case of compound interest easily. In the formula, A represents the final amount in the account after t years compounded 'n' times at interest rate 'r' with starting amount 'p'.<|endoftext|>
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Other Officers of the House
“The House of Representatives shall chuse their Speaker and other Officers; and shall have the sole Power of Impeachment.”
— U.S. Constitution, Article I, section 2, clause 5
When a quorum finally gathered on April 1, 1789, the House first elected its Speaker, Frederick Augustus Conrad Muhlenberg of Pennsylvania, and then its first Clerk, John Beckley of Virginia. While the Clerk’s title, like the Speaker’s, evolved from the British House of Commons, the duties of the office grew as the House wrote its rules. Fundamentally, the Clerk had an organizational function that included receiving the credentials of Members elected to the House, compiling the official list (or roll) of Members, and maintaining the journal of proceedings mandated by the Constitution—duties for which the Clerk’s office is still responsible. As the First Congress (1789–1791) established its rules, the Clerk’s duties multiplied from parliamentary functions like reading bills to ensuring that bills sent to the Senate (engrossed bills) were printed “in a fair round hand.” Clerks also managed the House’s finances and payroll until the Sergeant-at-Arms, and later, the Chief Administrative Officer assumed those duties.
A House of Representatives officer from 1789 to 1995, the Doorkeeper was elected by a resolution at the opening of each Congress. The Office of the Doorkeeper was based on precedent from the Continental Congresses, which controlled entrance to its largely private meetings. Without debate, the First Congress (1789–1791) created the Doorkeeper’s position by resolution on April 2, 1789. The Doorkeeper controlled access to the House Chamber and eventually oversaw the press in the gallery. A total of 34 individuals served in the Doorkeeper position until it was terminated at the opening of the 104th Congress (1995–1997). Many of its duties were transferred to the Sergeant-at-Arms, the Clerk of the House, and the newly created Chief Administrative Officer.
The House created the Sergeant-at-Arms position on April 14, 1789, and charged it with securing and keeping order within the Chamber, and serving warrants and arresting individuals at the direction of the House and the Speaker. The House also put the Sergeant-at-Arms in charge of the Mace—a symbol borrowed from British practice—to indicate the operating status of the chamber. The duties of the Sergeant-at-Arms have since expanded to include more administrative tasks. In addition to providing security for the chamber and the House side of the Capitol complex, the Sergeant-at-Arms also manages state visits, announces the arrival of the President and other dignitaries during Joint Sessions (a duty previously held by the Doorkeeper), administers parking, and issues identification for staff.
The election of the Rev. William Linn as Chaplain of the House on May 1, 1789, continued the tradition established by the Continental Congresses of opening each day's proceedings with a prayer. The early House Chaplains alternated duties with their Senate counterparts on a weekly basis. The two conducted Sunday services for the Washington community in the House Chamber every other week. In addition to opening proceedings with prayer, the Chaplain provides pastoral counseling to the House community, coordinates the scheduling of guest chaplains, and arranges memorial services for the House and its staff. In the past, Chaplains have performed marriage and funeral ceremonies for House Members.
The Chief Administrative Officer
The Chief Administrative Officer (CAO) position was created in 1995 to address a host of administrative functions previously performed by the Doorkeeper and the Postmaster. The office deals closely with the Speaker and the Committee on House Administration to assure the smooth operation of House finances as well as technology, human resources, procurement, and services within the House Office Buildings. The predecessor to the CAO was the Director of Non-Legislative and Financial Services. Established by H. Res. 423 in the 102nd Congress (1991–1993), this appointed position was eliminated at the end of the 103rd Congress (1993–1995).
An elected officer for nearly 160 years (24th through the 102nd Congresses), the Postmaster managed mail operations in the House. Initially, workers in the Doorkeeper’s office were paid additional compensation to perform mail duties, but a December 13, 1832, resolution made the Postmaster a distinct and permanent House employee. In 1834, William J. McCormick, a Doorkeeper’s employee, was named the first House Postmaster. Four years later, the House adopted a rule that charged the Postmaster with superintending the post office in the Capitol and overseeing the delivery of mail to Members and Officers. The House Reform Resolution of 1992 (H. Res. 423, April 9, 1992), abolished the Office of the Postmaster and reassigned mail handling procedures for the House of Representatives to other House Officers and private entities. A total of 21 Postmasters served in the House.
For Further Reading
Brudnick, Ida A. “Support Offices in the House of Representatives: Roles and Authorities.” 18 May 2011. Report RS33220. Congressional Research Service. (Washington, D.C.: Library of Congress).
Cannon’s Precedents of the House of Representatives of the United States. Six volumes. (Washington, D.C.: Government Printing Office, 1935–1941).
Constitution, Jefferson’s Manual and Rules of the House of Representatives. (Washington, D.C.: Government Printing Office, 1896 to present).
Deschler’s Precedents of the House of Representatives of the United States. Nine volumes. (Washington, D.C.: Government Printing Office, 1976–1977).
Hinds’ Precedents of the House of Representatives of the United States. Eight volumes. (Washington, D.C.: Government Printing Office, 1907–1908).
Sullivan, John. House Practice: A Guide to the Rules, Precedents, and Procedures of the House. (Washington: Government Printing Office, 2011).<|endoftext|>
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A recent high-profile study led by US climatologist James Hansen has warned that sea levels could rise by several metres by the end of this century. How realistic is this scenario?
We can certainly say that sea levels are rising at an accelerating rate, after several millennia of relative stability. The question is how far and how fast they will go, compared with Earth’s previous history of major sea-level changes.
Seas have already risen by more than 20 cm since 1880, affecting coastal environments around the world. Since 1993, sea level has been rising faster still (see chapter 3 here), at about 3 mm per year (30 cm per century).
One key to understanding future sea levels is to look to the past. The prehistoric record clearly shows that sea level was higher in past warmer climates. The best evidence comes from the most recent interglacial period (129,000 to 116,000 years ago), when sea level was 5-10 m higher than today, and high-latitude temperatures were at least 2℃ warmer than at present.
The two largest contributions to the observed rise since 1900 are thermal expansion of the oceans, and the loss of ice from glaciers. Water stored on land (in lakes, reservoirs and aquifers) has also made a small contribution. Satellite observations and models suggest that the amount of sea-level rise due to the Greenland and Antarctic ice sheets has increased since the early 1990s.
Before then, their contributions are not well known but they are unlikely to have contributed more than 20% of the observed rise.
Together, these contributions provide a reasonable explanation of the observed 20th-century sea-level rise.
The Intergovernmental Panel on Climate Change (IPCC) projections (see chapter 13 here) forecast a sea-level rise of 52-98 cm by 2100 if greenhouse emissions continue to grow, or of 28-61 cm if emissions are strongly curbed.
The majority of this rise is likely to come from three sources: increased ocean expansion; glacier melt; and surface melting from the Greenland ice sheet. These factors will probably be offset to an extent by a small increase in snowfall over Antarctica.
With continued emissions growth, it is entirely possible that the overall rate of sea-level rise could reach 1 m per century by 2100 – a rate not seen since the last global ice-sheet melting event, roughly 10,000 years ago.
Beyond 2100, seas will continue to rise for many centuries, perhaps even millennia. With continued growth in emissions, the IPCC has projected a rise of as much as 7 m by 2500, but also warned that the available ice-sheet models may underestimate Antarctica’s future contribution.
The joker in the pack is what could happen to the flow of ice from the Antarctic ice sheet directly into the ocean. The IPCC estimated that this could contribute about 20 cm of sea-level rise this century. But it also recognised the possibility of an additional rise of several tens of centimetres this century if the ice sheet became rapidly destabilised.
This could happen in West Antarctica and in parts of the East Antarctic ice sheets that are resting on ground below sea level, which gets deeper going inland from the coast. If relatively warm ocean water penetrates beneath the ice sheet and melts its base, this would cause the grounding line to move inland and ice to flow more rapidly into the ocean.
Several recently published studies have confirmed that parts of the West Antarctic ice sheet are already in potentially unstoppable retreat. But for these studies the additional rise above the IPCC projections of up to 98 cm by 2100 from marine ice sheet instability was more likely to be just one or two tenths of a metre by 2100, rather than several tenths of a metre allowed for in the IPCC report. This lower rise was a result of more rigorous ice-sheet modelling, compared with the results available at the time of the IPCC’s assessment.
How stable are ice sheets?
Ocean temperatures were thought to be the major control in triggering increased flow of the Antarctic ice sheet into the ocean. Now a new study published in Nature by US researchers Robert DeConto and David Pollard has modelled what would happen if you factor in increased surface melting of ice shelves due to warming air temperatures, as well as the marine melting.
Such an ice-shelf collapse has already been seen. In 2002, the Larsen-B Ice Shelf on the Antarctic Peninsula disintegrated into thousands of icebergs in a matter of weeks, allowing glaciers to flow more rapidly into the ocean. The IPCC’s predictions had considered such collapses unlikely to occur much before 2100, whereas the new study suggests that ice-sheet collapse could begin seriously affecting sea level as early as 2050.
With relatively high greenhouse emissions (a scenario referred to in the research literature as RCP8.5), the new study forecasts a rise of about 80 cm by 2100, although it also calculated that this eventuality could be almost totally averted with lower emissions. But when the model parameters were adjusted to simulate past climates, the Antarctic contribution was over 1 m by 2100 and as much as 15 m by 2500.
Greenland’s ice sheet is crucially important too. Above a certain threshold, warming air temperatures would cause surface melting to outstrip snow accumulation, leading to the ice sheet’s eventual collapse. That would add an extra 7 m to sea levels over a millennium or more.
The problem is that we don’t know where this threshold is. It could be as little as 1℃ above pre-industrial average temperatures or as high as 4℃. But given that present-day temperatures are already almost 1℃ above pre-industrial temperatures, it is possible we could cross this threshold this century, regardless of where exactly it is, particularly for high-emission scenarios.
Overall, then, it is clear that the seeds for a multi-metre sea-level rise could well be sown during this century. But in terms of the actual rises we will see in our lifetimes, the available literature suggests it will be much less than the 5 m by 2050 anticipated by Hansen and his colleagues.
The wider question is whether the ice-sheet disintegration modelled by DeConto and Pollard will indeed lead to rises of the order of 15 m over the coming four centuries, as their analysis and another recent paper suggest. Answering that question will require more studies, with a wider range of climate and ice-sheet models.<|endoftext|>
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__1664: __Thomas Newcomen, creator of the first practical atmospheric steam engine, is born in Devon, England.
Newcomen's exact birth date is a matter of debate. Some sources say "before Feb. 24," while others go as late as Feb. 28. What's more, some list 1663, because dates between Jan. 1 and March 25 (the old New Year's celebration) were sometimes listed as one year, sometimes the other and sometimes both, like this: 1663/64.
Whenever he came into this world, he certainly changed it before he left.
A devout Baptist, Newcomen worked as an ironmonger, fabricating and selling metal parts. Devon had plenty of tin mines around.
But flooding was a major problem in the mines, and water often had to be often pumped out of the mines – by human or animal power. It was for this task that Newcomen created the first practical steam engine. It was the prototype of the machinery that made the Industrial Revolution possible.
The initial design of Newcomen's engine used a vertical brass cylinder with a piston connected to a rocking beam. A copper boiler sat below the cylinder to heat the water to boiling. When the piston was at the top of its range of motion, water was sprayed into the cylinder. That cooled the insides, condensing the steam within. This formed a vacuum, pulling the piston down. The boiler, still on, then reheated the steam, driving the piston up again.
Repeating this process caused the rocker beam to rock up and down: a working steam engine. Newcomen's 1712 design attached the working end of the beam (opposite the piston) to chains that descended to pumps located deep in the mine.
The Newcomen engine and its successor, the more efficient Watt engine of the 1770s, are called "atmospheric engines" because the steam is under only "slight pressure."
Newcomen wasn't the only one to stumble on the basic principle driving his design. Thomas Savery had invented a device in 1698 that used a vacuum and atmospheric pressure to suck out water. High-pressure steam forced the water to the surface of the mine. It was a good idea in theory, but there were challenges in its construction, and the resulting equipment didn't work well enough when used in mines.
Newcomen's engine was easier to build and thus found greater application. But after Savery's death, a joint stock company was set up to issue licenses for both Newcomen and Savery engines.
The simplicity of the Newcomen design and its effectiveness helped it become popular quickly. The first successful Newcomen engine was installed at a coal works plant in 1712. More than 110 Newcomen engines were in operation in Britain and Europe by 1773.
The Newcomen steam-engine design eventually faded away, mainly because it was expensive to operate. (The Watt design that replaced it added a condenser cylinder to reduce heat loss and increase fuel efficiency.) But a full-sized replica of a Newcomen engine, based on a 1719 drawing, exists at the Black Country Living Museum in Dudley, England.
Newcomen himself finally ran out of steam and died at home in 1729.
Photo: The oldest surviving Newcomen engine, known as Fairbottom Bobs, was built around 1760 to pump water out of the Cannel coal pits in northwest England. The name arose from the bobbing motion of the wooden beam. Henry Ford moved the apparatus to his museum in Dearborn, Michigan, in 1929, four decades after this photo was taken.
- July 15, 1783: Marquis Invents Steamboat, Misses Esteem Boat
- Aug. 17, 1807: 'Fulton's Folly' Steams Up the Hudson
- June 1, 1849: Stanley Twins Steam Into History
- Oct. 9, 1855: Music-Making a Steampunk Can Love
- Steam-Driven Dreams: The Wondrously Whimsical World of Steampunk
- U.S. Team Hopes to Bring Steam Car Record Home
- Nov. 28, 1660: Hey, Guys, Let's Found Britain's Foremost Scientific Academy
- March 18, 1662: The Bus Starts Here ... in Paris
- Jan. 12, 1665: Fermat's Last Breath
- Nov. 14, 1666: Watching a Transfusion, and Taking Notes
- June 15, 1667: First Human Blood Transfusion Is Performed
- Feb. 24, 1582: Mark the Date
- Feb. 24, 1949: Piercing the Edge of the Final Frontier<|endoftext|>
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# Difference between revisions of "2012 AMC 10A Problems/Problem 23"
The following problem is from both the 2012 AMC 12A #19 and 2012 AMC 10A #23, so both problems redirect to this page.
## Problem
Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?
$\textbf{(A)}\ 60\qquad\textbf{(B)}\ 170\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 320\qquad\textbf{(E)}\ 660$
## Solution
Note that if $n$ is the number of friends each person has, then $n$ can be any integer from $1$ to $4$, inclusive.
Also note that the cases of $n=1$ and $n=4$ are the same, since a map showing a solution for $n=1$ can correspond one-to-one with a map of a solution for $n=4$ by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with $n=2$ when compared to configurations of $n=3$. Thus, we have two cases to examine, $n=1$ and $n=2$, and we count each of these combinations twice.
For $n=1$, if everyone has exactly one friend, that means there must be $3$ pairs of friends, with no other interconnections. The first person has $5$ choices for a friend. There are $4$ people left. The next person has $3$ choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are $15$ configurations with $n=1$.
For $n=2$, there are two possibilities. The group of $6$ can be split into two groups of $3$, with each group creating a friendship triangle. The first person has $\binom{5}{2} = 10$ ways to pick two friends from the other five, while the other three are forced together. Thus, there are $10$ triangular configurations.
However, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be seated anywhere on the hexagon Without Loss Of Generality. This person has $\binom{5}{2} = 10$ choices for the two friends on the adjoining vertices. Each of the three remaining people can be seated "across" from one of the original three people, forming a different configuration. Thus, there are $10 \cdot 3! = 60$ hexagonal configurations, and in total $70$ configurations for $n=2$.
As stated before, $n=3$ has $70$ configurations, and $n=4$ has $15$ configurations. This gives a total of $(70 + 15)\cdot 2 = 170$ configurations, which is option $\boxed{\textbf{(B)}\ 170}$.<|endoftext|>
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# Chapter 2: Inverse Trigonometric Functions EXERCISE 2.1
In this page we have NCERT Solutions for Class 12 Maths Chapter 2: Inverse Trigonometric Functions for EXERCISE 2.1 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Formula's used
Principle Value branch of Trigonometric Inverse Functions along with domain and range
### Find the principal values of the following
Question 1. $\sin^{-1}\left ( -\frac{1}{2} \right )$
Solution
Let $\sin^{-1}\left ( -\frac{1}{2} \right )$ = x, then
$\sin {x} = -\frac{1}{2} = - \sin\frac{ \pi}{6} = \sin ( - \frac{ \pi}{6} )$
We know from above table that
The principal value branch range for sin-1 is $\left [ -\frac{ \pi}{2}, \frac{ \pi}{2} \right ]$ and $\sin ( -\frac{ \pi}{6} ) = - \frac{1}{2}$
Therefore principal value for $\sin^{-1}\left ( -\frac{1}{2} \right ) \; is \; - \frac{ \pi}{6}$
Question 2. $\cos^{-1}\left ( - \frac{\sqrt{3}}{2} \right )$
Solution
Let $\cos^{-1}\left ( - \frac{\sqrt{3}}{2} \right )$ = x, then
$\cos x = \frac{\sqrt{3}}{2} = \cos (\frac{\pi}{6})$ We know from above table that
The principal value branch range for cos-1 is $\left [ 0 , \pi \right ]$ and $\cos (\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$
Therefore, principal value for $\cos^{-1}\left ( - \frac{\sqrt{3}}{2} \right ) \; is \; \frac{\pi}{6}$
Question 3. cosec-1 (2)
Solution
Let cosec-1 (2) = x. Then, cosec x = 2 = cosec $(\frac{\pi}{6} )$
We know from above table that
The principal value branch range for cosec-1 is $\left [ -\frac{\pi}{2}, \frac{\pi}{2}\right ] - {0}$ and cosec$(\frac{\pi}{6} )$ = 2
Therefore, principal value for cosec-1 (2) is $\frac{\pi}{6}$
Question 4. $\tan^{-1} \left ( - \sqrt{3} \right )$
Solution
Let $\tan^{-1} \left ( - \sqrt{3} \right ) = x$
Then, $\tan = - \sqrt{3} = - \tan \frac{\pi}{3} \tan (- \frac{\pi}{3})$
We know from above table that
The principal value branch range for $\tan^{-1} \; is \; \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ] \; and \; \tan\left ( -\frac{\pi}{3} \right ) = -\sqrt{3}$
Therefore, principal value for $\tan^{-1} \left ( - \sqrt{3} \right ) \; is \; -\frac{\pi}{3}$
Question 5. $\cos^{-1}\left ( -\frac{1}{2} \right )$
Solution
Let $\cos^{-1}\left ( -\frac{1}{2} \right )$ = x,
Then $\cos x = -\frac{1}{2} = -cos \frac{\pi}{3} = \cos ( \pi - \frac{\pi}{3} ) = \cos( \frac{2 \pi}{3} )$
We know from above table that
The principal value branch range for $\cos ^{-1} \; is \; \left [ 0 , \pi \right ] \; and \; \cos \left ( \frac{2 \pi}{3} \right ) = - \frac{1}{2}$
Therefore, principal value for $\cos^{-1}\left ( -\frac{1}{2} \right ) \; is \; \frac{2 \pi}{3}$
Question 6. $\tan^{-1} (-1)$
Solution
Let $\tan^{-1} (-1) = x$,
Then, tan x = -1 = $-\tan ( \frac{\pi}{4} ) = \tan ( - \frac{\pi}{4} )$
We know from above table that
The principal value branch range for $\tan^{-1} \; is \; \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ) \; and \; \tan ( - \frac{\pi}{4} ) = -1$
Therefore, principal value for $\tan^{-1} (-1) \; is \; -\frac{\pi}{4}$
Question 7. $\sec^{-1} \left ( \frac{2}{\sqrt{3}} \right )$
Solution
Let $\sec^{-1} \left ( \frac{2}{\sqrt{3}} \right ) = x$,
Then $\sec x = \frac{2}{\sqrt{3}} = \sec (\frac{\pi}{6})$
We know from above table that
The principal value branch range for $\sec^{-1} \; is \; \left [ 0 , \pi \right ] - \left \{ \frac{ \pi }{2} \right \} \; and \; \sec (\frac{\pi}{6}) = \frac{2}{\sqrt{3}}$
Therefore, principal value for $\sec^{-1} \left ( \frac{2}{\sqrt{3}} \right ) \; is ; \frac{\pi}{6}$
Question 8. $\cot^{-1} \sqrt{3}$
Solution
Let $\cot^{-1} \sqrt{3} = x$,
Then $\cot x = \sqrt {3} = \cot \left ( \frac{\pi}{6} \right )$
We know from above table that
The principal value branch range for cot-1 is $( 0 , \pi )$ and $\cot \left ( \frac{\pi}{6} \right ) = \sqrt{3}$
Therefore, principal value for $\cot^{-1} \sqrt{3} = \frac{\pi}{6}$
Question 9. Find principal value for $\cos^{-1} \left ( - \frac{1}{\sqrt{2}} \right )$
Solution
Let $\cos^{-1} \left ( - \frac{1}{\sqrt{2}} \right ) = a$
Then $\cos x = \frac{-1}{\sqrt{2}} = - \cos \left ( \frac{\pi}{4} \right ) = \cos \left ( \pi - \frac{\pi}{4} \right ) = \cos \left ( \frac{3 \pi}{4} \right )$
We know from above table that
The principal value branch range for cos-1 is $[0 , \pi] \; and \; \cos \left ( \frac{3 \pi}{4} \right ) = -\frac{1}{\sqrt{2}}$
Therefore, principal value for $\cos^{-1} \left ( - \frac{1}{\sqrt{2}} \right ) \; is \; \frac{3 \pi }{4}$
Question 10. Find principal value for cosec-1 $\left ( -\sqrt{2} \right )$
Solution
Let cosec-1$\left ( -\sqrt{2} \right )$ = x, Then
cosec x = $-\sqrt{2}$ = -cosec$\left ( \frac{\pi}{4}\right )$ = cosec $\left ( -\frac{\pi}{4}\right )$
We know,
The principal value branch range for cosec-1 is $\left [ -\frac{\pi}{2} , \frac{\pi}{2} \right ] - \left \{ 0 \right \}$ and cosec$\frac{-\pi}{4} = -\sqrt{2}$
Therefore, principal value for cosec-1 $\left ( -\sqrt{2} \right ) \; is \; -\frac{\pi}{4}$
### Find the values of the following
Question 11. Solve $\tan ^{-1}(1) + \cos^{-1} \left ( -\frac{1}{2} \right ) + \sin ^{-1}\left ( -\frac{1}{2} \right )$
Solution
Let $\tan ^{-1}(1) = x$, then
$\tan x = 1 = \tan \frac{\pi}{4}$ We know from above table that
The principal value branch range for $\tan ^{-1} \; is \; \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$
$\tan ^{-1}(1) = \frac{\pi}{4}$
Let $\cos^{-1} \left ( -\frac{1}{2} \right ) = y$, then
$\cos y = -\frac{1}{2} = -\cos \frac {\pi}{3} = \cos\left ( \pi - \frac{\pi}{3} \right ) = \cos\left ( \frac{2\pi}{3} \right )$
We know from above table that
The principal value branch range for cos-1 is $[0 , \pi]$
$\cos ^{-1} \left ( -\frac{1}{2} \right ) = \frac{2 \pi }{3}$
Let $\sin^{-1}\left ( -\frac{1}{2} \right ) = z$, then
$\sin z = - \frac{1}{2} = - \sin \frac{\pi}{6} = \sin \left ( -\frac{\pi}{6} \right )$
We know from above table that
The principal value branch range for $\sin ^{-1} \; is \; \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$
$\sin^{-1} \left ( -\frac{1}{2} \right ) = - \frac{\pi}{6}$ Now
$\tan ^{-1}(1) + \cos^{-1} \left ( -\frac{1}{2} \right ) + \sin ^{-1}\left ( -\frac{1}{2} \right )$ $= \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi}{4}$
Question 12. Solve $\cos ^{-1} \left ( \frac{1}{2} \right ) + 2 \sin ^{-1} \left ( \frac{1}{2} \right )$
Solution
Let $\cos ^{-1} \left ( \frac{1}{2} \right ) = x$, then $\cos x = \frac{1}{2} = \cos \frac{\pi}{3}$
We know from above table that
The principal value branch range for cos-1 is $\left [0 , \pi \right ]$
$\cos ^{-1} \left ( \frac{1}{2} \right ) = \frac{\pi}{3}$
Let $\sin ^{-1} \left (- \frac{1}{2} \right ) = y$, then $\sin y = \frac{1}{2} = \sin \frac{\pi}{6}$
We know,
The principal value branch range for $\sin ^{-1} \; is \; \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$
$\sin ^{-1} \left ( \frac{1}{2} \right ) = \frac{\pi}{6}$
Now,
$\cos ^{-1} \left ( \frac{1}{2} \right ) + 2 \sin ^{-1} \left ( \frac{1}{2} \right )$ $= \frac{\pi}{3} + 2 \times \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$
### Multiple Choice Questions
Question 13. If sin-1 a = b, then
(i) $0 \leq b \leq \pi$
(ii) $-\frac{\pi}{2} \leq b \leq \frac{\pi}{2}$
(iii) $0 < b < \pi$
(iv) $-\frac{\pi}{2} < b < \frac{\pi}{2}$
Solution
Given sin-1 a = b
We know from above table that
The principal value branch range for $\sin ^{-1} \; is \; \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$
Therefore, $-\frac{\pi}{2} \leq b \leq \frac{\pi}{2}$
Question 14. The value of $\tan ^{-1} \sqrt{3} - \sec ^{-1}(-2)$ is
(i) $\pi$
(ii) $- \frac{\pi}{3}$
(iii)$\frac{\pi}{3}$
(iv) $\frac{2 \pi}{3}$
Solution
Let $\tan ^{-1} \sqrt{3} = x$, then
$\tan x = \sqrt{3} = \tan \frac{\pi}{3}$
We know from above table that
The principal value branch range for $\tan ^{-1} \; is \; \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$
$\tan ^{-1}\sqrt{3} = \frac{\pi}{3}$
Let sec-1(-2) = y, then
sec y = -2 = $- \sec \frac{\pi}{3} = \sec \left ( \pi - \frac{\pi}{3} \right ) = \sec \left ( \frac{2 \pi}{3} \right )$
We know from above table that
The principal value branch range for sec-1 is $[0 , \pi] - \left \{ \frac{\pi}{2} \right \}$
$\sec ^{-1}(-2) = \frac{2 \pi}{3}$
Now,
$\tan ^{-1} \sqrt{3} - \sec ^{-1}(-2) = \frac{\pi}{3} - \frac{2 \pi}{3} = - \frac{\pi}{3}$
Hence option (ii) is correct
Go back to Class 12 Main Page using below links<|endoftext|>
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# How Do You Solve 2 Divided By 5?
