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# Slope-Intercept Form
## y = mx +b. Finding slope(m) and y-intercept(b) from equation or graph
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Finding the Cheapest Cell Phone Plan
Teacher Contributed
## Real World Applications – Algebra I
### Topic
How do you know which cell phone plan is the cheapest?
### Student Exploration
The student will investigate, compare and determine which cell phone plan is the least expensive over time.
1. Search for two different cell phone companies (i.e. AT&T, Verizon, Sprint, T-Mobile, Boost, Metro PCS).
2. The most organized way to list the cost of each cell phone plan is to make a table. Make a table for each cell phone company. The input \begin{align*}(x)\end{align*} values should be the # of months you have the phone, and the output \begin{align*}(y)\end{align*} values should be the total cost. “Month 0” should be the initial cost of purchasing the cell phone itself.
3. Create a graph for each table of points. What are realistic values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*} on this graph? Why?
4. Create an equation for each cell phone plan represented in slope-intercept form.
5. In each equation, what is the slope? What does this represent in the cell phone plan? What is the \begin{align*}y-\end{align*}intercept in each equation? What does this represent in the cell phone plan? Justify your thinking.
6. Which cell phone plan is the cheapest, and for how long?
7. When are both cell phone plans the same cost? Use the Substitution Method to solve this system of linear equations you created. What do “\begin{align*}x\end{align*}” and “\begin{align*}y\end{align*}” each represent in this solution?
### Notes/Highlights Having trouble? Report an issue.
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What does Square units mean in math?
What does Square units mean in math?
(sq or 2) These signify that two quantities measured in the same units have been multiplied together. For example, to find the area of a square or rectangle, length and breadth are multiplied together to give the area, which is measured in square units.
Why is area measured in square units?
Area is measured in “square” units. Since each side of a square is the same, it can simply be the length of one side squared. If a square has one side of 4 inches, the area would be 4 inches times 4 inches, or 16 square inches. (Square inches can also be written in2.)
Is circumference measured in square units?
It’s given in units of distance squared, such as cm2 or m2. The area of a circle is given by the formulas: A = πr2 (Area equals pi times the radius squared.) A = π(1/2 d)2 (Area equals pi times one-half the diameter squared.)
What is Square formula?
Area of square formula = s × s = s² Area of Square = s2. The Perimeter of a Square. The perimeter of the square is the total length of its boundary. The boundary of a square is represented by the sum of the length of all sides.The perimeter of the shape depends on the length of the shape.
What is perimeter formula?
The formula for the perimeter of a rectangle is often written as P = 2l + 2w, where l is the length of the rectangle and w is the width of the rectangle. The area of a two-dimensional figure describes the amount of surface the shape covers.
How do we calculate area?
For a square or rectangular room, you will first need to measure the length and then the width of the room. Then multiply the length and width. Length x Width = Area. So, if your room measures 11 feet wide x 15 feet long, your total area will be 165 square feet.
How do you calculate lot area in square meters?
Multiply the length and width together. Once both measurements are converted into metres, multiply them together to get the measurement of the area in square metres.
What is 3 meters by 3 meters in square meters?
How big is a 3 meter by 3 meter rectangle? Size of a 3 x 3 rectangle in meters squared….Area of a 3m x 3m Rectangle.
9 square meters
90,000 square centimeters
13,950 square inches
96.875 square feet
10.764 square yards
How many square meter is a plot?
An Acre is a propduct of any rectangular plot of land giving a total of 4,046sqm OR 43,560sq ft. An Acre consist of 6 plots each measuring 6 x 120ft. In Lagos State, the standard size of a plot is 60 x 120ft ( 18m x 36m ie 648sqm), while in some other cities of the country, plots are measured in 50 x100ft.
What does a square Metre look like?
A metre square is a square with sides one metre in length – it refers to the shape and the side length, not the area. By contrast, a square metre is an area and can be any shape….Updated 04/01/2020 (see below)
Area= Length x Breadth A=l × b
4 square metres A = 4 m2
How do you calculate cost per square meter?
Price per square metre is the amount of money a buyer is paying for each metre of land they are purchasing. It is one of many tools and measures that investors and buyers can use to compare properties and locations. To calculate it, take the cost of the house (e.g. \$750,000) and divide it by the land size (e.g. 150m2).
How do I calculate 2 square meters?
Multiply the length and width together. Once both measurements are converted into meters, multiply them together to get the measurement of the area in square meters. Use a calculator if necessary. For example: 2.35m x 1.08m = 2.538 square meters (m2).
How many meters is 100 square meters?
Square meter to Meter Calculator
1 m2 = 1 meter 1 m2
7 m2 = 2.6458 meter 49 m2
8 m2 = 2.8284 meter 64 m2
9 m2 = 3 meter 81 m2
10 m2 = 3.1623 meter 100 m2
How many square meters is a 2 bedroom apartment?
Ideally for mine, one-bedroom apartments should be no smaller that 60 to 65 square metres (including balcony) and two-bedroom apartments no smaller than 85 to 90 square metres gross.
How do I work out the square meter of a room?
To calculate the amount of m2 you will need for your floor, determine the area of your floor: measure the length of the room in meteres and the width of the room in metres, and multiply these (so length x width) to obtain the area of your floor.
What size is a square Metre?
The area equal to a square that is 1 meter on each side. Used for measuring areas of rooms, houses, blocks of land, etc. Example: A typical car parking space is about 12 square meters.
06/03/2021
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Section 4-5 "Greatest Common Factor"
Technology objective: Students will use the TI-34 calculators to find the greatest common factor of a pair of numbers.
Lesson: After a discussion of greatest common factor, ask the students if they can figure out how to use the TI-34 calculator to find the greatest common factor of a set of numbers. Put various suggestions on the board. Next, explain to the students how to use the TI-34 calculator to find the greatest common factor of a pair of numbers. They can do this by entering one number in the place of the numerator and one number in place of the denominator. Have them try to find the greatest common factor of 45 and 60. Have them enter the smaller number in for the numerator, 45, and the bigger number in for the denominator, 60. When the press enter, the calculator will simplify the fraction they entered. Ask the class to figure out what was factored out of the numerator and denominator. (This serves as a great mental exercise.) Once they have figured that number out, they know the greatest common factor. Have them try to find the greatest common factor of any two random numbers. Also discuss what happens when you enter two prime numbers. This can serve as a wonderful introduction to fractions.
Reinforcement:
Game
This game was found in the 1992 Yearbook Calculators in Mathematics Education in the article Statewide In-Service Programs on Calculators in Mathematics Teaching written by Bright, Lamphere, and Usnick. Students should be broken up into pairs. Each pair will be given a 6 x 6 grid with the following numbers inserted in no particular order:
25, 60, 45, 15, 10, 80, 48, 64, 36, 24, 65, 99, 27, 16, 42, 81, 75, 25, 200, 300, 500, 600, 800, 900, 360, 480, 640, 550, 270, 120, 144, 625, 525, 648, 864, and 468.
GCF game
1) Decide who plays first. Play then alternates.
2) On your turn, your opponent chooses an uncovered number on the grid.
3) You then choose a second uncovered number on the grid. Cover both choices.
4) Your score for the round is the GCF of the two numbers.
5) The winner is the player with the greatest cumulative score after five rounds.
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If GMAT probability questions are giving you a hard time this blog might just be all the help you need to get started. In this blog, we will learn how to approach and solve GMAT probability questions in the simplest way possible. So shall we?
## GMAT Probability Concepts
As we now move towards the mathematical aspects of the topic, one underlying factor that recurs in every question of probability is that whenever one is asked the question, what is the probability? The immediate question that arises or should arise in one’s mind is the probability of what?
The answer to this question is the probability of the EVENT. The EVENT is the cornerstone or the bottom line of probability. Hence, the first objective while trying to solve any question in probability is to define the event. The event whose probability is to be found out is described in the question and the task of the student in trying to solve the problem is to define it.
In general, the student can either define the event narrowly or broadly. Narrow definitions of events are the building blocks of any probability problem and whenever there is a doubt about a problem, you are advised to get into the narrowest form of the event definition. The difference between the narrow and broad definition of the event can be explained through an example:
GMAT Probability Questions:
1. What is the probability of getting a number greater than 2, in a throw of a normal unbiased dice having 6 faces?
The broad definition of the event here is getting a number greater than 2 and this probability is given by 4/6. However, this event can also be broken down into its more basic definitions as:
The event is defined as getting 3 or 4 or 5 or 6. The individual probabilities of each of these are 1/6, 1/6, 1/6, and 1/6 respectively.
Hence, the required probability is 1/6 + 1/6 + 1/6 + 1/6 = 4/6 = 2/3.
Although in this example it seems highly trivial, the narrow event-definition approach is very effective in solving difficult problems on probability.
In general, event definition means breaking up the event into the most basic building blocks, which have to be connected together through the two English conjunctions— ‘AND’ and ‘OR’.
## GMAT Probability Concepts: The Use of the Conjunction ‘AND’
Whenever we use AND as the natural conjunction joining two separate parts of the event definition, we replace the AND with the multiplication sign. Thus, if A AND B have to occur, and if the probability of their occurrence is P(A) and P(B) respectively, then the probability that A AND B occur is got by connecting P(A) AND P(B). Replacing the AND by multiplication sign, we get the required probability as P(A) X P(B)
GMAT Probability Questions:
1. If we have the probability of A hitting a target as 1/3 and that of B hitting the target as 1/2, then the probability that both hit the target if one shot is taken by both of them is got by?
Event Definition: A hits the target AND B hits the target.
P(A) x P(B) = 1/3 x 1/2 = 1/6
Note that since we use the conjunction AND in the definition of the event here, we multiply the individual probabilities that are connected through the conjunction AND.
## GMAT Probability Concepts: The Use of the Conjunction ‘OR’
Whenever we use OR as the natural conjunction joining two separate parts of the event definition, we replace the OR with the addition sign.
Thus, if A OR B has to occur, and if the probability of their occurrence is P(A) and P(B) respectively, then the probability that A OR B occurs is got by connecting P(A) OR P(B). Replacing the OR by addition sign, we get the required probability as P(A) + P(B).
GMAT Probability Questions:
1. If we have the probability of A winning a race as 1/3 and that of B winning the race as 1/2, then the probability that either A or B winning a race is got by?
Event Definition: A wins OR B wins.
P(A) + P(B) = 1/3 +1/2 = 5/6
Note that since we use the conjunction OR in the definition of the event here, we add the individual probabilities that are connected through the conjunction OR.
How can I improve GMAT Quant speed?
Have you taken the GMAT before?
## GMAT Probability Concepts: Combination of ‘AND’ and ‘OR’
If two dice are thrown, what is the chance that the sum of the numbers is not less than 10?
Event Definition: The sum of the numbers is not less than 10 if it is either 10 OR 11 OR 12. Which can be done by:
(6 AND 4) OR (4 AND 6) OR (5 AND 5) OR (6 AND 5) OR (5 AND 6) OR (6 AND 6)
that is, 1/6 x 1/6 + 1/6 x 1/6 + 1/6 x 1/6 + 1/6 x 1/6 + 1/6 x 1/6 + 1/6 x 1/6 = 6/36 = 1/6
The bottom line is that no matter how complicated the problem on probability is, it can be broken up into its narrower parts, which can be connected by ANDs and ORs to get the event definition.
Once the event is defined, the probability of each narrow event within the broad event is calculated and all the narrow events are connected by Multiplication (for OR) to get the final solution.
GMAT Probability Questions:
1. In a four-game match between Kasparov and Anand, the probability that Anand wins a particular game is 2/5 and that of Kasparov winning a game is 3/5. Assuming that there is no probability of a draw in an individual game, what is the chance that the match is drawn (Score – 2-2).
For the match to be drawn, 2 games have to be won by each of the players. If ‘A’ represents the event that Anand won a game, the event definition for the match to end in a draw can be described as [The student is advised to look at the use of narrow event definition.]
(A&A&K&K) OR (A&K&A&K) OR (A&K&K&A) OR (K&K&A&A) OR (K&A&K&A) OR (K&A&A&K)
This further translates into: (2/5)2 (3/5)2 + (2/5)2(3/5)2 + (2/5)2(3/5)2 + (2/5)2 (3/5)2 + (2/5)2(3/5)2 + (2/5)2(3/5)2 = (36/625) x 6 = 216/625
After a little bit of practice you can also think about this directly as:
4C2 x (2/5)2 x 2C2 x (3/5)2 = 6 x 1 x 36/625 = 216/625
Where, 4C2 gives us the number of ways in which Anand can win two games and 2C2 gives us the number of ways in which Kasparov can win the remaining 2 games (obviously, only one).
Have questions about GMAT Quant preparation or admissions? We can help! Connect with our counsellors for a free 15-minute consultation session and get all your doubts resolved!
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# Computing $999,999\cdot 222,222 + 333,333\cdot 333,334$ by hand.
I got this question from a last year's olympiad paper.
Compute $999,999\cdot 222,222 + 333,333\cdot 333,334$.
Is there an approach to this by using pen-and-paper?
EDIT Working through on paper made me figure out the answer. Posted below.
I'd now like to see other methods. Thank you.
• Place your calculator on a paper, and use your pen to hit the keys of the calculator... Have fun! Commented May 21, 2014 at 8:35
My observation suggests that we may simplify it somewhat like this assuming $x = 111,111$: \begin{align*} 9x\cdot2x+3x(3x + 1) &= 9x^2 +18x^2 + 3x \\ &= 27x^2 + 3x \\ &= 3x(9x + 1) \end{align*}
So this means: \begin{align*} 3x(9x + 1) &= 333,333\cdot (999,999+1) \\ &= 333,333\cdot1,000,000 \\ &= \boxed{333,333,000,000} \end{align*}
• Don't forget to accept this nice answer! :)
– SBF
Commented Jan 11, 2013 at 15:02
• @Ilya: I think it'd be rude to not entertain other solutions if posted. :)
– P.K.
Commented Jan 11, 2013 at 15:03
• Another potentially useful thing to remember (less so for this question) is that $111111 = \frac {10^6-1}{9}$, which could make the multiplication easier. See for example this. Commented Jan 11, 2013 at 15:22
\begin{aligned}&999,999 \cdot 222,222 + 333,333 \cdot 333,334 \\ = &333,333 \cdot 666,666 + 333,333 \cdot 333,334 \\ = &333,333 \cdot 1,000,000\end{aligned}
• @DumbCow more generally (2/3)A * (3/2)B = A * B Commented Jan 11, 2013 at 18:37
An algebra free way: the expression is $$(10^6 -1)\bigg({2 \over 9}(10^6 - 1)\bigg) + {10^6 - 1 \over 3}{10^6 + 2 \over 3}$$ $$= {10^6 - 1 \over 3}\bigg({2 \over 3}(10^6 - 1) + {10^6 + 2 \over 3}\bigg)$$ $$= \bigg({10^6 - 1 \over 3}\bigg)10^6$$ $$= 333,333*1,000,000$$ $$= 333,333,000,000$$
• I think that this is a nice answer. Commented Jan 12, 2013 at 4:11
My first (useful) though was factoring out 333,333:
\begin{align*} &999,999*222,222+333,333*333,334 =\\ &= 333,333*(3*222,222+333,334) =\\ &= 333,333*(666,666+333,334) =\\ &= 333,333*1,000,000 =\\ &= 333,333,000,000 \end{align*}
My immediate thought was to compute this as $$1,000,000\times222,222 - 222,222+111,111\times1,000,002$$
For what it's worth, my thought, different from the others so far:
$1000000\times(0.999999\times 222222 + 0.333333\times 333334)$
This directly leads to an approximation of the answer, if we pretend that 0.9999 is 1, and 0.3333 is 1/3.
That inspiration leads to (where M = $10^6$):
$(M-1)222222 + (M/3 - 1/3)33334$
Which rearranges to:
$222222M - 222222 + (333334M - 333334)/3$
(In the second term, we take the 1/3 out, and bring the 33334 in). The subtractions we can evaluate quite easily with mental calculations, because the minuends end with six zeros, and the digits are repeating. For instance by analogy with subtraction from 100 we know that 100 - 22 is 78, and so 100...00 - 22...22 is 77...78.
$222221777778 + 333333666666/3$
Of course the division by 3 is easy:
$222221777778 + 111111222222$
And the addition is also trivial. On the lower half, we are adding back the 2222222 digits that we subtracted in the first place to make 777778, which makes a million again, and the 1 which carries out of that bumps up the 222221 upper half to 222222, which adds with 111111 to make 333333, hence:
$333333000000$
$$999,999\cdot 222,222 + 333,333\cdot 333,334=333,333\cdot 666,666 + 333,333\cdot 333,334$$ $$=333,333 \cdot 1,000,000$$
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# JAC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1
Jharkhand Board JAC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Exercise Questions and Answers.
## JAC Board Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.1
Question 1.
Find the distance between the following pairs of points :
1. (2, 3), (4, 1)
2. (- 5, 7), (1, 3)
3. (a, b), (- a, – b)
1. Let A (2, 3) and B (4, 1) be the given points. Then,
AB = $$\sqrt{(2-4)^2+(3-1)^2}$$
= $$\sqrt{4+4}$$ = $$\sqrt{8}$$ = 2$$\sqrt{2}$$
Thus, the distance between the given points is 2$$\sqrt{2}$$.
2. Let A (- 5, 7) and B (- 1, 3) be the given points. Then,
AB = $$\sqrt{(-5+1)^2+(7-3)^2}$$
= $$\sqrt{16+16}$$ = $$\sqrt{32}$$ = 4$$\sqrt{2}$$
Thus, the distance between the given points is 4$$\sqrt{2}$$.
3. Let P (a, b) and Q(- a, – b) be the given points. Then,
PQ = $$\sqrt{(a+a)^2+(b+b)^2}$$
= $$\sqrt{4 a^2+4 b^2}$$
= 2$$\sqrt{a^2+b^2}$$
Thus, the distance between the given points is 2$$\sqrt{a^2+b^2}$$.
Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in section 7.2.
Let A (0, 0) and B (36, 15) be the given points.
Then,
AB = $$\sqrt{(0-36)^2+(0-15)^2}$$
= $$\sqrt{1296+225}$$ = $$\sqrt{1521}$$ = 39
Thus, the distance between the given points is 39.
Yes, now we can find the distance between the two towns A and B discussed in section 7.2 in the textbook.
Town B is located 36 km east and 15 km north of the town B. So, if we take town A to be situated at the origin with coordinates (0, 0) then the coordinates of town B become (36, 15). Now, as calculated above, the distance between A(0, 0) and B (36, 15) is 39. Hence, the distance between town A and town B is 39 km.
Question 3.
Determine if the points (1, 5), (2, 3) and (- 2, – 11) are collinear.
Let A(1, 5), B(2, 3) and C (- 2, – 11) be the given points. Then,
Now, 14.56 + 2.24 = 16.80 ≠ 16.28
Thus, AB + BC ≠ AC
Moreover, BC + AC ≠ AB and AC + AB ≠ BC are obvious.
Hence, the given points are not collinear.
Question 4.
Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Let the given points A (5, – 2), B (6, 4) and C(7, – 2) be the vertices of ΔABC. Then,
Here, in ΔABC, AB = BC ≠ AC.
Hence, ABC is an isosceles triangle.
Thus, the given points (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in the given figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Here A(3, 4), B(6, 7), C (9, 4) and D(6, 1) give the position of points where those four friends are seated. Then,
AB = BC = CD = DA and AC = BD.
In other words, all the sides of quadrilateral ABCD are equal and its diagonals are also equal.
Hence, ABCD is a square.
So. Champa is correct stating that ABCD is a square.
Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
1. (- 1, – 2), (1, 0), (1, 2), (- 3, 0)
2. (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
3. (4, 5), (7, 6), (4, 3), (1, 2)
1. If possible, let A (- 1, – 2), B(1, 0), C(- 1, 2) and D(- 3, 0) be the vertices of quadrilateral ABCD. Then,
AB = BC = CD = DA and AC = BD.
Hence, ABCD is a square.
2. If possible, let A (- 3, 5), B(3, 1), C(0, 3) and D(- 1, – 4) be the vertices of quadrilateral ABCD. Then,
Hence, A, B and C are collinear points in which C lies between A and B.
Hence, A, B, C and D do not form a quadrilateral.
3. If possible, let A (4, 5), B(7, 6), C(4, 3) and D (1, 2) be the vertices of quadrilateral ABCD. Then,
Thus, in quadrilateral ABCD, AB = CD, BC = DA, i.e., both the pairs of opposite sides are equal, but AC ≠ BD. i.e.. diagonals are not equal.
Hence, ABCD is a parallelogram.
Question 7.
Find the point on the x-axis which is equidistant from (2,-5) and (-2, 9).
Here, A (2, – 5) and B(- 2, 9) are two given points.
Coordinates of any point of the x-axis are of the form (x, 0).
Let P(x, 0) be the required point on the x-axis which is equidistant from A and B.
So, PA = PB
∴ PA² = PB²
∴ (x – 2)² + (0 + 5)² = (x + 2)² +(0 – 9)²
∴ x² – 4x + 4 + 25 = x² + 4x + 4 + 81
∴ -8x=56
∴ x = – 7
Thus, the required point on the x-axis which is equidistant from (2, – 5) and (- 2, 9) is (- 7, 0).
Question 8.
Find the values of y for which the distance between the points P (2, – 3) and Q(10, y) is 10 units.
The distance between P(2, – 3) and 9(10, y) is given to be 10.
∴ PQ = 10
∴ PQ² = 100
∴ (2 – 10)² + (- 3 – y)² = 100
∴ 64 + 9 + 6y + y² = 100
∴ y² + 6y – 27 = 0
∴ (y + 9) (y – 3) = 0
∴ y + 9 = 0 or y – 3 = 0
∴ y = – 9 or y = 3
Thus, the required values of y are – 9 and 3.
Question 9.
If Q(0, 1) is equidistant from P (5, – 3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Q(0, 1) is equidistant from P(5, -3) and R (x, 6).
∴ PQ = RQ
∴ PQ² = RQ²
∴ (5 – 0)² + (- 3 – 1)² = (x – 0)² + (6 – 1)²
∴ 25 + 16 = x² + 25
∴ x² = 16
∴ x = ± 4
∴ x = 4 or x = – 4
Thus, x = ± 4, QR = $$\sqrt{41}$$ and PR = $$\sqrt{82}$$ or 9$$\sqrt{2}$$.
Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
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Sketch the graphs of the following trigonometric functions:
Question:
Sketch the graphs of the following trigonometric functions:
(i) $f(x)=\cos \left(x-\frac{\pi}{4}\right)$
(ii) $g(x)=\cos \left(x+\frac{\pi}{4}\right)$
(iii) $h(x)=\cos ^{2} 2 x$
(iv) $\phi(x)=2 \cos \left(x-\frac{\pi}{6}\right)$
(v) $\Psi(x)=\cos 3 x$
(vi) $u(x)=\cos ^{2} \frac{x}{2}$
(vii) $f(x)=\cos \pi x$
(viii) $g(x)=\cos 2 \pi x$
Solution:
(i) $y=\cos \left(\mathrm{x}-\frac{\pi}{4}\right)$
$\Rightarrow y-0=\cos \left(x-\frac{\pi}{4}\right)$ …(i)
On shifting the origin at $\left(\frac{\pi}{4}, 0\right)$, we get :
$\mathrm{x}=\mathrm{X}+\frac{\pi}{4}$ and $\mathrm{y}=\mathrm{Y}+0$
On subsititut $i n g$ the values in (i) we get:
$\mathrm{Y}=\cos X$
Then, we draw the graph of $Y=\cos X$ and shift it by $\frac{\pi}{4}$ to the right.
Then, we obtain the following graph:
(ii) $y=\cos \left(\mathrm{x}+\frac{\pi}{4}\right)$
$\Rightarrow y-0=\cos \left(x+\frac{\pi}{4}\right)$ …(i)
On shifting the origin at $\left(-\frac{\pi}{4}, 0\right)$, we get:
$\mathrm{x}=\mathrm{X}-\frac{\pi}{4}$ and $\mathrm{y}=\mathrm{Y}+0$
On subsitituting the values in $(\mathrm{i})$, we get:
$\mathrm{Y}=\cos X$
Then, we draw the graph of $Y=\cos X$ and shift it by $\frac{\pi}{4}$ to the left.
Then, we obtain the following graph:
(iii) $y=\cos ^{2} 2 x$
The following graph is:
(iv) $y=2 \cos \left(x-\frac{\pi}{6}\right)$
$\Rightarrow y-0=2 \cos \left(x-\frac{\pi}{6}\right)$ ..(i)
On shifting the origin at $\left(\frac{\pi}{6}, 0\right)$, we get:
$x=X+\frac{\pi}{6}$ and $y=Y+0$
On subsitituting the values in $(\mathrm{i})$, we get:
$Y=2 \cos X$
Then, we draw the graph of $Y=\cos X$ and shift it by $\frac{\pi}{6}$ to the right.
Then, we obtain the following graph:
(v) $y=\cos 3 x$
The following graph is:
(vi) $y=\cos ^{2} \frac{x}{2}$
The following graph is:
(vii) $y=\cos \pi x$
The following graph is:
(viii)$y=\cos 2 \pi x$
The following graph is:
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Fortune Teller Math Games for Kids to Practice Math Skills
Today we are making a paper fortune teller math game for kids needing to practice their fractions and multiplication facts. Do you remember paper fortune tellers? They were *the* rage when I was a kid. We are going to use a paper fortune teller to introduce my kiddos to the joy of Cootie catchers while reviewing math lessons.
DIY Math Games for Kids to Make Math Fun
Math is one of those subjects that very few people enjoy. It is tedious and hard, but it doesn’t have to be if you make it a fun game.
We have come up with so many ways to make math fun:
Make a Math Fortune Teller Game
But today we made two paper games, one to review fractions and another to review times tables.
How to Make a Paper Fortune Teller
1. Basically, you start with a square piece of paper.
2. Fold the corner into the center.
3. Repeat for all four corners folding into the center.
4. Until you have all four corners folded to the center.
5. Flip it over and fold the corners into the center again.
6. Until you have all four corners folded to the center.
7. You then fold it like a hot dog – with the thumb flaps on the outside. Your kids stick their fingers into the flaps and move them to reveal the inside. Your kids can lift the inside flaps to see another message.
8. Write the messages or numbers on the flaps. We used math as our “messages” inside the classic fortune teller.
9. Place fingers into the paper fortune teller and move both in and out.
Related: Check out our reading fortune teller game: Learning to Read CVC words
Times Tables Game on Paper Fortune Teller
For our multiplication cootie catcher, we wrote each “family” of math problems on the outer flaps.
The tables we are working on with our second grader are 2, 3, 4 & 5s – so I wrote those numbers on the outside.
Inside the flaps we have the numbers written out by skip counting.
So as your kids move the paper game, they choose between the different “groups” of skip-count numbers. When they lift the flap, there are four multiplication problems for them to solve.
Paper Fractions Game on Paper Fortune Teller
For the fractions game, draw a circle on each of the four main sections. Break the circle apart into “fractions.
We did the following fractions on our catcher: 1/2, 1/3, 1/4, & 1/5.
The next level of flaps the kids had to figure out which “flap” matched the circle. Even though my kids are learning just the fractions, I wrote the decimal number beside “the answer” to try to help them begin matching the two numbers together.
When the kids pull up the flap, they see another “problem”. They had to color in the fraction amount on the bar using their finger.
I loved seeing the kids put their math lessons into their pockets and carry them around practicing them throughout the day!
More Math Games & Ideas from Kids Activities Blog
How did your kids like the paper fortune teller math game?