If you’ve ever pondered on the solution to the division of 2 by 5, then you’re in the correct location!
In this scenario, with 2 divided by 5, we are questioning “How frequently does five fit into two?”. When you perform division on one numerical value with another, it is akin to inquiring “How often does the second numerical value fit into the first?”. To begin, let us examine the significance of dividing one numerical value by another.
Let’s find out how we can break it down step by step. The answer to 0.4 (or 4/10ths) is obtained by dividing 5 by 2. The answer is not very many, but only once.
1 (since 2 is smaller than 5). – When we divide 5 by 2, the answer will be less than 1 (since 2 is smaller than 5). We can easily see that 2 is smaller than 5 by comparing their denominators. Now, for the second fraction: to get a numerator of 5 for 1/5 and 2/1, we just need to multiply 2 by 5.
In order to determine the precise value of our answer, let’s utilize long division.
When we divide 10 by 4 or 0.4, the result is 5 with a remainder of 0. The quotient, which is the answer, is 0. The dividend, which is the number being divided, is 2. Therefore, 5 divided by 2 equals 2.
You should be able to quickly and accurately figure out any division problem, and converting fractions into them with a common denominator is something you should remember to do – it’s not too difficult, as you can see from dividing these two numbers. Just practice patience and convert them with practice.
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The fraction value of 2/5 is obtained by dividing 2 by 5.
Table of Contents
## What Will Be The Remainder When Is Divided By 5?
When divided by 5, the remainder will be 3. When multiplied by 4, 20 divided by 5 leaves a remainder of 1. The remainder is 3 when 20 is divided by 5.
## What Can 5 Be Divided By?
5 can be divided by 1, 2, 3, 5, and 10.
## What Is The Answer For 2 Divided By 2?
The answer is obtained by dividing 2 by 2, because when you divide 2 into two equal parts, each part consists of either 1 orange or 1 apple.
## What Is 5 Divided By 2 Long?
2 and 1/2 is equivalent to the result of dividing 5 by 2. When you divide 2 by 5, you get a quotient of 2 and a remainder of 1. This can also be expressed as a fraction with a denominator of 2 and a remainder of 1.
## What Is The Remainder Of 2 Divided By 5?
The result of dividing 2 by 5 is 2.
## How Do We Divide Fractions?
Dividing fractions is necessary to obtain the reciprocal of the second fraction, which involves flipping the numerator and denominator. In this case, the reciprocal of 4/3 would be 3/4.
Then, we multiply the two top numbers. In this situation, it would be 3 x 4 = 12.
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Next, we multiply the two denominators. In this situation, it would be 4 x 3 = 12.
The solution is twelve-twelfths or one.
## Can 4 Be Divided By 2?
Yes, you can divide 2 groups of 2 into 4 evenly.
## How Do You Do 0.5 Divided By 2?
In this case, you will need to use the division algorithm to divide two numbers. The algorithm used for dividing two numbers that are less than 1 is called long division.
Begin by setting up the problem by drawing a line down the middle of a piece of paper. On the right side of the line, write the number 2. On the left side of the line, write the number 0.5.
To divide 0.5 by 2, you can determine how many times 5 can be divided into 2. After drawing a line to separate the numbers, underline the 2 on the right side and write 2 above the 5 on the left side. This indicates that there are two instances of 2 in 5, resulting in 0.5 being divided into two sections. The first section consists of 2, while the second section contains nothing. Moving forward.
Sketch a line across it and jot down -1.5 beneath 0.5 on the left-hand side of the line. Transfer this numerical value to the opposite side of the equality symbol as it denotes negativity. The solution equates to -1.5 when subtracting 2 from 0.5. Deduct 2 from 0.5, and now it is the appropriate moment.
The part is now completed, therefore the answer is -0.75. There is no number on the right side of the line when dividing by 2. There is no number on the left side of the line when dividing by 2. This is the final step.
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## Los 10 museos más famosos de Nueva York (y 10 alternativas)
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## Ratchet & Clank: Rift Apart is coming to PC—and it will be a technical showstopper
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## Satanic Panic Is Back: TikTok Calls To Boycott Coach Over Disney Villains Teddy Bear Collection
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## How Many Slices Are In A Costco Pizza? Get The Answer Now
Each pizza has its own unique flavor profile that will tantalize your taste buds. Costco offers a variety of pizzas, including gluten-free cheese pizzas and pesto chicken…
## Square Root of 361
In this article, we will also look at different methods for calculating the square root of 361 using a computer or calculator. We’ll answer some common questions…<|endoftext|>
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# How do you find the vertex of a quadratic equation y=(x+8)^2-2?
Nov 28, 2015
${x}_{\text{vertex}} = - 8$
I will let you find ${y}_{\text{vetex}}$ by sustitution
#### Explanation:
Expanding the brackets gives:
$y = {x}^{2} + 16 x + 64 - 2$
$y = {x}^{2} + 16 x + 64$
'----------------------------------------------------------------------
The equation given in the question is a quadratic that is already in vertex form (completing the square).
Consider the +8 from ${\left(x + 8\right)}^{2}$
The x_("vertex") " should be at "(-1)xx8=-8
Lets have a look!
Looks good!
Now all you have to do is substitute $x = \left(- 8\right)$ in the equation to find the value of y.
I will let you do that!<|endoftext|>
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GeoGebra Resources for Wayne Township
LAST UPDATED: 5/10/19
Most of these resources are designed to be used whole-class or small group, with a teacher-facilitated discussion. They are not designed to be used by individual students on a device, although that might be appropriate under certain conditions.
Note for non-Hoosiers: The strand names are taken from the Indiana Academic Standards, which are similar but not identical to Common Core State Standards.
Recently added materials
(3rd-4th grade) Use this simple applet to compare two fractions. Adjust the denominators, then drag the point to whatever fraction you're interested in. The points (and number text) will turn purple when the fractions are equivalent.
(3rd & 5th grade) Enter two whole numbers, then move the slider to see what happens when each is the dividend and divisor. (Works better with smaller numbers). Meant to help students understand the significance of dividend and divisor, including situations where the quotient would be a fraction.
(Game with 1st-4th grade math, depending on settings) A 2-player, GeoGebra version of Sara VanDerWerf's game. You will place (up to) 25 numbers in the grid. You score by placing (by clicking) matching numbers in adjacent squares. Decide whether you want to play a timed or untimed game, and whether you'd like the scores to be calculated by sums or products. See Sara's blog post for more information.
(Kindergarten) See numbers 0 - 20 expressed in ten frames, base 10 blocks, number line, and numerals. Intended to help students see the significance of a group of ten across representations. "How does this show us a group of ten? How does it show us the ones place?"
(K - 2nd grade) Enter any number of hundreds, tens, and ones to see that number drawn in any of three ways. One way of using this is to help students see the difference between a digit and its value. For example, a student might say the number 63 has 60 tens and 3 ones. If you enter that in this applet, you can see a number represented as...
- 60 tens and 3 ones ("How I typed it")
- 6 hundreds and 3 ones ("Standard place value")
- 603 ones ("All ones")
(1st - 2nd grade) This is meant to help students see the relationship between coins and numbers. Begin the discussion by asking them what they notice as you cycle through the different versions of splitting up the 100 (controlled by the "Switcher" slider). The work to make connections between the bundles/groups and various coins. See right for a potential prompt.
(3rd - 4th Grade) Use this resource to explore various fraction relationships, particularly how fractions can be composed and decomposed and the relationship between improper and mixed number forms. Make predictions about the result of a "little jump" (a unit fraction) and a "big jump" (a whole).
(3rd - 6th Grade) This resource is intended to get students to use proportional reasoning to estimate quotients. Use the sliders to generate random dividends and divisors, or input your own numbers. Start by dragging the numbers to the correct locations, then predict the number of "jumps" to land on the dividend.
(4th Grade) This resource uses an array approach to help students understand the relationship between the mixed number and improper forms of a fraction. Can students figure out the pattern?
(3rd - 5th Grade) This is designed to help students connect division algorithms to the "action" that is actually taking place. As students develop repeated subtraction or partial quotient approaches, use this to support them thinking "What is happening here?" as they represent the problem mathematically and attend to the meaning of the quantities.
(4th and 5th Grade) This uses an area model approach for multiplication. Use as a tool to support students as they learn multi-digit multiplication, or take an inquiry approach by clicking "Randomize" and working slowly, carefully to figure out the two factors. What's the fewest pieces of information you need?
(3rd - 7th Grade) Use the sliders to adjust the addends, then consider the size of the sum relative to the two bounds you are given. "Would the sum be closer to ___ or ___? How do you know?" As the level of precision increases, the level of mathematical reasoning increases with it. And the GRAND purpose behind this is to get students using an estimation strategy called front-end addition. Consider building this around a central question of "How do we make our estimates more precise?"<|endoftext|>
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4.10 Curve Sketching - Maple T.A. 10 Help
## 4.10 Curve Sketching
Your instructor can ask that you sketch a curve as part of an assignment. Some example curve sketching questions are:
• Graph one or multiple lines or curves.
• Indicate which region represents the solution set (cases: single line or curve; multiple lines or curves).
• Make the line solid or dashed to indicate whether the line is in the solution set.
• Remove the portion of the line or curve that does not belong.
• Use an open or closed circle to indicate whether the endpoints belong.
• Add asymptotes to logarithmic and exponential curves.
The sketch board is a set of tools that can be used to create and/or operate on various sketch types. With each question, Maple T.A. automatically provides the appropriate set and number of drawing tools (buttons).
#### Introduction
This section outlines the basic tasks you can perform to solve questions that involve curve sketching.
#### Graphing a Line
To graph a line:
1. Click Draw a Line (). This enables you to plot points on the coordinate grid.
1. Plot two points on the coordinate grid. After you plot two points on the grid, a line is drawn through these points.
1. To change the line, you can move either of the two points anywhere on the grid. You can also move the entire line by clicking and dragging on the line.
1. To delete the line, click Delete () and then click the object you want to delete.
#### Graphing a Parabola
To graph a parabola:
1. Click Draw Parabola (). This enables you to plot points on the coordinate grid that can be moved to shape and position the parabola.
1. On the coordinate grid, plot the vertex first followed by another point on the parabola.
1. You can change the parabola by dragging either of the two points anywhere on the grid. This will stretch, compress, reflect, or translate the parabola.
1. To delete the parabola, click Delete () and then click on the object you want to delete.
#### Indicating a Region of the Graph
To indicate which region on the graph represents the solution set:
1. Click Choose Region (). This enables you to select a region on either side of a single line or select a region bounded by two lines.
1. Click anywhere inside the region that represents the solution set to the problem. The region is automatically shaded.
1. To delete this shaded region, click Delete () and then click anywhere in the shaded area that you want to delete.
#### Making Solid or Dashed Lines for a Graph
To use solid or dashed lines for a graph:
1. Click Toggle Solid/Dashed (). This enables to you to switch between solid and dashed lines interchangeably.
1. Click on a line. If the line was solid, then the entire line becomes dashed. To change the line back to solid, click the line again and it will revert back to its original state.
1. To delete the solid or dashed line, click Delete () and then click on the line you want to delete.
#### Removing a Portion of a Line or Parabola
To remove a portion of a line or parabola:
1. Click Snip Left () or Snip Right (). These tools enable you to remove a portion to the left or right (respectively) of a location on the parabola.
2. Click on the portion of the parabola you want to keep. If you selected Snip Left in the previous step, the portion of the parabola to the left of your cursor is removed. Similarly, if you selected Snip Right in the previous step, the portion of the parabola to the right of your cursor is removed.
#### Removing a Portion of a Line or Parabola between Two Points
To remove a portion of a line or parabola between two points:
1. Click Snip Between (). This enables you to remove a portion of a parabola between two selected points.
2. Select two points on the parabola by clicking on them. A solid, green point indicates each selected location. The portion of the parabola between these two solid, green points is automatically removed.
#### Including or Excluding Endpoints
To indicate the position of endpoints on the graph:
1. Click Toggle Filled/Hollow (). This enables you to toggle back and forth between filled (included) and open (excluded) endpoints.
1. Click on the endpoint that you want to include or exclude. If the endpoint is filled (included), click on it to make it open (excluded). Similarly, if the endpoint is open (excluded), click on it to make it filled (included).
#### Graphing Absolute Values
To graph an absolute value:
1. Click Draw Absolute Value (). This enables you to plot the vertex and an additional point on the coordinate grid.
1. Plot the vertex first followed by an additional point on the grid. The absolute value graph is drawn automatically.
1. You can change the absolute value graph by dragging either of the two points anywhere on the grid. This will stretch, compress, reflect, or translate the absolute value.
1. To delete the absolute value graph, click Delete () and then click on the object you want to delete.
#### Graphing Exponential Curves
To graph an exponential curve:
1. Click Draw Exponential (). This enables you to plot two points and a horizontal asymptote.
1. Plot two points on the exponential curve.
1. Plot the horizontal asymptote for the curve: Click the location on the coordinate grid where you want to place the horizontal asymptote. A dashed, horizontal line is automatically drawn through the point on the grid where you clicked. If you would like to move this horizontal asymptote, simply drag it to its new location.
1. To delete the exponential curve, click Delete () and then click on the object you want to delete.
#### Graphing Logarithmic Curves
To graph a logarithmic curve:
1. Click Draw Logarithmic (). This enables you to plot two points and a vertical asymptote.
1. Plot two points on the logarithmic curve.
1. Plot the vertical asymptote for the curve: Click the location on the coordinate grid where you want to place the vertical asymptote. A dashed, vertical line is automatically drawn through the point on the grid where you clicked. If you would like to move this vertical asymptote, simply drag it to its new location.
1. To delete the logarithmic curve, click Delete () and then click on the object you want to delete.<|endoftext|>
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Hemophilia is a genetic disorder where protein factors that help blood clot are missing, or not fully functional. Those with hemophilia experience prolonged bleeding following injury, surgery or even minor trauma. Severe forms of the disease may cause spontaneous bleeding in the absence of any trauma to the body. Bleeding internally into the joints, brain or other organs can lead to serious complications when left untreated.
Types of Hemophilia:
There are three main types of hemophilia depending on which clotting factor is affected:
- Hemophilia A: deficiency in clotting factor VIII
- Hemophilia B: deficiency in clotting factor IX
- Hemophilia C: deficiency in clotting factor XI.
Symptoms associated with Hemophilia:
- Excessive bleeding following minor injuries.
- Large skin bruises from bleeding within the skin
- Uncontrolled bleeding after receiving a shot or pulling teeth
- Frequent nosebleeds that are hard to stop
- Blood in the urine or stool
- Pain and swelling in the joints due to internal bleeding
- Infantile bleeding after birth or circumcision
- Headaches and difficulties with vision.
Causes and risk factors of Hemophilia:
Hemophilia is an inherited disorder, which means the underlying genetic abnormalities are passed down from parents to their offspring. Hemophilia is the result of mutations (errors) in the genes that encode clotting factors VIII, IX and XI. These mutations lead to less effective versions of the clotting proteins, or their complete absence. When the critical concentration of clotting factors is not present in the body, clots form slower and bleeding is hard to control.
The mutations that are linked to hemophilia are on the X chromosome. Females receive two copies of the X chromosome (one from each parent), so gene products from one normal X chromosome often make up for any deficiencies present on the second X chromosome. Since inheriting two defective copies of the same gene is rare, the majority of females are carriers; they carry one hemophilia gene and can clot blood normally.
Males have a higher chance of suffering from hemophilia:
Males have a much higher chance of developing hemophilia because they only receive one copy of the X chromosome (from their mother). Instead of a second X chromosome, they inherit a Y chromosome (from their father), which does not duplicate the gene content of the X chromosome. Therefore, one defective X chromosome is sufficient for developing hemophilia. Hemophilia is called an X-linked trait because only mothers can pass it down to their sons.
Rare types of Hemophilia:
A rare form of hemophilia known as acquired hemophilia is not due to inheritance but instead happens when the body launches an immune attack against clotting factor VIII. This form of hemophilia begins in adulthood and leads to spontaneous internal bleeding into the joints, muscles and other tissue. Acquired hemophilia is not fully understood, but is linked to cancer, immune disease, allergies, and pregnancy.
Death due to bleeding-related causes is significantly higher with an inhibitor than without inhibitor. Inhibitors also increase the risk of bleeding and joint damage as well as the cross of care.<|endoftext|>
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## Pages
Showing posts with label distribute. Show all posts
Showing posts with label distribute. Show all posts
## Sunday, April 21, 2013
### Elementary Algebra Exam #3
Click on the 10 question exam covering topics in chapters 5 and 6. Give yourself one hour to try all of the problems and then come back and check your answers.
10. The length of a rectangle is 4 centimeters less than twice its width. The area is 96 square centimeters. Find the length and width. (Set up an algebraic equation then solve it)
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## Sunday, November 4, 2012
### Multiplying Polynomials
The distributive property and the product rule for exponents are the keys to multiplication of polynomials. Remember the product rule,
It says that when we multiply two numbers with the same base we must add the exponents.
Monomial x Polynomial
Multiply.
A common mistake is to distribute when working with multiplication. The distributive property only applies when addition or subtraction separates the terms.
Binomial x Polynomial: Use the distributive property to multiply polynomials by binomials.
We can save a step by distributing each term in the binomial individually. Sometimes this technique is referred to as the FOIL method. (FOIL – Multiply the First, Outer, Inner and then Last terms together.)
Multiply.
When combining like terms, be sure that the variable parts are exactly the same. Go slow and work in an organized fashion because it is easy to make an error when many terms are involved.
It is good practice to recheck your distributive step.
Trinomial x Polynomial
The following problem is one that we will have to be able to do before moving on to the next algebra course. These are tedious, time consuming, and often worked incorrectly. Use caution, take your time and work slowly.
Function notation for multiplication looks like
Given the functions f and g find (f g)(x).
The following special products will simplify things if we memorize them.
This special product is often called difference of squares. Notice that the middle terms cancel because one term will always be positive and the other will be negative.
We may use these formulas as templates when multiplying binomials. It is a good idea to memorize them.
Multiply.<|endoftext|>
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# Same Maths Coursework
Tags: Organizational Commitment Research PapersFree Homework PassesTypes Of Tones Used In EssaysCover Letter For Waitress With No ExperienceEnglish Iii Research PapersBenefits Of Team Sports Essay
Ex 1.2: Results for 2004 Months ( n ) Start day 1 Thursday 2 Sunday 3 Monday 4 Thursday 5 Saturday 6 Tuesday 7 Thursday 8 Sunday 9 Wednesday 10 Friday 11 Monday 12 Wednesday Therefore, we can see that the dates are thus: 1, 4, 7 = same 2, 8 = same 3, 11 = same 5 6 9, 12 = same 10 As you can see, the dates which fall on the same day are not the same as in the study sample Ex 1.1, so there is no pattern here.Ex 4.6 Starting July 4th n date 1 4 2 12 3 20 4 28 The above table gives a formula of 8n – 4.I have noticed that descending diagonally left is always n = 6n – x, where x depends on which date you choose to start with. Columns) is always n = 7n – x; and that descending diagonally right gives n = 8n – x. Studying relationships between adjacent numbers I drew a box around 4 numbers: Ex 5.1 Starting 2nd August 2 3 2 x 10 = 20 9 10 3 x 9 = 27 The difference is 7.Ex 3.2 The second row in June: n 1 6 3 7 4 8 5 9 6 11 7 12 As you can see from the table above, the formula is quite simple: n = n 5 Ex 3.3 The third row in June: n 1 13 2 14 3 15 4 16 5 17 6 18 7 19 The above table gives another formula: n = n 12 Ex 3.4, Finally the fourth row in June: n 1 20 2 21 3 22 4 23 5 24 6 25 7 26 The final formula is n = n 19 Ex 3.5: These are the formulae for the rows in the month of June: First row: n = n Second row: n = n 5 Third row: n = n 12 Fourth row: n = n 19 I didn’t draw the table for 5th, but predict it as: Fifth row: n = n 26 The general expression is n = n x depending on the start date. Studying diagonal relationships Ex 4.1 Diagonally descending to the left, starting with 2nd January.n date 1 2 2 8 3 14 4 20 5 26 The difference is always 6.n date 1 9 2 15 3 21 4 27 The above table gives the formula n = 6n 3 Ex 4.3: Diagonally descending to the left, starting with 5th June.n date 1 5 2 11 3 17 4 23 5 29 This table gives the formula n = 6n – 1 Now I will study Diagonals descending to the right, to examine whether there is any similar relationship between these numbers. Ex 4.4 Diagonally descending to the right, starting with 3rd February.Ex 1.4 Results for 2001 Months ( n ) Start day 1 Monday 2 Thursday 3 Thursday 4 Sunday 5 Tuesday 6 Friday 7 Sunday 8 Wednesday 9 Saturday 10 Monday 11 Thursday 12 Saturday The matching months are: 1, 10 = same 2, 3, 11 = same 4, 7 = same 5 7 8 9, 12 = same As you can see, once again the results for the months that start on the same day show the same pattern as in Ex1.1 and Ex1.3.This clearly shows that there is indeed a pattern between the start days of the months each non – leap year.This formula makes sense because when you descend diagonally to the left, you travel foreword in time by one week minus a day, i.e. I found that this applies to any box of 4 numbers on the calendar.Ex 5.2 Starting 6th October 6 7 6 x 14 = 84 13 14 7 x 13 = 91 The difference is 7 again, as in Ex 5.1.<|endoftext|>
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- Assess how Humanism gave rise to the art of the Renasissance
- Humanists reacted against the utilitarian approach to education, seeking to create a citizenry who were able to speak and write with eloquence and thus able to engage the civic life of their communities.
- The movement was largely founded on the ideals of Italian scholar and poet Francesco Petrarca, which were often centered around humanity’s potential for achievement.
- While Humanism initially began as a predominantly literary movement, its influence quickly pervaded the general culture of the time, reintroducing classical Greek and Roman art forms and leading to the Renaissance.
- Donatello became renowned as the greatest sculptor of the Early Renaissance, known especially for his Humanist, and unusually erotic, statue of David.
- While medieval society viewed artists as servants and craftspeople, Renaissance artists were trained intellectuals, and their art reflected this newfound point of view.
- In humanist painting, the treatment of the elements of perspective and depiction of light became of particular concern.
Humanism, also known as Renaissance Humanism, was an intellectual movement embraced by scholars, writers, and civic leaders in 14th- and early-15th-century Italy. The movement developed in response to the medieval scholastic conventions in education at the time, which emphasized practical, pre-professional, and scientific studies engaged in solely for job preparation, and typically by men alone. Humanists reacted against this utilitarian approach, seeking to create a citizenry who were able to speak and write with eloquence and thus able to engage the civic life of their communities. This was to be accomplished through the study of the “studia humanitatis,” known today as the humanities: grammar, rhetoric, history, poetry, and moral philosophy. Humanism introduced a program to revive the cultural—and particularly the literary—legacy and moral philosophy of classical antiquity. The movement was largely founded on the ideals of Italian scholar and poet Francesco Petrarca, which were often centered around humanity’s potential for achievement.
While Humanism initially began as a predominantly literary movement, its influence quickly pervaded the general culture of the time, re-introducing classical Greek and Roman art forms and contributing to the development of the Renaissance. Humanists considered the ancient world to be the pinnacle of human achievement, and thought its accomplishments should serve as the model for contemporary Europe. There were important centers of Humanism in Florence, Naples, Rome, Venice, Genoa, Mantua, Ferrara, and Urbino.
Humanism was an optimistic philosophy that saw man as a rational and sentient being, with the ability to decide and think for himself. It saw man as inherently good by nature, which was in tension with the Christian view of man as the original sinner needing redemption. It provoked fresh insight into the nature of reality, questioning beyond God and spirituality, and provided knowledge about history beyond Christian history.
Renaissance Humanists saw no conflict between their study of the Ancients and Christianity. The lack of perceived conflict allowed Early Renaissance artists to combine classical forms, classical themes, and Christian theology freely. Early Renaissance sculpture is a great vehicle to explore the emerging Renaissance style. The leading artists of this medium were Donatello, Filippo Brunelleschi, and Lorenzo Ghiberti. Donatello became renowned as the greatest sculptor of the Early Renaissance, known especially for his classical, and unusually erotic, statue of David, which became one of the icons of the Florentine republic.
Humanism affected the artistic community and how artists were perceived. While medieval society viewed artists as servants and craftspeople, Renaissance artists were trained intellectuals, and their art reflected this newfound point of view. Patronage of the arts became an important activity, and commissions included secular subject matter as well as religious. Important patrons, such as Cosimo de’ Medici, emerged and contributed largely to the expanding artistic production of the time.
In painting, the treatment of the elements of perspective and light became of particular concern. Paolo Uccello, for example, who is best known for “The Battle of San Romano,” was obsessed by his interest in perspective, and would stay up all night in his study trying to grasp the exact vanishing point. He used perspective in order to create a feeling of depth in his paintings. In addition, the use of oil paint had its beginnings in the early part of the 16th century, and its use continued to be explored extensively throughout the High Renaissance.
Some of the first Humanists were great collectors of antique manuscripts, including Petrarch, Giovanni Boccaccio, Coluccio Salutati, and Poggio Bracciolini. Of the three, Petrarch was dubbed the “Father of Humanism” because of his devotion to Greek and Roman scrolls. Many worked for the organized church and were in holy orders (like Petrarch), while others were lawyers and chancellors of Italian cities (such as Petrarch’s disciple Salutati, the Chancellor of Florence) and thus had access to book-copying workshops.
In Italy, the Humanist educational program won rapid acceptance and, by the mid-15th century, many of the upper classes had received Humanist educations, possibly in addition to traditional scholastic ones. Some of the highest officials of the church were Humanists with the resources to amass important libraries. Such was Cardinal Basilios Bessarion, a convert to the Latin church from Greek Orthodoxy, who was considered for the papacy and was one of the most learned scholars of his time.
Following the Crusader sacking of Constantinople and the end of the Byzantine Empire in 1453, the migration of Byzantine Greek scholars and émigrés, who had greater familiarity with ancient languages and works, furthered the revival of Greek and Roman literature and science.<|endoftext|>
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# CHAPTER 29 VOLUMES AND SURFACE AREAS OF COMMON SOLIDS
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1 CHAPTER 9 VOLUMES AND SURFACE AREAS OF COMMON EXERCISE 14 Page 9 SOLIDS 1. Change a volume of cm to cubic metres. 1m = 10 cm or 1cm = 10 6m 6 Hence, cm = m = 1. m. Change a volume of 5000 mm to cubic centimetres. 1cm = 10 mm or 1mm = 10 cm Hence, 5000 mm = mm = 5 cm. A metal cube has a surface area of 4 cm. Determine its volume. A cube had 6 sides. Area of each side = 4/6 = 4 cm Each side is a square hence the length of a side = 4 = cm Volume of cube = = 8 cm 4. A rectangular block of wood has dimensions of 40 mm by 1 mm by 8 mm. Determine (a) its volume in cubic millimetres and (b) its total surface area in square millimetres. (a) Volume of cuboid = l b h = = 840 mm (b) Surface area = (bh + hl + lb) = ( ) = ( ) = 896 = 179 mm , John Bird
2 5. Determine the capacity, in litres, of a fish tank measuring 90 cm by 60 cm by 1.8 m, given 1 litre = 1000 cm. Volume = ( ) cm Tank capacity = cm 1000 cm /litre = 97 litre 6. A rectangular block of metal has dimensions of 40 mm by 5 mm by 15 mm. Determine its volume in cm. Find also its mass if the metal has a density of 9 g/cm. Volume = length breadth width = = mm Mass = density volume = 9 g/cm 15cm = 15 g = cm = 15 cm 7. Determine the maximum capacity, in litres, of a fish tank measuring 50 cm by 40 cm by.5 m (1 litre = 1000 cm ). Volume = cm Tank capacity = cm 1000 cm /litre = 500 litre 8. Determine how many cubic metres of concrete are required for a 10 m long path, 150 mm wide and 80 mm deep. Width = 150 mm = 0.15 m and depth = 80 mm = m Hence, volume of path = length breadth width = = 1.44 m i.e. concrete required = 1.44 m , John Bird
3 9. A cylinder has a diameter 0 mm and height 50 mm. Calculate (a) its volume in cubic centimetres, correct to 1 decimal place, and (b) the total surface area in square centimetres, correct to 1 decimal place. Diameter = 0 mm = cm hence radius, r = / = 1.5 cm and height, h = 50 mm = 5 cm πrh = π = 11.5π = 5. cm, correct to 1 decimal place (a) Volume = ( ) (b) Total surface area = πrh + π r = ( π 1.5 5) + ( π 1.5 ) = 15π + 4.5π = 19.5π = 61. cm 10. Find (a) the volume and (b) the total surface area of a right-angled triangular prism of length 80 cm whose triangular end has a base of 1 cm and perpendicular height 5 cm. (a) Volume of right-angled triangular prism = 1 bhl = i.e. Volume = 400 cm (a) Total surface area = area of each end + area of three sides In triangle ABC, AC = AB + BC from which, AC = AB + BC = = 1 cm Hence, total surface area = 1 bh + (AC 80) + (BC 80) + (AB 80) , John Bird
4 i.e. total surface area = 460 cm = (1 5) + (1 80) + (1 80) + (5 80) = A steel ingot whose volume is m is rolled out into a plate which is 0 mm thick and 1.80 m wide. Calculate the length of the plate in metres. Volume of ingot = length breadth width i.e. = l from which, length = = 7.04 m 1. The volume of a cylinder is 75 cm. If its height is 9.0 cm, find its radius. Volume of cylinder, V = π rh i.e. 75 = π r (9.0) 75 from which, r = π 9.0 and radius, r = 75 π 9.0 = 1.6 cm 1. Calculate the volume of a metal tube whose outside diameter is 8 cm and whose inside diameter is 6 cm, if the length of the tube is 4 m. Outer diameter, D = 8 cm and inner diameter, d = 6 cm ( ) ( ) πd πd π 8 π 6 Area of cross-section of copper = = = = cm Hence, volume of metal tube = (cross-sectional area) length of pipe = = 8796 cm , John Bird
5 14. The volume of a cylinder is 400 cm. If its radius is 5.0 cm, find its height. Determine also its curved surface area. Volume of cylinder, V = π rh i.e. 400 = π (5.0) h from which, height, 400 h = π (5.0) = cm Curved surface area = πrh = ( π ) = 15.9 cm 15. A cylinder is cast from a rectangular piece of alloy 5 cm by 7 cm by 1 cm. If the length of the cylinder is to be 60 cm, find its diameter. Volume of rectangular piece of alloy = = 40 cm Volume of cylinder = π rh Hence, 40 = 40 7 π r (60) from which, r = π(60) = π and radius, r = 7 π = cm and diameter of cylinder, d = r = =.99 cm 16. Find the volume and the total surface area of a regular hexagonal bar of metal of length m if each side of the hexagon is 6 cm. A hexagon is shown below. In triangle 0BC, tan 0 = x from which, x = tan 0 = cm , John Bird
6 Hence, area of hexagon = = 9.5 cm and volume of hexagonal bar = = cm Surface area of bar = 6[ 0.06 ] + [ ] in metre units = m 17. A block of lead 1.5 m by 90 cm by 750 mm is hammered out to make a square sheet 15 mm thick. Determine the dimensions of the square sheet, correct to the nearest centimetre. Volume of block of lead = length breadth height = cm If length = breadth = x cm and height = 15/10 = 1.5 cm, then x ( 1.5) = from which, x = and x = = 81.6 cm = 8. m Hence, dimensions of square sheet are 8. m by 8. m 18. How long will it take a tap dripping at a rate of 800 mm /s to fill a three-litre can? litre = 000 cm = mm = 106mm Time to fill can = 10 mm 6 800mm /s = 750 s = = 6.5 minutes 19. A cylinder is cast from a rectangular piece of alloy 5.0 cm by 6.50 cm by 19. cm. If the height of the cylinder is to be 5.0 cm, determine its diameter, correct to the nearest centimetre. Volume of cylinder, V = π rh and volume of rectangular prism = cm i.e = π r (5.0) from which, r = π 5.0 and radius, 47 r = π 5.0 =.0 cm 014, John Bird
7 i.e. diameter = r =.0 = 4 cm 0. How much concrete is required for the construction of the path shown, if the path is 1 cm thick? 1 4 π Area of path = (8.5 ) + ( ) + (.1 ) + (.4 [ + 1.]) = = 4.0 m If thickness of path = 1 cm = 0.1 m then Concrete required = volume of path = = 4.08 m , John Bird
8 EXERCISE 15 Page If a cone has a diameter of 80 mm and a perpendicular height of 10 mm, calculate its volume in cm and its curved surface area. 1 1 = = mm = 01.1 cm Volume of cone = π r h π ( 40) ( 10) From diagram below, slant height, l = ( 1 4) + = cm Curved surface area = πrl = π(4)(1.649) = cm. A square pyramid has a perpendicular height of 4 cm. If a side of the base is.4 cm long find the volume and total surface area of the pyramid. A sketch of the pyramid is shown below = 7.68 cm Volume of pyramid = ( )( ) In the sketch, AB = 4 cm and BC =.4/ = 1. cm , John Bird
9 Length AC = ( 4 1.) + = cm 1 Hence, area of a side = (.4 )( ) = 5.01 cm Total surface area of pyramid = 4[ 5.01] + (.4) = 5.81 cm. A sphere has a diameter of 6 cm. Determine its volume and surface area. Volume of sphere = π π r = = 11.1 cm Surface are of sphere = 4πr = 4π = 11.1 cm 6 4. If the volume of a sphere is 566 cm, find its radius. 4 Volume of sphere = π r hence, 566 = 4 π r 566 from which, r = = π and radius, r = = 5.11 cm 5. A pyramid having a square base has a perpendicular height of 5 cm and a volume of 75 cm. Determine, in centimetres, the length of each side of the base. 1 1 If each side of base = x cm then volume of pyramid = A h= xh 1 i.e. 75 = (5) x and 75 x = = 9 5 from which, length of each side of base = 9 = cm 6. A cone has a base diameter of 16 mm and a perpendicular height of 40 mm. Find its volume correct to the nearest cubic millimetre , John Bird
10 π = π Volume of cone = rh ( 40) = 681 mm 7. Determine (a) the volume and (b) the surface area of a sphere of radius 40 mm. (a) Volume of sphere = ( ) (b) Surface are of sphere = ( ) 4 4 πr = π 40 = mm or cm 4πr = 4π 40 = mm or cm 8. The volume of a sphere is 5 cm. Determine its diameter. Volume of sphere = d π π 4 4 r = Hence, 5 = 4 d π from which, d 5 = 4π d and = 5 = 4.65 cm and diameter, d = 4.65 = 8.5 cm 4π 9. Given the radius of the Earth is 680 km, calculate, in engineering notation (a) its surface area in km and (b) its volume in km. 4πr = 4π 680 = km (a) Surface area of Earth = ( ) 4 4 πr = π 680 = km (b) Volume of earth = ( ) 10. An ingot whose volume is 1.5 m is to be made into ball bearings whose radii are 8.0 cm. How many bearings will be produced from the ingot, assuming 5% wastage? Volume of one ball bearing = ( ) Volume of x bearings = cm 4 4 πr = π 8 Let number of ball bearings = x , John Bird
11 Hence, [ x ] 10 = π ( ) from which, number of bearings, x = π 8 = A spherical chemical storage tank has an internal diameter of 5.6 m. Calculate the storage capacity of the tank, correct to the nearest cubic metre. If 1 litre = 1000 cm, determine the tank capacity in litres. Volume of storage tank = π π r = = = 9 m, correct to the nearest cubic metre Volume of tank = 9 106cm If 1 litre = 1000 cm, then capacity of storage tank = litres = litres , John Bird
12 EXERCISE 16 Page Find the total surface area of a hemisphere of diameter 50 mm. Total surface area = [ ] 1 π r + 4 π r = π r + π r = π r = 50 π = 5890 mm or cm. Find (a) the volume and (b) the total surface area of a hemisphere of diameter 6 cm. Volume of hemisphere = 1 (volume of sphere) = πr = π 6.0 = cm Total surface area = curved surface area + area of circle = 1 (surface area of sphere) + πr = 1 (4πr ) + πr = πr + πr = πr 6.0 = π = 84.8 cm. Determine the mass of a hemispherical copper container whose external and internal radii are 1 cm and 10 cm, assuming that 1 cm of copper weighs 8.9 g. Volume of hemisphere = π r = π[ ] 1 10 cm Mass of copper = volume density = [ ] π 1 10 cm 8.9 g/cm = 1570 g = 1.57 kg 4. A metal plumb bob comprises a hemisphere surmounted by a cone. If the diameter of the hemisphere and cone are each 4 cm and the total length is 5 cm, find its total volume. The plumb bob is shown sketched below , John Bird
13 1 1 Volume of bob = π rh+ π r = π ( ) ( 5 ) + π ( ) = 16 4π + π = 9. cm 5. A marquee is in the form of a cylinder surmounted by a cone. The total height is 6 m and the cylindrical portion has a height of.5 m, with a diameter of 15 m. Calculate the surface area of material needed to make the marquee, assuming 1% of the material is wasted in the process The marquee is shown sketched below. Surface area of material for marquee = πrl πrh Hence, surface area = π(7.5)(7.9057) + π(7.5)(.5) = = m + where l = ( 7.5.5) If 1% of material is wasted then amount required = = 9.4 m + = m 6. Determine (a) the volume and (b) the total surface area of the following solids: (i) a cone of radius 8.0 cm and perpendicular height 10 cm (ii) a sphere of diameter 7.0 cm (iii) a hemisphere of radius.0 cm , John Bird
14 (iv) a.5 cm by.5 cm square pyramid of perpendicular height 5.0 cm (v) a 4.0 cm by 6.0 cm rectangular pyramid of perpendicular height 1.0 cm (vi) a 4. cm by 4. cm square pyramid whose sloping edges are each 15.0 cm (vii) a pyramid having an octagonal base of side 5.0 cm and perpendicular height 0 cm (i) A sketch of the cone is shown below. 1 1 = = 670 cm (a) Volume of cone = π r h π ( 8.0) ( 10) (b) Total surface area = πr + πrl where l = ( ) = cm = π ( 8.0) + π ( 8.0)( ) = = 5 cm (ii) (a) Volume of sphere = π = 180 cm (b) Surface area = 7.0 4πr = 4π = 154 cm (iii) (a) Volume of hemisphere = ( ) πr = π.0 = 56.5 cm 1 (4 πr ) + πr = πr = π.0 = 84.8 cm (b) Surface area = ( ) (iv) A sketch of the square pyramid is shown below, where AB = 5.0 cm , John Bird
15 1 (a) Volume of pyramid = (.5 ) ( 5.0 ) = 10.4 cm (b) In the diagram, AC = ( AB BC ) ( ) Surface area = ( ) 1 + = + = = =.0 cm (v) A sketch of the rectangular pyramid is shown below = 96.0 cm (a) Volume of rectangular pyramid = ( )( ) (b) In the diagram, AC = ( 1.0.0) and AD = ( 1.0.0) + = 1.69 cm + = cm Hence, surface area = ( ) = = 146 cm (vi) The square pyramid is shown sketched below , John Bird
16 Diagonal on base = ( ) = cm hence, BC = = cm Hence, perpendicular height, h = ( ) = cm 1 (a) Volume of pyramid = ( 4. ) ( ) (b) AD = ( ) + = = 86.5 cm Hence, surface area = ( ) = = 14 cm (vii) A pyramid having an octagonal base is shown sketched below. One sector is shown in diagram (p) below, where from which, x =.5 tan.5 = cm.5 tan.5 = x 1 Hence, area of whole base = = cm 1 (a) Volume of pyramid = ( )( 0 ) = 805 cm (p) (q) (b) From diagram (q) above, y = ( ) + = cm , John Bird
17 Total surface area = = = 59 cm 7. A metal sphere weighing 4 kg is melted down and recast into a solid cone of base radius 8.0 cm. If the density of the metal is 8000 kg/m determine (a) the diameter of the metal sphere and (b) the perpendicular height of the cone, assuming that 15% of the metal is lost in the process. mass 4 kg Volume of sphere = = = 0.00m = cm = 000 cm density 8000 kg/m 4 (a) Volume of sphere = π r i.e. 000 = 4 π r and radius, r = 000 4π = cm Hence, the diameter of the sphere, d = r = = 17.9 cm (b) Volume of cone = = 550 from which, perpendicular height of cone, h = 1 1 cm = πrh = π ( 8.0) ( 8.0) h 550 = 8.0 cm π 8. Find the volume of a regular hexagonal pyramid if the perpendicular height is 16.0 cm and the side of base is.0 cm. The hexagonal base is shown sketched below. From the diagram, tan 0 = 1.5 h from which, h = 1.5 tan 0 =.598 cm , John Bird
18 Hence, area of hexagonal base = =.87 cm 1 and volume of hexagonal pyramid = (.87 )( 16.0 ) = 15 cm 9. A buoy consists of a hemisphere surmounted by a cone. The diameter of the cone and hemisphere is.5 m and the slant height of the cone is 4.0 m. Determine the volume and surface area of the buoy. The buoy is shown in the sketch below. Height of cone, h = ( ) =.80 m 1 1 Volume of buoy = π r + π rh= π ( 1.5) + π ( 1.5) (.80) = = 10. m πrl πr = π π 1.5 Surface area = ( ) ( )( ) ( ) = 5π +.15π = 8.15π = 5.5 m 10. A petrol container is in the form of a central cylindrical portion 5.0 m long with a hemispherical section surmounted on each end. If the diameters of the hemisphere and cylinder are both 1. m determine the capacity of the tank in litres (1 litre = 1000 cm ). The petrol container is shown sketched below , John Bird
19 4 + = + Volume of container = π r π rh π ( 0.6) π ( 0.6) ( 5.0) = 0.88π + 1.8π = m = cm and tank capacity = cm cm /litre = 6560 litres 11. The diagram below shows a metal rod section. Determine its volume and total surface area ( ) = ( ) Volume of rod = πrh l b w π ( ) ( ) 1 1 = 50π +500 = cm Surface area = ( π rh) + π r + (.50.0) + (.5 100) + (.0 100) = π(1.0)(100) + π( 1.0) = 107 cm , John Bird
20 1. Find the volume (in cm ) of the die-casting shown below. The dimensions are in millimetres. 1 π Volume = ( 0 50) = π = mm = cm = 0.7 cm 1. The cross-section of part of a circular ventilation shaft is shown below, ends AB and CD being open. Calculate (a) the volume of the air, correct to the nearest litre, contained in the part of the system shown, neglecting the sheet metal thickness, (given 1 litre = 1000 cm ), (b) the cross-sectional area of the sheet metal used to make the system, in square metres, and (c) the cost of the sheet metal if the material costs per square metre, assuming that 5% extra metal is required due to wastage , John Bird
21 (a) In cm, volume of air = π ( 00) + π + π ( 150) + π ( 150) = π π π π = π cm = π cm 1000 cm /litre = 1458 litres, correct to the nearest litre (b) In m, cross-sectional area of the sheet metal 1 4 = π ( 0.5)( ) + 4π ( 0.5 ) + π ( 0.5)( 1.5) + π ( 0.4)( 1.5) + π ( ) = π π π + 1.π π =.11π = m = 9.77 m correct to significant figures (c) Sheet metal required = m Cost of sheet metal = = , John Bird
22 EXERCISE 17 Page The radii of the faces of a frustum of a cone are.0 cm and 4.0 cm and the thickness of the frustum is 5.0 cm. Determine its volume and total surface area. A sketch of a side view of the frustum is shown below. 1 h R Rr r Volume of frustum = π ( + + ) = = 147 cm = π ( 5.0)( 4.0 (4.0)(.0).0 ) π ( 5.0)( 8.0) From the diagram below, slant length, l = ( ) = 9 Total surface area = ( ) πl R+ r + πr + πr = π ( 9 )( ) + π (.0) + π ( 4.0) =.1π + 4π + 16π = 164 cm. A frustum of a pyramid has square ends, the squares having sides 9.0 cm and 5.0 cm, respectively. Calculate the volume and total surface area of the frustum if the perpendicular distance between its ends is 8.0 cm. A side view of the frustum of the pyramid is shown below. By similar triangles: CG BG BH AH = from which, height, CG = ( ) (.5) BH 8.0 BG = = 10.0 cm AH , John Bird
23 Height of complete pyramid = = 18.0 cm 1 Volume of large pyramid = ( 9.0 ) ( 18.0 ) = 486 cm 1 Volume of small triangle cut off = ( 5.0 ) ( 10.0 ) Hence, volume of frustum = = 40 cm A cross-section of the frustum is shown below. = 8. cm BC = ( 8 ) + = 8.46 cm = cm Area of four trapeziums = ( )( ) Total surface area of frustum = = 7 cm. A cooling tower is in the form of a frustum of a cone. The base has a diameter of.0 m, the top has a diameter of 14.0 m and the vertical height is 4.0 m. Calculate the volume of the tower and the curved surface area. A sketch of the cooling tower is shown below , John Bird
24 1 h R Rr r Volume of frustum = π ( + + ) 1 π + + = π = m = ( 4.0)( 16.0 (16.0)(7.0) 7.0 ) 8 ( 417) ( ) Slant length, l = ( AB BC ) ( ) + = = 5.6 m Curved surface area = l R r ( )( ) π ( + ) = π = π = 185 m 4. A loudspeaker diaphragm is in the form of a frustum of a cone. If the end diameters are 8.0 cm and 6.00 cm and the vertical distance between the ends is 0.0 cm, find the area of material needed to cover the curved surface of the speaker. A sketch of the loudspeaker diaphragm is shown below. ( ) Slant length, l = ( AC AB ) ( ) + = = 1.95cm Curved surface area = πl (R + r) = π(1.95)( ) = 1707 cm , John Bird
25 5. A rectangular prism of metal having dimensions 4. cm by 7. cm by 1.4 cm is melted down and recast into a frustum of a square pyramid, 10% of the metal being lost in the process. If the ends of the frustum are squares of side cm and 8 cm respectively, find the thickness of the frustum. Volume of frustum of pyramid = 90% of volume of rectangular prism = 0.9( ) = cm. A cross-section of the frustum of the square pyramid is shown below (not to scale). By similar triangles: CG BG BH AH BH h BG = AH.5 = 0.6 h = from which, height, CG = ( ) ( 1.5) = 4.1h cm Volume of large pyramid = ( ) ( h h) h = 1.8 h cm Volume of small triangle cut off = ( ) ( ) Hence, = 4.1h 1.8h =.h Thus, thickness of frustum, h = = cm 6. Determine the volume and total surface area of a bucket consisting of an inverted frustum of a cone, of slant height 6.0 cm and end diameters 55.0 cm and 5.0 cm. A sketch of the bucket is shown below , John Bird
26 Thickness of frustum, h = ( 6.0 ( ) ) 1 h R Rr r Volume of frustum = π ( + + ) Total surface area = ( ) = 4.58 cm 1 π + + = cm correct to 4 = ( 4.58)( 7.5 (7.5)(17.5) 17.5 ) πl R+ r + π r π π 17.5 = ( )( ) ( ) = 160π π = 196.5π = 6051 cm significant figures 7. A cylindrical tank of diameter.0 m and perpendicular height.0 m is to be replaced by a tank of the same capacity but in the form of a frustum of a cone. If the diameters of the ends of the frustum are 1.0 m and.0 m, respectively, determine the vertical height required. Volume of cylinder = rh ( ) ( ) π = π = π m A sketch of the frustum of a cone is shown below , John Bird
27 1 h R Rr r Volume of frustum = π = π ( + + ) 1 1 = π ( h)( (1.0)(0.5) ) = π h( 1.75) from which, thickness of frustum = vertical height, h = π 9 = = 5.14 m 1 π 1.75 ( 1.75) , John Bird
28 EXERCISE 18 Page Determine the volume and surface area of a frustum of a sphere of diameter cm, if the radii of the ends of the frustum are 14.0 cm and.0 cm and the height of the frustum is 10.0 cm. π h h r r 6 Volume of frustum of sphere = ( ) = π ( 10.0) ( ( 14.0 ) + (.0 ) ) 6 = 1105 cm = cm correct to 4 significant figures Surface area of frustum = πrh = π ( ) = 150 cm. Determine the volume (in cm ) and the surface area (in cm ) of a frustum of a sphere if the diameter of the ends are 80.0 mm and 10.0 mm and the thickness is 0.0 mm. The frustum is shown shaded in the cross-section sketch below (in cm units). π h h r r 6 Volume of frustum of sphere = ( ) = ( ) π in cm units π = ( ) + + = 59. cm Surface area of frustum = πrh From the above diagram, r = 6 + OP (1) , John Bird
29 r = 4 + OQ Now OQ = + OP Hence, ( ) Equating equations (1) and () gives: r 6 OP = OP () + = 4 + (.0 + OP) i.e. 6 + OP = ( OP) + OP Thus, 6 = 5 + 6(OP) from which, OP = = 6 6 From equation (1), r 11 = and radius, r = = 6.74 cm Surface area of frustum = πrh = π(6.74)(.0) = 118. cm. A sphere has a radius of 6.50 cm. Determine its volume and surface area. A frustum of the sphere is formed by two parallel planes, one through the diameter and the other at a distance h from the diameter. If the curved surface area of the frustum is to be 1 5 of the surface area of the sphere, find the height h and the volume of the frustum. 4 4 πr = π 6.50 = 1150 cm Volume of sphere = ( ) 4πr = 4π 6.50 = 51 cm Surface area = ( ) The frustum is shown shaded in the sketch below , John Bird
30 1 Curved surface area = πrh = ( 51 ) cm 5 in this case 1 i.e. π(6.50)h = ( 51 ) 5 from which, height, h = 1 ( 51 ) 5 π 6.50 ( ) =.60 cm r = = cm From the diagram, 1 ( ) π h h r r 6 Volume of frustum of sphere = ( ) = π (.60 ) (.60 + ( 6.50 ) + ( ) ) 6 = 6.7 cm 4. A sphere has a diameter of.0 mm. Calculate the volume (in cm ) of the frustum of the sphere contained between two parallel planes distances 1.0 mm and mm from the centre and on opposite sides of it. A cross-section of the frustum is shown in the sketch below. From the diagram, 1 ( ) r = = mm and ( ) r = = mm π h h r r 6 Volume of frustum of sphere = ( ) , John Bird
31 = π (.0) (.0 + ( ) + ( ) ) 6 = 1487 mm = cm 5. A spherical storage tank is filled with liquid to a depth of 0.0 cm. If the inner diameter of the vessel is 45.0 cm, determine the number of litres of liquid in the container (1 litre = 1000 cm ). A cross-section of the storage tank is shown sketched below. Volume of water = π r + volume of frustum From the diagram, 1 ( ) r = = 1.1 cm π h h r r 6 Volume of frustum of sphere = ( ) = π ( 7.5 ) ( (.50 ) + ( 1.1 ) ) 6 = cm Hence, total volume of water = (.50) π = 5 41 cm Number of litres of water = 5 41cm 1000cm /litre = 5.4 litres , John Bird
32 EXERCISE 19 Page Use the prismoidal rule to find the volume of a frustum of a sphere contained between two parallel planes on opposite sides of the centre, each of radius 7.0 cm and each 4.0 cm from the centre. The frustum of the sphere is shown sketched in cross-section below. Radius, r = ( ) + = 8.06 cm x A A A 6 Using the prismoidal rule, volume of frustum = [ ] π π π = ( 7.0) + 4 ( 8.06) + ( 7.0) = 1500 cm. Determine the volume of a cone of perpendicular height 16.0 cm and base diameter 10.0 cm by using the prismoidal rule. x A A A 6 Using the prismoidal rule: Volume, V = [ ] Area, 10.0 A 1 = πr1 = π = 5π Area, A 5.0 = πr = π = 6.5π Area, ( ) A = πr = π 0 = 0 and x = 16.0 cm , John Bird
33 x 16.0 A A A π π 6 6 Hence, volume of cylinder, V = [ ] = [ 5 + 4(6.5 ) + 0] = π = cm. A bucket is in the form of a frustum of a cone. The diameter of the base is 8.0 cm and the diameter of the top is 4.0 cm. If the length is.0 cm, determine the capacity of the bucket (in litres) using the prismoidal rule (1 litre = 1000 cm ). The bucket is shown in the sketch below. The radius of the midpoint is = 17.5 cm x A A A 6 Using the prismoidal rule, volume of frustum = [ ].0 6 = π ( 1.0) + 4π ( 17.5) + π ( 14.0) = 1 00 cm Hence, capacity of bucket = 1 00cm 1000cm /litre = 1.0 litres 4. Determine the capacity of a water reservoir, in litres, the top being a 0.0 m by 1.0 m rectangle, the bottom being a 0.0 m by 8.0 m rectangle and the depth being 5.0 m (1 litre = 1000 cm ). The water reservoir is shown sketched below , John Bird
34 A mid-section will have dimensions of = 5 m by = 10 m x A A A 6 Using the prismoidal rule, volume of frustum = [ ] = ( 0 1) + 4( 5 10) + ( 0 8) = m = cm Hence, capacity of water reservoir = cm cm /litre = litre , John Bird
35 EXERCISE 10 Page The diameter of two spherical bearings are in the ratio :5. What is the ratio of their volumes? Diameters are in the ratio :5 Hence, ratio of their volumes = :5 i.e. 8:15. An engineering component has a mass of 400 g. If each of its dimensions are reduced by 0%, determine its new mass. New mass = ( 0.7) 400 = = 17. g , John Bird
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The ultraviolet (UV) region of the electromagnetic spectrum is intermediate in wavelength and frequency between the x-ray and the visible regions. UV radiation can produce damage to the eyes and skin. The wavelength of the radiation and the length of the exposure determine the type and extent of the damage. UV radiation is emitted when excited atoms make transitions from a higher to a lower energy state, thus releasing photons with energies in the UV range. The primary man-made method of generating UV radiation is to excite atoms via an electrical arc through a gas or a vapor and intense heat. Examples of sources of UV radiation are mercury vapor lamps, fluorescent lights, germicidal lamps, black light lamps, plasma torches, open arcs (such as those used in arc welding) and sunlamps used in the tanning salon industry.