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# Math 55: Discrete Mathematics
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1 Math 55: Discrete Mathematics UC Berkeley, Fall 2011 Homework # 5, due Wednesday, February Let P (n) be the statement that n 3 = (n(n + 1)/2) 2 for the positive integer n. a) What is the statement P (1)? b) Show that P (1) is true. c) What is the induction hypothesis? d) What do you need to prove in the inductive step? e) Complete the inductive step. f) Explain why these steps show that this formula is true for all positive integers n. a) P (1) is the statement 1 3 = ((1(1 + 1)/2) 2. b) This is true because both sides of the equation evaluate to 1. c) The induction hypothesis is the statement P (k) for some positive integer k, that is, the statement k 3 = (k(k +1)/2) 2. d) Assuming that P (k) holds, we need to show that P (k + 1) holds, that is, we need to derive the equation k 3 +(k+1) 3 = ((k + 1)(k + 2)/2) 2 from the equation in (c). e) We add (k + 1) 3 to both the left hand side and the right hand side of the equation in (c). This shows that the left hand side in (d) is equal to (k(k + 1)/2) 2 + (k + 1) 3. By expanding and factoring, we find that this expression equals ((k + 1)(k + 2)/2) 2. Hence we have shown that the left hand side of the equation in (d) equals the right hand side of the equation in (d). f) We have carried out both the basis step and the inductive step. The principle of mathematical induction now ensures that P (n) is true for all positive integers n. 1
2 5.1.6 Prove that 1 1! + 2 2! + + n n! = (n + 1)! 1 whenever n is a positive integer. We use mathematical induction. In fhe basis step, for n = 1, the equation states that 1 1! = (1 + 1)! 1, and this is true because both sides of the equation evaluate to 1. For the inductive step, we assume that 1 1! + 2 2! + + k k! = (k + 1)! 1 for some positive integer k. We add (k + 1)(k + 1)! to the left hand side to find that 1 1! + 2 2! + + (k + 1) (k + 1)! = (k + 1)! 1 + (k + 1)(k + 1)! The right hand side equals (k + 1)!(k + 2) 1 = (k + 2)! 1. This establishes the desired equation also for k + 1, and we are done by the principle of mathematical induction a) Find a formula for n(n + 1) by examining the values of this expression for small values of n. b) Prove the formula you conjectured in part (a). (a) By evaluating the sum for n = 1, 2, 3, 4, 5,..., we are led to conjecture that the following equation holds for all positive integers n: n(n + 1) = n n + 1. (1) (b) We use mathematical induction. The basis step is n = 1. Here both sides of the equation are equal to 1/2, so the claim holds. For the inductive step, we assume that (1) is true for n = k. We 1 add (k+1)(k+2) to both sides of this equation. Then the right hand side becomes k k (k + 1)(k + 2) = k(k + 2) + 1 (k + 1)(k + 2) = k + 1 k + 2. Hence the left hand side of (1) for n = k +1 equals the right hand side of (1) for n = k + 1. This completes the proof by induction Prove that n! < n n for all integers n 2, using the six suggested steps. Let P (n) be the propositional function n! < n n. 2
3 a) The statement P (2) says that 2! = 2 is less than 2 2 = 4. b) This statement is true because 4 is larger than 2. c) The inductive hypothesis states that P (k) holds for some integer k 2. d) We need to prove that k! < k k implies (k + 1)! < (k + 1) k+1. e) Given that k! < k k holds, easily seen inequalities imply (k + 1)! = k! (k + 1) < k k (k + 1) < (k + 1) k (k + 1) = (k + 1) k+1. f) We have carried out both the basis step and the inductive step. The principle of mathematical induction now ensures that P (n) is true for all integers n Prove that 3 divides n 3 + 2n whenever n is a positive integer. We use mathematical induction. For n = 1, the assertion says that 3 divides , which is indeed the case, so the basis step is fine. For the inductive step, we assume that 3 divides k 3 + 2k for some positive integer k. Hence there exists an integer l such that 3l = k 3 + 2k. A computation shows (k + 1) 3 + 2(k + 1) = (k 3 + 2k) + 3(k 2 + k + 1). The right hand is divisible by 3. This is evident for the second summand, and it is the induction hypothesis for the first summand. Hence we have proved that 3 divides (k + 1) 3 + 2(k + 1). This complete the inductive step, and hence the assertion follows Use mathematical induction to show that given a set of n + 1 positive integers, none exceeding 2n, there is at least one integer in this set that divides another integer in the set. Let P (n) be the following propositional function: given a set of n + 1 positive integers, none exceeding 2n, there is at least one integer in this set that divides another integer in the set. The proposition P (1) is true because there is only one set of positive integers none exceeding 2 1. This set is {1, 2} and it contains an integer, namely 1, that divides the other integer, namely 2. This verifies the basis step in our proof by mathematical induction. For the inductive step we assume that P (k) is true for some positive integer k. To prove P (k + 1), we consider a set S of k + 2 positive 3
4 integers none exceeding 2k + 2. If S {2k + 1, 2k + 2} has cardinality 0 or 1 then we apply the induction hypothesis to S\{2k + 1, 2k + 2} to conclude that this set contains a dividing pair of integers. Hence we are left with the case that 2k + 1 and 2k + 2 are both in S and S\{2k + 1, 2k + 2} consists of k positive integers of size at most 2k that pairwise don t divide each other. If k + 1 is in S then we are done because k + 1 divides 2k + 2. Suppose therefore that k + 1 S. Then we replace S by the set S = (S\{2k + 2}) {k + 1}. The new set S is covered by the previous case, so it contains a divisible pair. If that pair does not involve k + 1 then it is also in S. If it involves k + 1 then this means that some l S\{k + 1} divides l. That l must also divide 2k + 2 and hence S contains a divisible pair. This completes the inductive step and hence the proof Let P (n) be the statement that a postage of n cents can be formed using just 4-cent stamps and 7-cent stamps. Prove that P (n) is true for n 18, using the six suggested steps. We prove this using strong induction. The basis step is to check that P (18), P (19), P (20) and P (21) hold. This seen from the identities 18 = , 19 = , 20 = , 21 = For the inductive step, we assume that P (j) holds for all integers j with 18 j k where k 21. To realize k + 1 cents, we first realize k 3 cents using 4-cent stamps and 7-cent stamps. This is possible by the inductive hypothesis, since k Now add one more 4-cent stamp to realize k + 1 cents. This completes the induction step and it hence proves the assertion Assume that a chocolate bar consists of n squares arranged in a rectangular pattern. The entire bar, or a smaller rectangular piece of the bar, can be broken along a vertical or a horizontal line separating the squares. Assuming that only one piece can be broken at a time, determine how many breaks you must successively make to break the bar into n separate squares use strong induction to prove your answer. We claim that the number of needed breaks is n 1. We shall prove this for all positive integers n using strong induction. The basis step n = 1 is clear. In that case we don t need to break the chocolate at all, we can just eat it. Suppose now that n 2 and assume the assertion is true for all rectangular chocolate bars with fewer than n 4
5 squares. Then we break the chocolate into two pieces of size m and n m where 1 m < n. By the induction hypotheses, the bar with m pieces requires m 1 breaks and the bar with n m squares requires n m 1 breaks. Thus the original cholocate bar requires 1+(m 1) +(n m 1) breaks. This number equals n 1, as required Suppose that P (n) is a propositional function. Determine for which nonnegative integers n the statement P (n) must be true if a) P (0) is true; for all nonnegative integers n, if P (n) is true then P (n + 2) is true. b) P (0) is true; for all nonnegative integers n, if P (n) is true then P (n + 3) is true. c) P (0) and P (1) are true; for all non-negative integers n, if P (n) and P (n + 1) are true then P (n + 2) is true. d) P (0) is true; for all non-negative integers n, if P (n) is true then P (n + 2) and P (n + 3) are true. a) The statement P (n) is true for all nonnegative integers n that are even. b) The statement P (n) is true for all nonnegative integers n that are divisible by 3. c) The statement P (n) is true for all nonnegative integers n. d) The statement P (n) is true for all nonnegative integers n with n 1, since every such n is expressible as a sum of 2 s and 3 s Find f(2), f(3), f(4), and f(5) if f is defined recursively by f(0) = f(1) = 1 and for n = 1, 2,... a) f(n + 1) = f(n) f(n 1), b) f(n + 1) = f(n)f(n 1), c) f(n + 1) = f(n) 2 + f(n 1) 2, d) f(n + 1) = f(n)/f(n 1). a) 0, 1, 2, 3 b) 1, 1, 1, 1 c) 2, 5, 29, 866 d) 1, 1, 1, 1 5
6 5.3.6 Determine whether each of these proposed definitions is a valid recursive definition of a function f from the set of all nonnegative integers to the set of integers. If f is well defined, find a formula for f(n) when n is a nonnegative integer and prove that your formula is valid. a) f(0) = 1, f(n) = f(n 1) for n 1, b) f(0) = 1, f(1) = 0, f(2) = 2, f(n) = 2f(n 3) for n 3, c) f(0) = 0, f(1) = 1, f(n) = 2f(n + 1) for n 2, d) f(0) = 0, f(1) = 1, f(n) = 2f(n 1) for n 1, e) f(0) = 2, f(n) = f(n 1) if n is odd and n 1 and f(n) = 2f(n 2) if n 2. a) The function is well-defined and given by f(n) = ( 1) n. b) The function is well-defined and given by f(n) = 0 if n 1 (mod3) and f(n) = 2 (n 1)/2 otherwise. c) The function is not well-defined because the definition of f(n) involves the value at n + 1. d) The function is well-defined and given by f(n) = 2 (n+3)/ Give a recursive definition of the sequence [a n ], n = 1, 2, 3,... if a) a n = 6n, b) a n = 2n + 1, c) a n = 10 n, d) a n = 5. a) a 1 = 6 and a n = a n for n 2. b) a 1 = 3 and a n = a n for n 2. c) a 1 = 1 and a n = 10a n 1 for n 2. d) a 1 = 5 and a n = a n 1 for n Let f n denote the nth Fibonacci number. Prove that f 2 1 +f f 2 n = f n f n+1 when n is a positive integer. We prove this by induction on n. The statement is true for n = 1 because 1 2 = 1 1, it is true for n = 2 because = 1 2, and it is true for n = 3 because = 2 3. This takes care of the 6
7 basis step. For the inductive step, suppose it is true for n = k, and consider the left hand for n = k + 1. We find f f f 2 k + f 2 k+1 = f kf k+1 + f 2 k+1 = f k+1(f k + f k+1 ). The last expression equals f k+1 f k+2 by the recursive definition of the Fibonacci sequences. The verifies the equation for n = k + 1. We have thus completed both the basis step and the inductive step, and hence the claim follows by the principle of mathematical induction ** Give a recursive definition of the functions max and min, so that that max(a 1, a 2,..., a n ) and min(a 1, a 2,..., a n ) are the maximum and the minimum of the n numbers a 1, a 2,..., a n respectively. a 1 if n = 1 (base case) a n if n > 1 and max(a 1,..., a n ) = a n > max(a 1,..., a n 1 ) max(a 1,..., a n 1 ) otherwise min(a 1,..., a n ) = a 1 if n = 1 (base case) if n > 1 and a n min(a 1,..., a n 1 ) Find these values of Ackermann s function a) A(2, 3) b) A(3, 3) a n < min(a 1,..., a n 1 ) otherwise The Ackermann function is defined recursively by A(m, n) = 2n if m = 0; A(m, n) = 0 if m 1 and n = 0; A(m, n) = 2 if m 1 and n = 1; A(m, n) = A(m 1, A(m, n 1)) if m 1 and n 2. We have A(2, 2) = A(1, A(2, 1)) = A(1, 2) = A(0, A(1, 1)) = A(0, 2) = 4. Also, we have A(1, 1) = 2, A(1, 2) = A(0, A(1, 1)) = A(0, 2) = 4, A(1, 3) = A(0, A(1, 2)) = A(0, 4) = 8, and A(1, 4) = A(0, A(1, 3)) = A(0, 8) = 16. More generally, A(1, n) = 2 n. This is Exercise 50. 7
8 a) A(2, 3) = A(1, A(2, 2)) = A(1, 4) = 16. b) A(3, 3) = A(2, A(3, 2)) = A(2, A(1, A(3, 1))) = A(2, A(1, 2)) = A(2, 4) = A(1, A(2, 3)) = A(1, 16) = 2 16 =
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# Adding Fractions With Common Denominators
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Students must add and simplify fractions with common denominators.
• 1.
### Add and reduce to simplest term.5/8 + 1/8
Explanation
To add fractions, we need a common denominator. In this case, both fractions have a denominator of 8. So, we can add the numerators together and keep the denominator the same. 5/8 + 1/8 = (5+1)/8 = 6/8. To simplify the fraction, we can divide both the numerator and denominator by their greatest common divisor, which is 2. Dividing 6 by 2 and 8 by 2 gives us 3/4. Therefore, the simplest form of 5/8 + 1/8 is 3/4.
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• 2.
### Add and reduce to simplest term.1/7 + 2/7
Explanation
To add fractions, we need to have the same denominator. In this case, both fractions have a denominator of 7, so we can simply add the numerators. Adding 1 and 2 gives us 3, and the denominator remains 7. Therefore, the sum of 1/7 and 2/7 is 3/7.
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• 3.
### Add and reduce to simplest term.2/9 + 2/9
Explanation
The given question asks to add the fractions 2/9 and 2/9 and then reduce the result to its simplest form. When we add 2/9 and 2/9, we get 4/9. This fraction cannot be simplified any further, so the answer is 4/9.
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• 4.
### Add and reduce to simplest term.4/13 + 7/13
Explanation
The given question asks to add 4/13 and 7/13 and reduce the result to its simplest form. When we add the numerators (4+7), we get 11 as the numerator. Since the denominators are the same (13), we keep it the same in the result. Therefore, the sum of 4/13 and 7/13 is 11/13.
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• 5.
### Add and reduce to simplest term.7/15 + 2/15
Explanation
To add fractions, we need a common denominator. In this case, both fractions already have a denominator of 15, so we can simply add the numerators. 7/15 + 2/15 equals 9/15. To simplify this fraction, we can divide both the numerator and denominator by their greatest common factor, which is 3. So, 9/15 simplifies to 3/5.
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• 6.
### Add and reduce to simplest term.3/10 + 1/10
Explanation
To add fractions, we need to have a common denominator. In this case, both fractions have a denominator of 10, so we can simply add the numerators. 3/10 + 1/10 equals 4/10, which can be simplified to 2/5 by dividing both the numerator and denominator by 2. Therefore, the correct answer is 2/5.
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# How do you solve x^2<=4x-2 using a sign chart?
Apr 4, 2017
$2 - \sqrt{2} \le x \le 2 + \sqrt{2}$
#### Explanation:
We have ${x}^{2} \le 4 x - 2$ i.e. ${x}^{2} - 4 x + 2 \le 0$ and using the quadratic formula is
$x = \frac{4 \pm \sqrt{{4}^{2} - 4 \times 1 \times 2}}{2} = 2 \pm \sqrt{2}$
and our inequality is therefore
$\left(x - 2 + \sqrt{2}\right) \left(x - 2 - \sqrt{2}\right) \le 0$
From this, we know that the product $\left(x - 2 + \sqrt{2}\right) \left(x - 2 - \sqrt{2}\right)$ is negative or equal to $0$. It is apparent that sign of binomials $\left(x - 2 + \sqrt{2}\right)$ and $x - 2 - \sqrt{2}$ will change around the values $2 - \sqrt{2}$ and $2 + \sqrt{2}$ respectively. In a sign chart, we divide the real number line using these values, i.e. below $2 - \sqrt{2}$, between $2 - \sqrt{2}$ and $2 + \sqrt{2}$ and above $2 + \sqrt{2}$ and see how the sign of ${x}^{2} - 4 x + 2$ changes.
Sign Chart
$\textcolor{w h i t e}{X X X X X X X X X X X} 2 - \sqrt{2} \textcolor{w h i t e}{X X X X X} 2 + \sqrt{2}$
$\left(x - 2 + \sqrt{2}\right) \textcolor{w h i t e}{X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X X X} + i v e$
$\left(x - 2 - \sqrt{2}\right) \textcolor{w h i t e}{X} - i v e \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e$
$\left({x}^{2} - 4 x + 2\right) \textcolor{w h i t e}{X X} + i v e \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e$
It is observed that ${x}^{2} - 4 x + 2 \le 0$ when either $x \ge 2 - \sqrt{2}$ or $x \le 2 + \sqrt{2}$ i.e. $x$ lies between $2 - \sqrt{2}$ and $2 + \sqrt{2}$ including these numbers or $2 - \sqrt{2} \le x \le 2 + \sqrt{2}$, which is the solution for the inequality.
In interval form solution is $\left[2 - \sqrt{2} , 2 + \sqrt{2}\right]$
graph{x^2-4x+2 [-2.976, 7.024, -2.5, 2.5]}
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You can find out the tangent of an angle by using the angle() function. You will need a real number x and an angle to calculate the tangent of the angle. The function will return a real number in the interval -p 2 to p 2. When you have the tangent and the angle, you can use the angle() function to find the degree measure of the angle.
## tan
The tangent of an angle is a trigonometric measure that shows the ratio of the two sides of a right triangle. The tangent ratio is the same for any size of right triangle. You can think of the tangent as a function. However, the tangent ratio may take on different values depending on the degree measure or the radian measure of the angle. It is sometimes abbreviated as “cot.”
The tangent ratio is used in many math calculations. For example, let’s assume that an angle is one degree above or below zero. The tangent of an angle of one degree is equal to one half of the angle’s height. This angle is also known as a right-angled triangle.
A right triangle ABC is a right-angled triangle. Its tangent is at a point where the lines cross. This is the same as the tangent of an angle of the same size. If there are two tangent circles in the triangle ABC, one of them will be perpendicular to the circle’s radius. In the diagram below, the tangent circle is two-thirds of the radius of the circle AB.
## tan ratio
If you want to calculate the Tan ratio of an angle whose tangency is 1.19, then you have to know the sine of the angle and the cosine of the tangent. These two formulas are related to one another and it is very important to understand them properly. The sine of the angle equals 86.8 degrees, and the cosine of the angle equals 86.6 degrees.
In trigonometry, the tangent of an angle is the side of the right triangle adjacent to the opposite side. It is the same for right triangles of different sizes. It can also be thought of as a function and will have different values depending on the measure of the angle, which may be measured in degrees or radians. It is also abbreviated as cot.
The Tan ratio of an angle whose tangency is 1.19 is the ratio between the two sides of an angle. It can be calculated using a logarithmic scale. The two sides of a triangle are 90 degrees and c2, and their tangents are 90 degrees. If you’d like to calculate the Tan ratio of an angle whose tangent is 1.19, you can use the following formulas:
Arc cos x – cos a y – cos a tan ratio. If you want to find the tan ratio of an angle whose tangent is 1.19, you can use the cos x-cos -x.
The Cosine of an angle whose tangent is one-hundred degrees is 180 degrees. Therefore, it is 90 degrees plus A. Therefore, a cos x 90 degrees is 210 degrees. Moreover, the Cos x 90 degrees – x is 180 degrees is equal to 180 degrees cos x y.
The Tan ratio of an angle whose tangential is 1.19 is cos 2 495 deg, cos 215 deg, and cos x – y. Thus, an angle whose tangent is one hundred twenty-one degrees is cos x.
## tan value
The tangent function of an angle is defined as the ratio of x to y. It can be found by making a reference triangle from the terminal ray of th to the x-axis. This tangent function repeats every 180 degrees.
The tangent function can be written in radians instead of degrees. One full rotation of an angle is equal to 2 radians. For example, a thirty-degree angle is 1/60 of a full rotation of a circle, or 30 radians. In the same way, a one-sixth rotation of an angle of 180 degrees is equal to thirty radians. To determine the tangent value of an angle whose tangent value is 1.19, students need to plot the points on the y-axis that are equal to tan(th).
The angle whose tangent is 1.19 has a tan value of 0.86. If the angle is a right triangle, the angle will be 45 degrees. In the same way, the angle whose tangent is tan A = 2 will have a tan of 60deg.
You can use a calculator to find the tangent of an angle. The first step is to set your calculator to degrees. You may also want to refer to your calculator’s manual for more information. Then, you will be able to find the tangent of an angle with an unknown side.
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# Problem Solving Plan, Diagrams
## Develop selected strategies for problem solving.
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Problem Solving Plan, Diagrams
Credit: Andrea Goh
Source: https://www.flickr.com/photos/thebokeeffect/7308563880/
Katie is trying out a new cake recipe. The cake recipe said to add \begin{align*}2 \frac{1}{4}\end{align*} cups of flour. The cake turned out to be too moist. She tried the recipe again, but added \begin{align*}\frac{1}{3}\end{align*} cup of flour to the recipe. The cake was much better this time. How many cups of flour did Katie add in all?
In this concept, you will learn how to use the problem solving strategy: draw a diagram.
### Problem-Solving with Diagrams
Drawing a diagram or a picture is a strategy to help you solve many different problems involving fractions. The first thing that you should do when approaching a problem is to identify the given information and what the problem is asking for. Here is an example.
John ate \begin{align*}\frac{1}{5}\end{align*} of the cake. What fraction is left?
First, the problem gives you the amount of cake that John ate. The solution is how much cake he has left. This problem requires subtraction. Draw a diagram to show what is known about John and his cake.
This is a diagram of fraction bars to represent John’s cake. The blue section shows how much of the cake John has eaten. The white bars represent the amount of cake that is left. You can see that there are four-fifths left.
The answer to the problem is John has \begin{align*}\frac{4}{5}\end{align*} of the cake left.
You can use a diagram to solve a problem or create a numeric expression. Often times both ways will work, but one will make more sense than the other. Let's look at another problem.
Shannon jogged \begin{align*}1 \frac{3}{20}\end{align*} miles yesterday. Today, she jogged \begin{align*}\frac{1}{2}\end{align*} mile. How many total miles did Shannon jog?
Method 1 –– Draw a diagram:
One way to solve this problem is to draw a diagram. Look at the first distance that Shannon jogged. Draw two same-sized rectangles. Divide the rectangles into 20 equal-sized sections. Then shade \begin{align*}1 \frac{3}{20}\end{align*} of the diagram.
This represents the \begin{align*}1 \frac{3}{20}\end{align*} miles that Shannon jogged yesterday.
Shannon also jogged \begin{align*}\frac{1}{2}\end{align*} mile today.
So, shade \begin{align*}\frac{1}{2}\end{align*} of the partially filled rectangle to represent the distance she jogged today.
The diagram is \begin{align*}1 \frac{13}{20}\end{align*} shaded. Therefore, Shannon jogged a total of \begin{align*}1 \frac{13}{20}\end{align*} miles on those two days.
Method 2–– Set up an addition problem:
To find out how many miles she jogged all together, add \begin{align*}1 \frac{3}{20} + \frac{1}{2}\end{align*}. The fractional part of the mixed number has a different denominator than \begin{align*}\frac{1}{2}\end{align*}. Find the least common multiple (LCM) of both denominators. The least common multiple of 20 and 2 is 20. Next, rename the fractions with a common denominator.
\begin{align*}\frac{1}{2} = \frac{10}{20}\end{align*}
Now you can add the fractions together.
\begin{align*}1 \frac{3}{20} + \frac{10}{20} = 1 \frac{13}{20}\end{align*}
Notice that the answer is the same. Both methods will produce the same result. Choose the method that you find easiest when working on problems like this.
### Examples
#### Example 1
Earlier, you were given a problem about Katie's cake recipe.
Katie started with \begin{align*}2\frac{1}{4}\end{align*} cups of flour, but added \begin{align*}\frac{1}{3}\end{align*} cup of flour to the recipe. Add the flour quantities to find the total amount of flour used for the cake.
\begin{align*}2\frac{1}{4} + \frac{1}{3}\end{align*}
Look at the fractions. They do not have a common denominator.
First, rewrite the fractions with a common denominator. The LCM of 3 and 4 is 12.
\begin{align*}2\frac{1}{4} = 2\frac {3}{12} \quad \quad \frac{1}{3} = \frac {4}{12}\end{align*}
\begin{align*}2\frac{1}{4} + \frac{1}{3} =2\frac{3}{12} + \frac{4}{12}\end{align*}
Then, draw a diagram to find the sum
\begin{align*}2\frac{3}{12} + \frac{4}{12} = 2\frac{7}{12}\end{align*}
Katie used a total of \begin{align*}2\frac{7}{12}\end{align*} cups of flour.
#### Example 2
Use problem solving strategies-- make a diagram and set up an addition problem-- to solve.
Teri ran \begin{align*}1 \frac{1}{2}\end{align*} miles yesterday, and she ran \begin{align*}2 \frac{1}{2}\end{align*} miles today. How many miles did she run in all?
If John ran 7 miles, what is the difference between his total miles and Teri’s total miles? How many miles have they run altogether?
First, draw a diagram to find the total miles for Teri.
\begin{align*}1 \frac{1}{2} + 2 \frac{1}{2} = 4\end{align*}
Terri ran 4 miles.
Next, find the difference between Teri and John's total miles.
7 - 4 = 3
There is a difference of 3 miles.
Finally, find the sum of their total miles.
7 + 4 = 11
Together, they ran 11 miles.
#### Example 3
Draw a diagram to solve the problem: \begin{align*}2 \frac{1}{3} + 4\end{align*}.
The sum is \begin{align*}6 \frac{1}{3}\end{align*}.
#### Example 4
Draw a diagram to solve the problem: \begin{align*}\frac{4}{5} - \frac{1}{5}\end{align*}.
The difference is \begin{align*}\frac{3}{5}\end{align*}.
#### Example 5
Draw a diagram to solve the problem: \begin{align*}\frac{3}{4} + \frac{2}{4}\end{align*}.
The sum is \begin{align*}1 \frac{1}{4}\end{align*}.
### Review
Solve each of the following problems by using a problem solving strategy.
1. Tyler has eaten one-fifth of the pizza. If he eats another two-fifths of the pizza, what part of the pizza does he have left?
2. What part has he eaten in all?
3. How many parts of this pizza make a whole?
4. Maria decides to join Tyler in eating pizza. She orders a vegetarian pizza with six slices. If she eats two slices of pizza, what fraction has she eaten?
5. What fraction does she have left?
6. If Tyler was to eat half of Maria’s pizza, how many pieces would that be?
7. If Maria eats one-third, and Tyler eats half, what fraction of the pizza is left?
8. How much of the pizza have they eaten altogether?
9. John and Terri each ran 18 miles. If Kyle ran half the distance that both John and Teri ran, how many miles did he run?
10. If Jeff ran \begin{align*}3 \frac{1}{2}\end{align*} miles, how much did he and Kyle run altogether?
11. What is the distance between Jeff and Kyle’s combined mileage and John and Teri’s combined mileage?
12. Sarah gave Joey one-third of the pie. Kara gave him one-fourth of another pie. How much pie did Joey receive altogether?
13. Is this less than or more than one-half of a pie?
14. Who gave Joey a larger part of the pie, Kara or Sarah?
15. What is the difference between the two fractions of pie?
To see the Review answers, open this PDF file and look for section 6.18.
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Diagram A diagram is a drawing used to represent a mathematical problem.
Problem Solving Problem solving is using key words and operations to solve mathematical dilemmas written in verbal language.
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# Polls to probabilities
How likely is it that your favorite candidate will win the election? If your candidate is ahead of their opponent by 5%, are they certain to win? What about 10%? Or if they're down by 2%, are they out of the race? Victory probabilities are related to how far ahead or behind a candidate is in the polls, but the relationship isn't a simple one and has some surprising consequences as we'll see.
# Opinion poll example
Let's imagine there's a hard-fought election between candidates A and B. A newspaper publishes an opinion poll a few days before the election:
• Candidate A: 52%
• Candidate B: 48%
• Sample size: 1,000
Should candidate A's supporters pop the champagne and candidate B's supporters start crying?
# The spread and standard error
Let's use some standard notation. From the theory of proportions, the mean and standard error for the proportion of respondents who chose A is:
$p_a = {n_a \over n}$ $\sigma_a = { \sqrt {{p_a(1-p_a)} \over n}}$
where $$n_a$$ is the number of respondents who chose A and $$n$$ is the total number of respondents. If the proportion of people who answered candidate B is $$p_b$$, then obviously, $$p_a + p_b = 1$$.
Election probability theory usually uses the spread, $$d$$, which is the difference between the candidates: $d = p_a - p_b = 2p_a - 1$ From statistics theory, the standard error of $$d$$ is: $\sigma_d = 2\sigma_a$ (these relationships are easy to prove, but a bit tedious, if anyone asks, I'll show the proof.)
Obviously, for a candidate to win, their spread, $$d$$, must be > 0.
# Everything is normal
From the central limit theorem (CLT), we know $$p_a$$ and $$p_b$$ are normally distributed, and also from the CLT, we know $$d$$ is normally distributed. The next step to probability is viewing the normal distribution for candidate A's spread. The chart below shows the normal distribution with mean $$d$$ and standard error $$\sigma_d$$.
As with most things with the normal distribution, it's easier if we transform everything to the standard normal using the transformation: $z = {(x - d) \over \sigma_d}$ The chart below is the standard normal representation of the same data.
The standard normal form of this distribution is a probability density function. We want the probability that $$d>0$$ which is the light green shaded area, so it's time to turn to the cumulative distribution function (CDF), and its complement, the complementary cumulative distribution function (CCDF).
# CDF and CCDF
The CDF gives us the probability that we will get a result less than or equal to some value I'll label $$z_c$$. We can write this as: $P(z \leq z_c) = CDF(z_c) = \phi(z_c)$ The CCDF is defined so that: $1 = P(z \leq z_c) + P(z > z_c)= CDF(z_c) + CCDF(z_c) = \phi(z_c) + \phi_c(z_c)$ Which is a long-winded way of saying the CCDF is defined as: $CCDF(z_c) = P(z_c \gt 0) = \phi_c(z_c)$
The CDF is the integral of the PDF, and from standard textbooks: $\phi(z_c) = {1 \over 2} \left( 1 + erf\left( {z_c \over \sqrt2} \right) \right)$ We want the CCDF, $$P(z > z_c)$$, which is simply 1 - CDF.
Our critical value occurs when the spread is zero. The transformation to the standard normal in this case is: $z_c = {(x - d) \over \sigma_d} = {-d \over \sigma_d}$ We can write the CCDF as: $\phi_c(z_c) = 1 - \phi(z_c) = 1- {1 \over 2} \left( 1 + erf\left( {z_c \over \sqrt2} \right) \right)\$ $= 1 - {1 \over 2} \left( 1 + erf\left( {-d \over {\sigma_d\sqrt2}} \right) \right)$ We can easily show that: $erf(x) = -erf(-x)$ Using this relationship, we can rewrite the above equation as: $P(d > 0) = {1 \over 2} \left( 1 + erf\left( {d \over {\sigma_d\sqrt2}} \right) \right)$
What we have is an equation that takes data we've derived from an opinion poll and gives us a probability of a candidate winning.
# Probabilities for our example
For candidate A:
• $$n=1000$$
• $$p_a = {520 \over 1000} = 0.52$$
• $$\alpha_a = 0.016$$
• $$d = {{520 - 480} \over 1000} = 0.04$$
• $$\alpha_d = 0.032$$
• $$P(d > 0) = 90\%$$
For candidate B:
• $$n=1000$$
• $$p_b = {480 \over 1000} = 0.48$$
• $$\alpha_b = 0.016$$
• $$d = {{480 - 520} \over 1000} = -0.04$$
• $$\alpha_d = 0.032$$
• $$P(d > 0) = 10\%$$
Obviously, the two probabilities add up to 1. But note the probability for candidate A. Did you expect a number like this? A 4% point lead in the polls giving a 90% chance of victory?
# Some consequences
Because the probability is based on $$erf$$, you can quite quickly get to highly probable events as I'm going to show in an example. I've plotted the probability for candidate A for various leads (spreads) in the polls. Most polls nowadays tend to have about 800 or so respondents (some are more and some are a lot less), so I've taken 800 as my poll size. Obviously, if the spread is zero, the election is 50%:50%. Note how quickly the probability of victory increases as the spread increases.
What about the size of the poll, how does that change things? Let's fix the spread to 2% and vary the size of the poll from 200 to 2,000 (the usual upper and lower bounds on poll sizes). Here's how the probability varies with poll size for a spread of 2%.