The Agency currently regulates tanning facilities and mercury vapor lamps, see AAC Title 12, Chapter 1 Article 14 "The Control of nonionizing Radiation".
For more information on this topic, please visit the following sites:
EPA Ultraviolet Index
Ultraviolet Information Sheet (New Zealand National Institute of Water and Atmospheric Research at Lauder )
UV NASA (NASA)
The US National Institute of Health (Sunlight, Ultraviolet Radiation, and the Skin)
Ultraviolet Radiation DA UV-B Radiation Monitoring Program
USDA UV (USDA at Colorado State University)
MORE ON ULTRAVIOLET RADIATION
All energies that move at the speed of light are collectively referred to as electromagnetic radiation or 'light'. Various types of light differ in their wavelength, frequency and energy; higher energy waves have higher frequencies and shorter wavelengths. Pigments inside the retina of our eyes absorb wavelengths of light between 400nm-700nm, collectively referred to as 'visible light'. A "nm'' is a nanometer which is one billionth, or 10e-9, meters. Stratospheric Oxygen and Ozone molecules absorb 97-99% of the sun's high frequency Ultraviolet light, light with wavelengths between 150 and 300nm. Ultraviolet-B (UV-B) is a section of the UV spectrum, with wavelengths between 270 and 320nm.
The amount of UV-B light received by a location is strongly dependent on:
- Latitude and elevation of the location. At the high-latitude Polar Regions the sun is always low in the sky; because the sunlight passes through more atmosphere so more of the UV-B is absorbed. For this reason average UV-B exposure at the poles is over a thousand times lower than at the equator.
- Cloud cover; the reduction in UV-B exposure depends the cover's thickness.
- Proximity to an industrial area because of the protection offered by photochemical smog. Industrial processes produce ozone, one of the more irritating components of smog, which absorbs UV-B. This is thought to be one of the main reasons that significant ozone losses in the Southern Hemisphere have not been mirrored in the Northern Hemisphere.
HEALTH EFFECTS OF UV-B LIGHT
Genetic damage DNA absorbs UV-B light and the absorbed energy can break bonds in the DNA. Most of the DNA breakage’s are repaired by proteins present in the cell nucleus but unrepaired genetic damage of the DNA can lead to skin cancers. In fact one method that scientists use to analyze amounts of 'genetically-damaging UV-B is to expose samples of DNA to the light and then count the number of breaks in the DNA. For example J.Regan's work at the Florida Institute of Technology used human DNA to find that genetically significant doses of solar radiation could penetrate as far as 9 feet into non-turbulent ocean water.
The Cancer link. The principle danger of skin cancer is to light-skinned peoples. A 1% decrease in the ozone layer will cause a estimated 2% increase in UV-B irradiation; it is estimated that this will lead to a 4%increase in basal carcinomas and 6% increase in squamous-cell carcinomas.[Graedel&Crutzen]. 90% of the skin carcinomas are attributed to UV-B exposure [Wayne] and the chemical mechanism by which it causes skin cancer has been identified [Tevini]. The above named carcinomas are relatively easy to treat, if detected in time, and are rarely fatal. But the much more dangerous malignant melanoma is not as well understood. There appears to be a correlation between brief, high intensity exposures to UV and eventual appearance (as long as 10-20yrs!) of melanoma. Twice as many deaths due to melanomas are seen in the southern states of Texas and Florida, as in the northern states of Wisconsin and Montana, but there could be many other factors involved. One undisputed effect of long-term sun exposure is the premature aging of the skin due to both UV-A, UV-B and UV-C. Even careful tanning kills skin cells, damages DNA and causes permanent changes in skin connective tissue which leads to wrinkle formation in later life. There is no such thing as a safe tan.
Possible eye damage can result from high doses of UV light, particularly to the cornea, which is a good absorber of UV light. High doses of UV light can causes a temporary clouding of the cornea, called 'snow-blindness’ and chronic doses has been tentatively linked to the formation of cataracts. Higher incidences of cataracts are found at high elevations, Tibet and Bolivia; and higher incidences are seen at lower latitudes (approaching the equator).
Damage to marine life. The penetration of increased amounts of UV-B light has caused great concern over the health of marine plankton that densely populate the top 2 meters of ocean water. The natural protective-response of most chlorophyll containing cells to increased light-radiation is to produce more light-absorbing pigments but this protective response is not triggered by UV-B light. Another possible response of plankton is to sink deeper into the water but this reduces the amount of visible light they need for photosynthesis, and thereby reduces their growth and reproduction rate. In other words, the amount of food and oxygen produced by plankton could be reduced by UV exposure without killing individual organisms. There are several other considerations:
- Ultraviolet levels are over 1,000 times higher at the equator than at the polar regions so it is presumed that marine life at the equator is much better adapted to the higher environmental UV light than organisms in the polar regions. The current concern of marine biologists is mostly over the more sensitive Antarctic phytoplankton, which normally would receive very low doses of UV. Only one large-scale field survey of Antarctic phytoplankton has been carried out so far [Smith et.al _Science_1992]; they found a 6-12% drop in phytoplankton productivity once their ship entered the area of the springtime ozone hole. Since the hole only lasts from 10-12 weeks this translates into a 2-4% loss overall, a measurable but not yet catastrophic loss.
- Both plants and phytoplankton vary widely in their sensitivity to UV-B. When over 200 agricultural plants were tested, more than half showed sensitivity to UV-B light. Other plants showed negligible effects or even a small increase in vigor. Even within a species there were marked differences; for example one variety of soybean showed a 16% decrease in growth while another variety of the same soybean showed no effect [R.Parson]. An increase in UV-B could cause a shift in population rather than a large die-off of plants
- An increase in UV-B will cause increased amounts of Ozone to be produced at lower levels in the atmosphere. While some have hailed the protection offered by this 'pollution-shield' many plants have shown themselves to be very sensitive to photochemical smog.
R.Parson FAQ 111 ,UV and biological effects of UV
FDA Consumer Magazine and publications: FDA#87-8272, #81-8149 and #92-1146
M.Tevini, ed. UV-B Radiation and Ozone Depletion: Effects on humans, animals, plants, microorganisms and materials Lewis Pub. Boca Raton, 1993.
R.P.Wayne, Chemistry of the Atmospheres 2nd ed. Oxford 1991
R.Smith et al. "Ozone depletion: Ultraviolet radiation and phytoplankton biology in Antarctic waters"' Science, 255, 952. (1992)
Author: Brien Sparling
LINKS TO OTHER EMF AND RF INFORMATION SOURCES
Please read this Disclaimer prior to connecting to these web sites.
EMF Research Program (RAPID)
The Institute of Electrical and Electronics Engineers, Inc.
How did you get here?
Let us know how you found out about this site, it will help us to develop a more effective means of communicating information about Nonionizing Radiation to the public. Contact us<|endoftext|>
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Search by Topic
Resources tagged with Algebra - generally similar to More Mathematical Mysteries:
Filter by: Content type:
Stage:
Challenge level:
There are 31 results
Broad Topics > Algebra > Algebra - generally
More Mathematical Mysteries
Stage: 3 Challenge Level:
Write down a three-digit number Change the order of the digits to get a different number Find the difference between the two three digit numbers Follow the rest of the instructions then try. . . .
Is it Magic or Is it Maths?
Stage: 3 Challenge Level:
Here are three 'tricks' to amaze your friends. But the really clever trick is explaining to them why these 'tricks' are maths not magic. Like all good magicians, you should practice by trying. . . .
Stage: 3 Challenge Level:
Think of a number, add one, double it, take away 3, add the number you first thought of, add 7, divide by 3 and take away the number you first thought of. You should now be left with 2. How do I. . . .
Pyramids
Stage: 3 Challenge Level:
What are the missing numbers in the pyramids?
Stage: 3 Challenge Level:
A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . .
Why 8?
Stage: 3 Challenge Level:
Choose any four consecutive even numbers. Multiply the two middle numbers together. Multiply the first and last numbers. Now subtract your second answer from the first. Try it with your own. . . .
Stage: 3 Challenge Level:
Think of a number Multiply it by 3 Add 6 Take away your start number Divide by 2 Take away your number. (You have finished with 3!) HOW DOES THIS WORK?
Think of Two Numbers
Stage: 3 Challenge Level:
Think of two whole numbers under 10. Take one of them and add 1. Multiply by 5. Add 1 again. Double your answer. Subract 1. Add your second number. Add 2. Double again. Subtract 8. Halve this. . . .
Stage: 3 Challenge Level:
Replace the letters with numbers to make the addition work out correctly. R E A D + T H I S = P A G E
Harmonic Triangle
Stage: 3 Challenge Level:
Can you see how to build a harmonic triangle? Can you work out the next two rows?
Quick Times
Stage: 3 Challenge Level:
32 x 38 = 30 x 40 + 2 x 8; 34 x 36 = 30 x 40 + 4 x 6; 56 x 54 = 50 x 60 + 6 x 4; 73 x 77 = 70 x 80 + 3 x 7 Verify and generalise if possible.
AP Rectangles
Stage: 3 Challenge Level:
An AP rectangle is one whose area is numerically equal to its perimeter. If you are given the length of a side can you always find an AP rectangle with one side the given length?
Converse
Stage: 4 Challenge Level:
Clearly if a, b and c are the lengths of the sides of a triangle and the triangle is equilateral then a^2 + b^2 + c^2 = ab + bc + ca. Is the converse true, and if so can you prove it? That is if. . . .
DOTS Division
Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
Our Ages
Stage: 4 Challenge Level:
I am exactly n times my daughter's age. In m years I shall be exactly (n-1) times her age. In m2 years I shall be exactly (n-2) times her age. After that I shall never again be an exact multiple of. . . .
Euler's Squares
Stage: 4 Challenge Level:
Euler found four whole numbers such that the sum of any two of the numbers is a perfect square. Three of the numbers that he found are a = 18530, b=65570, c=45986. Find the fourth number, x. You. . . .
Square Mean
Stage: 4 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
Diophantine N-tuples
Stage: 4 Challenge Level:
Take any whole number q. Calculate q^2 - 1. Factorize q^2-1 to give two factors a and b (not necessarily q+1 and q-1). Put c = a + b + 2q . Then you will find that ab+1 , bc+1 and ca+1 are all. . . .
Stage: 4 Challenge Level:
Find all real solutions of the equation (x^2-7x+11)^(x^2-11x+30) = 1.
For What?
Stage: 4 Challenge Level:
Prove that if the integer n is divisible by 4 then it can be written as the difference of two squares.
Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
Stage: 3 Challenge Level:
Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
Bang's Theorem
Stage: 4 Challenge Level:
If all the faces of a tetrahedron have the same perimeter then show that they are all congruent.
Medallions
Stage: 4 Challenge Level:
I keep three circular medallions in a rectangular box in which they just fit with each one touching the other two. The smallest one has radius 4 cm and touches one side of the box, the middle sized. . . .
Loopy
Stage: 4 Challenge Level:
Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?
Novemberish
Stage: 4 Challenge Level:
a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.
Coffee
Stage: 4 Challenge Level:
To make 11 kilograms of this blend of coffee costs £15 per kilogram. The blend uses more Brazilian, Kenyan and Mocha coffee... How many kilograms of each type of coffee are used?
Plutarch's Boxes
Stage: 3 Challenge Level:
According to Plutarch, the Greeks found all the rectangles with integer sides, whose areas are equal to their perimeters. Can you find them? What rectangular boxes, with integer sides, have. . . .
Rudolff's Problem
Stage: 4 Challenge Level:
A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?
The Development of Algebra - 1
Stage: 3, 4 and 5
This is the first of a two part series of articles on the history of Algebra from about 2000 BCE to about 1000 CE.<|endoftext|>
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মূল বিষয়বস্তুতে এড়িয়ে যান
$\exponential{x}{2} - 10 x + 25 = 0$
Solve for x
গ্রাফ
## শেয়ার করুন
a+b=-10 ab=25
To solve the equation, factor x^{2}-10x+25 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
-1,-25 -5,-5
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 25.
-1-25=-26 -5-5=-10
Calculate the sum for each pair.
a=-5 b=-5
The solution is the pair that gives sum -10.
\left(x-5\right)\left(x-5\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
\left(x-5\right)^{2}
Rewrite as a binomial square.
x=5
To find equation solution, solve x-5=0.
a+b=-10 ab=1\times 25=25
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+25. To find a and b, set up a system to be solved.
-1,-25 -5,-5
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 25.
-1-25=-26 -5-5=-10
Calculate the sum for each pair.
a=-5 b=-5
The solution is the pair that gives sum -10.
\left(x^{2}-5x\right)+\left(-5x+25\right)
Rewrite x^{2}-10x+25 as \left(x^{2}-5x\right)+\left(-5x+25\right).
x\left(x-5\right)-5\left(x-5\right)
Factor out x in the first and -5 in the second group.
\left(x-5\right)\left(x-5\right)
Factor out common term x-5 by using distributive property.
\left(x-5\right)^{2}
Rewrite as a binomial square.
x=5
To find equation solution, solve x-5=0.
x^{2}-10x+25=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 25}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-10\right)±\sqrt{100-4\times 25}}{2}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-100}}{2}
Multiply -4 times 25.
x=\frac{-\left(-10\right)±\sqrt{0}}{2}
x=-\frac{-10}{2}
Take the square root of 0.
x=\frac{10}{2}
The opposite of -10 is 10.
x=5
Divide 10 by 2.
x^{2}-10x+25=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
\left(x-5\right)^{2}=0
Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-5\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
x-5=0 x-5=0
Simplify.
x=5 x=5
Add 5 to both sides of the equation.
x=5
The equation is now solved. Solutions are the same.<|endoftext|>
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# IB DP Math AA Topic: AHL 3.18 Intersections of: a line with a plane; two planes; three planes HL Paper 1
### Question
Consider the three planes
1 : 2x – y + z = 4
2 : x – 2y + 3z = 5
3 : -9x + 3y – 2z = 32
(a) Show that the three planes do not intersect.
(b) (i) Verify that the point P(1 , -2 , 0) lies on both ∏1 and ∏2 .
(ii) Find a vector equation of L, the line of intersection of ∏1 and ∏2 .
(c) Find the distance between L and ∏3 .
Ans:
(a) METHOD 1
attempt to eliminate a variable
obtain a pair of equations in two variables
EITHER
-3x + z = -3 and
-3x + z = 44
OR
-5x + y = -7 and
-5x + y = 40
OR
3x – z = 3 and
3x – z = $$\frac{79}{5}$$
THEN
the two lines are parallel ( – 3 ≠ 44 or – 7 ≠ 40 or 3 ≠ -$$\frac{79}{5}$$
Note: There are other possible pairs of equations in two variables.
To obtain the final R1, at least the initial M1 must have been awarded.
hence the three planes do not intersect
METHOD 2
vector product of the two normals = $$\begin{pmatrix}-1\\-5\\-3\end{pmatrix}$$ (or equivalent)
$$r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}$$ (or equivalent)
Note: Award A0 if “r =” is missing. Subsequent marks may still be awarded.
Attempt to substitute (1 + λ, -2 + 5λ, 3λ) in ∏3
-9(1 + λ) + 3(-2+5λ) – 2(3λ) = 32
− 15 = 32, a contradiction
hence the three planes do not intersect
(b) (i) ∏1 : 2 + 2 + 0 = 4 and ∏2 : 1 + 4 + 0 = 5
(ii) METHOD 1
attempt to find the vector product of the two normals
$$\begin{pmatrix}2\\-1 \\1 \end{pmatrix} \times \begin{pmatrix}1\\-2 \\3 \end{pmatrix}$$
$$=\begin{pmatrix}-1\\-5 \\-3 \end{pmatrix}$$
$$r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}$$
Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector.
Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize
lack of “r =” only once.
METHOD 2
attempt to eliminate a variable from ∏1 and ∏2
3x – z = 3 OR 3y – 5z = -6 OR 5x – y = 7
Let x = t
substituting x = t in 3x – z = 3 to obtain
z = -3 + 3t and y = 5t – 7 (for all three variables in parametric form)
$$r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}$$
Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector. Accept other position vectors
which satisfy both the planes ∏1 and ∏2 .
(c) METHOD 1
the line connecting L and ∏3 is given by L1
attempt to substitute position and direction vector to form L1
$$s = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + t \begin{pmatrix}-9\\3 \\-2 \end{pmatrix}$$
substitute (1 – 9t, – 2 + 3t, – 2t) in ∏3
-9(1-9t) + 3(-2+3t) – 2(-2t) = 32
$$94t = 47\Rightarrow t=\frac{1}{2}$$
attempt to find distance between (1, -2,0) and their point $$\begin{pmatrix}-\frac{7}{2}, & -\frac{1}{2}, & -1 \end{pmatrix}$$
$$=\left | \begin{pmatrix}1\\-2 \\0 \end{pmatrix} +\frac{1}{2}\begin{pmatrix}-9\\3 \\-2 \end{pmatrix}-\begin{pmatrix}1\\-2 \\0 \end{pmatrix}\right | = \frac{1}{2}\sqrt{(-9)^{2}+3^{2}+(-2)^{2}}$$
$$=\frac{\sqrt{94}}{2}$$
METHOD 2
unit normal vector equation of ∏3 is given by
$$=\frac{32}{\sqrt{94}}$$
### Question
Consider the three planes
1 : 2x – y + z = 4
2 : x – 2y + 3z = 5
3 : -9x + 3y – 2z = 32
(a) Show that the three planes do not intersect.
(b) (i) Verify that the point P(1 , -2 , 0) lies on both ∏1 and ∏2 .
(ii) Find a vector equation of L, the line of intersection of ∏1 and ∏2 .
(c) Find the distance between L and ∏3 .
Ans:
(a) METHOD 1
attempt to eliminate a variable
obtain a pair of equations in two variables
EITHER
-3x + z = -3 and
-3x + z = 44
OR
-5x + y = -7 and
-5x + y = 40
OR
3x – z = 3 and
3x – z = $$\frac{79}{5}$$
THEN
the two lines are parallel ( – 3 ≠ 44 or – 7 ≠ 40 or 3 ≠ -$$\frac{79}{5}$$
Note: There are other possible pairs of equations in two variables.
To obtain the final R1, at least the initial M1 must have been awarded.
hence the three planes do not intersect
METHOD 2
vector product of the two normals = $$\begin{pmatrix}-1\\-5\\-3\end{pmatrix}$$ (or equivalent)
$$r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}$$ (or equivalent)
Note: Award A0 if “r =” is missing. Subsequent marks may still be awarded.
Attempt to substitute (1 + λ, -2 + 5λ, 3λ) in ∏3
-9(1 + λ) + 3(-2+5λ) – 2(3λ) = 32
− 15 = 32, a contradiction
hence the three planes do not intersect
(b) (i) ∏1 : 2 + 2 + 0 = 4 and ∏2 : 1 + 4 + 0 = 5
(ii) METHOD 1
attempt to find the vector product of the two normals
$$\begin{pmatrix}2\\-1 \\1 \end{pmatrix} \times \begin{pmatrix}1\\-2 \\3 \end{pmatrix}$$
$$=\begin{pmatrix}-1\\-5 \\-3 \end{pmatrix}$$
$$r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}$$
Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector.
Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize
lack of “r =” only once.
METHOD 2
attempt to eliminate a variable from ∏1 and ∏2
3x – z = 3 OR 3y – 5z = -6 OR 5x – y = 7
Let x = t
substituting x = t in 3x – z = 3 to obtain
z = -3 + 3t and y = 5t – 7 (for all three variables in parametric form)
$$r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}$$
Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector. Accept other position vectors
which satisfy both the planes ∏1 and ∏2 .
(c) METHOD 1
the line connecting L and ∏3 is given by L1
attempt to substitute position and direction vector to form L1
$$s = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + t \begin{pmatrix}-9\\3 \\-2 \end{pmatrix}$$
substitute (1 – 9t, – 2 + 3t, – 2t) in ∏3
-9(1-9t) + 3(-2+3t) – 2(-2t) = 32
$$94t = 47\Rightarrow t=\frac{1}{2}$$
attempt to find distance between (1, -2,0) and their point $$\begin{pmatrix}-\frac{7}{2}, & -\frac{1}{2}, & -1 \end{pmatrix}$$
$$=\left | \begin{pmatrix}1\\-2 \\0 \end{pmatrix} +\frac{1}{2}\begin{pmatrix}-9\\3 \\-2 \end{pmatrix}-\begin{pmatrix}1\\-2 \\0 \end{pmatrix}\right | = \frac{1}{2}\sqrt{(-9)^{2}+3^{2}+(-2)^{2}}$$
$$=\frac{\sqrt{94}}{2}$$
METHOD 2
unit normal vector equation of ∏3 is given by
$$=\frac{32}{\sqrt{94}}$$
## Question
The points A, B, C have position vectors i + j + 2k , i + 2j + 3k , 3i + k respectively and lie in the plane $$\pi$$ .
(a) Find
(i) the area of the triangle ABC;
(ii) the shortest distance from C to the line AB;
(iii) the cartesian equation of the plane $$\pi$$ .
The line L passes through the origin and is normal to the plane $$\pi$$ , it intersects $$\pi$$ at the
point D.
(b) Find
(i) the coordinates of the point D;
(ii) the distance of $$\pi$$ from the origin.
## Markscheme
(a) (i) METHOD 1
$$\overrightarrow {{\text{AB}}} = \boldsymbol{b} – \boldsymbol{a} = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ 1 \end{array}} \right)$$ (A1)
$$\overrightarrow {{\text{AC}}} = \boldsymbol{c} – \boldsymbol{a} = \left( {\begin{array}{*{20}{c}} 3 \\ 0 \\ 1 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2 \\ { – 1} \\ { – 1} \end{array}} \right)$$ (A1)
$$\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\ 0&1&1 \\ 2&{ – 1}&{ – 1} \end{array}} \right|$$ M1
i(−1 + 1) − j(0 − 2) + k (0 − 2) (A1)
= 2j − 2k A1
Area of triangle ABC}} $$= \frac{1}{2}\left| {2{\mathbf{j}} – 2{\mathbf{k}}} \right| = {\text{ }}\frac{1}{2}\sqrt 8$$ $$( = \sqrt 2 )$$ sq. units M1A1
Note: Allow FT on final A1.
METHOD 2
$$\left| {{\text{AB}}} \right| = \sqrt 2 ,{\text{ }}\left| {{\text{BC}}} \right| = \sqrt {12} ,{\text{ }}\left| {{\text{AC}}} \right| = \sqrt 6$$ A1A1A1
Using cosine rule, e.g. on $${\hat C}$$ M1
$$\cos C = \frac{{6 + 12 – 2}}{{2\sqrt {72} }} = \frac{{2\sqrt 2 }}{3}$$ A1
$$\therefore {\text{Area }}\Delta {\text{ABC}} = \frac{1}{2}ab\sin C$$ M1
$$= \frac{1}{2}\sqrt {12} \sqrt 6 \sin \left( {\arccos \frac{{2\sqrt 2 }}{3}} \right)$$
$$= 3\sqrt 2 \sin \left( {\arccos \frac{{2\sqrt 2 }}{3}} \right){\text{ }}\left( { = \sqrt 2 } \right)$$ A1
Note: Allow FT on final A1.
(ii) $${\text{AB}} = \sqrt 2$$ A1
$$\sqrt 2 = \frac{1}{2}{\text{AB}} \times h = \frac{1}{2}\sqrt 2 \times h{\text{ , }}h$$ equals the shortest distance (M1)
$$\Rightarrow h = 2$$ A1
(iii) METHOD 1
(\pi \) has form $$r \cdot \left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ { – 2} \end{array}} \right) = d$$ (M1)
Since (1, 1, 2) is on the plane
$$d = \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 2 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ { – 2} \end{array}} \right) = 2 – 4 = -2$$ M1A1
Hence $$r \cdot \left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ { – 2} \end{array}} \right) = – 2$$
$$2y – 2z = – 2{\text{ (or }}y – z = – 1)$$ A1
METHOD 2
$$r = \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 2 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ 1 \end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}} 2 \\ { – 1} \\ { – 1} \end{array}} \right)$$ (M1)
$$x = 1 + 2\mu$$ (i)
$$y = 1 + \lambda – \mu$$ (ii)
$$z = 2 + \lambda – \mu$$ (iii) A1
Note: Award A1 for all three correct, A0 otherwise.
From (i) $$\mu = \frac{{x – 1}}{2}$$
substitute in (ii) $$y = 1 + \lambda – \left( {\frac{{x – 1}}{2}} \right)$$
$$\Rightarrow \lambda = y – 1 + \left( {\frac{{x – 1}}{2}} \right)$$
substitute $$\lambda$$ and $$\mu$$ in (iii) M1
$$\Rightarrow z = 2 + y – 1 + \left( {\frac{{x – 1}}{2}} \right) – \left( {\frac{{x – 1}}{2}} \right)$$
$$\Rightarrow y – z = – 1$$ A1
[14 marks]
(b) (i) The equation of OD is
$$r = \lambda \left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ { – 2} \end{array}} \right)$$, $$\left( {{\text{or }}r = \lambda \left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ { – 1} \end{array}} \right)} \right)$$ M1
This meets $$\pi$$ where
$$2\lambda + 2\lambda = – 1$$ (M1)
$$\lambda = – \frac{1}{4}$$ A1
Coordinates of D are $$\left( {0, – \frac{1}{2},\frac{1}{2}} \right)$$ A1
(ii) $$\left| {\overrightarrow {{\text{OD}}} } \right| = \sqrt {0 + {{\left( { – \frac{1}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} = \frac{1}{{\sqrt 2 }}$$ (M1)A1
[6 marks]
Total [20 marks]
## Examiners report
It was disappointing to see that a number of candidates did not appear to be well prepared for this question and made no progress at all. There were a number of schools where no candidate made any appreciable progress with the question. A good number of students, however, were successful with part (a) (i). A good number of candidates were also successful with part a (ii) but few realised that the shortest distance was the height of the triangle. Candidates used a variety of methods to answer (a) (iii) but again a reasonable number of correct answers were seen. Candidates also had a reasonable degree of success with part (b), with a respectable number of correct answers seen.
## Question
Consider the plane with equation $$4x – 2y – z = 1$$ and the line given by the parametric equations
$$x = 3 – 2\lambda$$
$$y = (2k – 1) + \lambda$$
$$z = – 1 + k\lambda .$$
Given that the line is perpendicular to the plane, find
(a) the value of k;
(b) the coordinates of the point of intersection of the line and the plane.
## Markscheme
(a) $$\boldsymbol{a} = \left( {\begin{array}{*{20}{c}} 4 \\ { – 2} \\ { – 1} \end{array}} \right) \bot$$ to the plane $$\boldsymbol{e} = \left( {\begin{array}{*{20}{c}} { – 2} \\ 1 \\ k \end{array}} \right)$$ is parallel to the line (A1)(A1)
Note: Award A1 for each correct vector written down, even if not identified.
line $$\bot$$ plane $$\Rightarrow$$ $$\boldsymbol{e}$$ parallel to $$\boldsymbol{a}$$
since $$\left( {\begin{array}{*{20}{c}} 4 \\ { – 2} \\ { – 1} \end{array}} \right) = t\left( {\begin{array}{*{20}{c}} { – 2} \\ 1 \\ k \end{array}} \right) \Rightarrow k = \frac{1}{2}$$ (M1)A1
(b) $$4(3 – 2\lambda ) – 2\lambda – \left( { – 1 + \frac{1}{2}\lambda } \right) = 1$$ (M1)(A1)
Note: FT their value of k as far as possible.
$$\lambda = \frac{8}{7}$$ A1
$${\text{P}}\left( {\frac{5}{7},\frac{8}{7}, – \frac{3}{7}} \right)$$ A1
[8 marks]
## Examiners report
Solutions to this question were often disappointing. In (a), some candidates found the value of k, incorrectly, by taking the scalar product of the normal vector to the plane and the direction of the line. Such candidates benefitted partially from follow through in (b) but not fully because their line turned out to be parallel to the plane and did not intersect it.
## Question
Consider the points $${\text{A(1, 0, 0)}}$$, $${\text{B(2, 2, 2)}}$$ and $${\text{C(0, 2, 1)}}$$.
A third plane $${\Pi _3}$$ is defined by the Cartesian equation $$16x + \alpha y – 3z = \beta$$.
a.Find the vector $$\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}}$$.[4]
b.Find an exact value for the area of the triangle ABC.[3]
c.Show that the Cartesian equation of the plane $${\Pi _1}$$, containing the triangle ABC, is $$2x + 3y – 4z = 2$$.[3]
d.A second plane $${\Pi _2}$$ is defined by the Cartesian equation $${\Pi _2}:4x – y – z = 4$$. $${L_1}$$ is the line of intersection of the planes $${\Pi _1}$$ and $${\Pi _2}$$.
Find a vector equation for $${L_1}$$.[5]
e.Find the value of $$\alpha$$ if all three planes contain $${L_1}$$.[3]
f.Find conditions on $$\alpha$$ and $$\beta$$ if the plane $${\Pi _3}$$ does not intersect with $${L_1}$$.[2]
## Markscheme
$$\overrightarrow {{\rm{CA}}} = \left( \begin{array}{c}1\\ – 2\\ – 1\end{array} \right)$$ (A1)
$$\overrightarrow {{\rm{CB}}} = \left( \begin{array}{c}2\\0\\1\end{array} \right)$$ (A1)
Note: If $$\overrightarrow {{\text{AC}}}$$ and $$\overrightarrow {{\text{BC}}}$$ found correctly award (A1) (A0).
$$\overrightarrow {{\rm{CA}}} \times \overrightarrow {{\rm{CB}}} = \left| {\begin{array}{*{20}{c}}i&j&k\\1&{ – 2}&{ – 1}\\2&0&1\end{array}} \right|$$ (M1)
$$\left( \begin{array}{c} – 2\\ – 3\\4\end{array} \right)$$ A1
[4 marks]
a.
METHOD 1
$$\frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}} } \right|$$ (M1)
$$= \frac{1}{2}\sqrt {{{( – 2)}^2} + {{( – 3)}^2} + {4^2}}$$ (A1)
$$= \frac{{\sqrt {29} }}{2}$$ A1
METHOD 2
attempt to apply $$\frac{1}{2}\left| {{\text{CA}}} \right|\left| {{\text{CB}}} \right|\sin C$$ (M1)
$${\text{CA.CB}} = \sqrt 5 .\sqrt 6 \cos C \Rightarrow \cos C = \frac{1}{{\sqrt {30} }} \Rightarrow \sin C = \frac{{\sqrt {29} }}{{\sqrt {30} }}$$ (A1)
$${\text{area}} = \frac{{\sqrt {29} }}{2}$$ A1
[3 marks]
b.
METHOD 1
r.$$\left( \begin{array}{c} – 2\\ – 3\\4\end{array} \right) = \left( \begin{array}{l}1\\0\\0\end{array} \right) \bullet \left( \begin{array}{c} – 2\\ – 3\\4\end{array} \right)$$ M1A1
$$\Rightarrow – 2x – 3y + 4z = – 2$$ A1
$$\Rightarrow 2x + 3y – 4z = 2$$ AG
METHOD 2
$$– 2x – 3y + 4z = d$$
substituting a point in the plane M1A1
$${\text{d}} = – 2$$ A1
$$\Rightarrow – 2x – 3y + 4z = – 2$$
$$\Rightarrow 2x + 3y – 4z = 2$$ AG
Note: Accept verification that all 3 vertices of the triangle lie on the given plane.
[3 marks]
c.
METHOD 1
$$\left| {\begin{array}{*{20}{c}}{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}}\\2&3&{ – 4}\\4&{ – 1}&{ – 1}\end{array}} \right| = \left( \begin{array}{c} – 7\\ – 14\\ – 14\end{array} \right)$$ M1A1
$${\mathbf{n}} = \left( \begin{array}{l}1\\2\\2\end{array} \right)$$
$$z = 0 \Rightarrow y = 0,{\text{ }}x = 1$$ (M1)(A1)
$${L_1}:{\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + \lambda \left( \begin{array}{l}1\\2\\2\end{array} \right)$$ A1
Note: Do not award the final A1 if $$\mathbf{r} =$$ is not seen.
METHOD 2
eliminate 1 of the variables, eg x M1
$$– 7y + 7z = 0$$ (A1)
introduce a parameter M1
$$\Rightarrow z = \lambda$$,
$$y = \lambda {\text{, }}x = 1 + \frac{\lambda }{2}$$ (A1)
$${\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + \lambda \left( \begin{array}{l}1\\2\\2\end{array} \right)$$ or equivalent A1
Note: Do not award the final A1 if $$\mathbf{r} =$$ is not seen.
METHOD 3
$$z = t$$ M1
write x and y in terms of $$t \Rightarrow 4x – y = 4 + t,{\text{ }}2x + 3y = 2 + 4t$$ or equivalent A1
attempt to eliminate x or y M1
$$x,{\text{ }}y,{\text{ }}z$$ expressed in parameters
$$\Rightarrow z = t$$,
$$y = t,{\text{ }}x = 1 + \frac{t}{2}$$ A1
$${\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + t \left( \begin{array}{l}1\\2\\2\end{array} \right)$$ or equivalent A1
Note: Do not award the final A1 if $$\mathbf{r} =$$ is not seen.
[5 marks]
d.
METHOD 1
direction of the line is perpendicular to the normal of the plane
$$\left( \begin{array}{c}16\\\alpha \\ – 3\end{array} \right) \bullet \left( \begin{array}{l}1\\2\\2\end{array} \right) = 0$$ M1A1
$$16 + 2\alpha – 6 = 0 \Rightarrow \alpha = – 5$$ A1
METHOD 2
solving line/plane simultaneously
$$16(1 + \lambda ) + 2\alpha \lambda – 6\lambda = \beta$$ M1A1
$$16 + (10 + 2\alpha )\lambda = \beta$$
$$\Rightarrow \alpha = – 5$$ A1
METHOD 3
$$\left| {\begin{array}{*{20}{c}}2&3&{ – 4}\\4&{ – 1}&{ – 1}\\{16}&\alpha &{ – 3}\end{array}} \right| = 0$$ M1
$$2(3 + \alpha ) – 3( – 12 + 16) – 4(4\alpha + 16) = 0$$ A1
$$\Rightarrow \alpha = – 5$$ A1
METHOD 4
attempt to use row reduction on augmented matrix M1
to obtain $$\left( {\begin{array}{*{20}{c}}2&3&{ – 4}\\0&{ – 1}&1\\0&0&{\alpha + 5}\end{array}\left| \begin{array}{c}2\\0\\\beta – 16\end{array} \right.} \right)$$ A1
$$\Rightarrow \alpha = – 5$$ A1
[3 marks]
e.
$$\alpha = – 5$$ A1
$$\beta \ne 16$$ A1
[2 marks]
f
## Question
Consider the points $${\text{A(1, 0, 0)}}$$, $${\text{B(2, 2, 2)}}$$ and $${\text{C(0, 2, 1)}}$$.
A third plane $${\Pi _3}$$ is defined by the Cartesian equation $$16x + \alpha y – 3z = \beta$$.
a.Find the vector $$\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}}$$.[4]
b.Find an exact value for the area of the triangle ABC.[3]
c.Show that the Cartesian equation of the plane $${\Pi _1}$$, containing the triangle ABC, is $$2x + 3y – 4z = 2$$.[3]
d.A second plane $${\Pi _2}$$ is defined by the Cartesian equation $${\Pi _2}:4x – y – z = 4$$. $${L_1}$$ is the line of intersection of the planes $${\Pi _1}$$ and $${\Pi _2}$$.
Find a vector equation for $${L_1}$$.[5]
e.Find the value of $$\alpha$$ if all three planes contain $${L_1}$$.[3]
f.Find conditions on $$\alpha$$ and $$\beta$$ if the plane $${\Pi _3}$$ does not intersect with $${L_1}$$.[2]
## Markscheme
$$\overrightarrow {{\rm{CA}}} = \left( \begin{array}{c}1\\ – 2\\ – 1\end{array} \right)$$ (A1)
$$\overrightarrow {{\rm{CB}}} = \left( \begin{array}{c}2\\0\\1\end{array} \right)$$ (A1)
Note: If $$\overrightarrow {{\text{AC}}}$$ and $$\overrightarrow {{\text{BC}}}$$ found correctly award (A1) (A0).
$$\overrightarrow {{\rm{CA}}} \times \overrightarrow {{\rm{CB}}} = \left| {\begin{array}{*{20}{c}}i&j&k\\1&{ – 2}&{ – 1}\\2&0&1\end{array}} \right|$$ (M1)
$$\left( \begin{array}{c} – 2\\ – 3\\4\end{array} \right)$$ A1
[4 marks]
a.
METHOD 1
$$\frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}} } \right|$$ (M1)
$$= \frac{1}{2}\sqrt {{{( – 2)}^2} + {{( – 3)}^2} + {4^2}}$$ (A1)
$$= \frac{{\sqrt {29} }}{2}$$ A1
METHOD 2
attempt to apply $$\frac{1}{2}\left| {{\text{CA}}} \right|\left| {{\text{CB}}} \right|\sin C$$ (M1)
$${\text{CA.CB}} = \sqrt 5 .\sqrt 6 \cos C \Rightarrow \cos C = \frac{1}{{\sqrt {30} }} \Rightarrow \sin C = \frac{{\sqrt {29} }}{{\sqrt {30} }}$$ (A1)
$${\text{area}} = \frac{{\sqrt {29} }}{2}$$ A1
[3 marks]
b.
METHOD 1
r.$$\left( \begin{array}{c} – 2\\ – 3\\4\end{array} \right) = \left( \begin{array}{l}1\\0\\0\end{array} \right) \bullet \left( \begin{array}{c} – 2\\ – 3\\4\end{array} \right)$$ M1A1
$$\Rightarrow – 2x – 3y + 4z = – 2$$ A1
$$\Rightarrow 2x + 3y – 4z = 2$$ AG
METHOD 2
$$– 2x – 3y + 4z = d$$
substituting a point in the plane M1A1
$${\text{d}} = – 2$$ A1
$$\Rightarrow – 2x – 3y + 4z = – 2$$
$$\Rightarrow 2x + 3y – 4z = 2$$ AG
Note: Accept verification that all 3 vertices of the triangle lie on the given plane.
[3 marks]
c.
METHOD 1
$$\left| {\begin{array}{*{20}{c}}{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}}\\2&3&{ – 4}\\4&{ – 1}&{ – 1}\end{array}} \right| = \left( \begin{array}{c} – 7\\ – 14\\ – 14\end{array} \right)$$ M1A1
$${\mathbf{n}} = \left( \begin{array}{l}1\\2\\2\end{array} \right)$$
$$z = 0 \Rightarrow y = 0,{\text{ }}x = 1$$ (M1)(A1)
$${L_1}:{\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + \lambda \left( \begin{array}{l}1\\2\\2\end{array} \right)$$ A1
Note: Do not award the final A1 if $$\mathbf{r} =$$ is not seen.
METHOD 2
eliminate 1 of the variables, eg x M1
$$– 7y + 7z = 0$$ (A1)
introduce a parameter M1
$$\Rightarrow z = \lambda$$,
$$y = \lambda {\text{, }}x = 1 + \frac{\lambda }{2}$$ (A1)
$${\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + \lambda \left( \begin{array}{l}1\\2\\2\end{array} \right)$$ or equivalent A1
Note: Do not award the final A1 if $$\mathbf{r} =$$ is not seen.
METHOD 3
$$z = t$$ M1
write x and y in terms of $$t \Rightarrow 4x – y = 4 + t,{\text{ }}2x + 3y = 2 + 4t$$ or equivalent A1
attempt to eliminate x or y M1
$$x,{\text{ }}y,{\text{ }}z$$ expressed in parameters
$$\Rightarrow z = t$$,
$$y = t,{\text{ }}x = 1 + \frac{t}{2}$$ A1
$${\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + t \left( \begin{array}{l}1\\2\\2\end{array} \right)$$ or equivalent A1
Note: Do not award the final A1 if $$\mathbf{r} =$$ is not seen.
[5 marks]
d.
METHOD 1
direction of the line is perpendicular to the normal of the plane
$$\left( \begin{array}{c}16\\\alpha \\ – 3\end{array} \right) \bullet \left( \begin{array}{l}1\\2\\2\end{array} \right) = 0$$ M1A1
$$16 + 2\alpha – 6 = 0 \Rightarrow \alpha = – 5$$ A1
METHOD 2
solving line/plane simultaneously
$$16(1 + \lambda ) + 2\alpha \lambda – 6\lambda = \beta$$ M1A1
$$16 + (10 + 2\alpha )\lambda = \beta$$
$$\Rightarrow \alpha = – 5$$ A1
METHOD 3
$$\left| {\begin{array}{*{20}{c}}2&3&{ – 4}\\4&{ – 1}&{ – 1}\\{16}&\alpha &{ – 3}\end{array}} \right| = 0$$ M1
$$2(3 + \alpha ) – 3( – 12 + 16) – 4(4\alpha + 16) = 0$$ A1
$$\Rightarrow \alpha = – 5$$ A1
METHOD 4
attempt to use row reduction on augmented matrix M1
to obtain $$\left( {\begin{array}{*{20}{c}}2&3&{ – 4}\\0&{ – 1}&1\\0&0&{\alpha + 5}\end{array}\left| \begin{array}{c}2\\0\\\beta – 16\end{array} \right.} \right)$$ A1
$$\Rightarrow \alpha = – 5$$ A1
[3 marks]
e.
$$\alpha = – 5$$ A1
$$\beta \ne 16$$ A1
[2 marks]
f.
## Question
a.Show that the points $${\text{O}}(0,{\text{ }}0,{\text{ }}0)$$, $${\text{ A}}(6,{\text{ }}0,{\text{ }}0)$$, $${\text{B}}({6,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12} })$$, $${\text{C}}({0,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12}})$$ form a square.
[3]
b.
Find the coordinates of M, the mid-point of [OB].[1]
c.
Show that an equation of the plane $${\mathit{\Pi }}$$, containing the square OABC, is $$y + \sqrt 2 z = 0$$.[3]
d.
Find a vector equation of the line $$L$$, through M, perpendicular to the plane $${\mathit{\Pi }}$$.[3]
e.
Find the coordinates of D, the point of intersection of the line $$L$$ with the plane whose equation is $$y = 0$$.[3]
f.
Find the coordinates of E, the reflection of the point D in the plane $${\mathit{\Pi }}$$.[3]
g.
(i) Find the angle $${\rm{O\hat DA}}$$.
(ii) State what this tells you about the solid OABCDE.[6]
## Markscheme
$$\left| {\overrightarrow {{\text{OA}}} } \right| = \left| {\overrightarrow {{\text{CB}}} } \right| = \left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| = 6$$ (therefore a rhombus) A1A1
Note: Award A1 for two correct lengths, A2 for all four.