Now imagine you're a cynical and seasoned poll analyst working on candidate A's campaign. The young and excitable intern comes rushing in, shouting to everyone that A is ahead in the polls! You ask the intern two questions, and then, like the Oracle at Delphi, you predict happiness or not. What two questions do you ask?
• What's the size of the poll?
# What's missing
There are two elephants in the room, and I've been avoiding talking about them. Can you guess what they are?
All of this analysis assumes the only source of error is random noise. In other words, there's no systemic bias. In the real world, that's not true. Polls aren't wholly based on random sampling, and the sampling method can introduce bias. I haven't modeled it at all in this analysis. There are at least two systemic biases:
• Pollster house effects arising from house sampling methods
• Election effects arising from different population groups voting in different ways compared to previous elections.
Understanding and allowing for bias is key to making a successful election forecast. This is an advanced topic for another blog post.
The other missing item is more subtle. It's undecided voters. Imagine there are two elections and two opinion polls. Both polls have 1,000 respondents.
Election 1:
• Candidate A chosen by 20%
• Candidate B chosen by 10%
• Undecided voters are 70%
Election 2:
• Candidate A chosen by 55%
• Candidate B chosen by 45%
• Undecided voters are 0%
In both elections, the spread from the polls is 10%, so candidate A has the same higher chance of winning in both elections, but this doesn't seem right. Intuitively, we should be less certain about an election with a high number of undecided voters. Modeling undecided voters is a topic for another blog post!
The best source of election analysis I've read is in the book "Introduction to data science" and the associated edX course "Inference and modeling", both by Rafael Irizarry. The analysis in this blog post was culled from multiple books and websites, each of which only gave part of the story.
# Serial numbers and losing business
Here's a story about how something innocuous and low-level like serial numbers can damage your reputation and lose you business. I have advice on how to avoid the problem too!
(Serial numbers can give away more than you think. Image source: Wikimedia Commons. License: Public Domain.)
# Numbered by design
Years ago, I worked for a specialty manufacturing company, its products were high precision, low-volume, and expensive. The industry was cut-throat competitive, and commentary in the press was that not every manufacturer would survive; as a consequence, customer confidence was critical.
An overseas customer team came to us to design a specialty item. The company spent a week training them and helping them design what they wanted. Of course, the design was all on a CAD system with some templated and automated features. That's where the trouble started.
One of the overseas engineers spotted that a customer-based serial number was automatically included in the design. Unfortunately, the serial number was 16, implying that the overseas team was only the 16th customer (which was true). This immediately set off their alarm bells - a company with only 16 customers was probably not going to survive the coming industry shake-out. The executive team had to smooth things over, which included lying about the serial numbers. As soon as the overseas team left, the company changed its system to start counting serial numbers from some high, but believable number (something like 857).
Here's the point: customers can infer a surprising amount from your serial numbers, especially your volume of business.
# Invoices
Years later, I was in a position where I was approving vendor invoices. Some of my vendors didn't realize what serial numbers could reveal, and I ended up gaining insight into their financial state. Here are the rules I used to figure out what was going on financially, which was very helpful when it came to negotiating contract renewals.
• If the invoice is unnumbered, the vendor is very small and they're likely to have only a handful of customers. All accounting systems offer invoice generation and they all number/identify individual invoices. If the invoice doesn't have a serial number, the vendor's business is probably too small to warrant buying an accounting system, which means a very small number of customers.
• Naive vendors will start invoice numbering from 1, or from a number like 1,000. You can infer size if they do this.
• Many accounting systems will increment invoice numbers by 1 by default. If you're receiving regular invoices from a vendor, you can use this to infer their size too. If this month's invoice is 123456 and next month's is 123466, this might indicate 10 invoices in a month and therefore 10 customers. You can do this for a while and spot trends in a vendor's customer base, for example, if you see invoices incrementing by 100 and later by 110, this may be because the vendor has added 10 customers.
The accounting tool suppliers are wise to this, and many tools offer options for invoice numbering that stop this kind of analysis (e.g. starting invoices from a random number, random invoice increments, etc.). But not all vendors use these features and serial number analysis works surprisingly often.
(Destroyed German Tank. Image source: Wikimedia Commons. License: Public Domain)
# The German tank problem
Serial number analysis has been used in wartime too. In World War II, the allied powers wanted to understand the capacity of Nazi industry to build tanks. Fortunately, German tanks were given consecutive serial numbers (this is a simplification, but it was mostly true). Allied troops were given the job of recording the serial numbers of captured or destroyed tanks which they reported back. Statisticians were able to infer changes in Nazi tank production capabilities through serial number analysis, which after the war was found to be mostly correct. This is known as the German tank problem and you can read a lot more about it online.
# Simple things say a lot
The bottom line is simple: serial numbers can give away more about your business than you think. They can tell your customers how big your customer base is, and whether it's expanding or contracting; crucial information when it comes to renegotiating contracts. Pay attention to your serial numbers and invoices!
# How opinion polls work on the ground
I worked as a street interviewer for an opinion polling organization and I know how opinion polls are made and executed. In this blog post, I'm going to explain how opinion polls were run on the ground, educate you on why polls can go wrong, and illustrate how difficult it is to run a valid poll. I'm also going to tell you why everything you learned from statistical textbooks about polling is wrong.
(Image Credit: Wikimedia Commons, License: Public Domain)
# Random sampling is impossible
In my experience, this is something that's almost never mentioned in statistics textbooks but is a huge issue in polling. If they talk about sampling at all, textbooks assume random sampling, but that's not what happens.
Random sampling sounds wonderful in theory, but in practice, it can be very hard; people aren't beads in an urn. How do you randomly select people on the street or on the phone - what's the selection mechanism? How do you guard against bias? Let me give you some real examples.
Imagine you're a street interviewer. Where do you stand to take your random sample? If you take your sample outside the library, you'll get a biased sample. If you take it outside the factory gates, or outside a school, or outside a large office complex, or outside a playground, you'll get another set of biases. What about time of day? The people out on the streets at 7am are different from the people at 10am and different from the people at 11pm.
Similar logic applies to phone polls. If you call landlines only, you'll get one set of biases. If you call people during working hours, your sample will be biased (is the mechanic fixing a car going to put down their power tool to talk to you?). But calling outside of office hours means you might not get shift workers or parents putting their kids to bed. The list goes on.
You might be tempted to say, do all the things: sample at 10am, 3pm, and 11pm; sample outside the library, factory, and school; call on landlines and mobile phones, and so on, but what about the cost? How can you keep opinion polls affordable? How do you balance calls at 10am with calls at 3pm?
Because there are very subtle biases in "random" samples, most of the time, polling organizations don't do wholly 'random' sampling.
# Sampling and quotas
If you can't get a random sample, you'd like your sample to be representative of a population. Here, representative means that it will behave in the same way as the population for the topics you're interested in, for example, voting in the same way or buying butter in the same way. The most obvious way of sampling is demographics: age and gender etc.
Let's say you were conducting a poll in a town to find out residents' views on a tax increase. You might find out the age and gender demographics of the town and sample people in a representative way so that the demographics of your sample match the demographics of the town. In other words, the proportion of men and women in your sample matches that of the town, the age distribution matches that of the town, and so on.
(US demographics. Image credit: Wikimedia Commons. License: Public domain)
In practice, polling organizations use a number of sampling factors depending on the survey. They might include sampling by:
• Gender
• Age
• Ethnicity
• Income
• Social class or employment category
• Education
but more likely, some combination of them.
In practice, interviewers may be given a sheet outlining the people they should interview, for example, so many women aged 45-50, so many people with degrees, so many people earning over $100,000, and so on. This is often called a quota. Phone interviews might be conducted on a pre-selected list of numbers, with guidance on how many times to call back, etc. Some groups of people can be very hard to reach, and of course, not everyone answers questions. When it comes to analysis time, the results are weighted to correct bias. For example, if the survey could only reach 75% of its target for men aged 20-25, the results for men in this category might be weighted by 4/3. # Who do you talk to? Let's imagine you're a street interviewer, you have your quota to fulfill, and you're interviewing people on the street, who do you talk to? Let me give you a real example from my polling days; I needed a man aged 20-25 for my quota. On the street, I saw what looked like a typical and innocuous student, but I also saw an aggressive-looking skinhead in full skinhead clothing and boots. Who would you choose to interview? (Image credit: XxxBaloooxxx via Wikimedia Commons. License: Creative Commons.) Most people would choose the innocuous student, but that's introducing bias. You can imagine multiple interviewers making similar decisions resulting in a heavily biased sample. To counter this problem, we were given guidance on who to select, for example, we were told to sample every seventh person or to take the first person who met our quota regardless of their appearance. This at least meant we were supposed to ask the skinhead, but of course, whether he chose to reply or not is another matter. The rules sometimes led to absurdity. I did a survey where I was supposed to interview every 10th person who passed by. One man volunteered, but I said no because he was the 5th person. He hung around so long that eventually, he became the 10th person to pass me by. Should I have interviewed him? He met the rules and he met my sampling quota. I came across a woman who was exactly what I needed for my quota. She was a care worker who had been on a day trip with severely mentally handicapped children and was in the process of moving them from the bus to the care home. Would you take her time to interview her? What about the young parent holding his child when I knocked on the door? The apartment was clearly used for recent drug-taking. Would you interview him? As you might expect, interviewers interpreted the rules more flexibly as the deadline approached and as it got later in the day. I once interviewed a very old man whose wife answered all the questions for him. This is against the rules, but he agreed with her answers, it was getting late, and I needed his gender/age group/employment status for my quota. The company sent out supervisors to check our work on the streets, but of course, supervisors weren't there all the time, and they tended to vanish after 5pm anyway. The point is, when it comes to it, there's no such thing as random sampling. Even with quotas and other guided selection methods, there are a thousand ways for bias to creep into sampling and the biases can be subtle. The sampling methodology one company uses will be different from another company's, which means their biases will not be the same. # What does the question mean? One of the biggest lessons I learned was the importance of clear and unambiguous questions, and the unfortunate creativity of the public. All of the surveys I worked on had clearly worded questions, and to me, they always seemed unambiguous. But once you hit the streets, it's a different world. I've had people answer questions with the most astonishing levels of interpretation and creativity; regrettably, their interpretations were almost never what the survey wanted. What surprised me was how willing people were to answer difficult questions about salary and other topics. If the question is worded well (and I know all the techniques now!), you can get strangers to tell you all kinds of things. In almost all cases, I got people to tell me their age, and when required, I got salary levels from almost everyone. A well-worded question led to a revelation that shocked me and shook me out of my complacency. A candidate had unexpectedly just lost an election in the East End of London and the polling organization I worked for had been contracted to find out why. To help people answer one of the questions, I had a card with a list of reasons why the candidate lost, including the option: "The candidate was not suitable for the area." A lot of people chose that as their reason. I was naive and didn't know what it meant, but at the end of the day, I interviewed a white man in pseudo-skinhead clothes, who told me exactly what it meant. He selected "not suitable for the area" as his answer and added: "She was black, weren't she?". The question setters weren't naive. They knew that people would hesitate before admitting racism was the cause, but by carefully wording the question and having people choose from options, they provided a socially acceptable way for people to answer the question. Question setting requires real skill and thought. (Oddly. there are very few technical resources on wording questions well. The best I've found is: "The Art of Asking Questions", by Stanley Le Baron Payne, but the book has been out of print for a long time.) # Order, order Question order isn't accidental either, you can bias a survey by the order you ask questions. Of course, you have to avoid leading questions. The textbook example is survey questions on gun control. Let's imagine there were two surveys with these questions: Survey 1: • Are you concerned about violent crime in your neighborhood? • Do you think people should be able to protect their families? • Do you believe in gun control? Survey 2: • Are you concerned about the number of weapons in society? • Do you think all gun owners secure their weapons? • Do you believe in gun control? What answers do you think you might get? As well as avoiding bias, question order is important to build trust, especially if the topic is a sensitive one. The political survey I did in the East End of London was very carefully constructed to build the respondent's trust to get to the key 'why' question. This was necessary for other surveys too. I did a survey on police recruitment, but as I'm sure you're aware, some people are very suspicious of the police. Once again, the survey was constructed so the questions that revealed it was about police recruitment came later on after the interviewer (me!) had built some trust with the respondent. # How long is the survey? This is my favorite story from my polling days. I was doing a survey on bus transport in London and I was asked to interview people waiting for a bus. The goal of the survey was to find out where people were going so London could plan for new or changed bus routes. For obvious reasons, the set of questions were shorter than usual, but in practice, not short enough; a big fraction of my interviews were cut short because the bus turned up! In several cases, I was asking questions as people were getting on the bus, and in a few cases, we had a shouted back and forth to finish the survey before their bus pulled off out of earshot. (Image credit: David McKay via Wikimedia Commons. License: Creative Commons) To avoid exactly this sort of problem, most polling organizations use pilot surveys. These are test surveys done on a handful of people to debug the survey. In this case, the pilot should have uncovered the fact that the survey was too long, but regrettably, it didn't. (Sometime later, I designed and executed a survey in Boston. I did a pilot survey and found that some of my questions were confusing and I could shorten the survey by using a freeform question rather than asking for people to choose from a list. In any survey of more than a handful of respondents, I strongly recommend running a pilot - especially if you don't have a background in polling.) The general lesson for any survey is to keep it as short as possible and understand the circumstances people will be in when you're asking them questions. # What it all means - advice for running surveys Surveys are hard. It's hard to sample right, it's hard to write questions well, and it's hard to order questions to avoid bias. Over the years, I've sat in meetings when someone has enthusiastically suggested a survey. The survey could be a HR survey of employees, or a marketing survey of customers, or something else. Usually, the level of enthusiasm is inversely related to survey experience. The most enthusiastic people are often very resistant to advice about question phrasing and order, and most resistant of all to the idea of a pilot survey. I've seen a lot of enthusiastic people come to grief because they didn't listen. If you're thinking about running a survey, here's my advice. • Make your questions as clear and unambiguous as you can. Get someone who will tell you you're wrong to review them. • Think about how you want the questions answered. Do you want freeform text, multiple choice, or a scale? Surprisingly, in some cases, free form can be faster than multiple choice. • Keep it short. • Always run a pilot survey. # What it means - understanding polling results Once you understand that polling organizations use customized sampling methodologies, you can understand why polling organizations can get the results wrong. To put it simply, if their sampling methodology misses a crucial factor, they'll get biased results. The most obvious example is state-level polling in the US 2016 Presidential Election, but there are a number of other polls that got very different results from the actual election. In a future blog post, I'll look at why the 2016 polls were so wrong and why polls were wrong in other cases too. # If you liked this post, you might like these ones ## Monday, July 27, 2020 ### The Electoral College for beginners # The (in)famous electoral college We're coming up to the US Presidential election so it's time for pundits and real people to discuss the electoral college. There's a lot of misunderstanding about what it is, its role, and its potential to undermine democracy. In this post, I'm going to tell you how it came to be, the role it serves, and some issues with it that may cause trouble. (Ohio Electoral College 2012. Image credit: Wikimedia Commons. Contributor: Ibagli. License: Creative Commons.) # How it came to be The thirteen original colonies had the desire for independence in common but had stridently different views on government. In the aftermath of independence, the US was a confederacy, a country with a limited and small (federal) government. After about ten years, it became obvious that this form of government wasn't working and something new was needed. So the states created a Constitutional Convention to discuss and decide on a new constitution and form of government. Remember, the thirteen states were the size of European countries and had very different views on issues like slavery. The states with smaller populations were afraid they would be dominated by the more populous states, which was a major stumbling block to agreements. The issue was resolved by the Great Compromise (or Connecticut Compromise if you come from Connecticut). The Convention created a two-chamber congress and a more powerful presidency than before. Here's how they were to be elected: • The lower house, the House of Representatives, was to have representatives elected in proportion to the population of the state (bigger states get more representatives). • The upper house, the Senate, was to have two Senators per state regardless of the population of the state • Presidents were to be elected through an electoral college, with each elector having one vote. Each state would be allocated a number of electors (and hence votes) based on their seats in congress. The electors would meet and vote for the President. For example in 1789, the state of New Hampshire had three representatives and two senators, which meant New Hampshire sent five electors (votes) to the electoral college. The states decided who the electoral college electors were. Think for a minute about why this solution worked. The states were huge geographically with low population densities and often poor communications. Travel was a big undertaking and mail was slow. It made sense to send voters to vote on your behalf at a college and these delegates may have to change their vote depending on circumstances. In short, the electoral college was a way of deciding the presidency in a big country with slow communications. Electoral college vote allocation is and was only partially representative of the underlying population size. Remember, each state gets two Senators (and therefore two electoral college votes) regardless of its population. This grants power disproportionately to lower-population states, which is a deliberate and intended feature of the system. # Early practice Prior to the formation of modern political parties, the President was the person who got the largest number of electoral college votes and the Vice-President was the person who got the next highest number of votes. For example, in 1792, George Washington was re-elected President with 132 votes, and the runner-up, John Adams, who got 77 votes, became Vice-President. This changed when political parties made this arrangement impractical, and by 1804, the President and the Vice-President were on the same ticket. Electoral college electors were originally selected by state legislators, not by the people. As time went on, more states started directly electing electoral college electors. In practice, this meant the people chose their Presidential candidate and the electoral college electors duly voted for them. By the late 19th century, all states were holding elections for the President and Vice-President through electoral college representation. # Modern practice Each state has the following representation in congress: • Two Senators • A number of House of Representative seats roughly related to the state's population. The size of each state's congressional delegation is their number of electoral college votes. For example, California has 53 Representatives and 2 Senators giving 55 electoral college electors and 55 electoral college votes. During a Presidential election, the people in each state vote for who they want for President (and by extension, Vice-President). Although it's a federal election, the voting is entirely conducted by each state; the ballot paper is different, the counting process is different, and the supervision is different. Most states allocate their electoral college votes on a winner-takes-all basis, the person with the largest share of the popular vote gets all the electoral college votes. For example, in 2016, the voting in Pennsylvania was: 2,926,441 votes for Hilary Clinton and 2,970,733 votes for Donald Trump, and Donald Trump was allocated all of Pennsylvania's electoral college votes. Two states do things a little differently. Maine and Nebraska use the Congressional District method. They allocate one of their electoral college votes to each district used to elect a member of the House of Representatives. The winner of the statewide vote is then allocated the other two electoral college votes. In Maine in 2016, Hilary Clinton won three electoral college votes and Donald Trump one. Washington D.C. isn't a state and doesn't have Senators; it has a non-voting delegate to the House of Representatives. However, it does have electoral college votes! Under the 23rd amendment to the Constitution, it has the same electoral college votes as the least populous state (currently 3). In total, there are 538 electoral college votes: • 100 Senators • 435 Representatives • 3 Electors for Washington D.C. The electoral college does not meet as one body in person. Electors meet in their respective state capitols and vote for President. # How electoral college votes are decided How are electoral college votes allocated to states? I've talked about the formula before, 2 Senators for every state plus the number of House of Representative seats. House of Representative seats are allocated on a population basis using census data. There are 435 seats that are reallocated every ten years based on census data. Growing states may get more seats and shrinking states fewer. This is why the census has been politicized from time to time - if you can influence the census you can gain a ten-year advantage for your party. # Faithless electors and the Supreme Court Remember, the electors meet and vote for President. Let's imagine we have two Presidential candidates, cat and dog, and that the people of the state vote for cat. What's to stop the electors voting for dog instead? Nothing at all. For many states, there's nothing to stop electors voting for anyone regardless of who won the election in the state. This can and does happen, even as recently as 2016. It happens so often that there's a name for them: faithless electors. In 2016, five electors who should have voted for Hilary Clinton didn't vote for her, and two who should have voted for Donald Trump didn't vote for him. These votes were officially accepted and counted. Several states have laws that mandate that electors vote as instructed or provide punishment for electors who do not vote as instructed. These laws were challenged in the Supreme Court, which voted to uphold them. On the face of it, faithless electors sound awful, but I do have to say a word in their defense. They do have some support from the original intent of the Constitutional Convention and they do have some support from the Federalist Papers. It's not entirely as black and white as it appears to be. Have faithless electors ever swayed a Presidential election? No. Could they? Yes. # Gerrymandering In principle, it's possible to gerrymander electoral college votes, but it hasn't been done in practice. Let me explain how a gerrymander could work. First off, you'd move to Congressional District representation. Because the shape of congressional districts are under state control, you could gerrymander these districts to your heart's content. Next, you'd base your senatorial electoral college votes on the congressional district winner on a winner-takes-all basis. Let's say you had 10 congressional districts and you'd gerrymandered them so your party could win 7. Because 7 of the 10 districts would be for one candidate, you'd award your other two votes to that candidate. In other words, a candidate could lose the popular vote but still gain the majority of the electoral college votes for a state. # The electoral college and representative democracy Everyone knows that Hilary Clinton won the popular vote but Donald Trump won the electoral college and became President. This was a close election, but it's theoretically possible for a candidate to lose the popular vote by a substantial margin, yet still win the presidency. Bear in mind what I said at the beginning of this piece, electoral college votes are not entirely representative of the population, by design. Here's a chart of electoral college votes per 1,000,000 population for 2020. Note how skewed it is in favor of low-population (and rural) states. If you live in Wyoming your vote is worth 5 times that of a voter in Texas. Obviously, some states are firmly Democratic and others firmly Republican. The distribution of electoral college votes pushes candidates to campaign more heavily in small swing states, giving them an outsize influence (for example, New Hampshire). Remember, your goal as a candidate is to win electoral college votes, your goal is not to win the popular vote. You need to focus your electoral spending so you get the biggest bang for your buck in terms of electoral college votes, which means small swing states. # Nightmare scenarios Here are two scenarios that are quite possible with the current system: Neither of these scenarios is good for democracy or stability. There is nothing to prevent them now. # Who else uses an electoral college? Given the problems with an electoral college, it's not surprising that there aren't many other cases in the world of its use. According to Wikipedia, there are several other countries that use it for various elections, but they are a minority. # Could the electoral college be changed for another system? Yes, but it would take a constitutional change, which is a major undertaking and would require widespread cross-party political support. Bear in mind, a more representative system (e.g. going with the popular vote) would increase the power of the more populous states and decrease the power of less populous states - which takes us all the way back to the Great Compromise and the Constitutional Convention. # What's next? I hope you enjoyed this article. I intend to write more election-based pieces as November comes closer. I'm not going to endorse or support any candidate or party; I'm only interested in the process of democracy! # If you liked this post, you might like these ones # Forecasting the 2020 election: a retrospectiveWhat do presidential approval polls really tell us?Fundamentally wrong? Using economic data as an election predictor - why I distrust forecasting models built on economic and other dataCan you believe the polls? - fake polls, leading questions, and other sins of opinion polling.President Hilary Clinton: what the polls got wrong in 2016 and why they got it wrong - why the polls said Clinton would win and why Trump did.Poll-axed: disastrously wrong opinion polls - a brief romp through some disastrously wrong opinion poll results.Who will win the election? Election victory probabilities from opinion pollsSampling the goods: how opinion polls are made - my experiences working for an opinion polling company as a street interviewer.The electoral college for beginners - how the electoral college works ## Monday, July 20, 2020 ### Sad! How not to create a career structure: good intentions gone wrong # Sad! I worked for a company that tried to do a good thing for people's careers and promotion prospects, but it back-fired horribly and ended up hurting more than it helped. I'm going to tell you the story and draw some lessons from it. Of course, I've changed and obscured some details to protect the guilty. (Sometimes doing the right thing badly can turn even placid software developers into a mob. Image source: Wikimedia Commons, License: Public Domain.) # The company The company was a large corporation with several hundred software developers working in different divisions on different projects. There was a formal grade level for developers: a developer fresh out of college might be level 20 and an experienced and senior developer might be level 30. Each level had a pay band with a rigid cap; once you reached the cap, you could only get more money by promotion. The pay bands overlapped and people knew them. Everyone was keenly aware of their level and there was open discussion of who was on what level. The problem was, the standards and levels were inconsistent across and within departments. Some departments and managers were known to be 'generous' and others were known to be 'mean'. Some developers moved departments as the only way to get promoted, but there were problems with this approach too. Some departments had one set of rules for pay and levels, while others had different rules. In some cases, developers tried to move to other groups to get a promotion, but they were offered jobs at the bottom of the pay band at that level, which unfortunately was less money than they were currently on. Managerial inconsistencies had bad consequences for individuals too. In one case, a person got a big promotion and their manager left not long after. Their new manager felt they were underperforming and that they had been over-promoted. Some managers promoted based on performance alone, but others insisted on time served on a level. The bottom line was, there were substantial inconsistencies and inequities and the software engineers were unhappy. # The learned organization Fortunately, there was an active learned organization that helped software companies. One of the things this very learned organization did was produce career guidance, specifically, it produced a career hierarchy showing what software developers would be expected to do at different levels of seniority. As with all management solutions, it was a matrix. There were different skills in different areas and some areas were only applicable to more senior positions. To put it simply, they'd mapped out a grading structure and the skills and responsibilities of each grade. On the face of it, this sounded like a perfect solution for my employer: why not roll out this grading structure and use it to decide people's levels? # The roll-out My employer took the skills/grades matrix from the learned society and mapped it to the internal levels 20-30. The idea was, you could see what was expected of you at your level and what skills you needed to develop and the responsibilities you needed to take on to get a promotion. Sounds great! They published this guidance to all developers. Light the blue touch paper... # The fall-out Here's what happened. The people on high grades (around 30) knew they weren't going to be demoted and their salary wasn't going to go down. Most of them noticed that they weren't doing many of the things the skills/grades matrix said they should be doing. They found the whole thing hilarious. They openly joked about the grade/skills matrix and freely talked about all the things they weren't doing that the matrix said they should be. The majority of people were on middling grades (23-27) and this was incendiary to them. Many of the people were doing more than what was expected of them at their grade level. In many cases, they were meeting the requirements two levels above their current level. The net results were urgent discussions with their managers about immediate promotions. The middling grades people were furious, while the high-grade people were laughing about it. It wasn't a happy mix and management had to step in to stifle debate to calm people down. Senior management soon realized this was explosive, so they backpedaled quickly. The roll-out program was canceled due to unexplained 'deficiencies in the process' and it 'not meeting business needs'. A review was promised, which as you might expect, never happened. The subject became taboo and senior management didn't attempt any further reform. Of course, this had its impact. The middling grades realized they had to leave to get career advancement and the senior people realized further advancement was unlikely. The people with get up and go got up and left. What about the people on the lower grades? When the roll-out happened, they were very happy. For the first time, they had clear goals for promotion, and consistency meant they wouldn't have to leave their group to get promoted. They were upset when it was canceled because it was back to the old opaque system. When the review never happened, it dawned on many of them that the company wasn't serious about their futures. Instead of being a positive change for the future, it set the company back several years. # Why did it go so badly wrong? ## Solving the wrong problem What problem was senior management trying to solve? There were two broad problems we faced as a company: • Current inequities and inconsistencies • Future inequities and inconsistencies The skill/grade matrix was a solution to the second problem, but not the first. The first problem was the most serious and the skills/grade matrix just made it more visible. The senior management team had no plan to address the first problem. The leadership applied the right solution to the wrong problem. ## Overnight fix to a long-standing problem Inconsistencies had built up over the course of about a decade, and there were reasons why things developed the way they did. Partly, it was company culture. As a general rule of thumb, entrenched problems take time to solve. Senior leadership plainly expected a magic wand, but that was never going to happen. They didn't put in enough effort post-launch to make the project successful. ## Didn't listen/lack of inclusion This was the main problem. Senior leadership should have talked to software developers to understand what their concerns really were. In short order, listening would have brought to the surface the issues with current inequities and it would have shown senior leadership that the plan wasn't going to work. The fact that developers weren't involved in discussions about their own future was telling. ## Pretend it never happened You can't put the genie back in the lamp. Senior management tried to brush the whole thing under the rug very quickly when it was apparent it was going wrong, which was the worst response they could have made. # How should they have dealt with it? Hindsight is a wonderful thing, but knowing what I know now, here's how I would have dealt with this situation differently. • Understand what the problem really is by talking to people. By people, I mean the people directly affected. I've learned never to rely on intermediaries who may be incentivized to mislead or tell you what you want to hear. • Deal with the worst cases up-front and as one-offs. A one-size fits all process can flounder when dealing with outliers, so it's better to deal with the most extreme cases individually. In this case, people who should be on much higher levels. • Trial the solution with a small group. Trying it on a group of about 20-30 software developers would have flushed out most of the problems. • Stick with it. Put the effort in to make it work, but have clear criteria for exiting if it doesn't work. More specifically, I might have followed these policies. • Assuming no or very little money for promotions, I would have extended the pay bands at the bottom end. This would allow me to promote people without incurring extra costs. In the ideal world, you promote everyone who deserves it and give them the appropriate pay rise, but the ideal world doesn't exist. • I might reform the whole 20-30 levels, for example, I might call the levels something else, and then I would map people to new levels, including mapping some people to higher levels. An exercise like this allows you to correct all kinds of oddities, but of course, some people will be upset. • For people who were on higher levels than they should be, I would leave them there for a while. It's too damaging for an organization to make corrections on this scale - there were just too many people who were over-promoted. But I wouldn't promote these people further until their performance warranted it. ## Saturday, July 11, 2020 ### The art of persuasion: pathos, logos, and ethos # The art of persuasion is ancient In this blog post, I'm going to talk about three high-level approaches to persuasion that people have used for thousands of years: pathos, logos, and ethos. Aristotle codified them, hence their Greek names, but despite their ancient origins, they pop up in modern sales methods, in the speeches given by politicians, and in preachers' sermons. Understanding these basics will let you craft more effective speeches and emails. (Martin Luther King was one of the best rhetoricians of the 20th century. King used pathos enormously effectively in his 'I have a dream speech'. Image source: Wikimedia Commons. Photographer: Rowland Scherman.) # Logos Sadly, this is the weakest of the three, it's the appeal to logic; the use of facts and figures, and inference and deduction to make an argument. For example, if I were trying to make the case that the economy was great under my leadership, I might talk about high employment numbers, pay rises, and business growth. My chain of reasoning would be: the numbers are great due to my leadership, so you should vote for me. Let's look at a couple of real examples. My first example is Winston Churchill's "Their finest hour speech" (you can read more analysis of this speech here): "‘Hitler knows that he will have to break us in this Island or lose the war. If we can stand up to him, all Europe may be free and the life of the world may move forward into broad, sunlit uplands.’" Note there are no numbers here, it's a chain of logic linking one idea to another, in this case, an if-then piece of logic. John F. Kennedy also used logos. Here's an excerpt from his famous "We choose to go to the moon..." speech: "Within these last 19 months at least 45 satellites have circled the earth. Some 40 of them were made in the United States of America and they were far more sophisticated and supplied far more knowledge to the people of the world than those of the Soviet Union." Note the use of numbers to make the point that the United States was ahead of, and more advanced than the Soviet Union. Logos isn't just about making cold scientific claims, it can be about bashing your opponents with logic. Margaret Thatcher was a master speaker, and she used logos to batter her opponents: "If the Labour Government is no longer able to act in the national interest, is there no alternative to the ruin of Britain? Yes, indeed there is - and that alternative is here at Brighton today." The point is, logos isn't about a mathematical deduction, it's about creating a chain of arguments to lead the audience to the conclusion that your point is true. In sales situations, logos is used in several ways, for example, a salesperson might say something like: "Our solution costs less on a yearly basis than competitor Y but has features A, B, and C that Y does not." # Ethos This is an appeal from authority, the idea is that the speaker has some kind of position that gives them special knowledge. The most obvious example is the old quote: "Trust me, I'm a doctor". Speakers use this device a lot to establish credibility, for example, you might hear people talk about years of government service, or their qualifications, or their awards. One of the best examples of ethos I've come across is Steve Jobs Stanford Commencement address of 2005. Pretty much, the whole speech is ethos, establishing his credibility with the audience. Here's a small extract: "I was lucky — I found what I loved to do early in life. Woz and I started Apple in my parents’ garage when I was 20. We worked hard, and in 10 years Apple had grown from just the two of us in a garage into a$2 billion company with over 4,000 employees. We had just released our finest creation — the Macintosh — a year earlier, and I had just turned 30."