Note: Award A1A0 for $$\overrightarrow {{\rm{OA}}} = \overrightarrow {{\rm{CB}}} = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{ or \,\,} } \overrightarrow {{\rm{OC}}} = \overrightarrow {A{\rm{B}}} = \left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right)$$ if no magnitudes are shown.
$$\overrightarrow {{\rm{OA}}}\,\, {\rm{ g}}\overrightarrow {{\rm{OC}}} = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{g}}\left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right) = 0$$ (therefore a square) A1
Note: Other arguments are possible with a minimum of three conditions.
[3 marks]
a.
$${\text{M}}\left( {3,{\text{ }} – \frac{{\sqrt {24} }}{2},{\text{ }}\frac{{\sqrt {12} }}{2}} \right)\left( { = \left( {3,{\text{ }} – \sqrt 6 ,{\text{ }}\sqrt 3 } \right)} \right)$$ A1
[1 mark]
b.
METHOD 1
$$\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OC}}} =$$$$\left( \begin{array}{l}6\\0\\0\end{array} \right) \times \left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right) = \left( \begin{array}{c}0\\ – 6\sqrt {12} \\ – 6\sqrt {24} \end{array} \right)\left( { = \left( \begin{array}{c}0\\ – 12\sqrt 3 \\ – 12\sqrt 6 \end{array} \right)} \right)$$ M1A1
Note: Candidates may use other pairs of vectors.
equation of plane is $$– 6\sqrt {12} y – 6\sqrt {24} z = d$$
any valid method showing that $$d = 0$$ M1
$$\mathit{\Pi} :y+\sqrt{2z}=0$$ AG
METHOD 2
equation of plane is $$ax + by + cz = d$$
substituting O to find $$d = 0$$ (M1)
substituting two points (A, B, C or M) M1
eg
$$6a = 0,{\text{ }} – \sqrt {24} b + \sqrt {12} c = 0$$ A1
$$\mathit{\Pi} :y+\sqrt{2z}=0$$ AG
[3 marks]
c.
$$\boldsymbol{r} = \left( \begin{array}{c}3\\ – \sqrt 6 \\\sqrt 3 \end{array} \right) + \lambda \left( \begin{array}{l}0\\1\\\sqrt 2 \end{array} \right)$$ A1A1A1
Note: Award A1 for r = , A1A1 for two correct vectors.
[3 marks]
d.
Using $$y = 0$$ to find $$\lambda$$ M1
Substitute their $$\lambda$$ into their equation from part (d) M1
D has coordinates $$\left( {{\text{3, 0, 3}}\sqrt 3 } \right)$$ A1
[3 marks]
e.
$$\lambda$$ for point E is the negative of the $$\lambda$$ for point D (M1)
Note: Other possible methods may be seen.
E has coordinates $$\left( {{\text{3, }} – 2\sqrt 6 ,{\text{ }} – \sqrt 3 } \right)$$ A1A1
Note: Award A1 for each of the y and z coordinates.
[3 marks]
f.
(i) $$\overrightarrow {{\text{DA}}} {\text{ g}}\overrightarrow {{\text{DO}}} =$$$$\left( \begin{array}{c}3\\0\\ – 3\sqrt 3 \end{array} \right){\rm{g}}\left( \begin{array}{c} – 3\\0\\ – 3\sqrt 3 \end{array} \right) = 18$$ M1A1
$$\cos {\rm{O\hat DA}} = \frac{{18}}{{\sqrt {36} \sqrt {36} }} = \frac{1}{2}$$ M1
hence $${\rm{O\hat DA}} = 60^\circ$$ A1
Note: Accept method showing OAD is equilateral.
(ii) OABCDE is a regular octahedron (accept equivalent description) A2
Note: A2 for saying it is made up of 8 equilateral triangles
Award A1 for two pyramids, A1 for equilateral triangles.
(can be either stated or shown in a sketch – but there must be clear indication the triangles are equilateral)
[6 marks]
## Question
a.Show that the points $${\text{O}}(0,{\text{ }}0,{\text{ }}0)$$, $${\text{ A}}(6,{\text{ }}0,{\text{ }}0)$$, $${\text{B}}({6,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12} })$$, $${\text{C}}({0,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12}})$$ form a square.[3]
b.
Find the coordinates of M, the mid-point of [OB].[1]
c.
Show that an equation of the plane $${\mathit{\Pi }}$$, containing the square OABC, is $$y + \sqrt 2 z = 0$$.[3]
d.
Find a vector equation of the line $$L$$, through M, perpendicular to the plane $${\mathit{\Pi }}$$.[3]
e.
Find the coordinates of D, the point of intersection of the line $$L$$ with the plane whose equation is $$y = 0$$.[3]
f.
Find the coordinates of E, the reflection of the point D in the plane $${\mathit{\Pi }}$$.[3]
g.
(i) Find the angle $${\rm{O\hat DA}}$$.
(ii) State what this tells you about the solid OABCDE.[6]
## Markscheme
$$\left| {\overrightarrow {{\text{OA}}} } \right| = \left| {\overrightarrow {{\text{CB}}} } \right| = \left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| = 6$$ (therefore a rhombus) A1A1
Note: Award A1 for two correct lengths, A2 for all four.
Note: Award A1A0 for $$\overrightarrow {{\rm{OA}}} = \overrightarrow {{\rm{CB}}} = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{ or \,\,} } \overrightarrow {{\rm{OC}}} = \overrightarrow {A{\rm{B}}} = \left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right)$$ if no magnitudes are shown.
$$\overrightarrow {{\rm{OA}}}\,\, {\rm{ g}}\overrightarrow {{\rm{OC}}} = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{g}}\left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right) = 0$$ (therefore a square) A1
Note: Other arguments are possible with a minimum of three conditions.
[3 marks]
a.
$${\text{M}}\left( {3,{\text{ }} – \frac{{\sqrt {24} }}{2},{\text{ }}\frac{{\sqrt {12} }}{2}} \right)\left( { = \left( {3,{\text{ }} – \sqrt 6 ,{\text{ }}\sqrt 3 } \right)} \right)$$ A1
[1 mark]
b.
METHOD 1
$$\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OC}}} =$$$$\left( \begin{array}{l}6\\0\\0\end{array} \right) \times \left( \begin{array}{c}0\\ – \sqrt {24} \\\sqrt {12} \end{array} \right) = \left( \begin{array}{c}0\\ – 6\sqrt {12} \\ – 6\sqrt {24} \end{array} \right)\left( { = \left( \begin{array}{c}0\\ – 12\sqrt 3 \\ – 12\sqrt 6 \end{array} \right)} \right)$$ M1A1
Note: Candidates may use other pairs of vectors.
equation of plane is $$– 6\sqrt {12} y – 6\sqrt {24} z = d$$
any valid method showing that $$d = 0$$ M1
$$\mathit{\Pi} :y+\sqrt{2z}=0$$ AG
METHOD 2
equation of plane is $$ax + by + cz = d$$
substituting O to find $$d = 0$$ (M1)
substituting two points (A, B, C or M) M1
eg
$$6a = 0,{\text{ }} – \sqrt {24} b + \sqrt {12} c = 0$$ A1
$$\mathit{\Pi} :y+\sqrt{2z}=0$$ AG
[3 marks]
c.
$$\boldsymbol{r} = \left( \begin{array}{c}3\\ – \sqrt 6 \\\sqrt 3 \end{array} \right) + \lambda \left( \begin{array}{l}0\\1\\\sqrt 2 \end{array} \right)$$ A1A1A1
Note: Award A1 for r = , A1A1 for two correct vectors.
[3 marks]
d.
Using $$y = 0$$ to find $$\lambda$$ M1
Substitute their $$\lambda$$ into their equation from part (d) M1
D has coordinates $$\left( {{\text{3, 0, 3}}\sqrt 3 } \right)$$ A1
[3 marks]
e.
$$\lambda$$ for point E is the negative of the $$\lambda$$ for point D (M1)
Note: Other possible methods may be seen.
E has coordinates $$\left( {{\text{3, }} – 2\sqrt 6 ,{\text{ }} – \sqrt 3 } \right)$$ A1A1
Note: Award A1 for each of the y and z coordinates.
[3 marks]
f.
(i) $$\overrightarrow {{\text{DA}}} {\text{ g}}\overrightarrow {{\text{DO}}} =$$$$\left( \begin{array}{c}3\\0\\ – 3\sqrt 3 \end{array} \right){\rm{g}}\left( \begin{array}{c} – 3\\0\\ – 3\sqrt 3 \end{array} \right) = 18$$ M1A1
$$\cos {\rm{O\hat DA}} = \frac{{18}}{{\sqrt {36} \sqrt {36} }} = \frac{1}{2}$$ M1
hence $${\rm{O\hat DA}} = 60^\circ$$ A1
Note: Accept method showing OAD is equilateral.
(ii) OABCDE is a regular octahedron (accept equivalent description) A2
Note: A2 for saying it is made up of 8 equilateral triangles
Award A1 for two pyramids, A1 for equilateral triangles.
(can be either stated or shown in a sketch – but there must be clear indication the triangles are equilateral)
[6 marks]
## Question
Consider the plane $${\mathit{\Pi} _1}$$, parallel to both lines $${L_1}$$ and $${L_2}$$. Point C lies in the plane $${\mathit{\Pi} _1}$$.
The line $${L_3}$$ has vector equation $$\boldsymbol{r} = \left( \begin{array}{l}3\\0\\1\end{array} \right) + \lambda \left( \begin{array}{c}k\\1\\ – 1\end{array} \right)$$.
The plane $${\mathit{\Pi} _2}$$ has Cartesian equation $$x + y = 12$$.
The angle between the line $${L_3}$$ and the plane $${\mathit{\Pi} _2}$$ is 60°.
a.Given the points A(1, 0, 4), B(2, 3, −1) and C(0, 1, − 2) , find the vector equation of the line $${L_1}$$ passing through the points A and B.[2]
b.
The line $${L_2}$$ has Cartesian equation $$\frac{{x – 1}}{3} = \frac{{y + 2}}{1} = \frac{{z – 1}}{{ – 2}}$$.
Show that $${L_1}$$ and $${L_2}$$ are skew lines.[5]
c.
Find the Cartesian equation of the plane $${\Pi _1}$$.[4]
d.
(i) Find the value of $$k$$.
(ii) Find the point of intersection P of the line $${L_3}$$ and the plane $${\mathit{\Pi} _2}$$.[7]
## Markscheme
direction vector $$\overrightarrow {{\rm{AB}}} = \left( \begin{array}{c}1\\3\\ – 5\end{array} \right)$$ or $$\overrightarrow {{\rm{BA}}} = \left( \begin{array}{c} – 1\\ – 3\\5\end{array} \right)$$ A1
$$\boldsymbol{r} = \left( \begin{array}{l}1\\0\\4\end{array} \right) + t\left( \begin{array}{c}1\\3\\ – 5\end{array} \right)$$ or $$\boldsymbol{r} = \left( \begin{array}{c}2\\3\\ – 1\end{array} \right) + t\left( \begin{array}{c}1\\3\\ – 5\end{array} \right)$$ or equivalent A1
Note: Do not award final A1 unless ‘$$\boldsymbol{r} = {\text{K}}$$’ (or equivalent) seen.
Allow FT on direction vector for final A1.
[2 marks]
a.
both lines expressed in parametric form:
$${L_1}$$:
$$x = 1 + t$$
$$y = 3t$$
$$z = 4 – 5t$$
$${L_2}$$:
$$x = 1 + 3s$$
$$y = – 2 + s$$ M1A1
$$z = – 2s + 1$$
Notes: Award M1 for an attempt to convert $${L_2}$$ from Cartesian to parametric form.
Award A1 for correct parametric equations for $${L_1}$$ and $${L_2}$$.
Allow M1A1 at this stage if same parameter is used in both lines.
attempt to solve simultaneously for x and y: M1
$$1 + t = 1 + 3s$$
$$3t = – 2 + s$$
$$t = – \frac{3}{4},{\text{ }}s = – \frac{1}{4}$$ A1
substituting both values back into z values respectively gives $$z = \frac{{31}}{4}$$
and $$z = \frac{3}{2}$$ so a contradiction R1
therefore $${L_1}$$ and $${L_1}$$ are skew lines AG
[5 marks]
b.
finding the cross product:
$$\left( \begin{array}{c}1\\3\\ – 5\end{array} \right) \times \left( \begin{array}{c}3\\1\\ – 2\end{array} \right)$$ (M1)
= – i – 13j – 8k A1
Note: Accept i + 13j + 8k
$$– 1(0) – 13(1) – 8( – 2) = 3$$ (M1)
$$\Rightarrow – x – 13y – 8z = 3$$ or equivalent A1
[4 marks]
c.
(i) $$(\cos \theta = )\frac{{\left( \begin{array}{c}k\\1\\ – 1\end{array} \right) \bullet \left( \begin{array}{l}1\\1\\0\end{array} \right)}}{{\sqrt {{k^2} + 1 + 1} \times \sqrt {1 + 1} }}$$ M1
Note: Award M1 for an attempt to use angle between two vectors formula.
$$\frac{{\sqrt 3 }}{2} = \frac{{k + 1}}{{\sqrt {2({k^2} + 2)} }}$$ A1
$$4{(k + 1)^2} = 6({k^2} + 2)$$ M1
$${k^2} – 4k + 4 = 0$$
$${(k – 2)^2} = 0$$
$$k = 2$$ A1
Note: Award M1A0M1A0 if $$\cos 60^\circ$$ is used $$(k = 0{\text{ or }}k = – 4)$$.
(ii) $$r = \left( \begin{array}{l}3\\0\\1\end{array} \right) + \lambda \left( \begin{array}{c}2\\1\\ – 1\end{array} \right)$$
substituting into the equation of the plane $${\Pi _2}$$:
$$3 + 2\lambda + \lambda = 12$$ M1
$$\lambda = 3$$ A1
point P has the coordinates:
(9, 3, –2) A1
Notes: Accept 9i + 3j – 2k and $$\left( \begin{array}{l}9\\3\\- 2\end{array} \right)$$.
Do not allow FT if two values found for k.
[7 marks]
d.
## Question
Two planes have equations
${\Pi _1}:{\text{ }}4x + y + z = 8{\text{ and }}{\Pi _2}:{\text{ }}4x + 3y – z = 0$
Let $$L$$ be the line of intersection of the two planes.
B is the point on $${\Pi _1}$$ with coordinates $$(a,{\text{ }}b,{\text{ }}1)$$.
The point P lies on $$L$$ and $${\rm{A\hat BP}} = 45^\circ$$.
a.Find the cosine of the angle between the two planes in the form $$\sqrt {\frac{p}{q}}$$ where $$p,{\text{ }}q \in \mathbb{Z}$$.[4]
b.
(i) Show that $$L$$ has direction $$\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)$$.
(ii) Show that the point $${\text{A }}(1,{\text{ }}0,{\text{ }}4)$$ lies on both planes.
(iii) Write down a vector equation of $$L$$.[6]
c.
Given the vector $$\overrightarrow {{\text{AB}}}$$ is perpendicular to $$L$$ find the value of $$a$$ and the value of $$b$$.[5]
d.
Show that $${\text{AB}} = 3\sqrt 2$$.[1]
e.
Find the coordinates of the two possible positions of $$P$$.[5]
## Markscheme
Note: Throughout the question condone vectors written horizontally.
angle between planes is equal to the angles between the normal to the planes (M1)
$$\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right) = 18$$ (A1)
let $$\theta$$ be the angle between the normal to the planes
$$\cos \theta = \frac{{18}}{{\sqrt {18} \sqrt {26} }} = \sqrt {\frac{{18}}{{26}}} {\text{ }}\left( {{\text{or equivalent, for example }}\sqrt {\frac{{324}}{{468}}} {\text{ or }}\sqrt {\frac{9}{{13}}} } \right)$$ M1A1
[4 marks]
a.
Note: Throughout the question condone vectors written horizontally.
(i) METHOD 1
$$\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 4} \\ 8 \\ 8 \end{array}} \right)$$ M1A1
which is a multiple of $$\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)$$ R1AG
Note: Allow any equivalent wording or $$\left( {\begin{array}{*{20}{c}} { – 4} \\ 8 \\ 8 \end{array}} \right) = 4\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)$$, do not allow $$\left( {\begin{array}{*{20}{c}} { – 4} \\ 8 \\ 8 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)$$.
METHOD 2
let $$z = t$$ (or equivalent)
solve simultaneously to get M1
$$y = t – 4,{\text{ }}x = 3 – 0.5t$$ A1
hence direction vector is $$\left( {\begin{array}{*{20}{c}} { – 0.5} \\ 1 \\ 1 \end{array}} \right)$$
which is a multiple of $$\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)$$ R1AG
METHOD 3
$$\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right) = – 4 + 2 + 2 = 0$$ M1A1
$$\left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right) = – 4 + 6 – 2 = 0$$ A1
Note: If only one scalar product is found award M0A0A0.
(ii) $${\Pi _1}:{\text{ }}4 + 0 + 4 = 8$$ and $${\Pi _2}:{\text{ }}4 + 0 – 4 = 0$$ R1
(iii) r $$= \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 4 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)$$ A1A1
Note: A1 for “r $$=$$” and a correct point on the line, A1 for a parameter and a correct direction vector.
[6 marks]
b.
Note: Throughout the question condone vectors written horizontally.
$$\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} a \\ b \\ 1 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {a – 1} \\ b \\ { – 3} \end{array}} \right)$$ (A1)
$$\left( {\begin{array}{*{20}{c}} {a – 1} \\ b \\ { – 3} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right) = 0$$ M1
Note: Award M0 for $$\left( {\begin{array}{*{20}{c}} a \\ b \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right) = 0$$.
$$– a + 1 + 2b – 6 = 0 \Rightarrow a – 2b = – 5$$ A1
lies on $${\Pi _1}$$ so $$4a + b + 1 = 8 \Rightarrow 4a + b = 7$$ M1
$$a = 1,{\text{ }}b = 3$$ A1
[5 marks]
c.
Note: Throughout the question condone vectors written horizontally.
$${\text{AB}} = \sqrt {{0^2} + {3^2} + {{( – 3)}^2}} = 3\sqrt 2$$ M1AG
[1 mark]
d.
Note: Throughout the question condone vectors written horizontally.
METHOD 1
$$\left| {\overrightarrow {{\text{AB}}} } \right| = \left| {\overrightarrow {{\text{AP}}} } \right| = 3\sqrt 2$$ (M1)
$$\overrightarrow {{\text{AP}}} = t\left( {\begin{array}{*{20}{c}} { – 1} \\ 2 \\ 2 \end{array}} \right)$$ (A1)
$$\left| {3t} \right| = 3\sqrt 2 \Rightarrow t = \pm \sqrt 2$$ (M1)A1
$${\text{P}}\left( {1 – \sqrt 2 ,{\text{ }}2\sqrt 2 ,{\text{ }}4 + 2\sqrt 2 } \right)$$ and $$\left( {1 + \sqrt 2 ,{\text{ }} – 2\sqrt 2 ,{\text{ }}4 – 2\sqrt 2 } \right)$$ A1
[5 marks]
METHOD 2
let P have coordinates $$(1 – \lambda ,{\text{ }}2\lambda ,{\text{ }}4 + 2\lambda )$$ M1
$$\overrightarrow {{\text{BA}}} = \left( {\begin{array}{*{20}{c}} 0 \\ { – 3} \\ 3 \end{array}} \right),{\text{ }}\overrightarrow {{\text{BP}}} = \left( {\begin{array}{*{20}{c}} { – \lambda } \\ {2\lambda – 3} \\ {3 + 2\lambda } \end{array}} \right)$$ A1
$$\cos 45^\circ = \frac{{\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BP}}} }}{{\left| {{\text{BA}}} \right|\left| {{\text{BP}}} \right|}}$$ M1
Note: Award M1 even if AB rather than BA is used in the scalar product.
$$\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BP}}} = 18$$
$$\frac{1}{{\sqrt 2 }} = \frac{{18}}{{\sqrt {18} \sqrt {9{\lambda ^2} + 18} }}$$
$$\lambda = \pm \sqrt 2$$ A1
$${\text{P}}\left( {1 – \sqrt 2 ,{\text{ }}2\sqrt 2 ,{\text{ }}4 + 2\sqrt 2 } \right)$$ and $$\left( {1 + \sqrt 2 ,{\text{ }} – 2\sqrt 2 ,{\text{ }}4 – 2\sqrt 2 } \right)$$ A1
Note: Accept answers given as position vectors.
[5 marks]
## Question
The following system of equations represents three planes in space.
$x + 3y + z = – 1$
$x + 2y – 2z = 15$
$2x + y – z = 6$
Find the coordinates of the point of intersection of the three planes.
## Markscheme
EITHER
eliminating a variable, $$x$$, for example to obtain $$y + 3z = – 16$$ and $$– 5y – 3z = 8$$ M1A1
attempting to find the value of one variable M1
point of intersection is $$( – 1,{\text{ }}2,{\text{ }} – 6)$$ A1A1A1
OR
attempting row reduction of relevant matrix, eg. M1
correct matrix with two zeroes in a column, eg. A1
further attempt at reduction M1
point of intersection is $$( – 1,{\text{ }}2,{\text{ }} – 6)$$ A1A1A1
Note: Allow solution expressed as $$x = – 1,{\text{ }}y = 2,{\text{ }}z = – 6$$ for final A marks.
[6 marks]
## Examiners report
This provided a generally easy start for many candidates. Most successful candidates obtained their answer through row reduction of a suitable matrix. Those choosing an alternative method often made slips in their algebra.
## Question
A line $$L$$ has equation $$\frac{{x – 2}}{p} = \frac{{y – q}}{2} = z – 1$$ where $$p,{\text{ }}q \in \mathbb{R}$$.
A plane $$\Pi$$ has equation $$x + y + 3z = 9$$.
Consider the different case where the acute angle between $$L$$ and $$\Pi$$ is $$\theta$$
where $$\theta = \arcsin \left( {\frac{1}{{\sqrt {11} }}} \right)$$.
Show that $$L$$ is not perpendicular to $$\Pi$$.
[3]
a.
Given that $$L$$ lies in the plane $$\Pi$$, find the value of $$p$$ and the value of $$q$$.
[4]
b.
(i) Show that $$p = – 2$$.
(ii) If $$L$$ intersects $$\Pi$$ at $$z = – 1$$, find the value of $$q$$.
[11]
c.
## Markscheme
EITHER
n $$= \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right)$$ and d $$= \left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)$$ A1A1
and n $$\ne$$ kd R1
OR
n $$\times$$ d $$= \left( {\begin{array}{*{20}{c}} { – 5} \\ {3p – 1} \\ {2 – p} \end{array}} \right)$$ M1A1
the vector product is non-zero for $$p \in \mathbb{R}$$ R1
THEN
$$L$$ is not perpendicular to $$\Pi$$ AG
[3 marks]
a.
METHOD 1
$$(2 + p\lambda ) + (q + 2\lambda ) + 3(1 + \lambda ) = 9$$ M1
$$(q + 5) + (p + 5)\lambda = 9$$ (A1)
$$p = – 5$$ and $$q = 4$$ A1A1
METHOD 2
direction vector of line is perpendicular to plane, so
$$\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) = 0$$ M1
$$p = – 5$$ A1
$$(2,{\text{ }}q,{\text{ }}1)$$ is common to both $$L$$ and $$\Pi$$
either $$\left( {\begin{array}{*{20}{c}} 2 \\ q \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) = 9$$ or by substituting into $$x + y + 3z = 9$$ M1
$$q = 4$$ A1
[4 marks]
b.
(i) METHOD 1
$$\alpha$$ is the acute angle between n and L
if $$\sin \theta = \frac{1}{{\sqrt {11} }}$$ then $$\cos \alpha = \frac{1}{{\sqrt {11} }}$$ (M1)(A1)
attempting to use $$\cos \alpha = \frac{{n \bullet d}}{{\left| n \right|\left| d \right|}}$$ or $$\sin \alpha = \frac{{n \bullet d}}{{\left| n \right|\left| d \right|}}$$ M1
$$\frac{{p + 5}}{{\sqrt {11} \times \sqrt {{p^2} + 5} }} = \frac{1}{{\sqrt {11} }}$$ A1A1
$${(p + 5)^2} = {p^2} + 5$$ M1
$$10p = – 20$$ (or equivalent) A1
$$p = – 2$$ AG
METHOD 2
$$\alpha$$ is the angle between n and L
if $$\sin \theta = \frac{1}{{\sqrt {11} }}$$ then $$\sin \alpha = \frac{{\sqrt {10} }}{{\sqrt {11} }}$$ (M1)A1
attempting to use $$\sin \alpha = \frac{{\left| {n \times d} \right|}}{{\left| n \right|\left| d \right|}}$$ M1
$$\frac{{\sqrt {{{( – 5)}^2} + {{(3p – 1)}^2} + {{(2 – p)}^2}} }}{{\sqrt {11} \times \sqrt {{p^2} + 5} }} = \frac{{\sqrt {10} }}{{\sqrt {11} }}$$ A1A1
$${p^2} – p + 3 = {p^2} + 5$$ M1
$$– p + 3 = 5$$ (or equivalent) A1
$$p = – 2$$ AG
(ii) $$p = – 2$$ and $$z = – 1 \Rightarrow \frac{{x – 2}}{{ – 2}} = \frac{{y – q}}{2} = – 2$$ (A1)
$$x = 6$$ and $$y = q – 4$$ (A1)
this satisfies $$\Pi$$ so $$6 + q – 4 – 3 = 9$$ M1
$$q = 10$$ A1
[11 marks]
c.
## Examiners report
Parts (a) and (b) were often well done, though a small number of candidates were clearly puzzled when trying to demonstrate $$\left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) \ne k\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)$$, with some scripts seen involving needlessly convoluted arguments.
a.
Parts (a) and (b) were often well done, though a small number of candidates were clearly puzzled when trying to demonstrate $$\left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) \ne k\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)$$, with some scripts seen involving needlessly convoluted arguments.
b.
Part (c) often proved problematic, as some candidates unsurprisingly used the sine (or cosine) of an incorrect angle, and few consequent marks were then available. Some good clear solutions were seen, occasionally complete with diagrams in the cases of the thoughtful candidates who were able to ‘work through’ the question rather than just apply a standard vector result.
c.
## Question
Find the coordinates of the point of intersection of the planes defined by the equations $$x + y + z = 3,{\text{ }}x – y + z = 5$$ and $$x + y + 2z = 6$$.
## Markscheme
METHOD 1
for eliminating one variable from two equations (M1)
eg, $$\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ {2x + 2z = 8} \\ {2x + 3z = 11} \end{array}} \right.$$ A1A1
for finding correctly one coordinate
eg, $$\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ {(2x + 2z = 8)} \\ {z = 3} \end{array}} \right.$$ A1
for finding correctly the other two coordinates A1
$$\Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 1} \\ {y = – 1} \\ {z = 3} \end{array}} \right.$$
the intersection point has coordinates $$(1,{\text{ }} – 1,{\text{ }}3)$$
METHOD 2
for eliminating two variables from two equations or using row reduction (M1)
eg, $$\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ { – 2 = 2} \\ {z = 3} \end{array}} \right.$$ or $$\left( {\begin{array}{*{20}{c}} 1&1&1 \\ 0&{ – 2}&0 \\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 3 \\ 2 \\ 3 \end{array}} \right.} \right)$$ A1A1
for finding correctly the other coordinates A1A1
$$\Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 1} \\ {y = – 1} \\ {(z = 3)} \end{array}} \right.$$ or $$\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ 3 \end{array}} \right.} \right)$$
the intersection point has coordinates $$(1,{\text{ }} – 1,{\text{ }}3)$$
METHOD 3
$$\left| {\begin{array}{*{20}{c}} 1&1&1 \\ 1&{ – 1}&1 \\ 1&1&2 \end{array}} \right| = – 2$$ (A1)
attempt to use Cramer’s rule M1
$$x = \frac{{\left| {\begin{array}{*{20}{c}} 3&1&1 \\ 5&{ – 1}&1 \\ 6&1&2 \end{array}} \right|}}{{ – 2}} = \frac{{ – 2}}{{ – 2}} = 1$$ A1
$$y = \frac{{\left| {\begin{array}{*{20}{c}} 1&3&1 \\ 1&5&1 \\ 1&6&2 \end{array}} \right|}}{{ – 2}} = \frac{2}{{ – 2}} = – 1$$ A1
$$z = \frac{{\left| {\begin{array}{*{20}{c}} 1&1&3 \\ 1&{ – 1}&5 \\ 1&1&6 \end{array}} \right|}}{{ – 2}} = \frac{{ – 6}}{{ – 2}} = 3$$ A1
Note: Award M1 only if candidate attempts to determine at least one of the variables using this method.
[5 marks]
[N/A]
## Question
Consider the lines $${l_1}$$ and $${l_2}$$ defined by
$${l_1}:$$ r $$= \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ a \end{array}} \right) + \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)$$ and $${l_2}:\frac{{6 – x}}{3} = \frac{{y – 2}}{4} = 1 – z$$ where $$a$$ is a constant.
Given that the lines $${l_1}$$ and $${l_2}$$ intersect at a point P,
a.find the value of $$a$$;[4]
b.determine the coordinates of the point of intersection P.[2]
## Markscheme
METHOD 1
$${l_1}:$$r $$= \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ a \end{array}} \right) = \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right) \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = – 3 + \beta } \\ {y = – 2 + 4\beta } \\ {z = a + 2\beta } \end{array}} \right.$$ M1
$$\frac{{6 – ( – 3 + \beta )}}{3} = \frac{{( – 2 + 4\beta ) – 2}}{4} \Rightarrow 4 = \frac{{4\beta }}{3} \Rightarrow \beta = 3$$ M1A1
$$\frac{{6 – ( – 3 + \beta )}}{3} = 1 – (a + 2\beta ) \Rightarrow 2 = – 5 – a \Rightarrow a = – 7$$ A1
METHOD 2
$$\left\{ {\begin{array}{*{20}{l}} { – 3 + \beta = 6 – 3\lambda } \\ { – 2 + 4\beta = 4\lambda + 2} \\ {a + 2\beta = 1 – \lambda } \end{array}} \right.$$ M1
attempt to solve M1
$$\lambda = 2,{\text{ }}\beta = 3$$ A1
$$a = 1 – \lambda – 2\beta = – 7$$ A1
[4 marks]
a.
$$\overrightarrow {{\text{OP}}} = \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ { – 7} \end{array}} \right) + 3 \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)$$ (M1)
$$= \left( {\begin{array}{*{20}{c}} 0 \\ {10} \\ { – 1} \end{array}} \right)$$ A1
$$\therefore {\text{P}}(0,{\text{ 10, }} – 1)$$
[2 marks]
b.
## Question
The points A and B are given by $${\text{A}}(0,{\text{ }}3,{\text{ }} – 6)$$ and $${\text{B}}(6,{\text{ }} – 5,{\text{ }}11)$$.
The plane Π is defined by the equation $$4x – 3y + 2z = 20$$.
a.Find a vector equation of the line L passing through the points A and B.[3]
b.Find the coordinates of the point of intersection of the line L with the plane Π.[3]
## Markscheme
$$\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 6 \\ { – 8} \\ {17} \end{array}} \right)$$ (A1)
r = $$\left( {\begin{array}{*{20}{c}} 0 \\ 3 \\ { – 6} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 6 \\ { – 8} \\ {17} \end{array}} \right)$$ or r = $$\left( {\begin{array}{*{20}{c}} 6 \\ { – 5} \\ {11} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 6 \\ { – 8} \\ {17} \end{array}} \right)$$ M1A1
Note: Award M1A0 if r = is not seen (or equivalent).
[3 marks]
a.
substitute line L in $$\Pi :4(6\lambda ) – 3(3 – 8\lambda ) + 2( – 6 + 17\lambda ) = 20$$ M1
$$82\lambda = 41$$
$$\lambda = \frac{1}{2}$$ (A1)
r = $$\left( {\begin{array}{*{20}{c}} 0 \\ 3 \\ { – 6} \end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}} 6 \\ { – 8} \\ {17} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ { – 1} \\ {\frac{5}{2}} \end{array}} \right)$$
so coordinate is $$\left( {3,{\text{ }} – 1,{\text{ }}\frac{5}{2}} \right)$$ A1
Note: Accept coordinate expressed as position vector $$\left( {\begin{array}{*{20}{c}} 3 \\ { – 1} \\ {\frac{5}{2}} \end{array}} \right)$$.
[3 marks]
## Question
The following figure shows a square based pyramid with vertices at O(0, 0, 0), A(1, 0, 0), B(1, 1, 0), C(0, 1, 0) and D(0, 0, 1).
The Cartesian equation of the plane $${\Pi _2}$$, passing through the points B , C and D , is $$y + z = 1$$.
The plane $${\Pi _3}$$ passes through O and is normal to the line BD.
$${\Pi _3}$$ cuts AD and BD at the points P and Q respectively.
a.Find the Cartesian equation of the plane $${\Pi _1}$$, passing through the points A , B and D.[3]
b.Find the angle between the faces ABD and BCD.[4]
c.Find the Cartesian equation of $${\Pi _3}$$.[3]
d.Show that P is the midpoint of AD.[4]
e.Find the area of the triangle OPQ.[5]
## Markscheme
recognising normal to plane or attempting to find cross product of two vectors lying in the plane (M1)
for example, $$\mathop {{\text{AB}}}\limits^ \to \,\, \times \mathop {{\text{AD}}}\limits^ \to = \left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 0 \hfill \\ \end{gathered} \right) \times \left( \begin{gathered} – 1 \hfill \\ \,0 \hfill \\ \,1 \hfill \\ \end{gathered} \right) = \left( \begin{gathered} 1 \hfill \\ 0 \hfill \\ 1 \hfill \\ \end{gathered} \right)$$ (A1)
$${\Pi _1}\,{\text{:}}\,\,x + z = 1$$ A1
[3 marks]
a.
EITHER
$$\left( \begin{gathered} 1 \hfill \\ 0 \hfill \\ 1 \hfill \\ \end{gathered} \right) \bullet \left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 1 \hfill \\ \end{gathered} \right) = 1 = \sqrt 2 \sqrt 2 \,{\text{cos}}\,\theta$$ M1A1
OR
$$\left| {\left( \begin{gathered} 1 \hfill \\ 0 \hfill \\ 1 \hfill \\ \end{gathered} \right) \times \left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 1 \hfill \\ \end{gathered} \right)} \right| = \sqrt 3 = \sqrt 2 \sqrt 2 \,{\text{sin}}\,\theta$$ M1A1
Note: M1 is for an attempt to find the scalar or vector product of the two normal vectors.
$$\Rightarrow \theta = 60^\circ \left( { = \frac{\pi }{3}} \right)$$ A1
angle between faces is $$20^\circ \left( { = \frac{{2\pi }}{3}} \right)$$ A1
[4 marks]
b.
$$\mathop {{\text{DB}}}\limits^ \to = \left( \begin{gathered} \,1 \hfill \\ \,1 \hfill \\ – 1 \hfill \\ \end{gathered} \right)$$ or $$\mathop {{\text{BD}}}\limits^ \to = \left( \begin{gathered} – 1 \hfill \\ – 1 \hfill \\ \,1 \hfill \\ \end{gathered} \right)$$ (A1)
$${\Pi _3}\,{\text{:}}\,\,x + y – z = k$$ (M1)
$${\Pi _3}\,{\text{:}}\,\,x + y – z = 0$$ A1
[3 marks]
c.
METHOD 1
line AD : (r =)$$\left( \begin{gathered} 0 \hfill \\ 0 \hfill \\ 1 \hfill \\ \end{gathered} \right) + \lambda \left( \begin{gathered} \,1 \hfill \\ \,0 \hfill \\ – 1 \hfill \\ \end{gathered} \right)$$ M1A1
intersects $${\Pi _3}$$ when $$\lambda – \left( {1 – \lambda } \right) = 0$$ M1
so $$\lambda = \frac{1}{2}$$ A1
hence P is the midpoint of AD AG
METHOD 2
midpoint of AD is (0.5, 0, 0.5) (M1)A1
substitute into $$x + y – z = 0$$ M1
0.5 + 0.5 − 0.5 = 0 A1
hence P is the midpoint of AD AG
[4 marks]
d.
METHOD 1
$${\text{OP}} = \frac{1}{{\sqrt 2 }},\,\,{\text{O}}\mathop {\text{P}}\limits^ \wedge {\text{Q}} = 90^\circ ,\,\,{\text{O}}\mathop {\text{Q}}\limits^ \wedge {\text{P}} = 60^\circ$$ A1A1A1
$${\text{PQ}} = \frac{1}{{\sqrt 6 }}$$ A1
area $$= \frac{1}{{2\sqrt {12} }} = \frac{1}{{4\sqrt 3 }} = \frac{{\sqrt 3 }}{{12}}$$ A1
METHOD 2
line BD : ( =)$$\left( \begin{gathered} 1 \hfill \\ 1 \hfill \\ 0 \hfill \\ \end{gathered} \right) + \lambda \left( \begin{gathered} – 1 \hfill \\ – 1 \hfill \\ \,1 \hfill \\ \end{gathered} \right)$$
$$\Rightarrow \lambda = \frac{2}{3}$$ (A1)
$$\mathop {{\text{OQ}}}\limits^ \to = \left( \begin{gathered} \frac{1}{3} \hfill \\ \frac{1}{3} \hfill \\ \frac{2}{3} \hfill \\ \end{gathered} \right)$$ A1
area = $$\frac{1}{2}\left| {\mathop {{\text{OP}}}\limits^ \to \, \times \mathop {{\text{OQ}}}\limits^ \to } \right|$$ M1
$$\mathop {{\text{OP}}}\limits^ \to = \left( \begin{gathered} \frac{1}{2} \hfill \\ 0 \hfill \\ \frac{1}{2} \hfill \\ \end{gathered} \right)$$ A1
Note: This A1 is dependent on M1.
area = $$\frac{{\sqrt 3 }}{{12}}$$ A1
[5 marks
## Question
The points A, B, C and D have position vectors a, b, c and d, relative to the origin O.
It is given that $$\mathop {{\text{AB}}}\limits^ \to = \mathop {{\text{DC}}}\limits^ \to$$.
The position vectors $$\mathop {{\text{OA}}}\limits^ \to$$, $$\mathop {{\text{OB}}}\limits^ \to$$, $$\mathop {{\text{OC}}}\limits^ \to$$ and $$\mathop {{\text{OD}}}\limits^ \to$$ are given by
a = i + 2j − 3k
b = 3ij + pk
c = qi + j + 2k
d = −i + rj − 2k
where p , q and r are constants.
The point where the diagonals of ABCD intersect is denoted by M.
The plane $$\Pi$$ cuts the x, y and z axes at X , Y and Z respectively.
a.i.Explain why ABCD is a parallelogram.[1]
a.ii.Using vector algebra, show that $$\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to$$.[3]
b.Show that p = 1, q = 1 and r = 4.[5]
c.Find the area of the parallelogram ABCD.[4]
d.Find the vector equation of the straight line passing through M and normal to the plane $$\Pi$$ containing ABCD.[4]
e.Find the Cartesian equation of $$\Pi$$.[3]
f.i.Find the coordinates of X, Y and Z.[2]
f.ii.Find YZ.[2]
## Markscheme
a pair of opposite sides have equal length and are parallel R1
hence ABCD is a parallelogram AG
[1 mark]
a.i.
attempt to rewrite the given information in vector form M1
ba = cd A1
rearranging d − a = c − b M1
hence $$\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to$$ AG
Note: Candidates may correctly answer part i) by answering part ii) correctly and then deducing there
are two pairs of parallel sides.
[3 marks]
a.ii.
EITHER
use of $$\mathop {{\text{AB}}}\limits^ \to = \mathop {{\text{DC}}}\limits^ \to$$ (M1)
$$\left( \begin{gathered} 2 \hfill \\ – 3 \hfill \\ p + 3 \hfill \\ \end{gathered} \right) = \left( \begin{gathered} q + 1 \hfill \\ 1 – r \hfill \\ 4 \hfill \\ \end{gathered} \right)$$ A1A1
OR
use of $$\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to$$ (M1)
$$\left( \begin{gathered} – 2 \hfill \\ r – 2 \hfill \\ 1 \hfill \\ \end{gathered} \right) = \left( \begin{gathered} q – 3 \hfill \\ 2 \hfill \\ 2 – p \hfill \\ \end{gathered} \right)$$ A1A1
THEN
attempt to compare coefficients of i, j, and k in their equation or statement to that effect M1
clear demonstration that the given values satisfy their equation A1
p = 1, q = 1, r = 4 AG
[5 marks]
b.
attempt at computing $$\mathop {{\text{AB}}}\limits^ \to \, \times \mathop {{\text{AD}}}\limits^ \to$$ (or equivalent) M1
$$\left( \begin{gathered} – 11 \hfill \\ – 10 \hfill \\ – 2 \hfill \\ \end{gathered} \right)$$ A1
area $$= \left| {\mathop {{\text{AB}}}\limits^ \to \, \times \mathop {{\text{AD}}}\limits^ \to } \right|\left( { = \sqrt {225} } \right)$$ (M1)
= 15 A1
[4 marks]
c.
valid attempt to find $$\mathop {{\text{OM}}}\limits^ \to = \left( {\frac{1}{2}\left( {a + c} \right)} \right)$$ (M1)
$$\left( \begin{gathered} 1 \hfill \\ \frac{3}{2} \hfill \\ – \frac{1}{2} \hfill \\ \end{gathered} \right)$$ A1
the equation is
r = $$\left( \begin{gathered} 1 \hfill \\ \frac{3}{2} \hfill \\ – \frac{1}{2} \hfill \\ \end{gathered} \right) + t\left( \begin{gathered} 11 \hfill \\ 10 \hfill \\ 2 \hfill \\ \end{gathered} \right)$$ or equivalent M1A1
Note: Award maximum M1A0 if ‘r = …’ (or equivalent) is not seen.
[4 marks]
d.
attempt to obtain the equation of the plane in the form ax + by + cz = d M1
11x + 10y + 2z = 25 A1A1
Note: A1 for right hand side, A1 for left hand side.
[3 marks]
e.
putting two coordinates equal to zero (M1)
$${\text{X}}\left( {\frac{{25}}{{11}},\,0,\,0} \right),\,\,{\text{Y}}\left( {0,\,\frac{5}{2},\,0} \right),\,\,{\text{Z}}\left( {0,\,0,\,\frac{{25}}{2}} \right)$$ A1
[2 marks]
f.i.
$${\text{YZ}} = \sqrt {{{\left( {\frac{5}{2}} \right)}^2} + {{\left( {\frac{{25}}{2}} \right)}^2}}$$ M1
$$= \sqrt {\frac{{325}}{2}} \left( { = \frac{{5\sqrt {104} }}{4} = \frac{{5\sqrt {26} }}{2}} \right)$$ A1
[4 marks]
f.ii.
### Question
Find a vector that is normal to the plane containing the lines $$L_1$$ , and $$L_2$$ , whose equations are:
$$L_1 : r = i + k + \lambda (2i + j – 2k)$$ $$L_2: r = 3i + 2j + 2k + \mu (j + 3k)$$
A vector that is normal to the plane is given by the vector product $$d1 × d2$$ where $$d1$$ and $$d2$$ are the direction vectors of the lines $$L_1$$ and $$L_2$$ respectively.
$$d1 × d2 = \begin{vmatrix} i& j& k\\ 2 & 1 & -2\\ 0 & 1 & 3 \end{vmatrix}$$
$$= 5i – 6j + 2k$$ (or any multiple)<|endoftext|>
| 4.625 |
1,485 |
## Surface Area and Volumes – Exercise 13.7 – Class IX
Assume π = 22/7 , unless stated otherwise.
1. Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm
Solution:
(i) r = 6 cm; h = 7cm
Volume of the cone = 1/3πr2h
= 1/3 x 22/7 x 62 x 7
= 264 cm3
(ii) r = 3.5 cm ; h = 12 cm
Volume of the cone = 1/3πr2h
= 1/3 x 22/7 x 3.52 x 12
= 154 cm3
1. Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm
Solution:
(i) r = 7cm , l = 25 cm
h2 = l2 – r2
= 252 – 72
= 576
h = 24cm
Volume of conical vessel = 1/3πr2h
= 1/3 x 22/7 x 72 x 24
= 1232 cm3
The capacity of conical vessel in litres = 1232/1000 = 1.232 litres
(ii)h = 12 cm , l = 13 cm
h2 = l2 – r2
r2 = l2 – h2
= 132 – 122
= 25
r = 5
Volume of conical vessel = 1/3πr2h
= 1/3 x 22/7 x 52 x 12
= 314.29 cm3
The capacity of conical vessel in litres = 314.29/1000 = 0.31429 litres
1. The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base.(Use π = 3.14)
Solution:
h = 15 cm
Volume of cone = 1570 cm3
We know, volume of the cone = 1/3πr2h
1570 = 1/3 x 3.14 x r2 x 15
1570 = 15.7 x r2
r2 = 1570/15.7 = 100
r = 10 cm
1. If the volume of a right circular cone of height 9 cm is 48 π cm3, find the diameter of its base.
Solution:
h = 9 cm
Volume of the right circular cone = 48π cm3
We know, volume of the cone = 1/3πr2h
48 π =1/3 x π x r2 x 9
48 = 1/3 x r2 x 9
r2 = 48/3 = 16
r = 4 cm
Therefore, diameter of right circular cone, d = 2r = 2 x 4 = 8 cm
1. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Solution:
d = 3.5 m ⇒ r = d/2 = 3.5/2 = 7/4 m
h = 12 m
Volume of conical pit = 1/3πr2h
= 1/3 x 22/7 x (7/4)2 x 12
= 38.5 m3
Therefore, the capacity of conical pit in litres = 38.5/1000 = 0.0385 litres
1. The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone
Solution:
(i) Volume of right circular cone = 9856 cm3
d = 28 cm ⇒r = 28/2 = 14 cm
Volume of conical pit = 1/3πr2h
9856 = 1/3 x 22/7 x 142 x h
h = 9856x3x7/22×142 = 48 cm
(ii) h2 = l2 – r2
482 = l2 – 142
l2 = 482 + 142
l2 = 2500
l = 50 cm
(iii) Curved surface area of the cone = πrl
= 22/7 x 14 x 50
= 2200 cm2
1. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Solution:
r = 5 cm, h = 12 cm , l = 13 cm
Volume of the cone = 1/3 πr2h
= 1/3 x π x 52 x 12
= 100 π cm3
1. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Solution:
r = 12 cm, h = 5 cm
Volume of the cone = 1/3 πr2h
= 1/3 x π x 122 x 5
= 240 π cm3
Required ratio = 100π/240π= 5/12 = 5 : 12
1. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Solution:
d = 10.5 m
r = d/2 = 10.5/2 = 5.25 m
h = 3 m
Volume of the cone = 1/3 πr2h
= 1/3 x π x (5.25)2 x 3
= 86.625 cm3
l = √(h2 + r2) = √[32+(5.25)2] = √[9+27.5625] = √(36.5625) = 6.05 m
Curved surface area of the cone = πrl
= 22/7 x 5.25 x 6.05 m2
= 99.825 m2
Therefore, 99.825m2 of canvas is needed.<|endoftext|>
| 4.46875 |
1,242 |
BRINGING IN THE HOME CULTURE AND LANGUAGE
As educators prepare their programs, they need to constantly be thinking of how to enhance the children’s language. One way to do this is to incorporate the home languages into the setting (e.g., by adding posters or alphabets in various languages or by using music from various cultures). Families are a great resource for this. Libraries also provide some variety. When children recognize the script or sound of their first language, it can increase their comfort. It also shows parents that their home language is valued and provides a way for educators to demonstrate their openness to learning a few words or phrases in a child’s language. Cultural and religious celebrations can also provide educators with opportunities to learn about and share the families’ cultures. Frequently celebrated days in Canada can also be celebrated (e.g., Family Day, Canada Day). This can provide new language stimulation but should not be the core for program planning.
The program set up can be adapted to add a variety of play spaces where children can interact in small groups, as well as areas where they can retreat from language overload and engage in play without the risk of unwanted interference. Activities where chil- dren work together, arranging chairs in a cozy cor- ner for two, or bringing in special objects grouped in unique ways for discussion (e.g., a collection of different footwear) can also encourage language development.
LANGUAGE ADAPTATIONS FOR ESL LEARNERS
When activities are based on children’s interests, there is more likelihood of language being used and retained. For example, children may want to know how to say “fast” because they are excited about which car could go fastest down various ramps. When they notice a missing wheel, they learn the word “broken.” With younger children, try labeling actions (e.g., “Your car went fast.”) If this is accompanied by a zooming sound or arm action, it is even more appealing. Make the word or phrase roll off your tongue in an inviting way. “Bounce the ball,” can be said in a bouncy voice, for example.
When children feel listened to, they are more likely to listen to others. Model listening by giving your full attention to a child. Talk with children having them face you, away from distractions. Do a full squat to get down to their level and establish eye contact if the child is comfortable with this. Repeat some of what you heard to make sure you got it right.
Listening activities might include:
- games where the name of an object is important (e.g., hiding games, “Where is…?”);
- sound effects recordings where children match sounds to pictures of objects;
- soothing music at specific times of the day;
- singing—making up songs using a child’s name, for example.
Here are some other tips for helping children learn language:
Simplify your speech and add gestures.
When introducing a new language, simplify your speech; provide one concept at a time; add gesture, animation and props to make your message clearer.
Use consistent words and teach children important phrases.