There are several ways speakers establish ethos. Sometimes, they talk about their education, or they talk about their experiences, or they talk about their awards. But how do you establish ethos if you have none of those things? The answer is, you can co-opt it from other people and organizations.
A salesperson might co-opt ethos by talking about the awards and success of their company or product, for example, "Product X was the biggest selling product in category Y, it's the industry standard." They also use external references from trusted sources. Gartner's magic quadrant analysis is widely used for a reason, it lends external credibility and is a great source of ethos. Success stories serve similar purposes too; you can use a quote from an industry figure to boost your credibility.
# Pathos
This is an emotional appeal to do something. It's is one of the most effective persuasive techniques, but it has to be used carefully; it's a blunt instrument that can undermine your argument. One of the clearest examples of pathos also illustrates the problem: "do as I ask or I kill this kitten".
I'm going to return to Winston Churchill and another of his famous speeches for my first example, here's a famous extract:
"I would say to the House, as I said to those who have joined this government: “I have nothing to offer but blood, toil, tears and sweat.”"
Churchill's goal was to get the politicians and the people to join him in an existential struggle against fascism. He used vivid imagery to illustrate the struggle ahead. If logos is the use of facts and figures, pathos is the use of adjectives and adverbs.
One of my other favorite speakers is Martin Luther King, here's an extract from his most famous speech:
"I have a dream that one day even the state of Mississippi, a state sweltering with the heat of injustice, sweltering with the heat of oppression, will be transformed into an oasis of freedom and justice."
King is using evocative language to appeal to his audience, to motivate them to appeal for change.
I've seen salespeople use pathos in a clever way when they ask a prospect what the success of product would mean for them. The crude example here is telling someone they would look good in a sports car. Less crudely, the success of a product could lead to promotions. This has to be artfully done, but I've seen it used very effectively.
# Using them in a speech
Not every speech uses all the elements of persuasion, but many of them do. Let's imagine you're promoting your company's product at a convention and had a short slot to do it. Here's how you might use pathos, logos, and ethos.
Speech Commentary In 2020, Gartner placed us in the leader quadrant of their X magic quadrant for the fifth year running. Ms A, Vice-President for Important Things at Big Company B said that Product Y was a game-changer for them. Ethos. She said it reduced design costs by 20% and reduced time-to-market by 15%. Her experience is common, which is why we have the largest market share and we're growing more rapidly than any other company in the sector. We're becoming the industry standard. Logos. But it's not just about money. Because we reduce design flow issues, we reduce low-level re-work and rote work, so we free up engineers' time to be more creative and use their skills in more engaging and interesting ways. Our software enabled Dr S to design H, which won the industry design award and got him on the cover of Trade Publication. Pathos
This is a little crude, but you get the general idea.
Churchill, Kennedy, and King did not just make up their speeches. They knew very well what they were doing. Churchill and King in particular were master rhetoricians, adept at using words to change the world. Knowing a little of rhetoric can pay dividends if you're trying to be more persuasive.
A lot of books on rhetoric are unreadable, which is odd given that rhetoric is about communication. Here are some of the ones I like:
• Thank You for Arguing, Fourth Edition: What Aristotle, Lincoln, and Homer Simpson Can Teach Us About the Art of Persuasion - Jay Heinrichs
• Words Like Loaded Pistols: Rhetoric from Aristotle to Obama - Sam Leith
# Rhetoric for managers series
This post is part of a series of posts on rhetoric for managers. Here are the other posts in the series:
# Managers need to actively listen
Almost every MBA or business school degree has units on communications, including spoken communication. A great deal of attention is lavished on learning how to give speeches and how to communicate ideas, but no time is given to a more important management skill: how to actively listen.
(If a dog can listen attentively, why can't you? Image credit: Wikimedia Commons. License: Public Domain.)
If you're thinking of listening as a passive skill, you've got it wrong. If you think of it just as letting someone speak, you're mistaken. If you think of it as something you can just do, you've totally misunderstood it.
By actively listening, I mean understanding the true meaning of what someone says and not just the superficial content. I mean actively probing for the key underlying messages and feelings. I mean communicating back to the speaker that you've heard and understand what they have to say by your choice of gestures, noises, and responses.
By actively listening, you can understand what's really going on and earn trust as a manager. You can demonstrate empathy, respect, and the value you place on people.
In the last few years, the 'fake it until you make it' business meme has done the rounds, meaning pretending to be something you're not so you can get promoted to be what you were pretending to be. Active listening is the antithesis of this approach; it's all about communicating genuine human warmth. You can't fake caring.
# Listen without distraction
Our first listening lesson is the simplest: listen with undivided attention and communicate that your attention is undivided. This means no ringing phones and no beeping devices. Turn off or silence your phone or watch. Turn off or silence notifications on your computer - better still, close up your laptop or turn your screen away.
If at all possible, you should sit close to someone without distractions or obstructions. If you have an office, get up from behind your desk so you can sit with the person without diversion. Sitting behind your desk communicates power, and it's even worse if your monitor even partially obscures your view. If you don't have an office, go to a meeting room where you won't be distracted. On Zoom or video conferencing, close down any distracting windows and look at the person only (not other computer windows or something else in your home office).
These things sound simple, but it's surprising how many people don't understand the basics of turning off distractions. If you want people to talk, you need to give them the right environment - and that means you have to take action.
Let's imagine you wanted to talk to your manager about a difficult situation at work. Perhaps you suspect you're the victim of sexual or racial harassment, or you've seen something in another department you think you should report, or you're unhappy with a colleague. Each of these situations is difficult for you to talk about and requires you to trust your manager. Now imagine when you finally screw up the courage to see them, they sit behind their desk fiddling with their phone, their computer pings every minute, and they keep on turning away to check their monitor. Do you think you would tell a manager who behaves like this something that's risky and that requires trust? Do you want to be the kind of manager your team trusts, or do you just want to just fake it? Being authentic requires effort on your part.
Just making time for people isn't enough. You have to understand what's expected of you and what you mustn't do.
# Listening isn't problem-solving
One of the big mistakes managers make is slipping into problem-solving mode as soon as someone starts discussing a problem or issue. For people from a technical background, this is a comforting response to a difficult situation, after all, solving problems is what technical people are trained to do. But this might be the exact opposite of what the person wants. They might just want to talk through a difficult or disturbing situation and be heard. The other risk is, by offering solutions too soon, you might block more serious content. Sometimes, someone might come to see you with a 'presenting problem' that's relatively innocuous; they're seeing if they can trust you before disclosing the bigger issue. Before suggesting any solution, you should make sure you've heard the totality of what someone has to say. Your first goal should be making sure the person feels heard.
# The biggest risk of active listening for managers
I do have to address one of the big risks of listening: role confusion. You are a manager, not a counselor. The art of listening as a manager is to know your limits and not try to be a cheap therapist. Remember, as a manager, you may have to use the performance management process on someone which may be next to impossible if the person is relying on you as a counselor. If you've allowed an unhealthy pseudo-counseling relationship to develop, you've put yourself in a career-limiting situation. At all times, you must remember you are a manager.
# Listening leads to acting - sometimes
There's another key difference between a therapist and a manager. As a manager, you need to act on what people tell you. You don't need to act all the time, sometimes people need to vent and your role is solely to listen and communicate you've heard what they have to say. But on other occasions, you have to act - in fact, acting is the ultimate in active listening: 'I've heard what you have to say and I'm going to do something about it'. If you do nothing in response to what you hear, ultimately people will do something themselves: they'll leave and go somewhere else.
# What's next?
I'm going to blog a bit more about listening. I'm going to focus on some micro-skills you can use to communicate you're listening, and provide examples and exercises you can follow. These blog posts are not about faking listening, they're about being human and about your role for your team - demonstrating why you're a manager.
# References
The best single-volume book I've come across that covers the skills of active listening is "Swift to hear" by Michael Jacobs (ISBN: 978-0281052608). The book is aimed at people involved in pastoral care, but it focuses on technical skills, making it a great resource for anyone who needs to listen attentively.
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# Algebra Notes: Gilbert&Gilbert
Lillian Ratliff
July 18, 2012
## Preliminaries and Notation
Sets and Intervals
An interval with round brackets such as (a, b) means all the numbers between a and b not including a
and b. We call this interval an open interval.
An interval with square brackets such as [a, b] means all the numbers between a and b including a and
b. We call this interval a closed interval.
An interval with one square bracket and one closed bracket is called a clopen interval.
(i) [a, b): means all numbers between a and b including a and not b.
(ii) (a, b] means all the numbers between a and b including b and not a.
## General Mathematics Notation
The symbol means for all
The symbol means there exists
The symbol means in. We use it to say things like the number number 1 is in the interval (0, 2)
and we write it mathematically as 1 (0, 2).
R is the set of all real numbers, i.e. all numbers in the interval (, ).
is used to denote a general operation (i.e. it could be multiplication or addition +). Make sure to
look at the context. Here we will use when we mean multiplication and + when we mean addition.
We will use to denote a dummy binary operation (meaning you can replace it with any operation
that you want and the statement should hold true).
## 1.4 Binary Operations
Definition 1.17 A Binary Operation on a nonempty set A is a mapping f from A A to A. We use
the following notation
f :AA A
## Examples of binary operations:
Let f : R R R be defined as follows: (x, y) R R
f (x, y) = x + y
## Let f : R R R be defined as follows: (x, y) R R
f (x, y) = x y
1
Let f : Z Z Z be defined as follows: (x, y) Z Z
f (x, y) = x y + 1
## Definition 1.18 We define the following:
We say a binary operation is commutative if the following relation holds:
xy =yx
## We say a binary operation is associative if the following relation holds:
(x y) z = x (y z)
Examples:
(i) Let be defined as follows: x y = x + y 1. Then, the operation is commutative and associative
since
xy =x+y1=y+x1=yx (commutativity)
and
(x y) z = (x + y 1) z = (x + y 1) + z 1 = x 1 + y + z 1
= x + (y z) 1 = x (y z) (associativity)
x, y B, x y B.
## For the following definitions let A be a nonempty set.
Definition 1.20 We call e A an identity element with respect to the binary operation if the following
holds
ex=xe=x
for all x A.
## Definition 1.21 Let e be the identity element of A with respect to . Let a, b A.
(i) right inverse: b is a right inverse of a if a b = e.
(ii) left inverse: b is a left inverse of a if b a = e.
(iii) inverse: b is an inverse of a if a b = b a = e. That is to say that b is both a right and left inverse
of a.
Definition 1.22 A permutation is a one-to-one correspondence from a set to itself. For any nonempty
set A we denote the set of all permutations on A as S(A). The set of all mappings from A to A is denoted
M(A).
Definition 1.15 The composite mapping f g where f : A B and g : B A is the mapping from A
to C defined by
(f g)(x) = f (g(x))
Example: Let f : [0, 1] [1, 2] be defined by f (x) = x + 1 and let g : [1, 2] [1, 4] be defined by g(x) = x2 .
Then,
(g f )(x) = g(f (x)) = (x + 1)2
and it maps from [0, 1] to [1, 4].
## Let us now define the identity map.
2
Definition (Idenity Map) The map IA (x) is called the identity map on the set A. It maps A onto A
and it is a one-to-one correspondence. Hence, it is a permutation of A and as such it lives in S(A), i.e.
IA A(A). It is defined as follows:
x A, IA (x) = x
In addition, for any f A(A) (which means f : A A so that f (A) A) we have the following:
## (IA f )(x) = IA (f (x)) = f (x)
and
f IA (x) = f (IA (x)) = f (x).
Consider Example 12 from the book. Let f : Z Z be defined by f (z) = 2z and let g : Z Z be defined
by z
g(z) = 2, z E
4, z O
Now, since f (z) is always even, (g f )(z) = z for all z Z. Hence, g f = IZ which means that g is a left
inverse of f (similarly, f is a right inverse for g). Now,
z, z E
f g(z) =
8, z O
## Hence, f g 6= IZ so that g is not a right inverse for f .
1.4.a Proofs
Now, lets talk a little bit about constructing a proof. Say that you want to prove a statement like p implies
q (mathematically we can denote this as p q). This may also be stated as If p, then q. There are a few
ways to do this.
(1) Direct proof: You just assume p and then argue that q is true.
(2) Contradiction: You assume p is true and that q is false. Then you argue until you get a contradiction.
An example of this type of proof method is as follows: Statement: If a, b Z, then a2 4b 6= 2. We
start the proof by supposing the q portion of the statement is false. Which is to say that if a, b Z,
then a2 4b = 2. Thus,
a2 = 2 + 4b = 2(2b + 1) a2 E a E
Thus, we can write a = 2c for some number c. Plug 2c in for a in the equation a2 4b = 2 to get
## 4c2 4b = 2 4(c2 b) = 2 2(c2 b) = 1
Thus, since c2 b is an integer, 2(c2 b) = 1 implies 1 E. But, this is false. Hence we have proven
the orginal statement: If a, b Z, then a2 4b 6= 2.
(3) Contrapositive: For this proof method we prove that the negation of the statement holds. To negate a
logical statement we do the following:
(p q) = ( q p)
So, let us try to prove the following claim: For a, b Z, a + b 15 implies a 8 or b 8. Here
a + b 15 is our p statement
and
a 8 or b 8 is our q statement
So to form the negation we want to do
## (a + b 15 a 8 b 8) = (a < 8 b < 8 a + b < 15)
So, we now try to prove the following statement directly: a < 8 b < 8 a + b < 15.
3
Proof. Assume that a < 8 and b < 8 for a, b Z. Since a, b Z
## We can now add the two inequalities a 7 and b 7 to get
a + b 14
This implies that a + b < 15. Hence we have proven that a < 8 b < 8 a + b < 15 which is equivalent
to our original statement.
Now, lets talk about proving an if and only if statement (note that we can abbreviate if and only
if as iff. First, when we make the statement p iff q we mean both if p, then q and if q, then p
must hold. So, as is stated in the book, to prove the statement p iff q we must prove the following two
statements:
(1) (p q): This is the if part of the proof and it is often referred to as the sufficient condition. Here,
we assume q is true and we prove that p must then be true.
(2) (p q): This is the only if part of the proof and it is often referred to as the necessary condition.
Here, we assume p is true and then we prove that q must hold.
## 1.6. Equivalence Relations
Definition 1.35 (Relation) A binary relation on a nonempty set A is a set R of pairs (x, y) for x A
and y A. Note that R A A. If two elements x and y are related, i.e. (x, y) R, then we write xRy.
The symbol is often also used to denote a relation. Using this notation we write x y. We will use the
notation to denote the actual relation (so when we want to say x is related to y, we write x y) and we
will use R to denote the set of ordered pairs. Namely,
## To clarify this definition, let us take an example.
Example 1
Let A be the set of integers between 1 and 5. Consider the relation defined by a is related to b if a mod b = 2.
Then, we may construct the set R. In the set A, we have the following:
1 mod 2 = 1, 1 mod 3 = 1, 1 mod 4 = 1, and 1 mod 5 = 1.
2 mod 1 = 0, 2 mod 3 = 2, 2 mod 4 = 2, and 2 mod 5 = 2.
3 mod 1 = 0, 3 mod 2 = 1, 3 mod 4 = 3, and 3 mod 5 = 3.
4 mod 1 = 0, 4 mod 2 = 0, 4 mod 3 = 1, 4 mod 5 = 4.
5 mod 1 = 0, 5 mod 2 = 1, 5 mod 3 = 2, 5 mod 4 = 1.
Hence,
R = {(x, y) A A|x mod y = 2} = {(2, 3), (2, 4), (2, 5), (5, 3)}.
## Definition 1.36 (Equivalence Relation) A relation on A is an equivalence relation on A if the
following are satisfied for all x, y, z A:
(i) x x, x A. (reflexive)
(ii) x y = y x. (symmetric)
(iii) x y and y z = x z. (transitive)
Note that the relation in example (1) is not an equivalence relation since 2 3 but 3 2, i.e. the relation
is not reflexive. ( means not related to).
4
Example 2
Consider the set of all integers, Z. We define to be congruence modulo 4. This is to say that a number
x is congruent modulo 4 iff x y is a multiple of 4, i.e. (x y) mod 4 = 0. We write x y( mod 4). Note
that for (x y) mod 4 to be zero this means that 4 divides (x y) some number of times (let us say k
times) with zero remainder. We can write this as 4|(x y) = k remainder 0. Hence, (x y) is a multiple of
4. Specifically, (x y) = 4k.
We can show that the relation x y( mod 4) in the set of integers Z is an equivalence relation by checking
the three properties in the definition above. Let x, y andz be arbitrary integers in Z.
(reflexive): We want to show that x x( mod 4). Equivalently, we must show that x x = 4k for
some k.
xx= 0= 40
Hence, x x = 4k for k = 0 which implies that x x( mod 4).
(symmetric): Assume that x y( mod 4). We want to show that y x( mod 4).
## x y( mod 4) = (x y) = 4k for some k = (y x) = 4(k) = y x( mod 4).
(transitive): Assume that x y( mod 4) and y z( mod 4). We want to show x z( mod 4).
## y z( mod 4) = (y z) = 4j for some j
So, we have
(x y) (y z) = 4k 4j = x z = 4(k j)
Hence, x z( mod 4). Since x y( mod 4) in the set of integers Z satisfies the three properties of
an equivalence relation, it is an equivalence relation.
One nice application of equivalence relations is to define quotient sets. We may define an equivalence class
with respect to an element in a set A. Again, let be an equivalence relation. Then, for an element x A,
we define the equivalence class of x as the following:
## [a] = {x Z|a x( mod 4)}
We call the quotient set the set of all equivalence classes. We denote it by A/ . As an example, consider
the congruence modulo 2 relation on the integers. That is x y( mod 2) or (xy) = 2k for some integer k.
Then, the equivalence class of any even number (say 2) is the set of all even numbers. And, the equivalence
class of any odd number (say 3) is the set of all odd numbers. So, Z/ where is the congruence modulo
2 equivalence relation is the set
Z/ = {[2], [3]}
where [2] is the equivalence class of all even numbers and [3] is the equivalence class of all odd numbers.
2. Integers
2.1 Postulates for the Integers
(1) Addition postulates: Note that (Z, +) forms an abelian group. Hence, the addition postulates are just
the properties of an abelian group.
(a) Closure: Z is closed under +.
(c) Identity element: 0 is the identity element for (Z, +).
5
(d) Inverse: For a Z, a is the additive inverse since a + (a) = 0
(e) Commutative: a + b = b + a.
(2) Multiplicative postulates: Note that (Z, ) forms a commutative monoid . A commutative monoid
has all the properties of an abelian group except for the inverse.
(a) Closure: Z is closed under .
(b) Associativity: multiplication is associative.
(c) 1 is the identity element: z 1 = z.
(d) Multiplication is commutative.
(3) Distribution Law : Multiplication distributes over addition.
(i) (left distributive law): x (y + z) = x y + x z
(4) Positive Integers: Z+ Z where Z+ denotes the set of positive integers (it does not contain 0). Z+
has the following properties:
(a) Z+ is closed under addition.
(b) Z+ is closed under multiplication.
(c) Law of Trichotomy : For x Z one and only one of the following is true:
(i) x Z+
(ii) x = 0
(iii) x Z+ .
(5) Induction Postulate: If S Z+ such that 1 S and x S x + 1 S, then S = Z+ .
Note: Since Z satisfies postulates (1)-(3), Z forms a commutative ring . We may prove that the integers
also satisfy something called the right distributive law.
Theorem 2.1 The equality
## holds for all x, y, z Z.
Proof. Let x, y, z be arbitrary elements in Z. Then, since Z is closed under addition, y + z Z. We also
have that multiplication is commutative so that
(y + z) x = x (y + z)
## Since (Z, +, ) satisfies the left distributive law, we have
(y + z) x = x (y + z) = x y + x z = y x + z x
where the last equality holds because multiplication is commutative. Hence, we have shown that
(y + z) x = y x + z x.
## Now, we state a lemma such that will be used to prove a theorem.
Lemma 2.3 (Cancellation Law for Addition)
If a, b, c Z and a + b = a + c, then b = c.
6
Proof. We prove this lemma directly. Let a, b, c be integers and suppose that a + b = a + c. Since (Z, +, )
is a commutative ring, it contains the additive inverse of every element in Z. Hence, a Z. We may use
the other postulates of the integers to get the following implications:
a + b = a + c (a) + (a + b) = (a) + (a + c)
(a + a) + b = (a + a) + c (by postulate 1-b)
0+b=0+c (by postulate 1-d)
b=c (by postulate 1-c)
## Now, we state the theorem we are interested in proving.
Theorem 2.2 (Additive Inverse of a Product) For arbitrary x, y Z,
(x) y = (x y)
Proof.
## x y + (x) y = (x + (x)) y (by theorem 2.1)
= 0y (by postulate 1-d)
= 0y+0 (by postulate 1-c)
=0 y + (0 y + ((0 y))) (by postulate 1-d)
=(0 y + 0 y) + ((0 y)) (by postulate 1-b)
= (0 + 0) y + ((0 y)) (by theorem 2.1)
= 0 y + (0 y) (by postulate 1-c)
= 0 (by postulate 1-d)
Hence, x y + (x) y = 0. So, we know that x y + (x y) = 0 by postulate 1-d. And, we have shown that
x y + (x) y = 0. Thus, x y + (x y) = x y + (x) y. So, by the lemma, (x) y = (x y).
## Definition 2.4 (Order Relation <) For x, y Z
x < y y x Z+
where y x = y + (x).
Since we have an order relation < meaning less than, we can state the well ordering axiom.
Well ordering axiom: Every nonempty subset of the set of nonnegative integers contains a smallest ele-
ment.
The well ordering axiom does not hold on the set of all integers Z, i.e. there is no smallest negative integer.
## We may now define powers of elements in Z.
Definition (powers) Let x Z. xn := |x {z
x}. Note that this is well defined since we take x1 = x and
n times
inductively we define xk+1 = xk x.
## Definition (multiples) Let x Z. Then we define 1x = x and (k + 1)x = kx + x.
7
2.2 Mathematical Induction
Often we want to prove a statement Pn for every n Z+ . Logically the way we can think about induction
is that if P1 is true and if Pk is true implies that Pk+1 is true for arbitrary k Z+ , Pn must be true for all
n Z+ .
Proof by Mathematical Induction We want to prove Pn is true for all n Z+ . We take the following
three steps:
1. Verify the statement for n = 1.
2. (inductive assumption or hypothesis): Assume the statement holds for n = k.
3. Under the assumption in 2., prove the statement hold for n = k + 1.
## Let us take an example to get the gist of proof by induction.
Example 3
Prove
n(n + 1)
1 + 2 + n =
2
Proof. Lets go through the three steps above.
(a) Check for n = 1:
1(1 + 1) 2
= =1
2 2
(b) Assume n = k holds:
k(k + 1)
1 + 2 + + k = (Induction Hypothesis)
2
(c) Show n = k + 1 holds: So, we must show that
(k + 1)(k + 1 + 1)
1 + 2 + + (k + 1) = .
2
k(k + 1)
1 + 2 + + k + (k + 1) = +k+1 (by induction hypothesis)
2
k(k + 1) 2(k + 1)
= +
2 2
k(k + 1) + 2(k + 1)
=
2
(k + 1)((k + 1) + 1)
=
2
## Since we showed the three steps,
n(n + 1)
1 + 2 + n =
2
holds for all n Z+ .
Theorem 2.6 (Least Positive Integer) The integer 1 is the least positive integer, i.e. 1 n for all
n Z+ .
Proof. Let S be the set of all positive integers greater than or equal to 1, i.e.
S := {n Z|1 n}
Now, we want to use induction to show that S contains all the positive integers. If we can show this, then
we have shown that all positive integers satisfy the inequality 1 n and hence 1 is the least positive integer.
8
1. Check for n = 1: 1 S since 1 1.
2. Assume for n = k: assume that 1 k holds.
3. Show for n = k + 1 that 1 k + 1: By definition 2.4, we know that x < y y x Z+ . So,
1 0 = 1 + (0) = 1 Z+
## Hence, 0 < 1. Now, 0 < 1 implies that k < k + 1 since
k + 1 (k + 0) = k + 1 k = 1 Z+
so again by definition 2.4 k < k + 1. By the induction hypothesis in 2. above we have 1 k. We just
showed that k < k + 1. Hence,
1 k < k + 1 = 1 k + 1
Thus, k S implies k + 1 S.
Using induction we have shown that all the positive integers are contained in S. Therefore, 1 is the least
positive integer.
2.3 Divisibility
Definition 2.8 (Divisor, Multiple) Let a, b Z. We say a divides b if there is an integer c such that
b = ac. We often denote a divides b as a|b. We may also say that b is a multiple of a, a is a factor of b,
or that a is a divisor of b.
Theorem 2.10 (Division Algorithm) Let a, b Z such that b > 0. Then, unique integers q and r such
that
a = bq + r with 0 r < b
We call q the quotient and we call r the remainder . Another form of the theorem conclusion is
a r
=q+
b b
Proof. We prove this theorem in two parts. We first prove existence and then we prove uniqueness.
(existence): We must show that there exists q and r such that a = bq + r with 0 r < b. Consider
a, b Z such that b > 0. Let S be the set defined by
S := {x Z|x = a bn, n Z}
Let S := {x S|x 0}. The set S 6= since if a = 0, then b S with n = 1 and if a 6= 0, then
a + 2b|a| S with n = 2|a|. Now, if 0 S , we have 0 = a bq for some q. Hence, a = bq + 0 so
that r = 0. If 0 / S , then S contains a least element r by the well ordering axiom. Since r S we
can write it as r = a bq for some q Z. Hence, a = bq + r with r 0. Now,
r b = a bq b = a b(q + 1)
so that r b S. Since r is the least element in S and b > 0 so that r b < r, we must have that
r b < 0 because S contains all nonnegative integers in S. Hence, r < b. Thus, we have
a = bq + r with 0r<b
## And we are done.
(uniqueness): We show uniqueness by contradiction. We assume that q and r are not unique. That is
to says there exists q1 , q2 , r1 , r2 such that for a, b Z
## a = bq1 + r1 and a = bq2 + r2
9
where 0 r1 < b and 0 r2 < b. Without loss of generality we may assume r1 < r2 so that we have
0 r2 r1 r2 < b.
We also have
0 r2 r1 = (a bq2 ) (a bq1 ) = b(q1 q2 )
so that along with 0 r2 r1 r2 < b we have
0 r2 r1 = b(q1 q2 ) < b
But, the last inequality only holds when q1 q2 0 since b > 0. Thus, 0 b(q1 q2 ) 0 which implies
that q1 q2 = 0. Further, this implies that r1 = r2 . So, q and r are in fact unique.
## 2.4 Prime Factors and Greatest Common Divisor
Definition 2.11 (Greatest Common Divisor or gcd) An integer d is a greatest common divisor
(gcd) of a and b if the following hold:
1. d Z+
2. d|a and d|b (i.e. a = dq1 and b = dq2 )
3. c|a and c|b imply c|d. (a = cq3 and b = cq4 , then d = cq5 ).
Theorem 2.12 (gcd) Let a, b Z and let at least one of them be nonzero. Then, there exists a gcd d of a
and b. Moreover, we can write d as
d = am + bn
for m, n Z and d is the smallest positive integer that can be written in this form.
Proof. If b = 0 so that a 6= 0 or equivalently |a| > 0. Then, d = |a| is a gcd for a and b since d = |a| Z+ ,
a = |a| (1) if a < 0 or a = |a| 1 if a > 0 so that d|a. Similarly, b = |a| 0 so that d|b. Also, c|a and c|b
imply c|d since a = cq1 and b = cq2 imply d = |a| = |cq1 | which implies that c|d. Thus, either d = a 1 + b 0
or d = a (1) + b 0.