Instead of using different words to describe the same thing (e.g., toilet, bathroom or washroom) pick one and use it consistently. It also helps to provide children with simple phrases that can help them get their needs met (e.g., “I want____” or “Can I have _____?”) and phrases that help them to socialize (e.g., “No,” “Mine,” “Don’t touch,” and “Can I play?”)
Be concrete and model simple speech.
Be concrete with young children. Phrases like “use your words” are confusing, especially when the child may not know the words. Model ways children can enter play or share a toy to encourage them to try these strategies. If a child comes to you for help because someone grabbed their toy, show them how to get it back (e.g., by saying “It’s my toy” or “Mine.”) When you add gestures, such as the stopping gesture or pointing to yourself when saying “Mine,” the meaning becomes clearer.
Match your language to the child.
Observe the child interacting with the parent and look for the tone of voice, the rhythm of the home language, the volume and any non-verbal communication. By altering your volume, tone and pitch to more closely match the home language, you can make it easier for the child to listen.
Increase language requirements gradually and read children’s cues.
At first it is good to respond to children’s nonverbal cues, such as grunting or pointing, but over time, you can model the appropriate language. Eventually you can ask open-ended questions like, “What do you want?” Children show understanding in different ways including doing the action, nodding or smiling. It is important to look for their readiness through their responses. If children watch you as you are speaking with others or if you hear them singing songs from group time, then they are ready for more language.
Help children communicate with one another.
It is surprising how children can carry on conversations when they don’t share the same language. They use gestures and animation, model their play and use shared understandings. It is good to let children communicate in this way; however, when children get frustrated that another child can’t understand them, provide the English words to help.
Julie Dotsch is an ECE Diversity consultant for her company One World. She is well known in the community for her interactive workshops and her specialized knowledge of immigrant preschoolers and their families. Julie can be contacted at [email protected].<|endoftext|>
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# Simple Thevenin and Norton Equivalent AC Circuits
Thevenin’s and Norton’s theorem are applied to ac circuits in the same way as they are to dc circuit.
Contents
The only additional effort arises from the need to manipulate complex numbers.
Make sure to read what is ac circuit first.
Learning Kichhoff’s laws for AC circuit will lead us to:
## Thevenin’s and Norton’s Theorem for AC Circuit
The frequency-domain version of a Thevenin equivalent circuit is drawn in Figure.(1), where a linear circuit is replaced by a voltage source in series with an impedance.
The Norton equivalent circuit is depicted in Figure.(2), where a linear circuit is replaced by a current source in parallel with an impedance.
Figure 1. Thevenin equivalent
Figure 2. Norton equivalent
Keep in mind that the two equivalent circuits are related as
just as in source transformation, VTh is the open-circuit voltage while IN is the short circuit current.
If the circuit has sources operating at different frequencies (will be shown in the example below), the Thevenin or Norton equivalent circuit has to be determined at each frequency.
This leads to entirely different equivalent circuits, one for each frequency, not one equivalent circuit with equivalent sources and equivalent impedances.
We will not cover the ‘step-by-step’ using these methods, make sure you learn it from the previous posts about Thevenin’s Theorem and Norton’s Theorem for dc circuit.
It is not that different from the ac circuit.
## Thevenin and Norton Equivalent AC Circuit Examples
For a better understanding let us review the examples below.
1. Obtain the Thevenin equivalent at terminals a-b of the circuit in Figure.(3).
Figure 3
Solution :
We find ZTh by setting the voltage source to zero. As shown in Figure.(4a), the 8 Ω resistance is now in parallel with -j6 reactance, so that their combination gives
Similarly, the 4 Ω resistance is in parallel with the j12 reactance, and their combination gives
Figure 4. Solution for Figure.(3) : (a) finding ZTh, (b) finding VTh
The Thevenin impedance is the series combination of Z1 and Z2; that is,
To find VTh, consider the circuit in Figure.(4b). Currents I1 and I2 are obtained as
Applying KVL around the loop bcdeab in Figure.(4b) gives
or
2. Find the Thevenin equivalent of the circuit in Figure.(5) as seen from terminals a-b.
Figure 5
Solution :
To find VTh, we apply KCL at node 1 in Figure.(6a).
Applying KVL to the loop on the right-hand side in Figure.(6a), we obtain
or
Hence, the Thevenin voltage is
Figure 6. The solution of Figure.(5) : (a) finding VTh, (b) finding ZTh
To obtain ZTh, we remove the independent source.
Due to the presence of the dependent current source, we connect a 3 A current source (3 is an arbitrary value chosen for convenience here, a number divisible by the sum of currents leaving the node) to terminals a-b as shown in Figure.(6b). At the node, KCL gives
Applying KVL to the outer loop in Figure.(6b) gives
The Thevenin impedance is
3. Obtain current Io in Figure.(7) using Norton’s theorem.
Figure 7
Solution :
Our first objective is to find the Norton equivalent at terminals a-bZN is found in the same way as ZTh.
We set the sources to zero as shown in Figure.(8a). As evident from the figure, the (8 – j2) and (10 + j4) impedances are short-circuited, so that
To get IN, we short-circuit terminals a-b as in Figure.(8b) and apply mesh analysis.
Notice that meshes 2 and 3 form a supermesh because of the current source linking them. For mesh 1,
(3.1)
Figure 8. Solution of the circuit in Figure.(7) : (a) finding ZN, (b) finding VN, (c) calculating Io
For the supermesh,
(3.2)
At node a, due to the current source between meshes 2 and 3,
(3.3)<|endoftext|>
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Estimating Sheet
# Stopping Sight Distance Calculation
Construction Software
In this civil engineering article, you will learn the calculation process for sight distance in a highway as highway geometric design.
Sight distance belongs to the distance of highway a driver requires that can be clearly visible to a driver.
The sight distance is also described as the length of road or street that is far ahead to be viewed by the driver.
While making design of a highway or road, the following types of sight distances are commonly found :-
1) Stopping sight distance
2) Overtaking sight distance
3) Intersection sight distance
Stopping sight distance – Stopping sight distance is the sum of reaction and braking distance.
It is defined as follow :-
D = D1 + D2
Here D denotes stopping, D1 reaction sight distance and D2 is the braking distance.
The value of D1 and D2 can be measured with the following formula :-
D1 = 0.28Vt
D2 = 0.01V2
or, D2 = (0.28V)2 / 2gɳ
V refers to the speed or velocity of the vehicle. It is taken as design speed in KM/hrs.
T refers to the time in seconds.
G refers to the gravitational acceleration.
ɳ refers to the brake efficiency.
### Also Read: How to measure different types of curves in roads
The braking sight can also be measured with another formula given below :-
D2 = V2 / 100 ± 2.5 S; S denotes the gradient of road.
To clearly explain it, the solution is given for the following problem :-
Calculate the stopping sight distance of a highway for which the design speed is 65 km/hrs. Suppose, the brake efficiently is 40% and the driver reaction time is 3 seconds.
We know, stopping sight distance = Reaction distance + braking distance
D = D1 + D2
D = 0.28Vt + (0.28 V)2 / 2gɳ
D = 0.28(65)(3) + (0.28 x 65)2 / (2)(9.81)(0.40)[velocity = 65 km/hour; t = 3 seconds; brake efficiency = 40%]
D = 54.6 + 42.2 = 96.8 meters
So, 96.8 meters should always be noticeable to the driver for risk-free travel.
Video Source: SL Khan<|endoftext|>
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# 28 Creative Area and Perimeter Activities
From making a graph paper “person” to making your own mosaic.
As young kids learn about shapes and measurement, they soon enough come to measure the perimeter and area of those shapes. There are shapes all around us. In fact, most of us live in homes that are made of a lot of rectangles. These rectangles, like rooms, are often measured and perimeters and areas are calculated for buying rugs or flooring. The same is true for painting those rectangular walls. So, calculating area and perimeter is a pretty practical skill. Here are some interesting area and perimeter activities and explorations to help students learn about perimeter and area in hands-on, motivating ways.
# Area and Perimeter Activities
## 1. Tiling Area and Perimeter Chart
As a first exploration with both perimeter and area, put out a set of square counting tiles or even real tiles. Circle students up, give them six tiles each, and have them build along with you. First, put out four tiles and explain how this is a model of a floor. Explain how we count the outside of the tile’s edges to determine the perimeter. Then show how we count the tiles inside to determine the area. Draw a picture of this on chart paper and label “area” and “perimeter.”
Let students experiment with their tiles, making different kinds of rectangles and figuring out their areas and perimeters. Have them share back what they find, and add these ideas to the chart. This can become a good class reference as you continue working with these concepts.
## 2. Graph Paper Person
Graph paper is great for exploring area and perimeter because the squares are easily counted. Give students colored pencils and/or crayons and a piece of graph paper. Have them create a “drawing” of a person using only colored-in squares on the graph paper. Then ask them to determine the perimeter and area of their person and write both on the paper along with the calculations.
## 3. Sticky Note Measurement
Provide students with pads of colorful sticky notes and drawing paper. Ask them to create interesting designs or shapes using this colorful material. After some free exploration, direct them to clear the paper and create a design according to the perimeter and area you give them. You might ask for a rectangle that has a perimeter of 12 or a square that has an area of 16. After a while, let students make their own and then calculate their area and perimeter. Then take turns sharing the area and perimeter and see if the rest of the class can make the same thing.
## 4. Math Letter Mosaic
Break out the sticky notes again. Explain to students that they can create a mosaic using this material on a piece of large drawing paper. However, you will give them a capital letter to create, like an L or R. Once they do this, they measure and record the perimeter and area on a separate sheet of paper. It’s great if you can assign letters that will spell out a sentence. These make a great bulletin board or class sign.
## 5. LEGO Land
It seems like everyone loves LEGO and has for decades! Provide a set of LEGO to students and have them freely explore building flat designs. Then in a box or bag, put a set of cards with areas and perimeters written on them. Students take turns pulling a card, and using their LEGO bricks, they build a flat design with that exact perimeter and area.
## 6. Floor Tile–Style Field Trip
Take a mini field trip to another part of the school. Bring along yardsticks, rulers, measuring tapes, and painter’s tape. Many school floors in classrooms, cafeterias, and hallways are made of linoleum tiles. Partner students up and give them a roll of painter’s tape. Take the class to a large area to work together. Partners should create an interesting shape with the tiles and tape. Have them then calculate the area and perimeter with each floor tile being a unit of measure. You might not need the measuring tools, but you have them just in case.
## 7. Dream House Room Design
Barbie had a dream house! Maybe you do too. Provide each student with graph paper, pencil, scissors, a glue stick, and colored paper. Have them each draw a rectangle that represents a room in their dream house. Then cut out colored paper to represent furniture like tables, beds, sofas, and chairs. Arrange these in the room and glue them down. On each piece, write the perimeter and area of that piece as well as the perimeter and area of the room. You can add more rooms each session to build a total dream house floor plan.
## 8. Perimeter and Area Museum
Using tiles, cubes, sticky notes, attribute blocks, and any other manipulatives, explain to students that they will be building an area-and-shape museum on their desks. They should have at least five exhibits at the museum, each one showing a different perimeter or area. Use small pieces of construction paper and write the measurements on them. Fold them in half and stand them up as exhibit signs. Finally, have students take a museum walk and visit all the exhibits.
## 9. Perimeter and Area Hunt
Kids love scavenger hunts. Here’s one that helps them practice perimeter and area measures. Have each student make a flat model using multilink cubes according to a perimeter and area you give them individually. Then take these models and place them around the room, labeling each one with a letter. Students get a piece of lined paper, pencil, and clipboard and go around the room determining the perimeter and area of each model. They write it next to the corresponding letter on their paper. When done, review them together while holding the model up as a reference.
## 10. Vegetable Garden Planner
Give each student a piece of graph paper, colored pencils or crayons, and a pencil. Discuss what kinds of vegetables are grown in gardens. Ask students to use their graph paper to plan out a garden with at least five different things growing there. Determine how much of the garden they want to have for each vegetable, and draw and color in the areas for each. At the bottom of the plan, have a key that tells what each color means and how big that part of the garden is in terms of area and perimeter.
## 11. Geo Board Explore
Geo boards are a great way to explore shapes and measurements. Give each student a geo board and a pack of rubber bands. Begin by allowing students free exploration for five minutes. Then ask them to create a shape and calculate the perimeter and area. Finally, students take turns giving measurements to the class of the shape they made, like, “Make a triangle with a perimeter of 16,” or “Make a rectangle with an area of 8.” Everyone tries it on their geo boards and holds them up to show their answers.
## 12. How Many Rectangles Can You Make?
Here’s one of the area and perimeter activities that can stretch some thinking. Pair up students and give them 16 tiles. Explain, “You have 16 tiles. I want you to find how many different rectangles you can make using no more or no less than 16 tiles each time.” Have partners record these on a piece of graph paper, numbering them, and writing the perimeter and area of each. Share together at the end of the session.
## 13. Draw a Card, Draw a Rectangle
On a set of blank index cards, write a series of task cards. Each card should list a perimeter, an area, or an area and a perimeter. Give each student a piece of graph paper and a pencil. Ask them to choose a card from the deck. They should draw whatever the card specifies on the paper. When done, have it checked, and then students can trade cards to try some others.
## 14. How Big Are the Books?
Classrooms are full of books. Most are rectangles. Give each student lined paper, a pencil, and a ruler. For these area and perimeter activities, explain that they are to go around the room and select three books. They then write the titles down and measure the perimeter and area of each, recording it next to the title. Share back to see which book had the biggest and smallest perimeter and area. To extend this, send students two at a time to the library to find even bigger or smaller books to measure.
## 15. Tangram Time
Tangrams are always fun to play with. For these area and perimeter activities, print them out and let students explore and create for five minutes. Then focus back and have students create a tangram design on a piece of graph paper. Trace it, color it, and then find its perimeter and area. Suggest that students create at least two designs that are very different from each other.
## 16. The Area of My Pizza
Draw a couple of different-size circles on the board. Explain how you can find the area of a circle using the formula Area = 3.14 (pi) x radius (r)squared. Demonstrate how to measure the radius, and then use a calculator to do the operations in the formula. Using lids from cans or a compass, have students make at least two circular pizzas on a piece of drawing paper and with crayons “top” their pizzas with things like sausage, peppers, and so on. After decorating the pizzas, they should use a ruler and calculator to determine the area of each one and write it underneath the pizzas.
## 17. Graph Paper Names
Kids are always interested in their own names. Give students graph paper, colored pencils, and/or crayons. Explain that they should write their names in block letters using graph paper and color them in. Once each student completes their area and perimeter activities, ask them to calculate the area and perimeter of each letter and write that information under each letter.
## 18. Mystery Box
Put a collection of flat objects like books, notebooks, file folders, brochures, maps, box lids, stamps, stickers, and so on in a cardboard box. Students close their eyes and reach into the box and pull out an object. They take the object to their desks, write down the name of the object on a piece of lined paper, and then measure its perimeter and area, recording it next to its name. When done, return the object to the box and pick a new one. At the end, have students share their measurements for each object as you name them one by one.
## 19. Same/Same
Give each student a piece of graph paper and a pencil. The simple challenge is this: Can you draw two shapes that have the same area and perimeter but are different? Is it possible? How many can you make? Students draw these and share them on a bulletin board at the end. This is a great exploration no matter the results. Everyone will be thinking hard and measuring away!
## 20. Number Card Design
Have students pick two number cards from a deck of cards and then multiply them to determine the area. Now they take a piece of graph paper and try to draw a shape with that exact area. You can do the same with perimeter by having them pick two cards, one of which will show width and the other length.
## 21. Run Out of Room
Pair students up and give them crayons, graph paper to share, a pencil, and a pair of dice or number cubes. Each player chooses a color crayon to use. Players take turns rolling the dice. The two numbers that come up must be multiplied to create an area. The player then takes the color they chose and draws and colors in a shape of that area on the graph paper. The next player does the same thing. Players keep repeating this procedure until they can no longer fit any shapes on the paper. They add up the areas of their shapes and the player with the most area covered is the winner.
## 22. Land Rush
Stake your claim! There’s a land rush out in Area Acres. It’s a small place—in fact, only ants live there. You can stake your claim, but you can’t have a parcel of land with an area of more than 24 square centimeters or less than 12 square centimeters. Those are the rules. Using centimeter graph paper, have students draw two parcels of land that meet the requirements, and record their areas and perimeters.
## 23. Cheez-It Designs
Cheez-Its are tasty and they are also perfect little squares, which means you can tile with them and create some rectangles or other shapes. Give each student a couple of handfuls of crackers and a paper towel to build on. How motivating, especially if after you determine the perimeter and area, you get to eat them!
## 24. BIG Rectangles
Pair students up and give them some colored chalk. Have them draw a fairly big rectangle on the playground. Provide measuring tapes and ask them to work together to measure the perimeter and write it under the rectangle. Strategize on how they might determine the area. They could draw a grid on top and try to count squares and partial squares, adding them up to get a final answer, or just multiply the length by the width.
## 25. Magic Carpets
Here’s an opportunity to measure something a little larger. Bring in some rugs or carpet samples from a local carpet store. Pair up students and give them rulers, yardsticks, and tape measures. Tape a piece of paper with a letter on each rug. Spread these out around the room. Assign a letter for each pair to start at. They should use their measuring tools to discover the perimeter and area of each rug and record it on a piece of lined paper. When done with one, they rotate to the next until they’ve measured all the rugs and shared their findings.
## 26. Self-Portrait Picture Frame Perimeter
Bring in a framed picture or poster. Show students how you would measure the perimeter using a ruler or measuring tape. Give students a piece of drawing paper and ask them to draw a perimeter frame with a pattern design on it. Inside the frame, they should draw a self-portrait or a portrait of someone they admire. They then measure the perimeter of the frame and record it on the back of the paper.
## 27. Centimeters and Inches at the Pool
Give each student a ruler that has centimeters and inches on it or two rulers, one with centimeters and the other with inches. Ask students to draw a bird’s-eye view of a swimming pool on a piece of drawing paper and color it in. They can add deck chairs, umbrellas, tables, and so on around the pool. When finished, have them measure the perimeter and area in both inches and centimeters and record it on the same paper. Have students compare and discuss the differences in the numbers.
## 28. Design Estimation
We rarely carry measurement tools around with us—instead we mostly we estimate. Here’s one of the easy area and perimeter activities to get some practice with that.
Draw a fairly large, closed shape with straight lines on the board. Ask the class to take out whiteboards and markers, estimate the perimeter and area of your drawing, and record it. They can come up to the board to get a closer look, but they can’t use a measuring tool or touch the board. Once they have an estimate, they should share it back and explain their thinking.
Next, take a ruler and measure a couple of the lines of your drawing and tell the class the measurements. Based on the new information, they can revise their estimates or leave as is. Finally, ask students to take turns coming up to help measure the rest of the design and get a final answer.<|endoftext|>
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Introduction to Lightning
Lightning is one of the oldest observed natural phenomena on earth. At the same time, it also is one of the least understood. While lightning is simply a gigantic spark of static electricity (the same kind of electricity that sometimes shocks you when you touch a doorknob), scientists do not have a complete grasp on how it works, or how it interacts with solar flares impacting the upper atmosphere or the earth's electromagnetic field.
Lightning has been seen in volcanic eruptions, extremely intense forest fires, surface nuclear detonations, heavy snowstorms, and in large hurricanes. However, it is most often seen in thunderstorms. In fact, lightning (and the resulting thunder) is what makes a storm a thunderstorm.
At any given moment, there can be as many as 2,000 thunderstorms occurring across the globe. This translates to more than 14.5 MILLION storms each year. NASA satellite research indicated these storms produce lightning flashes about 40 times a second worldwide.
This is a change from the commonly accepted value of 100 flashes per second which was an estimate from 1925. Whether it is 40, 100, or somewhere in between, we live on an electrified planet.<|endoftext|>
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Distributive property of Multiplication over Addition
Formula
$a \times (b+c)$ $\,=\,$ $a \times b + a \times c$
An arithmetic property that distributes the multiplication across the addition is called the distributive property of multiplication over addition.
Introduction
$a$, $b$ and $c$ are three literals and represent three terms.
The product of the term $a$ and the sum of the terms $b$ and $c$ is written in mathematical form as follows.
$a \times (b+c)$
The product of them can be evaluated by distributing the multiplication over the addition.
$\implies$ $a \times (b+c)$ $\,=\,$ $a \times b + a \times c$
This distributive property can also be used to distribute the multiplication of a term over the sum of two or more terms.
$\implies$ $a \times (b+c+d+\ldots)$ $\,=\,$ $a \times b + a \times c + a \times d + \ldots$
Proof
Learn how to prove the distributive property of multiplication across addition in algebraic form by geometric method.
Verification
$2$, $3$ and $4$ are three numbers. Find the product of number $2$ and sum of the numbers of $3$ and $4$.
$2 \times (3+4)$
Find the value of this arithmetic expression.
$\implies$ $2 \times (3+4)$ $\,=\,$ $2 \times 7$
$\implies$ $2 \times (3+4) \,=\, 14$
Now, find the sum of the products of $2$ and $3$, and $2$ and $4$.
$2 \times 3 + 2 \times 4$ $\,=\,$ $6+8$
$\implies$ $2 \times 3 + 2 \times 4$ $\,=\,$ $14$
Now, compare the results of both expressions. They are equal.
$\,\,\, \therefore \,\,\,\,\,\,$ $2 \times (3+4)$ $\,=\,$ $2 \times 3 + 2 \times 4$ $\,=\,$ $14$
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Until 1883 a Chicagoan asked to tell what time it was could give more than one answer and still be correct. There was local time, determined by the position of the sun at high noon at a centrally located spot in town, usually City Hall. There was also railroad time, which put Columbus, Ohio, six minutes faster than Cincinnati and 19 minutes faster than Chicago. Scattered across the country were 100 different local time zones, and the railroads had some 53 zones of their own. Typically, a traveler journeying from one end of the U.S. to another would have to change his watch at least 20 times.
To do away with the inevitable confusion, the railroads took the matter into their own hands, holding a General Time Convention in the fall of 1883 at the Grand Pacific Hotel at LaSalle Street and Jackson Boulevard, the site of the present Continental Bank Building. Its purpose: to develop a better and more uniform system of railroad scheduling. The convention secretary, William F. Allen, editor of the Official Railway Guide in New York, proposed that four equal time zones be established across the country. Five were actually adopted-Intercolonial (now Atlantic, including Nova Scotia and New Brunswick in Canada), Eastern, Central, Mountain and Pacific.
The public, for the most part, welcomed the change, drastic as it was. Professional organizations, including the American Meteorological Society, American Geographical Society, Canadian Institute, International Geographical Congress in Venice, Italy, and Imperial Academy of Sciences in St. Petersburg, Russia, approved the plan as a necessary and long-overdue step to relieve confusion and restore efficiency to the country`s massive network of railroads.
The Standard Time System-based on the mean solar time at the central meridian of each time zone-was formally inaugurated on Sunday, Nov. 18, 1883, a day that came to be known as the ''Day of Two Noons.'' At 11:45 a.m., Chicago time, conductors and engineers in the West Side offices of the Pennsylvania, the Burlington, the Panhandle and the Alton rail companies pulled out their timepieces and watched intently as the minutes ticked away. Every railroad clock across the country was to be synchronized that day at high noon. At exactly noon Standard Time, or 9 minutes and 32 seconds past 12 (old) Chicago time, the Allegheny Observatory at the University of Pittsburgh, which kept the ''official'' clock for many of the railroad lines, transmitted a telegraph signal by which railroad clocks across the land were reset according to their particular time zones.
Almost immediately, federal, state and local governments adopted the new system. In Chicago, City Hall quickly conformed to the Standard Time system, as did the post office, public schools, police stations, firehouses and the Chicago Board of Trade. The Western Electric Manufacturing Co. on East Kinzie Street began furnishing Standard Time over private wires to jewelers and railroad lines. Still unsure about the new system, the company sent out inquiries to subscribers to see how they felt about it. The response was overwhelmingly positive.
Jewelers and watchmakers, upon whom corporations and private citizens depended for accurate timekeeping, were the most supportive, placing signs in their store windows to remind passersby to ''set back nine minutes.'' Groups of people would gather in front of jewelers` windows, stop, look at the clocks and correct their watches. All along State Street, Chicagoans greeted one another with, ''Have you the new time?''
There were a few holdouts, however. The North Chicago City Railway Co.`s clock still displayed local time, which was 9 1/2 minutes faster than Standard Time. To show their disapproval, a delegation of North Siders good naturedly presented one of the company officials with a badge depicting a street car running wildly out of control.
Not everyone was happy with the changes. The Indianapolis Sentinel bitterly opposed the new system, suggesting that Standard Time went against nature. ''People-55 million of them-must eat, sleep and work as well as travel by railroad time. The Sun will be requested to rise and set by railroad time. People will have to marry by railroad time and die by railroad time.''
But Standard Time took, and, what was most remarkable, it was done completely without legislation and on a voluntary basis. Congress didn`t pass the Standard Time Act until March 19, 1918.<|endoftext|>
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