## S := {z Z|z = ax + by for some x, y Z}
and define
S + := {z S|z > 0}
The set S contains b = a 0 + b 1 and b = a 0 + b (1) so S + 6= . By the well ordering axiom, S + has a
least element d and since d S + , d is positive and we write d = am + bn. Now, what is left to show is that
d is a gcd of a and b. First, we already have that d Z+ . By theorem 2.10 (division algorithm), since d > 0
and a, d Z, there exists integers q and r such that
a = dq + r with 0 r < d
Hence,
r = a dq
= a (am + bn)q
= a(1 mq) + b(nq)
which shows that r S. Since 0 r < d and since d is the least element in S + , it must be the case that
r = 0. Hence,
0 = a dq a = dq d|a
10
We may make a similar argument to show that d|b. Now, we check that last point. That is we must show
that if c|a and c|b then c|d.
c|a a = ck and c|b b = cj
Thus,
d = am + bn
= ckm + cjn
= c(km + jn)
so that c|d. We have shown that d satisfies 1-3 of definition 2.11 so that we may conclude that d is a gcd of
a and b.
## 2.7 Introduction to Coding Theory
3. Groups
Now, we may define a group.
Definition (Group) A group is a set G together with a binary operation such that the following hold:
(i) Closure: a, b G a b G.
(ii) Associativity: a, b, c G (a b) c = a (b c).
(iii) Identity element: e G such that a G, a e = e a = a.
(iv) Inverse Element: For each a G, an element b G such that a b = b a = e.
We sometimes denote the group as (G, ) or G .
## Definition (Commutative Group or Abelian Group) A commutative (abelian) group is a set G
with operation such that (G, ) is a group (see definition above, i.e. all the properties above hold) and the
following holds:
a, b G, a b = b a.
## Some examples of groups:
(i) The set of all non-zero rational numbers (denoted as Q\{0} which means the rational numbers Q with
the zero element removed) is an abelian group under the operation (multiplication). We denote this
group as (Q\{0}, ).
1 is the identity element: r 1 = r.
r1 is the inverse of r Q\{0}.
(ii) Q under addition forms an abelian group.
(commutativity): a + b = b + a.
(inverse): a Q. a Q
(identity): a + 0 = a.
(closure): a + b Q.
(associativity): (a + b) + c Q and a + (b + c) Q.
1
(iii) Z is not a group under multiplication. 2 Z but 21 = 2
/ Z.
(iv) (Cyclic group): A cyclic group is a group whose elements are generated by a single element, i.e.
the elements are powers of a single element. Consider a cyclic group (G, ). a is the generator for
this group if for each element g G we can write g = ak for some k Z. For notation, we mean
the following a3 = a1 a1 a1 . a3 = a a a. As an example, consider the multiplicative group
11
Z11 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and let the operation be multiplication, . Then, we claim that 2 is a
generator for this group.
21 ( mod 11) = 2, 22 ( mod 11) = 4( mod 11) = 4, 23 ( mod 11) = 8( mod 11) = 8,
24 = 16( mod 11) = 5, 25 ( mod 11) = 32( mod 11) = 10, 26 ( mod 11) = 64( mod 11) = 9,
7 8
2 ( mod 11) = 128( mod 11) = 7, 2 ( mod 11) = 256( mod 11) = 3
29 ( mod 11) = 512( mod 11) = 6, 210 ( mod 11) = 1024( mod 11) = 1
12
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# Ratio Ratio A ratio compares the sizes of
• Slides: 11
Ratio
Ratio A ratio compares the sizes of parts or quantities to each other. For example, What is the ratio of red counters to blue counters? red : blue =9: 3 =3: 1 For every three red counters there is one blue counter.
Ratio A ratio compares the sizes of parts or quantities to each other. For example, The ratio of blue counters to red counters is not the same as the ratio of red What is the ratio of blue counters to red counters to blue counters? blue : red =3: 9 =1: 3 For every blue counter there are three red counters.
Ratio What is the ratio of red counters to yellow counters to blue counters? red : yellow : blue = 12 : 4 : 8 = 3 : 1 : 2 For every three red counters there is one yellow counter and two blue counters.
Simplifying ratios Ratios can be simplified like fractions by dividing each part by the highest common factor. For example, 21 : 35 ÷ 7 =3: 5 For a three-part ratio all three parts must be divided by the same number. For example, 6 : 12 : 9 ÷ 3 =2: 4: 3 ÷ 3
Equivalent ratio spider diagrams
Simplifying ratios with units When a ratio is expressed in different units, we must write the ratio in the same units before simplifying. Simplify the ratio 90 p : £ 3 First, write the ratio using the same units. 90 p : 300 p When the units are the same we don’t need to write them in the ratio. 90 : 300 ÷ 30 = 3 : 10 ÷ 30
Simplifying ratios with units Simplify the ratio 0. 6 m : 30 cm : 450 mm First, write the ratio using the same units. 60 cm : 30 cm : 45 cm 60 : 30 : 45 ÷ 15 =4: 2: 3
Simplifying ratios containing decimals When a ratio is expressed using fractions or decimals we can simplify it by writing it in whole-number form. Simplify the ratio 0. 8 : 2 We can write this ratio in whole-number form by multiplying both parts by 10. 0. 8 : 2 × 10 = 8 : 20 ÷ 4 =2: 5
Simplifying ratios containing fractions Simplify the ratio : 4 2 3 We can write this ratio in whole-number form by multiplying both parts by 3. 2 3 : 4 × 3 = 2 : 12 ÷ 2 =1: 6
Ratio 1. 2. 3. 4. 5. 6. 7. 8. ANSWERS There are 10 girls and 15 boys in a class, what is the ratio of girls to boys in its simplest 1. 2: 3 2. 7: 8 form? There are 14 cats and 16 dogs in an animal shelter, what is the ratio of cats to dogs in its 3. 2: 5 4. simplest form? a) There 22 caramels and 55 fudges in a bag of sweets, what is the ratio of caramels to fudges b) in its simplest form? c) Simplify these ratio to their simplest forms: d) e) a) 48: 60 5. b) 45: 75 a) c) 63: 108 b) d) 25: 40: 80 c) e) 24: 56: 96 6. f) 120: 180: 600 a) b) g) 320: 400: 440 c) Archie and Charlie share their Thomas the tank engine toys in the ratio 1: 4, how many do 7. they each get if they have: a) a. 10 toys b. 30 toys c. 45 toys b) Tom and Jerry share sweets in the ratio 2: 3, how many do they each get if they share: c) a. 20 sweets b. 30 sweets c. 55 sweets 8. a) Sue and Linda share some money in the ratio 3: 7, how many do they each get if they share: b) a. £ 30 b. £ 60 c. £ 90 c) Mike, Dave and Henry share some little bits of blue tack in the ratio 1: 2: 3, how many do they each get if they share: a. 60 pieces b. 72 pieces c. 300 pieces Home 4: 5 3: 5 7: 12 5: 8: 16 3: 7: 12 2 and 8 6 and 24 9 and 36 8 and 12 12 and 18 22 and 33 9 and 21 18 and 42 27 and 63 10 and 20 and 30 12 and 24 and 36 50 and 100 and 150
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# Elementary Number Theory Problems 4.3 Solution (David M. Burton's 7th Edition) - Q7
Establish the following divisibility criteria: (a) An integer is divisible by $2$ if and only if its units digit is $0, 2, 4, 6,$ or $8$. (b) An integer is divisible by $3$ if and only if the sum of its digits is divisible by $3$. (c) An integer is divisible by $4$ if and only if the number...
## Background
All theorems, corollaries, and definitions listed in the book's order:
I will only use theorems or facts that are proved before this question. So, you will not see that I quote theorems or facts from the later chapters.
## Question
Establish the following divisibility criteria:
(a) An integer is divisible by $2$ if and only if its units digit is $0, 2, 4, 6,$ or $8$.
(b) An integer is divisible by $3$ if and only if the sum of its digits is divisible by $3$.
(c) An integer is divisible by $4$ if and only if the number formed by its tens and units digits is divisible by $4$.
[Hint: $10^{k} \equiv 0 \pmod {4}$ for $k \geq 2$.]
(d) An integer is divisible by $5$ if and only if its units digit is $0$ or $5$.
## Solution
### (a)
Let $N$ be the integer and we can write $N = a_{m}10^{m} + a_{m - 1}10^{m - 1} + \cdots + a_{2}10^{2} + a_{1}10 + a_{0}$. By Theorem 4.4, we have $10 \equiv 0 \pmod {2}$, so $N = P(10) \equiv P(0) \equiv a_{0} \pmod {2}$. It follows that $N \equiv 0 \pmod {2}$ if and only if $a_{0} \equiv 0 \pmod {2}$. As $a_{0}$ is the units digit, it will only be $0, 2, 4, 6,$ or $8$ when it is divisible by $2$.
Therefore, an integer is divisible by $2$ if and only if its units digit is $0, 2, 4, 6,$ or $8$.
### (b)
Let $N$ be the integer and we can write $N = a_{m}10^{m} + a_{m - 1}10^{m - 1} + \cdots + a_{2}10^{2} + a_{1}10 + a_{0}$. By Theorem 4.4, we have $10 \equiv 1 \pmod {3}$, so $N = P(10) \equiv P(1) \equiv a_{m} + a_{m - 1} + \cdots + a_{2} + a_{1} + a_{0} \pmod {3}$. It follows that $N \equiv 0 \pmod {3}$ if and only if $a_{m} + a_{m - 1} + \cdots + a_{2} + a_{1} + a_{0} \equiv 0 \pmod {3}$.
Therefore, an integer is divisible by $3$ if and only if the sum of its digits is divisible by $3$.
### (c)
Let $N$ be the integer and we can write $N = a_{m}10^{m} + a_{m - 1}10^{m - 1} + \cdots + a_{2}10^{2} + a_{1}10 + a_{0}$. We know that $10^{k} \equiv 0 \pmod {4}$ for $k \geq 2$. Thus $N \equiv a_{1}10 + a_{0} \pmod {4}$. It follows that $N \equiv 0 \pmod {4}$ if and only if $a_{1}10 + a_{0} \equiv 0 \pmod {4}$.
Therefore, an integer is divisible by $4$ if and only if the number formed by its tens and units digits is divisible by $4$.
### (d)
Let $N$ be the integer and we can write $N = a_{m}10^{m} + a_{m - 1}10^{m - 1} + \cdots + a_{2}10^{2} + a_{1}10 + a_{0}$. By Theorem 4.4, we have $10 \equiv 0 \pmod {5}$, so $N = P(10) \equiv P(0) \equiv a_{0} \pmod {5}$. It follows that $N \equiv 0 \pmod {5}$ if and only if $a_{0} \equiv 0 \pmod {5}$. As $a_{0}$ is the units digit, it will only be $0$ or $5$ when it is divisible by $5$.
Therefore, an integer is divisible by $5$ if and only if its units digit is $0$ or $5$.
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Teacher Notes
CONCEPT 1 – Sample Spaces as Sets
While probability has not typically been connected to geometry, the common core restructuring has placed it within the geometry curriculum. Probability and statistics have a direct application to events we see in everyday life and as such they are a very important branch of mathematics for all students to understand.
Probability is often defined as how likely something is to happen. In determining how likely something is to happen we must first determine what the total number of things that could happen for any given event or experiment. The list of all possible outcomes is called the sample space.
A LIST -- A sample space is often organized into a set of elements. For example the sample space for rolling a die is {1, 2, 3, 4, 5, 6} or the sample space for flipping a coin is {Head, Tail}. These two sample spaces represent uniform probability. Uniform probability is where each element of the set has the same chance of happening or in other words, each element is equally likely to happen. In a uniform probability, the number of elements in the set is the total number of outcomes possible for the sample space.
Some sample spaces are not uniform, such as a bag of marbles with 2 red and 1 green, the sample space is {red, green} because only a red or a green marble can be chosen. But in this sample space each element is NOT equally likely to happen because there are 2 ways to pick a red marble and only 1 way to pick a green. While there are only two elements listed in the sample space, the total number of outcomes of this sample space is 3, written n(S) = 3, because red has 2 ways of obtaining it and green has 1. The notation n(S) refers to the total number of outcomes possible in the set. Also notice when we list the elements of a set we use the set brackets {} and each element in the set is separated by a comma. {12, 34} is a set containing only two elements, whereas the set {1, 2, 3, 4} is a set containing 4 elements.
Some typical sample spaces written in set notation are:
Some sample spaces can be quite large and difficult to list such as the 52 cards in a standard deck.
Large sample spaces can make listing them a daunting task and often quite inefficient. For this reason we have a few other ways to organize a sample space.
A TREE DIAGRAM. The first tree diagram below shows two flips of a coin. The first flip could be a head or a tail and then the second flip could produce a head or a tail from each of the previous options. The tree displays the sample space for flipping a coin twice as {HH, HT, TH, TT}. We also notice the number of elements of the sample space is (2)(2) = 4. In the second tree diagram below we see the sample space for selecting two scoops of ice cream where the choices are vanilla, chocolate, and strawberry. There are three choices for the first scoop and then the second choice gives us three more options for each previous option. The tree diagram displays the sample space for choosing a double scoop as {VV, VC, VS, CV, CC, CS, SV, SC, SS}. Again we notice an easy method for determining the number of elements in this sample space is to multiply 3 by 3 to get 9.
In this case, getting a head is as likely as getting a tail. Thus this tree diagram represents a uniform probability. In this sample space each event is equally likely – getting a {Tail, Tail} is as likely as getting a {Head, Tail} or any of the other events. In this case, the choice of flavors might not be equally likely. If chocolate was the most popular flavor it would have a greater chance of being selected. The fact that each choice would not be equally likely would make the probabilities non-uniform. For example, getting {Chocolate, Chocolate} would be more likely to occur than {Strawberry, Strawberry}.
A TABLE OR CHART. A tree diagram would probably not be the best way to organize the sample space for all possible outcomes when rolling two dice. The tree would have 6 initial branches and then for each of those branches there would 6 more branches – in just two levels of the tree we would already have 36 branches. This would create a very large (hard to draw) tree diagram. A better way to organize this data might be to create a table.
When dealing with the occurrence of more than one event or activity such as this one, it is important to be able to quickly determine how many possible outcomes exist without listing all of the possible events. In this case to determine the total number of outcomes we could simply multiply 6 times 6 to get 36. This simple multiplication process is known as the Fundamental Counting Principle.
The number of ways in which a series of successive things can occur is found by multiplying the number of ways in which each thing can occur.
Ex. #1 - The Club Restaurant offers 6 appetizers, and 14 main courses. What is the sample space if a person orders one appetizer & one main course? (6)(14) = 84 Ex. #2 – A coin if flipped, a dice is rolled and a card is picked. What is the sample space for these events? (2)(6)(52) = 624 Ex. #3 - The pass lock on a iPad requires a 4 digit password. What is the sample space for all 4 digit passcodes? (10)(10)(10)(10) = 10,000 Ex. #4 - Jeff has 5 t-shirts, 3 pants, and 4 pairs of shoes. What is the sample space for all possible outfits? (5)(3)(4) = 60
The fundamental counting principle handles the calculation for most sample spaces. We will look at the use of permutations and combinations to handle other types of situations at a later time.
CONCEPT 2 – Intuitive Idea of Probability
We use ratios to show how likely, or unlikely, an outcome might be. This ratio is called the probability of the event. A probability is expressed as:
Ratios are most often expressed in the form of a fraction. Fractions, however, can be expressed as decimals or percent. So probabilities may be expressed as fractions, decimals or percent.
The range of the values for a probability is from 0 to 1. To have a probability of 0 means that it is impossible for the event to happen such as picking a red marble from a bag of 7 blue and 3 white marbles. To have a probability of 1 means that it is a certain event such as picking an even numbered tile from a bag that contain tiles with the numbers 2, 4, 6 and 12.
The big focus for this objective are the concepts that follow this point. Much of the earlier stuff is simply to set the table for the Venn diagramming.
CONCEPT 3 – Diagramming the Sample Space using Venn Diagrams
A sample space represents all things that could occur for a given event. In set theory language this would be known as the Universal Set – all elements defined by that set. We often use Venn diagrams to display the relationships within sets and sample spaces.
Diagramming the Universal Set (Sample Space)
The universal set is usually diagrammed as a rectangle. The set name which is being used as the universal set is usually placed in the upper left hand corner of the shape. Depending on the size of the set you do not have to include all elements of the set in the diagram, usually a few are provided to give an image of some of the values of the set. If the set is small, then all elements should be listed.
To diagram our sample space, the set M, a bag of marbles with 4 red marbles (solid) and 6 white marbles (empty) we create the rectangle, label it the universal set M, and then list out the elements of the set. In this case because there are only 10 elements it is easy to list them all out in the diagram.
CONCEPT 4 – Define and Diagram Outcomes (Subsets) of the Sample Space (Universal Set)
As stated earlier, a probability has two components, the sample space, which represents all possible things that could happen, and the defined successful outcomes, which represents the number of times a particular event occurs in that sample space. The outcome could be picking a heart from a deck of cards, rolling an even number on a dice, spinning a spinner and getting blue….. an outcome is simply a subset of the universal set. A subset is a collection of elements that all exist within another set. If all elements of set X belong to set Y, then it is said that set X is a subset of set Y. Any set formed with elements of the universal set is a subset of that universal set.
For example if the sample space was rolling a D12 (a 12 sided dice) some subsets might be:
Rolling a prime number, Set P = {2, 3, 5, 7, 11} is a subset of Set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
Rolling an even number less than 5, Set E = {2, 4} is a subset of Set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
Rolling a number greater than 12, Set B = {} is a subset of Set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
When determining the probability of a complement it is usually simplest to calculate the probability of the event and then subtract it from 1.
CONCEPT 6 – Mutually Exclusive or Disjoint Sets
More than one subset can be defined at a time from a universal set, so for example we could define the set of all red marbles, or the set of all even numbers, or the set of red marbles with numbers greater than 3 - the list seems like it could go on forever. Sometimes when we define more than one set at a time they have no elements in common. This is known as being mutually exclusive or disjoint. Two events are mutually exclusive events if the events
cannot both occur in the same trial of an experiment, for example the flip of a coin cannot be both heads and tails and thus those two events are mutually exclusive.
In both of these cases you cannot be both red and white or even and odd, thus they are mutually exclusive.
Understanding the intersection is so so important!! The intersection is critical to independence... it is critical to understanding the addition and multiplication rules for probabilities. Make sure this is understood.
CONCEPT 7 – The Intersection, “AND”
Union isn't quite as critical as intersection but this is also very important to understand. Take time to focus on the visual representations and how the uniion is determined.
CONCEPT 8 – The Union, “OR”
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## Intermediate Algebra (12th Edition)
$\sqrt[12]{6,912}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\sqrt[4]{3}\cdot\sqrt[3]{4} ,$ express the radicals as radicals with same indices by finding the $LCD$ of the indices. Once the indices are the same, use the laws of radicals to simplify the expression. $\bf{\text{Solution Details:}}$ The $LCD$ of the indices, $4$ and $3 ,$ is $12$ since it is the lowest number that can be divided exactly by both indices. Multiplying the index by a number to make it equal to the $LCD$ and raising the radicand by the same multiplier results to \begin{array}{l}\require{cancel} \sqrt[4(3)]{3^3}\cdot\sqrt[3(4)]{4^4} \\\\= \sqrt[12]{27}\cdot\sqrt[12]{256} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to\begin{array}{l}\require{cancel} \sqrt[12]{27(256)} \\\\= \sqrt[12]{6,912} .\end{array}
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The Improving Mathematics Education in Schools (TIMES) Project
Further trigonometry
Measurement and Geometry : Module 24Year : 10
June 2011
Assumed Knowledge
• Familiarity with the content of the module,
• Familiarity with basic coordinate geometry.
• Facility with simple algebra, formulas and equations.
• Familiarity with surds.
Motivation
In the module, we showed that if we know the angles and one side in a right-angled triangle we can find the other sides using the trigonometric ratios sine, cosine and tangent. Similarly, knowing any two of the sides in a right-angled triangle enables us to find all the angles.
Not all triangles contain a right-angle. We can relate sides and angles in an arbitrary triangle using two basic formulas known as the sine rule and the cosine rule.
Armed with these we can solve a greater range of problems in two dimensions and extend these ideas to three-dimensional problems as well. This is an essential tool for surveyors and civil engineers.
It soon becomes apparent that in some cases we need to be able to define the trigonometric ratio of an obtuse angle. This allows us to deal with a broader range of problems and applications. It will also provide the model for extending the definition of the trigonometric ratios to any angle. This idea will be picked up in the module,
Content
In the module, Introductory Trigonometry − Years 9-10, we defined the three standard trigonometric ratios sine, cosine and tangent of an angle θ, called the reference angle,
in a right-angled triangle.
These are defined by:
sin θ = , cos θ = , tan θ = , where 0° < θ < 90°.
Students should learn these ratios thoroughly. One simple mnemonic that might assist them is SOH CAH TOA, consisting of the first letter of each ratio and the first letter of the sides making up that ratio.
In a right-angled triangle, the other two angles are complements of each other. As the diagram below shows, the side opposite one of these angles is adjacent to the other.
Thus, it can be seen that,
sin θ = cos (90° − θ) and cos θ = sin (90° − θ) if 0° < θ < 90°
The cosine (co-sine) is so named since the cosine of an angle is the sine of its complement.
These ratios can be used to find sides and angles in right-angled triangles.
EXAMPLE
Find, correct to two decimal places, the value of the pronumeral in each triangle.
a b
Solution
a sin 15° = = x = 8 × sin 15° ≈ 2.07 (correct to two decimal places) b cos 28° = = a = 12.2 × cos 28° ≈ 10.77 (correct to two decimal places)
EXAMPLE
Calculate the value of θ, correct to one decimal place.
a b
Solution
a cos θ = So, θ = cos−1 ≈ 54.1° (correct to one decimal place) b tan θ = So, θ = tan−1 ≈ 30.3° (correct to one decimal place)
Note that for 0 < x < 1 the statement sin-1 x = θ means that sin θ = x. This notation is standard but it is essential that students do not confuse the inverse notation with the usual meaning of the index -1 used in algebra. To help avoid this confusion, it is best to always read sin-1 x as inverse sine of x and tan-1 x as inverse tangent of x.
Exact values
The trigonometric ratios for the angles 30°, 45° and 60° can be expressed using surds and occur very frequently in introductory trigonometry, in senior mathematics and in calculus. It is thus important for students to become familiar with them.
One way to find them quickly is to sketch the following triangles and then simply write down the ratios.
A right-angled triangle containing a 45° angle will be isosceles, so we choose the two shorter sides to be 1 unit in length and use Pythagoras’ theorem to find the hypotenuse.
For the angles 30° and 60°, we start with an equilateral triangle of side 2 units in length and drop a perpendicular as shown. Simple geometry and Pythagoras’ theorem gives the remaining information as shown in the diagram.
The table of values can now be completed from these diagrams.
θ sin θ cos θ tan θ 30° 45° 1 60°
Index notation
There are several deviations from the usual index notation that arise in trigonometry. Students may initially find them confusing.
We write, for example (tan θ) 2 as tan2 θ, (sin θ)3 as sin3 θ and so on. This must not be confused with the inverse notation discussed above. We do not write, for example,
sin-2 θ for (sin θ)-2, since this would confuse the usual meaning of indices with inverses.
EXERCISE 1
aSimplify tan2 30 − .
b The top of a tower has an angle of elevation of 45° from a point on the ground. From a point 100m further on, the angle of elevation is 30°, as shown in the diagram. Find the exact value, x, of the height of the tower.
Three-dimensional problems
We can use our knowledge of trigonometry to solve problems in three dimensions.
EXAMPLE
In the triangular prism opposite, find:
a the length CF
b the length BF
c the angle BFC, correct to one decimal place.
Solution
a Applying Pythagoras’ theorem to CEF, CF2 = 42 + 52 = 41 Hence CF = cm
b Applying Pythagoras’ theorem to BCF, BF2 = 32 + ()2 = 50 So BF = 5 cm
c To find the angle BFC, draw BCF and let BFC = θ. Now tan θ = giving θ = tan−1 So BFC ≈ 25.1° (to one decimal place)
EXERCISE 2
Find CEG in the cube shown below.
The Sine Rule
In many applications we encounter triangles that are not right-angled. We can extend our knowledge of trigonometry to deal with these triangles. This is done using two basic formulas, the first of which is called the sine rule.
We will assume, for the moment, that we are dealing with an acute-angled triangle ABC.
As shown in the diagram, we drop a perpendicular CP of length h from C to AB.
Then in APC we have sin A = , so h = b sin A.
Similarly, in CPB we have sin B = , so h = a sin B.
Equating these two expressions for h we have b sin A = a sin B which we can write as
= .
The same result holds for the side and the angle , so we can write
This is known as the sine rule. In words it says: any side of a triangle over the sine of the opposite angle equals any other side of the triangle over the sine of its opposite angle.
We will soon see how to extend this result to obtuse-angled triangles.
EXAMPLE
In ABC, AB = 9 cm. ABC = 76° and ACB = 58°.
Find, correct to two decimal places:
a AC b BC
Solution
a Apply the sine rule: = and so AC = ≈ 10.30 cm (to two decimal places) b To find BC, we need the angle CAB opposite it. CAB = 180° − 58° − 76° = 46° Thus, by the sine rule: = BC = ≈ 7.63 cm (to two decimal places)
Bearings
True bearings were covered in the module, Introductory Trigonometry.
We can now use the sine rule to solve simple surveying problems involving non-right-angled triangles.
EXAMPLE
From the points A and B, 800 metres apart, on a straight North-South road, the bearing of a house is 125°T and 050°T respectively. Find how far each point is from the house, correct to the nearest metre.
Solution
We draw a diagram to represent the information.
We can find the angles in AHB.
HAB = 180° − 125° = 55° and AHB = 180° − 50° − 55° = 75°
Apply the sine rule to ABH:
= and so BH = ≈ 678.44 m (to two decimal places)
Thus B is approximately 678 metres from the house.
Similarly, = and so AH = ≈ 634.45 m (to two decimal places)
Thus A is approximately 634 metres from the house.
EXERCISE 3
From a point at P, west of a building OA, the angle of elevation of the top A of the building OA is 28°. From a point Q 10m further west from P the angle of elevation is 20°. Draw a diagram and then use the sine rule to find the distance AP and hence the exact height of the building. Finally, evaluate the height OA to the nearest centimetre.
Finding angles
The sine rule can be used to find angles as well as sides in a triangle. One of the known sides, however, must be opposite one of the known angles.
EXAMPLE
Assuming that all the angles are acute.
Find the angle θ in the triangle FGH, to the nearest degree.
Solution
Apply the sine rule to FGH, = To make the algebra easier, take the reciprocal of both sides. = Hence sin θ = = 0.644... Hence θ = 40° (to the nearest degree)
As seen in the example above, it is easier when finding angles, to write the sine rule as = before substituting in the given information.
Dealing with obtuse angles
Both the sine rule and the cosine rule are used to find angles and sides in triangles. What happens when one of the angles is obtuse? To deal with this we need to extend the definition of the basic trigonometric ratios from acute to obtuse angles. We use coordinate geometry to motivate the extended definitions as follows.
We draw the unit circle centre the origin in the Cartesian plane and mark the point on the circle in the first quadrant.
In the diagram shown, since = cos θ, we can
see that the x-coordinate of P is cos θ. Similarly,
the y-coordinate of P is sin θ.
Hence the coordinates of P are (cos θ, sin θ).
We can now turn this idea around and say that
if θ is the angle between OP and the positive x-axis, then:
• the cosine of θ is defined to be the x-coordinate of
the point P on the unit circle and
• the sine of θ is defined to be the y-coordinate of the
point P on the unit circle.
This definition can be applied to all angles, both positive and negative, but in this module we will restrict the angle to be between 0° and 180°.
Consistency of the definitions
In the module, Introduction to Trigonometry, we defined sin θ = and cos θ = , where 0° < θ < 90°. In the previous section we defined cos θ = OQ and sin θ = PQ. We must show that the two definitions agree.
The diagram below shows the right-angled triangle OAB, and the triangle OPQ both containing the angle θ. Triangle OPQ has its vertex P on the unit circle. These triangles are similar and so the ratio = = PQ, which is the y-coordinate of the point P. Similarly = = 1, which is the x-coordinate of P.
So we have shown the two definitions agree.
As an example, let us take θ to be 30°, so has coordinates (cos 30°, sin 30°).
Now move the point P around the circle to P′ so that OP′ makes an angle of 150° with the positive x-axis. Note that 30° and 150° are supplementary angles.
The coordinates of P′ are (cos 150°, sin 150°).
But we can see that the triangles OPQ and OP′Q′ are congruent, so the y-coordinates
of P and P′ are the same. Thus, sin 150° = sin 30°.
Also, the x-coordinates of P and P′ have the same magnitude but opposite sign,
so cos 150° = −cos 30°.
From this typical example, we see that if θ is any obtuse angle, then its supplement,
180°− θ is acute, and the sine of θ is given by
sin θ = sin (180° − θ), where 90° < θ < 180°.
Similarly, if θ is any obtuse angle then the cosine of θ is given by
cos θ = −cos (180° − θ), where 90° < θ < 180°.
In words this says:
• the sine of an obtuse angle equals the sine of its supplement,
• the cosine of an obtuse angle equals minus the cosine of its supplement.
EXAMPLE
Find the exact value of:
a sin 150° b cos 150° c sin 120° d cos 120°
Solution
a sin 150° = sin (180 − 150)° b cos 150° = −cos (180 − 150)° = sin 30° = −cos 30° = = − c sin 120° = sin (180 − 120)° d cos 120° = −cos (180 − 120)° = sin 60° = −cos 60° = = −
Note: You can verify these results using your calculator.
The sine rule is also valid for obtuse-angled triangles.
EXERCISE 4
Reprove the sine rule = for a triangle in which angle A is obtus.
The Angles 0°, 90°, 180°
We can use the extended definition of the trigonometric functions to find the sine and cosine of the angles 0°, 90°, 180°.
EXERCISE 5
Draw a diagram showing the point on the unit circle at each of the above angles. Use the coordinates of to complete the entries in the table below.
θ 0° 90° 180° sin θ cos θ
The tangent of an obtuse angle
For θ in the range 0° < θ < 90° or 90° < θ < 180° we define the tangent of an angle θ by
tan θ = , for cos θ ≠ 0.
In the case when cos θ = 0, the tangent ratio is undefined. This will happen, when θ = 90°.
If θ is in the range 0° < θ < 90°, this definition agrees with the usual definition of
tan θ =
Hence, if θ is an obtuse angle, then
tan θ = (from the definition) = (since θ is obtuse) = −tan (180° − θ) (from the definition).
Hence the tangent of an obtuse angle is the negative of the tangent of its supplement.
Note that tan 0° = 0 and tan 180°= 0 since the sine of these angles is 0 and that tan 90° is undefined since cos 90°= 0.
EXERCISE 6
Find the exact values of tan 150° and tan 120°.
The Ambiguous Case
In our work on congruence, it was emphasized that when applying the SAS congruence test, the angle in question had to be the angle included between the two sides. Thus, the following diagram shows two non-congruent triangles ABC and ABC′ with two pairs of matching sides sharing a common (non-included) angle.
Suppose we are told that a triangle PQR has PQ = 9, PQR = 45°, and PR = 7. Then the angle opposite PQ is not uniquely determined. There are two non-congruent triangles that satisfy the given data.
Applying the sine rule to triangle we have
=
and so sin θ = ≈ 0.9091.
Thus θ ≈ 65°, assuming that θ is acute. But the supplementary angle θ′ = 115°. The angle PR′Q also satisfies the given data. This situation is sometimes referred to as the ambiguous case.
Since the angle sum of a triangle is 180°, in some circumstances only one of the two angles calculated is geometrically valid.
EXERCISE 7
Find the value of θ in the following diagram,
explaining why the answer is unique.
The cosine rule
We know from the SAS congruence test, that a triangle is completely determined if we are given two sides and the included angle. However, if we know two sides and the included angle in a triangle, the sine rule does not help us determine the remaining side.
The second important formula for general triangles is the cosine rule.
Suppose ABC is a triangle and that the angles A and C are acute. Drop a perpendicular from B to AC and mark the lengths as shown in the diagram.
In BDA, Pythagoras’ theorem gives
c2 = h2 + (bx)2.
Also in CBD, another application of Pythagoras’ theorem gives
h2 = a2x2.
Substituting this expression for into the first equation and expanding,
c2 = a2 − x2 + (b − x)2 = a2 − x2 + b2 − 2bx + x2 = a2 + b2 − 2bx.
Finally, from CBD, we have x = a cos C and so
c2 = a2 + b2 − 2abcos C
This last formula is known as the cosine rule. By relabeling the sides and angle, we can also write a2 = b2 + c2 − 2bc cos A, and b2 = a2 + c2 − 2ac cos B.
Notice that if C = 90° then, since cos C = 0, we obtain Pythagoras’ theorem, and so we can regard the cosine rule as Pythagoras’ theorem with a correction term.
The cosine rule is also true when C is obtuse, but note that in this case the final term in the formula will produce a positive number, because the cosine of an obtuse angle is negative. Some care must be taken in this instance.
EXAMPLE
Find the value of x to one decimal place.
solution
Applying the cosine rule:
x2 = 72 + 82 − 2 × 7 × 8 cos 110° = 151.30... so x = 12.3 (to one decimal place)
EXERCISE 8
Prove that the cosine rule also holds in the case when C is obtuse.
Finding angles
We know from the SSS congruence test that if the three sides of a triangle are known then the three angles are uniquely determined. Again, the sine rule is of no help in finding them since it requires the knowledge of (at least) one angle, but we can use the cosine rule instead.
We can substitute the three side lengths a, b and c into the formula c2 = a2 + b2 − 2ab cos C where C is the angle opposite the side c, and then re-arrange to find cos C and hence C.
Alternatively, we can re-arrange the formula to obtain
cos C =
and then substitute. Students may care to rearrange the cosine rule or learn a further formula. Using this form of the cosine rule often reduces arithmetical errors.
Recall that in any triangle ABC, if a > b then A > B.
EXAMPLE
A triangle has side lengths 6 cm, 8 cm and 11 cm. Find the smallest angle in the triangle.
Solution
The smallest angle in the triangle is opposite the smallest side.
Applying the cosine rule:
62 = 82 + 112 − 2 × 8 × 11 × cos θ cos θ = = and so θ ≈ 32.2 (correct to one decimal place)
Extension — The longest side and the largest angle of a triangle
In the module Congruence, we proved an important relationship between the relative sizes of the angles of a triangle and the relative lengths of its sides: The angle of a triangle opposite a longer side is larger than the angle opposite a shorter side.
For scalene triangles, this can be restated in terms of inequalities of all three sides as follows:
If ABC is a triangle in which a > b > c, then A > B > C.
This result can be proved in an interesting way using either the sine rule or the cosine rule.
The longest side and the sine rule
The following exercise uses the fact that sin θ increases from 0 to 1 as θ increases from
0° to 90°.
EXERCISE 9
Let ABC be a triangle in which a > b > c.
1. What can you conclude about the relative sizes of sin A, sin B and sin C using the sine rule?
2. If no angle is obtuse, what can you conclude about the relative sizes of A, B and C?
3. If a triangle PQR has an obtuse angle P = 180° − θ, where θ is acute, use the identity sin (180°− θ) = sin θ to explain why sin P is larger than sin Q and sin R.
4. Hence prove that if the triangle ABC has an obtuse angle, then A > B > C.
The longest side and the cosine rule
This exercise uses the fact that cos θ decreases from 1 to −1 as θ increases from 0° to 180°.
EXERCISE 10
Let ABC be a triangle in which a > b > c.
a
Write down cos A and cos B in terms of a, b and c, and express each in terms of their common denominator 2abc.
b
Show that cos B − cos A =
= .
c
Hence explain why cos B > cos A.
d
Similarly explain why cos C > cos B, and hence show that A > B > C.
EXERCISE 11
a
Use Pythagoras’ theorem to show that the hypotenuse is the longest side of a right-angled triangle
b
Why is the result of part a special case of the theorem above?
Area of a Triangle
We saw in the module, Introductory Trigonometry that if we take any triangle with two given sides and about a given (acute) angle θ, then the area of the triangle is given by
Area = ab sin θ.
EXERCISE 12
Derive this formula in the case when θ is obtuse.
EXERCISE 13
A triangle has two sides of length 5 cm and 4 cm containing an angle θ. Its area is 5 cm2. Find the two possible (exact) values of θ and draw the two triangles that satisfy the
given information.
EXERCISE 14
Write down two expressions for the area of triangle and derive the sine rule from them.
The sine and cosine rules can be used to solve a range of practical problems in surveying and navigation.
Three-dimensional problems
EXAMPLE
The point M is directly across the river from the base B of a tree AB. From a point 7 metres upstream from M, the angle of elevation of the top A of the tree is 17°. From a point Q, 5 metres downstream from M the angle of elevation of the top of the tree is 19°. Assuming that PMQ is a straight line and that the tree is on the edge of the river,
we wish to find the width w metres of the river.
Solution
In problems such as these, it is imperative to draw a careful diagram.
Let BP = a, BQ = b, and AB = h, and then applying Pythagoras’ theorem to triangles
BMP and BMQ, we have a = , b = .
From triangles ABP and ABQ we have h = a sin 17°, h = b sin 19°.
Equating these, substituting in the values of a and b and squaring we arrive at
(49 + w2)sin2 17° = (25 + w2)sin2 19°
We can now make w2 the subject and obtain w2 = , and so
w ≈ 8.66 m correct to 2 decimal places.
Note: Do not evaluate until the last step to retain full calculator accuracy.
Civil engineers analyzing forces and stresses in buildings and other structures often use vectors to represent the direction and magnitude of these forces. A vector is an arrow which has both direction and magnitude. The sine and cosine rules are used in vector diagrams to find resultant forces and stresses. This is an important application.
Trigonometric Identities
As well as having practical uses, the sine and cosine rules can be used to derive theoretical results, known as trigonometric identities that have important implications and applications in later work. Among these are the double angle results which we will describe below.
Using the area formula, A = ab sin C, for a triangle with two sides a and b, containing an angle C, we can do the following:
Fix acute angles a and β and let angle C = a + β. From the point C draw CD of length y and construct triangle ABC as shown in the diagram, where BA is perpendicular to CD.
From the diagram, we have
= cos α => y = a cos α (1) and = cos β => y = b cos β (2).
Comparing areas,
ab sin (α + β) = ay sin α + by sin β.
Substituting in the value of y from (2) into the first term and that from (1) into the second, we have, after some simplification,
sin (α + β) = sin α cos β + cos α sin β.
In the discussion above we assumed that α, β were acute angles. This identity holds for all
α and β but to show this requires a different approach.
Note that sin (α + β) ≠ sin α + sin β. For example, sin(60°+30°) = sin 90° = 1, whereas
sin 60° + sin 30° = ≠ 1.
EXERCISE 15
Use the above formula to show that the exact value of sin 75° is .
Putting α = β = θ in the above formula, we obtain the double angle formula
for sine, namely
sin 2θ = 2sin θ cos θ.
There is a similar double angle formula for cosine,
cos 2θ = cos2 θ − sin2 θ.
Both formulas are extremely useful when calculus is applied to the trigonometric functions.
Angles of any magnitude and the trigonometric functions
We saw in this module how to use the unit circle to give meaning to the sine and cosine of an obtuse angle. This definition can be extended to include angles greater than 180° and also to negative angles.
Thus, for example, if θ is between 180° and 270°, then sin θ = −sin (θ − 180°) and
cos θ = −cos (θ − 180°).
Once we can find the values of sin θ and cos θ for values of θ, we can plot graphs of the functions y = sin θ, y = cos θ.
These ideas will be developed in the module, Trigonometric Functions.
The graphs of the sine and cosine functions are used to model wave motion and electrical signals. They are an essential part of modern signal processing and telecommunications. This provides a breathtaking example of how a simple idea involving geometry and ratio was abstracted and developed into a remarkably powerful tool that has changed the world.
History
In the module, we mentioned that the Greeks had a version of trigonometry involving chords. This is shown in the diagram below.
In the diagram, the chord of the angle is the length of the chord that subtends an angle α at the centre in the circle of radius R.
Ptolemy (85-185AD), who lived and worked in Alexandria, wrote an extremely influential book called the Mathematical Syntaxis. It was translated into Arabic and given the Arabic title of Almagest.
Ptolemy considered chords subtending an angle α on the circumference. Using modern notation, if we take a diameter of AB length 1 unit as shown and a chord AC subtending an angle α at the circumference, then the lengths AC and BC are respectively sin α and cos α.
Since angles on the circumference subtended by the same arc are equal, if α is the angle subtended by any chord in this circle, then the length of that chord is always sin α.
Ptolemy also showed that if ABCD is a cyclic quadrilateral, then
AB.CD + BC.DA = AC.BD.
That is, the sum of the products of the opposite sides of a cyclic quadrilateral is equal to the product of the diagonals.
This result is known as Ptolemy’s theorem.
Applying Ptolemy’s theorem in the diagram below, where the circle has diameter 1, we obtain the result sin (α + β) = sin α cos β + cos α sin β, that we derived above.
Using this, and other formulas, Ptolemy was able to construct a detailed table of chords of angles. Since chords are closely related to the sine ratio, he essentially had a table of sines.
The sine rule in the circle
Given a triangle ABC, we can draw its circumcircle with diameter BD as shown. Let 2R be the diameter of the circumcircle.
Then for α acute, the sine of angle BDC is given by
sin BDC = sin α = .
Re-arranging gives = 2R.
Thus the quantities , and in the sine rule are all equal to the diameter of the circumcircle of the triangle ABC.
Regiomontanus (1436-1476), who wrote the first modern European book on trigonometry, included the sine rule and its derivation in his work.
The regular pentagram and regular pentagon
The regular pentagram and the regular pentagon have always been a source of fascination and is often used in astrology. It is based on a triangle whose properties are investigated in the following exercise.
EXERCISE 16
We begin with an isosceles triangle with angles 72°, 72°, 36°, as shown. This triangle
occurs naturally inside both the regular pentagon and the regular pentagram.
For ease of calculation we take the two equal sides to be 4 units in length.
We take the point D on AB such that BDC = 72°. Finally, we let BC = 2x.
a Show that the information marked on the diagram is correct.
b Prove that the triangles ABC and CDB are similar.
c Deduce that = and solve this equation to obtain x = − 1.
d By dropping a perpendicular from A to BC show that cos 72° = .
e Use the identity cos2 θ + sin2 θ = 1 to show that sin 72° = .
Exercise 1
a −1 b 50( + 1)
Exercise 2
tan-1 ≈ 35.26°, correct to two decimal places.
Exercise 3
AP = 24.57… m and OA = 11.54 m to the nearest cm.
Exercise 4
Let CAM = θ
Therefore CAB = 180° − θ
h = b sin θ (triangle CAM) and h = a sin B (triangle CAB)
Therefore a sin B = b sinθ = b sin (180° − A) = b sin A so =
hence the result.
Exercise 5
sin 0° = 0, sin 90° = 1, sin 180° = 0
cos 0° = 1, cos 90° = 0, cos 180° = −1
Exercise 6
, −
Exercise 7
sin−1 ≈ 22.62° (correct to two decimal places). The two sides and the right angle define a unique triangle (RHS congruence).
Exercise 8
In triangle BCM, h2 = a2 − (c + MA)2
In triangle CMA, h2 = b2MA2
Therefore, a2 − (c + MA)2= b2MA2
a2 = b2 + c2 + 2c × MA
But MA = b cos (180° − A) = − b cos A
Hence a2 = b2 + c2 − 2bc cos A
Exercise 9
a sin A > sin B > sin C (if a > b and a sin B = b sin A then sin A > sin B
b By the remark at the beginning of this paragraph, A > B > C.
c The angles P and Q add to θ, because the angle sum of the triangle is 180°.
Hence P and Q are smaller than θ, so sin P and sin Q are less than sin θ = sin P.
d From part c, the obtuse angle is A, which is therefore the largest of the three angles.
Hence sin B > sin C, where B and C are acute, so A > B > C.
Exercise 10
a cos A = and cos B =
=
=
b Grouping terms, cos B − cos A = = =
c From a > b > c > 0, it follows that ab > c2 and a3 > b3. Hence cos B > cos A.
d A similar argument proves that cos C > cos B, so cos C > cos B > cos A.
By the remark at the start of this paragraph, A > B > C.
Exercise 11
a Let ABC be right-angled at C. Then AB2 = AC2 + BC2. Hence AB2 is larger than both AC2 and BC2, so AB is longer than both AC and BC.
b In a right-handed triangle, the other two angles are acute, because the angle sum of the triangle is 180°. Hence the right angle is the largest angle.
Exercise 12
We use the same diagram as that used for the obtuse angle case proof of the sine and cosine rule.
Area = c × h = cb sin θ = cb sin A
The stated result is obtained by symmetry of argument.
Exercise 13
θ = 30° or 150°
Exercise 14
For a given triangle ABC, cb sin A = ca sin B. Therefore bsin A = asin B and = .
Exercise 15
sin 75° = sin (45° + 30°) = sin 45° cos 30° + sin 30° cos 45° = × + × =
Exercise 16
a DCB = 36° (angle sum of triangle), ACD = 36° (DCA = 36° and ACD = 72°) CD = 2x (triangle CBD is isosceles), DA = 2x (triangle CDA is isosceles), DB = AB − AD = 4 − 2x.
bTriangle ABC is similar to triangle CDB (AA)
c = d cos B = e sin2 B = 1 − =
hence = = x2 − 2x = 4 x2 + 2x + 1 = 5 (x + 1)2 = 5 x = −1 + as x > 0
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# The cross product of two vectors gives zero when the vectors enclose an angle of A. ${90^0}$B. ${180^0}$C. ${45^0}$D. ${120^0}$
Last updated date: 21st Jul 2024
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Hint: To answer this question, we first need to understand what is a vector. A vector is a two-dimensional object with both magnitude and direction. A vector can be visualized geometrically as a guided line segment with an arrow indicating the direction and a length equal to the magnitude of the vector.
Cross product: The cross product a$\times$b is defined as a vector c that is perpendicular (orthogonal) to both a and b, with a magnitude equal to the area of the parallelogram that the vectors span and a direction given by the right-hand law.Cross product formula of two vectors,
$\overrightarrow a \times \overrightarrow b = a.b.\sin \theta$
Here $\overrightarrow a$ and $\overrightarrow b$ are the two vectors and $\theta$ is the angle between two vectors. Here $a$ and $b$ are the magnitudes of both vectors
As given in the question, the cross product is zero. Therefore,
$a.b.\sin \theta = 0$
Now as we know that magnitude can’t be zero
So, to make this product zero $\sin \theta$must be zero
So, $\sin \theta = 0$
As $\sin \theta$=0 so the angle must be ${0^0}$ or ${180^0}$.
As given in this question, the option available is ${180^0}$.
Hence, the correct answer is option B.
Note: In three-dimensional spaces, the cross product, area product, or vector product of two vectors is a binary operation on two vectors. It is denoted by the symbol ($\times$). A vector is the cross product of two vectors.
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## How do you work out the interquartile range from a frequency table?
Steps:
1. Step 1: Put the numbers in order.
2. Step 2: Find the median.
3. Step 3: Place parentheses around the numbers above and below the median. Not necessary statistically, but it makes Q1 and Q3 easier to spot.
4. Step 4: Find Q1 and Q3.
5. Step 5: Subtract Q1 from Q3 to find the interquartile range.
How do you calculate the interquartile range?
To find the interquartile range (IQR), first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.
### How do we calculate cumulative frequency?
The cumulative frequency is calculated by adding each frequency from a frequency distribution table to the sum of its predecessors. The last value will always be equal to the total for all observations, since all frequencies will already have been added to the previous total.
What is cumulative frequency example?
The cumulative frequency of a value of a variable is the number of values in the collection of data less than or equal to the value of the variable. For example: Let the raw data be 2, 10, 18, 25, 15, 16, 15, 3, 27, 17, 15, 16. The cumulative frequency of 15 = 6 (Since, values ≤ 15 are 2, 10, 15, 15, 3, 15).
#### What is a cumulative frequency histogram?
It is an estimate of the probability distribution of a continuous variable. For a histogram In order to calculate the frequency density, we use. Cumulative frequency is accumulation of the frequencies. First plot the graph and then join up the points to make a cumulative curve.
How do you solve for quartiles?
The quartile formula helps in calculating the value that divides a list of numbers into quarters….Quartile Formula
1. First Quartile(Q1) = ((n + 1)/4)th Term.
2. Second Quartile(Q2) = ((n + 1)/2)th Term.
3. Third Quartile(Q3) = (3(n + 1)/4)th Term.
## What is the formula of lower quartile?
When the set of observations are arranged in ascending order the quartiles are represented as, First Quartile(Q1)=((n+1)/4)th Term also known as the lower quartile. The second quartile or the 50th percentile or the Median is given as: Second Quartile(Q2)=((n+1)/2)th Term.
What is the difference between cumulative frequency and frequency?
Frequency means how many times a particular datum(may be of any series) is repeated in general whereas cumulative frequency is the addition of the frequency of previous class to that of frequency of next class.
### What is cumulative frequency diagram?
A cumulative frequency graph shows the total number of values that fall below the upper boundary of each variable. All this means is that it represents the running-total of frequencies.
How do you get the cumulative frequency polygon?
To obtain the cumulative frequency polygon, we draw straight line sections to join these points in sequence. Estimate the median from the graph. There are 119 values, so the median will be the 119 + 1 2 = 60th value. This can be read from the graph as shown above. Estimate the interquartile range from the graph.
#### How to find the interquartile range from cumulative frequency?
41. 3 4 5 6 7 8 9 10 0 5 10 15 20 25 30 x c.f Upper Quartile: 8 Lower Quartile: 5.5 So, the Interquartile Range is: 8 – 5.5 = 2.5 Our Final Answer! 54. 3 4 5 6 7 8 9 10 0 5 10 15 20 25 30 x c.f The Median is just below 7, by estimation it is 6.8
How to calculate the interquartile range in unit 16?
Estimate the median from the graph. There are 119 values, so the median will be the 119 + 1 2 = 60th value. This can be read from the graph as shown above. Estimate the interquartile range from the graph. The lower quartile will be given by the 119 + 1 4 th value. The upper quartile will be given by the 3 (119 + 1) 4 th value.
## How to calculate cumulative frequency in unit 16?
Unit 16 Section 3 : Cumulative Frequency Lower quartile = n + 1 4 th value Median = n + 1 2 th value Upper quartile = 3 (n + 1) 4 th value Interquartile range = upper quartile – lower quartile Semi-interquartile range = interquartile range 2
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# How do you determine the height for Part B of this question?
Jul 7, 2016
$14$
$47.68 m {s}^{-} 1$, rounded to two decimal places.
$142 J$
#### Explanation:
Let height of the building be $= h \text{ } m$
After the wrench is dropped from the roof top, the kinematic equation is given by
${v}^{2} - {u}^{2} = 2 g s$ ......(1)
where $v , u \mathmr{and} g$ are final velocity after dropping distance $s$, initial velocity and acceleration due to gravity respectively. Let $g = 9.8 m {s}^{-} 2$
Let us find out the distance dropped when workman standing at the eighth floor observes the wrench. (for simplicity eyes of worker assumed at the floor level)
${\left(33.1\right)}^{2} - {0}^{2} = 2 \times 9.8 \times s$
$\implies s = {\left(33.1\right)}^{2} / \left(2 \times 9.8\right)$
$\implies s \approx 55.9 m$
Since the floors above the first are of height $8 m$ each
Hence number of floors above the eighth floor$= \frac{55.9}{8}$
$= 7$, rounded to nearest digit as number of floors can not be a fraction.
Total number of floors of the building $= 7 + 7 = 14$
We need to remember that the workman standing on the eighth floor has only seven floors below him.
Height of the building $h = 12.0 + 13 \times 8.00 = 116.00 m$
(First floor is of $12.0 m$ and all other floors are of $8.00 m$ height)
From (1) velocity $v$ when the wrench hits the ground
${v}^{2} = 2 \times 9.8 \times 118.00$
$v = 47.68 m {s}^{-} 1$, rounded to two decimal places.
Kinetic energy of wrench when it hits the ground$= m g h$
$= 0.125 \times 9.8 \times 116 = 142 J$
(All its potential energy while at the roof gets converted into its kinetic energy as it hits the ground.
It can also be calculated using the expression $K E = \frac{1}{2} m {v}^{2}$)
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# How do you solve 1/x + 1 = x/2?
Feb 27, 2016
$x = 1 \pm \sqrt{3}$
#### Explanation:
$1$. In order to solve this equation, each fraction must have the same denominator. To do this, multiply every term by $\textcolor{b l u e}{2 x}$ since the equation consists of fractions with denominators $\textcolor{\mathmr{and} a n \ge}{x}$ and $\textcolor{\mathmr{and} a n \ge}{2}$. When you multiply the whole equation by $\textcolor{b l u e}{2 x}$, you can get rid of the denominators.
$\frac{1}{\textcolor{\mathmr{and} a n \ge}{x}} + 1 = \frac{x}{\textcolor{\mathmr{and} a n \ge}{2}}$
$\textcolor{b l u e}{2 x} \left(\frac{1}{x} + 1\right) = \textcolor{b l u e}{2 x} \left(\frac{x}{2}\right)$
$2$. Simplify.
$\frac{2 x}{x} + 2 x = \frac{2 {x}^{2}}{2}$
$\frac{2 \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}} + 2 x = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}$
$2 + 2 x = {x}^{2}$
$3$. Rewrite the equation in the form, $a {x}^{2} + b x + c = 0$.
${x}^{2} - 2 x - 2 = 0$
$4$. Solve for $x$ using the quadratic formula.
$\textcolor{p u r p \le}{a = 1} , \textcolor{t u r q u o i s e}{b = - 2} , \textcolor{b r o w n}{c = - 2}$
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(1\right) \left(- 2\right)}}{2 \left(1\right)}$
$x = \frac{2 \pm \sqrt{4 + 8}}{2}$
$x = \frac{2 \pm \sqrt{12}}{2}$
$x = \frac{2 \pm 2 \sqrt{3}}{2}$
$x = \frac{2 \left(1 \pm \sqrt{3}\right)}{2 \left(1\right)}$
$x = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \left(1 \pm \sqrt{3}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \left(1\right)}$
$\textcolor{g r e e n}{x = 1 \pm \sqrt{3}}$
$\therefore$, $x = 1 \pm \sqrt{3}$.
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# Derivative
In mathematics, the derivative is one of the two central concepts of calculus. (The other is the integral; the two are related via the fundamental theorem of calculus.)
The simplest type of derivative is the derivative of a real-valued function of a single real variable. It has several interpretations:
• The derivative gives the slope of a tangent to the graph of the function at a point. In this way, derivatives can be used to determine many geometrical properties of the graph, such as concavity or convexity.
• The derivative provides a mathematical formulation of rate of change; it measures the rate at which the function's value changes as the function's argument changes.
This derivative is the kind usually encountered in a first course on calculus, and historically was the first to be discovered. However, there are also many generalizations of the derivative.
The remainder of this article discusses only the simplest case (real-valued functions of real numbers).
## Differentiation and differentiability
In physical terms, differentiation expresses the rate at which a quantity, y, changes with respect to the change in another quantity, x, on which it has a functional relationship. Using the symbol Δ to refer to change in a quantity, this rate is defined as a limit of difference quotients
${\displaystyle {\frac {\Delta y}{\Delta x}}}$
as Δx approaches 0. In Leibniz's notation for derivatives, the derivative of y with respect to x is written
${\displaystyle {\frac {dy}{dx}}}$
suggesting the ratio of two infinitesimal quantities. The above expression is pronounced in various ways such as "dy by dx" or "dy over dx". The form "dy dx" is also used conversationally, although it may be confused with the notation for element of area.
Modern mathematicians do not bother with "dependent quantities", but simply state that differentiation is a mathematical operation on functions. The precise definition of this operation (which therefore need not deal with infinitesimal quantities) is given as:
${\displaystyle \lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}.}$
A function is differentiable at a point x if its derivative exists at that point; a function is differentiable on an interval if it is differentiable at every x within the interval. If a function is not continuous at x, then there is no tangent line and the function is therefore not differentiable at x; however, even if a function is continuous at x, it may not be differentiable there. In other words, differentiability implies continuity, but not vice versa. One famous example of a function that is continuous everywhere but differentiable nowhere is the Weierstrass function.
The derivative of a differentiable function can itself be differentiable. The derivative of a derivative is called a second derivative. Similarly, the derivative of a second derivative is a third derivative, and so on.
## Newton's difference quotient
The derivative of a function f at x is geometrically the slope of the tangent line to the graph of f at x. Without the concept which we are about to define, it is impossible to directly find the slope of the tangent line to a given function, because we only know one point on the tangent line, namely (x, f(x)). Instead, we will approximate the tangent line with multiple secant lines that have progressively shorter distances between the two intersecting points. When we take the limit of the slopes of the nearby secant lines in this progression, we will get the slope of the tangent line. The derivative is then defined by taking the limit of the slope of secant lines as they approach the tangent line.
File:Tangent-calculus.png
Tangent line at (x, f(x))
File:Secant-calculus.png
Secant to curve y= f(x) determined by points (x, f(x)) and (x+h, f(x+h)).
To find the slopes of the nearby secant lines, choose a small number h. h represents a small change in x, and it can be either positive or negative. The slope of the line through the points (x,f(x)) and (x+h,f(x+h)) is
${\displaystyle {f(x+h)-f(x) \over h}.}$
This expression is Newton's difference quotient. The derivative of f at x is the limit of the value of the difference quotient as the secant lines get closer and closer to being a tangent line:
${\displaystyle f'(x)=\lim _{h\to 0}{f(x+h)-f(x) \over h}.}$
File:Lim-secant.png
Tangent line as limit of secants.
If the derivative of f exists at every point x in the domain, we can define the derivative of f to be the function whose value at a point x is the derivative of f at x.
Since immediately substituting 0 for h results in division by zero, calculating the derivative directly can be unintuitive. One technique is to simplify the numerator so that the h in the denominator can be cancelled. This happens easily for polynomials; see calculus with polynomials. For almost all functions however, the result is a mess. Fortunately, many guidelines exist.
## Notations for differentiation
### Lagrange's notation
The simplest notation for differentiation that is in current use is due to Joseph Louis Lagrange and uses the prime mark:
${\displaystyle f'(x)\;}$ for the first derivative, ${\displaystyle f''(x)\;}$ for the second derivative, ${\displaystyle f'''(x)\;}$ for the third derivative, and ${\displaystyle f^{(n)}(x)\;}$ for the nth derivative, provided n > 3
### Leibniz's notation
The other common notation is Leibniz's notation for differentiation which is named after Leibniz. For the function whose value at x is the derivative of f at x, we write:
${\displaystyle {\frac {d\left(f(x)\right)}{dx}}.}$
We can write the derivative of f at the point a in two different ways:
${\displaystyle {\frac {d\left(f(x)\right)}{dx}}\left.{\!\!{\frac {}{}}}\right|_{x=a}=\left({\frac {d\left(f(x)\right)}{dx}}\right)(a).}$
If the output of f(x) is another variable, for example, if y=f(x), we can write the derivative as:
${\displaystyle {\frac {dy}{dx}}.}$
Higher derivatives are expressed as
${\displaystyle {\frac {d^{n}\left(f(x)\right)}{dx^{n}}}}$ or ${\displaystyle {\frac {d^{n}y}{dx^{n}}}}$
for the n-th derivative of f(x) or y respectively. Historically, this came from the fact that, for example, the 3rd derivative is:
${\displaystyle {\frac {d\left({\frac {d\left({\frac {d\left(f(x)\right)}{dx}}\right)}{dx}}\right)}{dx}}}$
which we can loosely write as:
${\displaystyle \left({\frac {d}{dx}}\right)^{3}\left(f(x)\right)={\frac {d^{3}}{\left(dx\right)^{3}}}\left(f(x)\right).}$
Dropping brackets gives the notation above.
Leibniz's notation allows one to specify the variable for differentiation (in the denominator). This is especially relevant for partial differentiation. It also makes the chain rule easy to remember, because the "du" terms appear symbolically to cancel:
${\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}.}$
(In the popular formulation of calculus in terms of limits, the "du" terms cannot literally cancel, because on their own they are undefined; they are only defined when used together to express a derivative. In nonstandard analysis, however, they can be viewed as infinitesimal numbers that cancel.)
### Newton's notation
Newton's notation for differentiation (also called the dot notation for differentiation) requires placing a dot over the function name:
${\displaystyle {\dot {x}}={\frac {dx}{dt}}=x'(t)}$
${\displaystyle {\ddot {x}}=x''(t)}$
and so on.
Newton's notation is mainly used in mechanics, normally for time derivatives such as velocity and acceleration, and in ODE theory. It is usually only used for first and second derivatives.
### Euler's notation
Euler's notation uses a differential operator, denoted as D, which is prefixed to the function with the variable as a subscript of the operator:
${\displaystyle D_{x}f(x)\;}$ for the first derivative, ${\displaystyle {D_{x}}^{2}f(x)\;}$ for the second derivative, and ${\displaystyle {D_{x}}^{n}f(x)\;}$ for the nth derivative, provided n > 1
This notation can also be abbreviated when taking derivatives of expressions that contain a single variable. The subscript to the operator is dropped and is assumed to be the only variable present in the expression. In the following examples, u represents any expression of a single variable:
${\displaystyle Du\;}$ for the first derivative, ${\displaystyle D^{2}u\;}$ for the second derivative, and ${\displaystyle D^{n}u\;}$ for the nth derivative, provided n > 1
Euler's notation is useful for stating and solving linear differential equations.
## Critical points
Points on the graph of a function where the derivative is undefined or equals zero are called critical points or sometimes stationary points (in the case where the derivative equals zero). If the second derivative is positive at a critical point, that point is a local minimum; if negative, it is a local maximum; if zero, it may or may not be a local minimum or local maximum. Taking derivatives and solving for critical points is often a simple way to find local minima or maxima, which can be useful in optimization. In fact, local minima and maxima can only occur at critical points. This is related to the extreme value theorem.
## Physics
Arguably the most important application of calculus to physics is the concept of the "time derivative"—the rate of change over time—which is required for the precise definition of several important concepts. In particular, the time derivatives of an object's position are significant in Newtonian physics:
• Velocity (instantaneous velocity; the concept of average velocity predates calculus) is the derivative (with respect to time) of an object's position.
• Acceleration is the derivative (with respect to time) of an object's velocity.
• Jerk is the derivative (with respect to time) of an object's acceleration.
For example, if an object's position ${\displaystyle p(t)=-16t^{2}+16t+32}$; then, the object's velocity is ${\displaystyle {\dot {p}}(t)=p'(t)=-32t+16}$; the object's acceleration is ${\displaystyle {\ddot {p}}(t)=p''(t)=-32}$; and the object's jerk is ${\displaystyle p'''(t)=0.}$
If the velocity of a car is given, as a function of time, then, the derivative of said function with respect to time describes the acceleration of said car, as a function of time.
## Algebraic manipulation
Messy limit calculations can be avoided, in certain cases, because of differentiation rules which allow one to find derivatives via algebraic manipulation; rather than by direct application of Newton's difference quotient. One should not infer that the definition of derivatives, in terms of limits, is unnecessary. Rather, that definition is the means of proving the following "powerful differentiation rules"; these rules are derived from the difference quotient.
• Constant rule: The derivative of any constant is zero.
• Constant multiple rule: If c is some real number; then, the derivative of ${\displaystyle cf(x)}$ equals c multiplied by the derivative of f(x) (a consequence of linearity below)
• Linearity: (af + bg)' = af ' + bg' for all functions f and g and all real numbers a and b.
• General power rule (Polynomial rule): If ${\displaystyle f(x)=x^{r}}$, for some real number r; ${\displaystyle f'(x)=rx^{r-1}.}$
• Product rule: ${\displaystyle (fg)'=f'g+fg'}$ for all functions f and g.
• Quotient rule: ${\displaystyle (f/g)'=(f'g-fg')/(g^{2})}$ unless g is zero.
• Chain rule: If ${\displaystyle f(x)=h(g(x))}$, then ${\displaystyle f'(x)=h'[g(x)]*g'(x)}$.
• Inverse functions and differentiation: If ${\displaystyle y=f(x)}$, ${\displaystyle x=f^{-1}(y)}$, and f(x) and its inverse are differentiable, with ${\displaystyle dy/dx}$ non-zero, then ${\displaystyle dx/dy=1/(dy/dx).}$
• Derivative of one variable with respect to another when both are functions of a third variable: Let ${\displaystyle x=f(t)}$ and ${\displaystyle y=g(t)}$. Now ${\displaystyle dy/dx=(dy/dt)/(dx/dt).}$
• Implicit differentiation: If ${\displaystyle f(x,y)=0}$ is an implicit function, we have: dy/dx = - (∂f / ∂x) / (∂f / ∂y).
In addition, the derivatives of some common functions are useful to know. See the table of derivatives.
As an example, the derivative of
${\displaystyle f(x)=2x^{4}+\sin(x^{2})-\ln(x)\;e^{x}+7}$
is
${\displaystyle f'(x)=8x^{3}+2x\cos(x^{2})-{\frac {1}{x}}\;e^{x}-\ln(x)\;e^{x}.}$
## Using derivatives to graph functions
Derivatives are a useful tool for examining the graphs of functions. In particular, the points in the interior of the domain of a real-valued function which take that function to local extrema will all have a first derivative of zero. However, not all critical points are local extrema; for example, f(x)=x3 has a critical point at x=0, but it has neither a maximum nor a minimum there. The first derivative test and the second derivative test provide ways to determine if the critical points are maxima, minima or neither.
In the case of multidimensional domains, the function will have a partial derivative of zero with respect to each dimension at local extrema. In this case, the Second Derivative Test can still be used to characterize critical points, by considering the eigenvalues of the Hessian matrix of second partial derivatives of the function at the critical point. If all of the eigenvalues are positive, then the point is a local minimum; if all are negative, it is a local maximum. If there are some positive and some negative eigenvalues, then the critical point is a saddle point, and if none of these cases hold then the test is inconclusive (e.g., eigenvalues of 0 and 3).
Once the local extrema have been found, it is usually rather easy to get a rough idea of the general graph of the function, since (in the single-dimensional domain case) it will be uniformly increasing or decreasing except at critical points, and hence (assuming it is continuous) will have values in between its values at the critical points on either side.
## Generalizations
Where a function depends on more than one variable, the concept of a partial derivative is used. Partial derivatives can be thought of informally as taking the derivative of the function with all but one variable held temporarily constant near a point. Partial derivatives are represented as ∂/∂x (where ∂ is a rounded 'd' known as the 'partial derivative symbol'). Some people pronounce the partial derivative symbol as 'der' rather than the 'dee' used for the standard derivative symbol, 'd'.
The concept of derivative can be extended to more general settings. The common thread is that the derivative at a point serves as a linear approximation of the function at that point. Perhaps the most natural situation is that of functions between differentiable manifolds; the derivative at a certain point then becomes a linear transformation between the corresponding tangent spaces and the derivative function becomes a map between the tangent bundles.
In order to differentiate all continuous functions and much more, one defines the concept of distribution.
For complex functions of a complex variable differentiability is a much stronger condition than that the real and imaginary part of the function are differentiable with respect to the real and imaginary part of the argument. For example, the function f(x + iy) = x + 2iy satisfies the latter, but not the first. See also Holomorphic function.
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# Tae Sung Ye Rin. Order of Operations For one month Tae Sung gets \$30 for his job, and he uses \$2 every weekend and gains \$5 a week in his another job.
## Presentation on theme: "Tae Sung Ye Rin. Order of Operations For one month Tae Sung gets \$30 for his job, and he uses \$2 every weekend and gains \$5 a week in his another job."— Presentation transcript:
Tae Sung Ye Rin
Order of Operations For one month Tae Sung gets \$30 for his job, and he uses \$2 every weekend and gains \$5 a week in his another job. How much will he have after 3 months? (count one month for 4 weeks) (30 –2 x2 +5 x7)3=(30 – 4 +35)3 =(30 +31)3 =(61)3 =183
Solving proportions we ordered 2 pizzas. sung min said he will divide it into 11 peaces and eat 7 slices. Yeo chang said he will same as sung min but eating 42 slices. How many does he need to cut to eat equally to sung min?? We can have a proportion 42/x = 7/11. First, you multiple the cross ones. 7x = 462 Then divide 462 by 7 which is x = 66 Yeo chang needs to cut the pizza with 66 slices.
Combine like terms to solve an equation One day, yeo chang went out and bought 7 CDs. And he gave 100 us dollars. Then he got back 30 dollars. What is the price of each CD? We can have a equation 100- 7x = 30 When we move 100 to the right, it is -7x= -70 When we divide -70 by -7, we have x = 10 The price of a CD is \$10.
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# Appendix A
## Introduction
This appendix to the elementary arithmetic course provides proofs, additional explanations and derivations of some of the statements, theorems and rules used in the course.
## Why a negative times a negative is a positive
$−1 × 0 = 0 ⇔ −1 × (1 + −1) = 0 ⇔ (−1 × 1)+(−1 × −1)=0 ⇔ −1 + (−1 × −1) = 0 ⇔ −1 = −(−1 × −1) ⇔ 1 = −1 × −1 ☐.$
In the second line, 0 in the left side of the equation is replaced by (1+ −1). The distributive rule applied to the second line result in the third line. Subtracting (−1 × −1) from both sides of the equation in the fourth line result in the fifth line. Finally, negating both sides result in what was to be demonstrated: −1 × −1 = 1.
## Proof that the set of prime numbers is infinite
We use a proof by contradiction. The set of prime numbers is either finite or infinite. Let's assume it is finite and try to show that this assumption leads to a contradiction, which proves that it must be infinite.
So, let the finite number of primes all be collected in set $P = {p1, p2, p3, ... , pn}$ and let number $N = p1 × p2 × p3 × ... × pn + 1$.
Now, N is either prime, in which case we have a new prime number other than the original ones, which contradicts our assumption, or N is composite, in which case $N = q1 × q2 × q3 × ... × qm$, where $q1, q2, q3, ... , qm$ are N's prime factors. But no q can be a prime number from the original set P, because dividing N by any original prime or product of original primes leaves a remainder of 1 (N = p1 × p2 × p3 × ... × pn + 1). Thus $q1, q2, q3, ... , qm$ are new primes, other than the original ones, which also contradicts our start assumption. So, the set of prime numbers cannot be a finite; it must be infinite ☐.
## Why the divisibility rules work
### Divisibility by 2 or 5
$4578 = (4 × 1000) + (5 × 100) + (7 × 10) + 8$
Powers of 10 i.e. 10, 100, 1000, etc. are always divisible by 2. So, if and only if the last digit, the ones digit, is divisible by 2, the whole number is divisible by 2. Divisibility by 5 works the same.
### Divisibility by 3
$4578 = (4 × (999+1)) + (5 × (99+1)) + (7 × (9+1)) + 8 = (4 × 999) + 4 + (5 × 99) + 5 + (7 × 9) + 7 + 8 = (4 × 999)+ (5 × 99) + (7 × 9) + 4 + 5 + 7 + 8.$
9, 99, 999, etc. are always divisible by 9 and thus also by 3. So, if and only if the addition of all digits is divisible by 3, the whole number is divisible by 3.
### Divisibility by 7
$4578 = (457 × 10) + 8$
If 4578 is divisible by 7, then 2 × 4578 is as well, and vice versa:
$2 × 4578 = 457 × 20 + (2 × 8) = 457 × (21 - 1) + (2 × 8) = (457 × 21) - 457 + (2 × 8).$
(457 × 21) is divisible by 7 (because 21 is divisible by 7). If $−457 + (2 × 8)$ is divisible by 7, then $457 - (2 × 8)$ is divisible by 7 and 4578 is divisible by 7 (and vice versa).
So, a number is divisible by 7 if and only if the subtraction of 2 times the last digit from the number formed by the rest of the digits is divisible by 7.
### Divisibility by 11
$4578 = 457 × (11 - 1) + 8 = (457 × 11) - 457 + 8.$
If 4578 is divisible by 11, then −457 + 8 is divisible by 11, then 457 − 8 is divisible by 11 (and vice versa).
So, a number is divisible by 11 if and only if the subtraction of the last digit from the number formed by the rest of the digits is divisible by 11.
Or:
$4578 = (4 × (999+1)) + (5 × (99+1)) + (7 × (9+1)) + 8 = (4 × 999) + 4 + (5 × 99) + 5 + (7 × 9) + 7 + 8 = (4 × 999)+ (5 × 99) + (7 × 9) + 4 + 5 + 7 + 8.$
If and only if the alternating subtraction and addition of all digits is divisible by 11, the whole number is divisible by 11.
### Divisibility by 13
$9 × 4578 = 457 × 90 + (9 × 8) = 457 × (91 - 1) + (9 × 8) = (457 × 91) - 457 + (9 × 8).$
If and only if the subtraction of 9 times the last digit from the number formed by the rest of the digits is divisible by 13, the whole number is divisible by 13.
## Derivation of some of the arithmetic rules for fractions
$(0/ a) = 0 × (1/a) = 0$
$(a/ b) = 1 × (a/ b) = (c/c) × (a/ b) = (a×c)/(b×c)$
$(a/ b) ± (c/ b) = a×(1/ b) ± c×(1/ b) = (1/ b)×(a±c) = (a±c)/b$
$(a/ b) ± (c/d) = (a×d)/(b×d) ± (c×b)/(b×d) = (a×d ± c×b)/(b×d)$
$(a/ b) / (c/d) = ((a/ b)×(d/c)) / ((c/d)×(d/c)) = ((a/ b)×(d/c)) / 1 = (a/ b) × (d/c)$
$−(a/ b) = −1 × (a/ b) = (-1 × a)/ b = (-a/ b) = (-1/-1) × (-a/ b) = (-1 × -a)/(-a × b) = a/−b$
## Why does the Euclidean algorithm work?
Let:
• $gcd(a=px,b=py) = p, a > b, b > 0$, a > b, b > 0,
• t be any positive integer,
• qk be a quotient and rk be a remainder:
$a-tb = px-tpy = p(x-ty)$
Thus,
$gcd(a,b) = gcd(a,b,a-tb)$
$a/b = q_0 + (r_0/b) = (bq_0)/b + (r_0/b).$
$a/b = (bq_0 + r_0)/b ⇔ a = bq_0 + r_0 ⇔ a - bq_0 = r_0$
Thus,
$gcd(a,b) = gcd(b,a-bq_0) = gcd(b,r_0), (b>r_0)$
$\begin{array}{lr}\mathrm{gcd}\left(a,b\right)=\mathrm{gcd}\left(b,{r}_{0}\right)& b>{r}_{0}\\ \mathrm{gcd}\left(b,{r}_{0}\right)=\mathrm{gcd}\left({r}_{0},{r}_{1}\right)& {r}_{0}>{r}_{1}\\ \mathrm{gcd}\left({r}_{0},{r}_{1}\right)=\mathrm{gcd}\left({r}_{1},{r}_{2}\right)& {r}_{1}>{r}_{2}\\ \mathrm{\dots }& \\ \mathrm{gcd}\left({r}_{k},{r}_{k-1}\right)=\mathrm{gcd}\left({r}_{n},{r}_{n-1}\right)& {r}_{n-1}>{r}_{n}\end{array}$
The remainder rk (always an integer) is always smaller than the remainder in the step before. So, the remainder must finally be: $r_n = 0$. Thus we can write:
$gcd(a,b) = gcd(r_k,r_(k-1)) = gcd(0,r_(n-1)) = r_(n-1)$
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# Math
• ### Eureka Math – Kindergarten – Lesson 14
Solving number sentences – Are you ready to solve math stories with number sentences? Join Mr. Hammer for this lesson using drawings and number sentences to solve math stories with totals up to 8! For this lesson, you’ll need a pencil and paper or whiteboard and dry erase marker along with some markers or crayons for the problem set.
• ### Eureka Math – Kindergarten – Lesson 13
Number sentences that make 8 – How many addition number sentences can you write with a total of 8? Join Mr. Hammer for this lesson as we use space rocks to write number sentences with a total of 8! For this lesson, you’ll need eight objects that students can move around and count, along with a pencil and paper or whiteboard and dry erase marker.
• ### Eureka Math – 1st – Lesson 12
Make the next 10 to add 2 digits – How can you make the next ten? Join Mrs. Coleman for this lesson that encourages students to decompose one addend to make the next ten, then add the remaining amount to find the total. Students will need up to 40 small counting objects, as well as a whiteboard and marker or pencil and paper.
• ### Eureka Math – 1st – Lesson 11
Making 10’s – How many do you need to make a new ten? Join Mrs. Coleman for this lesson that encourages students to apply the strategies of counting on and making ten to larger numbers.
• ### Eureka Math – 1st – Lesson 10
Add 10’s to a 2-digit number – How can you add tens to a 2-digit number? Join Ms. Lassiter for this lesson as students learn to recognize that when tens are added to a number, the number of ones remain the same.
• ### Eureka Math – 3rd – Lesson 12
Fractions on a number line – How can fractions be placed on a number line? Join Ms. Roose as she partitions and labels number lines to represent any fraction between 0 and 1. Students will need a whiteboard and marker or a pencil and paper.
• ### Eureka Math – 3rd – Lesson 11
Comparing fractions – How can number lines represent fractions? In this lesson, Ms. Roose uses fraction strips to partition and label a number line. Students will need their fraction strips as well as a whiteboard and marker or a pencil and paper.
• ### Eureka Math – 2nd – Lesson 10
More rectangle building – Let’s draw! In this lesson with Mr. Waldorf, students build on prior skills with concrete square tiles to draw rectangles composed of squares.
• ### Eureka Math – 2nd – Lesson 11
Remove rows from an array – What happens when you remove rows or columns from an array? Join Mr. Waldorf for this lesson that encourages students to decompose rectangles and analyze the remaining rows or columns.
• ### Eureka Math – 2nd – Lesson 12
Splitting up rectangles – Let’s make rectangles! Join Mr. Waldorf for this lesson as students explore different ways to make rectangles using the same number of squares each time. Students will need a pair of scissors.
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## AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers InText Questions
AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers InText Questions and Answers.
### 8th Class Maths 1st Lesson Rational Numbers InText Questions and Answers
Do this
Question 1.
Consider the following collection of numbers 1, $$\frac{1}{2}$$, -2, 0.5, 4$$\frac{1}{2}$$, $$\frac{-33}{7}$$, 0, $$\frac{4}{7}$$, $$0 . \overline{3}$$, 22, -5, $$\frac{2}{19}$$, 0.125. Write these numbers under the appropriate category. [A number can be written in more than one group] (Page No. 2)
i) Natural numbers 1, 22
ii) Whole numbers 0, 1, 22
iii) Integers 0, 1, 22, -5, -2
iv) Rational numbers 1, $$\frac{1}{2}$$, -2, 0.5, 4$$\frac{1}{2}$$, $$\frac{-33}{7}$$, 0, $$\frac{4}{7}$$, $$0 . \overline{3}$$, 22, -5, $$\frac{2}{19}$$, 0.125 etc.
Would you leave out any of the given numbers from rational numbers? No
Is every natural number, whole number and integer is a rational number? Yes
Question 2.
Fill the blanks in the table. (Page No. 6)
Question 3.
Complete the following table. (Page No. 9)
Question 4.
Complete the following table. (Page No. 13)
Question 5.
Complete the following table. (Page No. 16)
Question 6.
Complete the following table. (Page No. 17)
Question 7.
Represent – $$\frac{13}{5}$$ on the number line. (Page No. 22)
Representing – $$\frac{13}{5}$$ on the number line.
Try These
Question 1.
Hamid says $$\frac{5}{3}$$ is a rational number and 5 is only a natural number. Shikha says both are rational numbers. With whom do you agree? (Page No. 3)
I would not agree with Hamid’s argument. Since $$\frac{5}{3}$$ is a rational number. But ‘5’ is not only
a natural number, it is also a rational number.
Since every natural number is a rational number,
According to Shikha’s opinion $$\frac{5}{3}$$, 5 are rational numbers.
∴ I agree with Shikha’s opinion.
Question 2.
Give an example to satisfy the following statements. (Page No.3)
i) All natural numbers are whole numbers but all whole numbers need not be natural numbers.
ii) All whole numbers are integers but all integers are not whole numbers.
iii) All integers are rational numbers but all rational numbers need not be integers.
i) ‘0’ is not a natural number.
∴ Every whole number is not a natural number. (∵ N ⊂ W)
ii) -2, -3, -4 are not whole numbers.
∴ All integers are not whole numbers. (∵ W ⊂ Z)
iii) $$\frac{2}{3}$$, $$\frac{7}{4}$$ are not integers.
∴ Every rational number is not an integer. (∵ Z ⊂ Q)
Question 3.
If we exclude zero from the set of integers is it closed under division? Check the same for natural numbers. (Page No. 6)
If ‘0’ is subtracted from the set of integers then it becomes Z – {0}.
Closure property under division on integers.
Ex: -4 ÷ 2 = -2 is an integer.
3 ÷ 5 = $$\frac{3}{5}$$ is not an integer.
∴ Set of integers doesn’t satisfy closure property under division.
Closure property under division on natural numbers.
Ex: 2 ÷ 4 = $$\frac{1}{2}$$ is not a natural number.
∴ Set of natural numbers doesn’t satisfy closure property under division.
Question 4.
Find using distributivity. (Page No. 16)
A) $$\left\{\frac{7}{5} \times\left(\frac{-3}{10}\right)\right\}+\left\{\frac{7}{5} \times\left(\frac{9}{10}\right)\right\}$$
B) $$\left\{\frac{9}{16} \times 3\right\}+\left\{\frac{9}{16} \times-19\right\}$$
Distributive law: a × (b + c) = ab + ac
A)
B)
Question 5.
Write the rational number for the points labelled with letters, on the number line. (Page No. 22)
i)
ii)
i) A = $$\frac{1}{5}$$, B = $$\frac{4}{5}$$, C = $$\frac{5}{5}$$ = 1, D = $$\frac{7}{5}$$, E = $$\frac{8}{5}$$, F = $$\frac{10}{5}$$ = 2.
ii) S = $$\frac{-6}{4}$$, R = $$\frac{-6}{4}$$, Q = $$\frac{-3}{4}$$, P = $$\frac{-1}{4}$$
Think, discuss and write
Question 1.
If a property holds good with respect to addition for rational numbers, whether it holds good for integers? And for whole numbers? Which one holds good and which doesn’t hold good? (Page No. 15)
Under addition the properties which are followed by set of rational numbers are also followed by integers.
Question 2.
Write the numbers whose multiplicative inverses are the numbers themselves. (Page No. 15)
The number T is multiplicative inverse of itself.
∵ 1 × $$\frac{1}{1}$$ = 1 ⇒ 1 × 1 = 1
∴ The multiplicative inverse of 1 is 1.
Question 3.
Can you find the reciprocal of ‘0’ (zero)? Is there any rational number such that when it is multiplied by ‘0’ gives ‘1’?
(Page No. 15)
The reciprocal of ‘0’ is $$\frac{1}{0}$$.
But the value of $$\frac{1}{0}$$ is not defined.
∴ There is no number is found when it is multiplied ‘0’ gives 1.
∵ 0 × (Any number) = 0
∴ No, there is no number is found in place of ‘A’.
Question 4.
Express the following in decimal form. (Page No. 28)
i) $$\frac{7}{5}$$, $$\frac{3}{4}$$, $$\frac{23}{10}$$, $$\frac{5}{3}$$,$$\frac{17}{6}$$,$$\frac{22}{7}$$
ii) Which of the above are terminating and which are non-terminating decimals?
iii) Write the denominators of above rational numbers as the product of primes.
iv) If the denominators of the above simplest rational numbers has no prime divisors other than 2 and 5 what do you observe?
i) $$\frac{7}{5}$$ = 0.4,
$$\frac{3}{4}$$ = 0.75,
$$\frac{23}{10}$$ = 2.3,
$$\frac{5}{3}$$ = 1.66… = $$1 . \overline{6}$$,
$$\frac{17}{6}$$ = 2.833… = $$2.8 \overline{3}$$,
$$\frac{22}{7}$$ = 3.142
ii) From the above decimals $$\frac{7}{5}$$, $$\frac{3}{4}$$, $$\frac{23}{10}$$ are terminating decimals.
While $$\frac{5}{3}$$,$$\frac{17}{6}$$,$$\frac{22}{7}$$ are non-terminating decimals
iii) By writing the denominators of above decimals as a product of primes is
iv) If the denominators of integers doesn’t have factors other than 2 or 5 and both are called terminating decimals.
Question 5.
Convert the decimals $$0 . \overline{9}$$, $$14 . \overline{5}$$ and $$1.2 \overline{4}$$ to rational form. Can you find any easy method other than formal method? (Page No. 31)
Let x = $$0 . \overline{9}$$
⇒ x = 0.999 ……. (1)
The periodicity of the above equation is ‘1’. So it is to be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 0.999
10x = 9.999 …….. (2)
From (1) & (2)
∴ x = 1 or $$0 . \overline{9}$$ = 1
Second Method:
$$0 . \overline{9}$$ = 0 + $$0 . \overline{9}$$
= 0 + $$\frac{9}{9}$$
= 0 + 1
= 1
Let x = $$14 . \overline{5}$$
⇒ x = 14.55 …….. (1)
The periodicity of the equation (1) is 1.
So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 14.55
10x = 145.55 …….. (2)
Second Method:
Let x = $$1.2 \overline{4}$$
⇒ x= 1.244 …….. (1)
Here periodicity of equation (1) is 1. So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 1.244
10 x = 12.44 …….. (2)
Second Method:
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# How do you know if a graph is proportional?
Oralia Micallef asked, updated on December 1st, 2021; Topic: graph
π 408 π 9 β
β
β
β
β4.9
##To determine whether x and y have a proportional relationship, see if the line through these points passes through the origin (0, 0). The points are on a line that passes through the origin. So, x and y have a proportional relationship.
Additional, what two things make a graph proportional?
A graph of a proportional relationship is a straight line that passes through the origin. The constant of a proportionality in a graph of a proportional relationship is the constant ratio of y to x (the slope of the line).
Whatever the case, what does proportional mean in graphs? About Transcript. Proportional relationships are relationships between two variables where their ratios are equivalent. Another way to think about them is that, in a proportional relationship, one variable is always a constant value times the other. That constant is know as the "constant of proportionality".
So, what makes a graph non proportional?
A non-proportional graph is a straight line that does not go through the origin. How to tell the difference: A proportional table has a constant of proportionality in that y divided by x always equals the same value. A non-proportional table will have different values when y is divided by x.
What does a proportional graph look like?
The graph of the proportional relationship equation is a straight line through the origin. Example 1: Given that y varies proportionally with x , with a constant of proportionality k=13 , find y when x=12 .
### Does proportional mean equal?
A proportional relationship is states that they are the same. For example, 1/2 and 6/12 have a proportional relationship, which means they are the same.
### What is a proportion in math?
Proportion says that two ratios (or fractions) are equal. Example: 1/3 = 2/6. See: Equivalent Fractions. Proportions.
### Can a proportional graph be negative?
Negative proportional relationships are special types of linear relationships. Because their slopes are negative, the x-value increases and goes to the right as the y-value decreases and goes down.
### What is an inversely proportional graph?
Hyperbola graphs, like the one immediately below, show that the quantities on the graph are in inverse proportion. This graph states, therefore, that A is inversely proportional to B. ... It means: By whatever factor A changes, B changes by the inverse of that factor. (Or you could say, βby the reciprocal of that factorβ.)
### What is an example of directly proportional?
Examples. If an object travels at a constant speed, then the distance traveled is directly proportional to the time spent traveling, with the speed being the constant of proportionality. The circumference of a circle is directly proportional to its diameter, with the constant of proportionality equal to Ο.
### How do you know if something is proportional?
Ratios are proportional if they represent the same relationship. One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional.
### What does a non proportional equation look like?
The graph of a linear equation is a line. If b = 0 in a linear equation (so y = mx), then the equation is a proportional linear relationship between y and x. If b β 0, then y = mx + b is a non-proportional linear relationship between y and x.
### Is Y proportional or Nonproportional?
If the ratios are equivalent then the relationship is proportional. With the graph, the origin is 0,0 (the starting point). A straight line through the origin = proportional. equation is of the form y=kx (y=2x)....x(lbs)y(cost)
1β
2β
3β
4β
### What does proportion mean?
1 : harmonious relation of parts to each other or to the whole : balance, symmetry. 2a : proper or equal share each did her proportion of the work. b : quota, percentage. 3 : the relation of one part to another or to the whole with respect to magnitude, quantity, or degree : ratio.
### How do you write an equation for a proportional graph?
The equation that represents a proportional relationship, or a line, is y = k x , where is the constant of proportionality. Use k = y x from either a table or a graph to find k and create the equation. Proportional relationships can be represented by tables, graphs and equations.
### What does a inverse graph look like?
So if you're asked to graph a function and its inverse, all you have to do is graph the function and then switch all x and y values in each point to graph the inverse. Just look at all those values switching places from the f(x) function to its inverse g(x) (and back again), reflected over the line y = x.
### What is the difference between proportional and equal?
As adjectives the difference between proportional and equal is that proportional is at a constant ratio (to) two magnitudes (numbers) are said to be proportional if the second varies in a direct relation arithmetically to the first while equal is (label) the same in all respects.
### Is proportional to symbol?
The symbol used to denote the proportionality is'β'. For example, if we say, a is proportional to b, then it is represented as 'aβb' and if we say, a is inversely proportional to b, then it is denoted as 'aβ1/b'.
### What is difference between direct and inverse proportion?
In a direct proportion, the ratio between matching quantities stays the same if they are divided. (They form equivalent fractions). In an indirect (or inverse) proportion, as one quantity increases, the other decreases. ... In an inverse proportion, the product of the matching quantities stays the same.
### What is proportion formula?
A proportion is simply a statement that two ratios are equal. It can be written in two ways: as two equal fractions a/b = c/d; or using a colon, a:b = c:d. ... To find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.
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# Word Problem Solver
Word problems in math test students on their ability to decipher and use mathematical information when it is presented verbally. Word problems often appear complicated and lengthy, and contain a lot of information which makes it confusing. However, to solve a problem we only need the relevant points, and the first step to solving word problems successfully is to identify and separate the required information.
## Math Word Problems
Like any other type of math problem, word problems get easier with practice. An easy way to understand the relation between the given parameters is to learn the vocabulary that is used to denote the different operations. There are stock phrases and words which repeat themselves in math word problems, so once students get familiar with these, their work is half done. Making formulas to solve the problem also becomes simpler.
Solved Example
Question: The sum of the square of 3 consecutive numbers is 110. Find the numbers.
Solution:
Let the smallest integer be m, then 3 consecutive integers are m, m + 1, m + 2.
From the statement :
$m^2 + (m + 1)^2 + (m + 2)^2 = 110$
$m^2 + m^2 + 1 + 2m + m^2 + 4 + 4m = 110$
$3m^2 + 6m + 5 = 110$
$3m^2 + 6m - 105 = 0$
Solve for m:
$3m^2 + 6m - 105 = 0$
$3m^2 + 21m - 15m - 105 = 0$ (By factorization)
$3m(m + 7) - 15(m + 7) = 0$
$(3m - 15)(m + 7) = 0$
$3m - 15 = 0$ or $m + 7 = 0$
=> $m = 5$ or $m = -7$
Therefore the three consecutive integers are 5, 6, 7 or -7, -6, -5.
## Algebra Word Problems
There are plenty of algebraic solved examples online which can help students when they get stuck or are unsure about the solution. The entire solution is included so that students can learn and practice form the beginning to the end.
## Solved Examples
Question 1: Sujen leaves 15420 dollars behind. According to her wish, the money is to be divided between her daughter and son in the ratio 2 : 3. Find the sum received by her daughter.
Solution:
We know that, if a quantity x is divided in the ratio m : n then two parts are $\frac{mx}{m+n}$ and $\frac{nx}{m+n}$.
The sum received by the daughter = $\frac{2 * 15420}{2 + 3}$
= 6168
$\therefore$ The sum received by her daughter is $\$$6168. Question 2: The age of a father is twice the square of the age of his daughter. Four years hence, the age of the father will be one year more than the three times the age of the daughter. Find the present age of his daughter. Solution: Let the present age of the daughter be$x$years. From the statement: The present age of the father is 2$x^2$years. 4 years hence: The age of the daughter will be$(x + 4)$and that of the father, (2$x^2$+$4$)years. From the question:$2x^2 + 4 = 1 + 3(x + 4)2x^2 + 4 = 1 + 3x + 122x^2 + 4 = 3x + 13$2$x^2 - 3x - 9 = 0$2$x^2 - 6x + 3x - 9 = 02x(x - 3) + 3(x - 3) = 0(2x + 3)(x - 3) = 0$=>$x$=$\frac{-3}{2}$or$x$= 3 But age cannot be negative. So$x\neq\frac{-3}{2}$. Therefore,$x$= 3.$\therefore\$ The present age of the daughter is 3 years.
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## 4(3-2x)=-44
HHHHHHHHHEEEEEEEEELLLLLLLLLLPPPPPPPPPPP????????????
4(3 - 2x) = -44
I'm assuming you are looking to solve for the only unknown variable, that being 'x'. To do so, you need to isolate this variable to one side of the equation.
First, divide both sides of the equation by 4:
(4(3 - 2x))/4 = (-44)/4
(4/4)(3 - 2x) = -(44/4)
(3 - 2x) = -(11)
3 - 2x = -11
Next, subtract 3 from both sides of the equation and combine like terms:
3 - 2x - 3 = -11 - 3
3 - 3 - 2x = -11 - 3
-2x = -14
Then, to solve for x, divide both sides of the equation by -2:
(-2x)/-2 = (-14)/-2
(-2/-2)x = (-/-)(14/2)
x = 7
You can check to see if this is the correct answer by plugging in x = 7 into the original equation:
4(3 - 2x) = 4(3 - 2(7)) = 4(3 - 14) = 4(-11) = -44
First, let's distribute the left hand side:
4*3 - 4*2x = -44
12 - 8x = -44
Next, isolate our variable. We're going to do this by subtracting 12 from both sides of the equation.
-8x = -44 - 12 = -56.
Lastly, divide by -8 on both sides to get x completely alone.
x = -56/-8 = 7.
You can check your answer by plugging 7 in for x and seeing what comes out!
4(3-2*7) = 4(3-14) = 4(-11) = -44.
4(3 - 2x) = -44
use distributive law of multiplication by multiplying 4 and 3 and 4 and -2x:
12 -8x = -44
-44 -12 = -8x
-56 = -8x
x = 7
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Properties of real numbersPage 1
WATCH ALL SLIDES
Slide 1
THE NUMBER LINE
AND
NUMBER SYSTEMS
Standards 6, 25
PROPERTIES OF REAL NUMBERS
PROPERTIES OF REAL NUMBERS
EXAMPLES OF PROPERTIES
OF REAL NUMBERS
SIMPLIFYING EXPRESSIONS
APPLYING PROPERTIES
END SHOW
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Slide 2
Standard 6:
Students add, subtract, multiply, and divide complex numbers.
Estándar 6:
Los estudiantes suman, restan, multiplican, y dividen números complejos.
Standard 25:
Students use properties from number systems to justify steps in combining and simplifying functions.
Estándar 25:
Los estudiantes usan propiedades de los sistemas numéricos para justificar pasos al combinar y simplificar funciones.
ALGEBRA II STANDARDS THIS LESSON AIMS:
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Slide 3
The number line
Standards 6, 25
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Slide 4
Standards 6, 25
RATIONAL NUMBERS:
where:
m and n are integers.
0.25
0.75
7
10
IRRATIONAL NUMBERS
-8
0
None can be expressed as a fraction!
The following numbers can be expressed as fractions and therefore they are Rational numbers:
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Slide 5
Standards 6, 25
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Slide 6
PROPERTIES OF REAL NUMBERS
COMMUTATIVE PROPERTY:
Addition:
a + b = b + a
5 + 7
= 7 + 5
1 + 6
= 6 + 1
3.6 + 1.1
= 1.1 + 3.6
Multiplication:
For any real numbers a, b, and c:
Standards 6, 25
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Slide 7
PROPERTIES OF REAL NUMBERS
ASSOCIATIVE PROPERTY:
Addition:
(a + b) + c = a + (b + c)
(3 + 4) +1
= 3 + (4 + 1)
(2 + 5) + 7
= 2 + (5 + 7)
(6.2 + 4.1) +3.3
= 6.2 + (4.1 + 3.3)
Multiplication:
For any real numbers a, b, and c:
Standards 6, 25
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Slide 8
PROPERTIES OF REAL NUMBERS
IDENTITY PROPERTY:
Addition:
a + 0 = 0 + a=a
5 + 0
= 0 + 5
1 + 0
= 0 + 1
Go to page:
1 2
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## SUBTRACTING FRACTIONS
• Same denominators
• Different denominators
• Mixed numbers
Type 1: Subtracting fractions with same denominators
For subtracting fractions having same denominators, subtract numerators and just write the common denominator in the final difference.
Example:
Subtract 2/5 from 3/5 Or Find 3/5 – 2/5
Solution:
Since denominator 5, is same in both 2/5 and 3/5, therefore:
In the final answer, i.e the difference of the two fractions, write 5 in denominator and 3 – 2 = 1 in numerator.
So, 3/5 – 2/5 = (3 – 2)/5
Note:
To subtract 3/5 from 2/5, i.e. to find 2/5 – 3/5
So, 2/5 – 3/5 = (2 – 3)/5 = -1/5
Type 2: Subtracting fractions having different denominators
For subtracting fractions having different denominators, follow the below steps
Step 1: Find LCM of the various denominators in the fractions.
Step 2: Write equivalent fractions of the original fractions with LCM in denominator. (to make denominators same)
Step 3: Add the equivalent fractions.
Example:
Subtract 3/4 from 5/6
Solution:
Step 1:
First, what is the LCM of the denominators 4 and 6?
4 = 2 × 2, and 6 = 2 × 3
So, LCM of 4 and 6 = 2 × 2 × 3 = 12
Step 2:
Now, write equivalent fractions for 3/4 and 5/6 having LCM 12 as the common denominator.
Equivalent fraction for 3/4
Now, 4 × ? = 12, it is
12 ÷ 4 = 3
So, multiply 3 to 1 and 4 in ¼ to get 12 in denominator.
3/4 = (3 × 3)/ (3 × 4) = 9/12
Equivalent fraction for 5/6
6 × ? = 12, it’s
12 ÷ 6 = 2
So, multiply 2 to both 5 and 6 in 5/6 to get 12 in denominator
5/6 = (5 × 2)/ (6 × 2) = 10/12
Step 3:
Finally,
To find the difference: 10/12 – 9/12
Subtract the equivalent fraction 9/12 from 10/12
10/12 – 9/12 = 1/12
Therefore,
(5/6) – (3/4) = 1/12
Short-cut for subtracting fractions having different denominators
To find (a/b) – (c/d)
First, find (a × d – b × c)/(b × d)
Next, reduce the fraction so obtained into simplest term.
Example:
Find (5/6) – (3/4)
• First, (5/6) – (3/4)
= (5 × 4 – 3 × 6)/6 × 4
= (20 – 18)/24
= 2/24
2. Next, reduce 2/24, by cancelling the common factor 2, as below:
2/24 = (2 × 1)/ (2 × 12) = 1/12
Type 3: Subtracting fractions which are mixed numbers
Subtract mixed numbers in two different methods. They are:
1st method:
• First, subtract the whole numbers and the fractions separately.
• Next, add the differences of the whole numbers and the fractions obtained in the first step
2nd method:
• First, convert the mixed fractions into improper fractions
• Next, subtract the two improper fractions.
Example:
Find the difference: 32/3 – 23/4
Below, let us apply both of the above methods for subtracting fractions.
Method 1:
Step 1:
Subtract the whole numbers: 3 – 2 = 1, and
Subtract the fractions: 2/3 – 3/4
Now, lcm of 3 and 4 is 3 × 4 = 12
{If two numbers do not have any common factor, then their LCM is their product}
2/3 = (2 × 4)/ (3 × 4) = 8/12, and
3/4 = (3 × 3)/ (4 × 3) = 9/12
So, 2/3 – 3/4 = (8/12) – (9/12) = -1/12
Step 2:
Next, add the fractions’ difference and the whole numbers’ difference
1 + (-1/12) = 1 – 1/12 = (12/12) – (1/12) = 11/12
Method 2 is
Subtracting fractions by converting them into improper fractions.
The mixed fraction 32/3 = (3 × 3 + 2)/3 = 11/3, and
The mixed fraction 23/4 = (4 × 2 + 3)/4 = 11/4.
Step 2:
Next, subtract the above two improper fractions, for subtracting fractions
32/3 and 23/4
i.e. 32/3 – 23/4 = (11/3) – (11/4)
= (11× 4)/(3 × 4) – (11 × 3)/(4 × 3)
{converting into equivalent fractions}
= (44/12) – (33/12) = 11/12
32/3 – 23/4
Or, using the above short-cut for subtracting fractions, we have
(11/3) – (11/4)
= (11× 4 – 11× 3)/ ( 3 × 4) =
= (44 – 33)/12 = 11/12
RELATED MATH LESSONS:
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# Calculating the Volume of a Pyramid
When calculating the volume of a pyramid we can substitute the values of the length, width and perpendicular height into the formula V = 1/3 lwh. In my experience this is often provided for the students with little explanation as to why a volume of a pyramid is exactly one third the volume of a cuboid.
## Volume of a pyramid is one-third the volume of a cuboid
I wanted my class of 14 and 15-year-olds to derive the formula for the volume of a pyramid for themselves. This way they will be less likely to forget the formula in future as it is something they created through their own knowledge of cuboids and algebraic manipulation.
## Why is a cuboid special?
I start the lesson by asking students to discuss in pairs how they would describe the method for finding the volume of a cuboid to a ten-year-old. To encourage them to think about the different cross-sections students they are not allowed to describe the volume as simply the product of length, width and height. The volume of a cuboid is later described as product of cross-sectional area of a face and its depth. The formula V = lwh works because each of the faces of a cuboid can be its cross-section.
At this point students are asked to calculate the volume and surface area of the three cuboids below. This is done in their books so the method can be referred to later if needed.
## Calculating the Volume of a Pyramid
To introduce the volume of a pyramid I present the diagram below.
I explain the height of the square based pyramid is exactly one half the height of the cube. Students are to consider how many of these pyramids would fit perfectly inside the cube. I encourage them to discuss this in pairs and use mini-whiteboards to aid their processing.
The almost immediate response from about a quarter of the class is that four pyramids would fit inside. I encourage students to take more time to consider the problem before they commit to a solution.
## Visualising the pyramids inside a cube
Some students realised there must be a pyramid on each square face of the cube. However, drawing this in 3D proved difficult to visualise so some students decided to sketch the net.
## Deriving the formula for calculating the volume of a pyramid
I now ask the class to write on their whiteboards the volume of the pyramid as a fraction of the volume of the cube. All students wrote 1/6. I write on the main whiteboard our findings so far.
Volume of a Pyramid = 1/6 lwh
We discuss that the pyramid shares the same base length and width of the cube but not the perpendicular height. I ask the students to write the height of the cube in terms of the height of the pyramid and simplify the result.
h = 2x
Volume of Pyramid = 1/6 lw2x
Volume of Pyramid = 1/3 lwx
x is the perpendicular height of the pyramid.
## Finding the volume of a pyramid to solve more complex problems
As we progress through the remainder of the lesson, we apply the formula for calculating the volume of a pyramid and other composite solids. The questions become more challenging as students need to apply Pythagoras’ Theorem to calculate the perpendicular height as well as calculating the total volume of composite solids.
## Calculating the Volume of a Frustrum
Calculating the volume of a frustum is the final problem which we do in the plenary. Click here to view the video.
About a third of the class found it difficult to visualise the frustum as the difference between two similar rectangular-based pyramids. I help by sketching the big pyramid and smaller top pyramid as two separate shapes on the main board. It is now clear to the students the difference between the two forms the volume of the frustum
## Revising Volume of Prisms
Students revise how to calculate the volume of prisms as...
## Surface Area of Square Based Pyramids
Students learn how to use net drawings and apply Pythagoras'...
## Metric Units of Area and Volume
Students learn how to convert between metric units of area...
## Total Surface Area of Cylinders
Students learn how to calculate the total surface area of...
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### Mr Mathematics Blog
#### Getting Ready for a New School Year
When getting ready for a new school year I have a list of priorities to work through. Knowing my team have all the information and resources they need to teach their students gives me confidence we will start the term in the best possible way. Mathematics Teaching and Learning Folder All teachers receive a folder […]
#### Mathematics OFSTED Inspection – The Deep Dive
Earlier this week, my school took part in a trial OFSTED inspection as part of getting ready for the new inspection framework in September 2019. This involved three Lead Inspectors visiting our school over the course of two days. The first day involved a ‘deep dive’ by each of the Lead Inspectors into Mathematics, English […]
#### How to Solve Quadratics by Factorising
The method of how to solve quadratics by factorising is now part of the foundational knowledge students aiming for higher exam grades are expected to have. Here is an example of such a question. Solve x2 + 7x – 18 = 0 In my experience of teaching and marking exam papers students often struggle with […]
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# What is 201/266 as a decimal?
## Solution and how to convert 201 / 266 into a decimal
201 / 266 = 0.756
Fraction conversions explained:
• 201 divided by 266
• Numerator: 201
• Denominator: 266
• Decimal: 0.756
• Percentage: 0.756%
201/266 or 0.756 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. Depending on the situation, decimals can be more clear. We don't say 1 and 1/2 dollar. We use the decimal version of \$1.50. Same goes for fractions. We will say 'the student got 2 of 3 questions correct'. If we need to convert a fraction quickly, let's find out how and when we should.
201 / 266 as a percentage 201 / 266 as a fraction 201 / 266 as a decimal
0.756% - Convert percentages 201 / 266 201 / 266 = 0.756
## 201/266 is 201 divided by 266
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 201 is being divided into 266. Think of this as our directions and now we just need to be able to assemble the project! The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! Now we divide 201 (the numerator) into 266 (the denominator) to discover how many whole parts we have. This is our equation:
### Numerator: 201
• Numerators are the portion of total parts, showed at the top of the fraction. 201 is one of the largest two-digit numbers you'll have to convert. 201 is an odd number so it might be harder to convert without a calculator. Values closer to one-hundred make converting to fractions more complex. Now let's explore X, the denominator.
### Denominator: 266
• Denominators represent the total parts, located at the bottom of the fraction. 266 is one of the largest two-digit numbers to deal with. And it is nice having an even denominator like 266. It simplifies some equations for us. Overall, two-digit denominators are no problem with long division. Now it's time to learn how to convert 201/266 to a decimal.
## How to convert 201/266 to 0.756
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 266 \enclose{longdiv}{ 201 }$$
Use long division to solve step one. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 266 \enclose{longdiv}{ 201.0 }$$
Uh oh. 266 cannot be divided into 201. So that means we must add a decimal point and extend our equation with a zero. Now 266 will be able to divide into 2010.
### Step 3: Solve for how many whole groups you can divide 266 into 2010
$$\require{enclose} 00.7 \\ 266 \enclose{longdiv}{ 201.0 }$$
How many whole groups of 266 can you pull from 2010? 1862 Multiply by the left of our equation (266) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.7 \\ 266 \enclose{longdiv}{ 201.0 } \\ \underline{ 1862 \phantom{00} } \\ 148 \phantom{0}$$
If there is no remainder, you’re done! If you still have numbers left over, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. This is also true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But 201/266 and 0.756 bring clarity and value to numbers in every day life. Here are examples of when we should use each.
### When you should convert 201/266 into a decimal
Dining - We don't give a tip of 201/266 of the bill (technically we do, but that sounds weird doesn't it?). We give a 75% tip or 0.756 of the entire bill.
### When to convert 0.756 to 201/266 as a fraction
Carpentry - To build a table, you must have the right measurements. When you stretch the tape measure across the piece of wood, you won't see 10.6 inches. You'll see a tick mark at 10 and 3/5 inches.
### Practice Decimal Conversion with your Classroom
• If 201/266 = 0.756 what would it be as a percentage?
• What is 1 + 201/266 in decimal form?
• What is 1 - 201/266 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.756 + 1/2?
### Convert more fractions to decimals
From 201 Numerator From 266 Denominator What is 201/256 as a decimal? What is 191/266 as a decimal? What is 201/257 as a decimal? What is 192/266 as a decimal? What is 201/258 as a decimal? What is 193/266 as a decimal? What is 201/259 as a decimal? What is 194/266 as a decimal? What is 201/260 as a decimal? What is 195/266 as a decimal? What is 201/261 as a decimal? What is 196/266 as a decimal? What is 201/262 as a decimal? What is 197/266 as a decimal? What is 201/263 as a decimal? What is 198/266 as a decimal? What is 201/264 as a decimal? What is 199/266 as a decimal? What is 201/265 as a decimal? What is 200/266 as a decimal? What is 201/266 as a decimal? What is 201/266 as a decimal? What is 201/267 as a decimal? What is 202/266 as a decimal? What is 201/268 as a decimal? What is 203/266 as a decimal? What is 201/269 as a decimal? What is 204/266 as a decimal? What is 201/270 as a decimal? What is 205/266 as a decimal? What is 201/271 as a decimal? What is 206/266 as a decimal? What is 201/272 as a decimal? What is 207/266 as a decimal? What is 201/273 as a decimal? What is 208/266 as a decimal? What is 201/274 as a decimal? What is 209/266 as a decimal? What is 201/275 as a decimal? What is 210/266 as a decimal? What is 201/276 as a decimal? What is 211/266 as a decimal?
### Convert similar fractions to percentages
From 201 Numerator From 266 Denominator 202/266 as a percentage 201/267 as a percentage 203/266 as a percentage 201/268 as a percentage 204/266 as a percentage 201/269 as a percentage 205/266 as a percentage 201/270 as a percentage 206/266 as a percentage 201/271 as a percentage 207/266 as a percentage 201/272 as a percentage 208/266 as a percentage 201/273 as a percentage 209/266 as a percentage 201/274 as a percentage 210/266 as a percentage 201/275 as a percentage 211/266 as a percentage 201/276 as a percentage
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## Adding and Subtracting within 20
1.OA.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8. Add and subtract within 20.
1.OA.5 Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
1.OA.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
## Adding and Subtracting within 20
###### 1.OA.4 1.OA.5 1.OA.6
Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
## Addition and Subtraction Word Problems
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
## Number Sentences
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.
## Properties of Operations
Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
## Counting to 120
Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.
## Counting to 120
###### 1.NBT.1
Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.
## Subtract Multiples of 10
Subtract multiples of 10 in the range 10-90 from multiples of 10 in the range 10-90 (positive or zero differences), using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.
## Tens & Ones
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
## Mental Math
Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.
## Comparing Numbers
Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <.
## Ordering Objects by Length
Order three objects by length; compare the lengths of two objects indirectly by using a third object.
## Tell and Write Time
Tell and write time in hours and half-hours using analog and digital clocks.
## Interpreting Data
Organize, represent, and interpret data with up to three categories; ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another.
## Length of an Object
Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps.
## Shape Attributes
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes.
## Composite Shapes
Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1
## Partitioning Into Equal Shares
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Conversion of Decimals, Fractions, and Percent
## Convert back and forth between decimals, fractions, and percents.
0%
Progress
Practice Conversion of Decimals, Fractions, and Percent
Progress
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Financing a Fiat
Teacher Contributed
## Real World Applications – Algebra I
Financing a Fiat
### Student Exploration
In 2012, one of the most popular cars to purchase in California was a Fiat. They advertised on their commercials their cheap monthly payments. Let’s look into what a potential car-owner would look forward to, and use two-step equations to find out the best deal.
In April, a quote was given to a potential Fiat owner based on how much they could first pay up front (called the down payment). The total cost of a Fiat is $18,408.06, including all taxes and fees. All of the following calculations are based on a 60-month pay period. If there is “zero down,” or no money paid as a down payment, based on a 60-month pay period, we can figure out how much the owner would pay per month. Let’s use the formula, $y = mx + b$, and let’s assume that $y$ represents the total amount paid for the car, $m$ represents the monthly payment, $x$ represents the number of months the owner is paying, and $b$ represents the initial down payment. If there is no money down, then the equation is $18408.06 = (m)(60) + 0$. This equation can be solved in one step. We divide both sides by 60, and find that the monthly payment is$306.80. Not bad, but the potential owner really wants to pay as close to $200 a month, as in what the commercial on television advertised. To lower the monthly payment, the owner would have to initially put down money. What if the down payment was$2,000? If the owner took 60 months to pay off the car, how much is the monthly payment? Let’s use the formula to find out:
$Y &= mx + b\\18,408.06 &= (m)(60) + 2000\\16,408.06 &= 60m\\\273.47 &= m$
What if the down payment was $5,000? How much would this car owner pay every month for 60 months? We can also apply our understanding of percent equations to find out the percent tax on the car. We set the dollar amount of tax equal to the original sticker price of the car multiplied by the percent (which is the unknown). On the printout that a car salesperson gave, we know that the tax is$1,466.06 and the sticker price is \$16,700. Our equation is,
$1466.06 - 16700(p)$, and $p$ represents the sales tax. We solve this equation by dividing both sides by 16700, and we get $p = 0.0878$.
We use our knowledge of converting this decimal to percent by multiplying this by 100. The sales tax is 8.78%.
### Extension Investigation
Try looking up a car that you would potentially purchase (either new or used). Try to find out how much you could pay every month for 60 months, knowing the total price of the car.
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Question Video: Finding a Probability for a Continuous Random Variable | Nagwa Question Video: Finding a Probability for a Continuous Random Variable | Nagwa
# Question Video: Finding a Probability for a Continuous Random Variable Mathematics • Third Year of Secondary School
## Join Nagwa Classes
Let π be a continuous random variable with the probability density function π(π₯), represented by the following graph. Find the value of π that makes π(5 < π < π) = 1/3.
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### Video Transcript
Let π be a continuous random variable with the probability density function π of π₯ represented by the following graph. Find the value of π that makes the probability that π is greater than five but less than π equal to one-third.
We recall first that for a continuous random variable π, the probability that π₯ lies in a given interval is equivalent to the area under the graph of its probability density function π of π₯ between the endpoints of that interval. Weβre looking to find the probability that π is greater than five and less than some value π. So this will correspond to the orange area shown.
We know what we want the area to be. Itβs one-third. We also know the width of this rectangle. From the graph of the probability density function, we can see that the height is one-sixth. The base or length of the rectangle is from five to the unknown value π. So an expression for the rectangleβs length is π minus five. Recalling that the area of a rectangle is found by multiplying its length and width together, we can therefore form an equation. One-sixth multiplied by π minus five is equal to one-third. We can then solve this equation to determine the value of π.
First, we multiply both sides of the equation by six. On the left-hand side, we have π minus five, and on the right-hand side, six multiplied by a third or six over three, which is equal to two. We then add five to each side of the equation, giving π equals seven.
So by recalling that for a continuous random variable π, the probability that π lies in a given interval is equal to the area under the graph of its probability density function between the endpoints of that interval, we found the value of π such that the probability π is greater than five but less than π is one-third is seven.
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## New Maths Curriculum: Year 1 Multiplication and Division
Let’s look at the new targets for year 1 multiplication and division. A lot of the work this year should be practically based.
The statutory requirements for Year 1 Multiplication and Division are:
Multiplication and division
Pupils should be taught to:
• solve one-step problems involving multiplication and division, by calculating the answer using concrete objects, pictorial representations and arrays with the support of the teacher.
Main changes
Multiplication and division problems including arrays are now included, whilst previously these were expectations for Year 2 onwards.
Grouping and sharing small quantities forms the major part of the work.
Doubling small numbers is an important part of this year’s work. Plenty of opportunity should also be given to counting, using 2p, 5p and 10p pieces, both counting up and down. Similar activities can be carried out with pairs of socks, gloves etc. Once again, it is better to use real life objects when beginning to combine groups.
Practice should be given at counting up in twos before moving on to fives or tens and can be shown as arrays or number patterns.
Remember, division can be understood in two ways:
1. Sharing equally e.g. 6 bars of chocolate are shared between 2 people.
How many bars does each person get?
This concept is best introduced with practical apparatus – bowls of sweets, counters, buttons etc. are ideal for sharing into equal groups, and then counting the number in the group.
2. Grouping, or repeated subtraction e.g. 8 divided by 2
can be seen as how many lots of two can I take from eight?
Again, this is best done with apparatus at first and it is a slightly different process to sharing equally (where one item is given to each person in turn).
This can be answered by repeatedly taking two from the pile and then counting how many lots of two have been taken altogether.
Multiplication and division in Year 1
## New Maths Curriculum: Year 1 Addition and Subtraction
Let’s look at the new targets for year 1 addition and subtraction; children will be expected to work with larger numbers!
Pupils should be taught to:
• read, write and interpret mathematical statements involving addition (+), subtraction (-) and equals (=) signs
• represent and use number bonds and related subtraction facts within 20
• add and subtract one-digit and two-digit numbers to 20, including zero
• solve one-step problems that involve addition and subtraction, using concrete objects and pictorial representations, and missing number problems
such as 4 = ? – 5.
Again, there does not seem to be much new here, but it must be remembered that these are end of year targets and much needs to be done during the year to reach them.
Children will be expected to make real progress with knowing, off by heart, all addition facts for each number totalling up to 20 (e.g. know that 9 + 5 = 14). They should also understand the effect of adding zero.
The key here is ‘knowing off by heart’, or as the new programme of study puts it in the corresponding notes:
‘memorise and reason with number bonds to 10 and 20…’
Not only should they know these number bonds, they should be able to reason using that knowledge.
Subtraction is much easier once it is realised that it is the inverse (or opposite) of addition. Many subtraction questions can be answered by adding on, but children are also expected to make progress with knowing subtraction facts (e.g. know that 9 – 4 = 5). In future years knowing these facts will be just as important as knowing ‘tables’.
New programme of Study: Year 1 Addition and Subtractionddition and subtraction
## New Maths Curriculum: Year 1 Number and Place Value
It’s a whole new set of targets for Year 1 from 2014. Let’s begin by having a look at Number and Place Value.
The statutory requirements for Year 1 Number and Place Value are:
Number and place value
Pupils should be taught to:
• count to and across 100, forwards and backwards, beginning with 0 or 1, or from any given number
• count, read and write numbers to 100 in numerals; count in multiples of twos, fives and tens
• given a number, identify one more and one less
• identify and represent numbers using objects and pictorial representations including the number line, and use the language of: equal to, more than, less than (fewer), most, least
• read and write numbers from 1 to 20 in numerals and words.
Main changes
The new maths programme of study is much shorter than previous models with much broader statements. This makes it tricky to know what has been left out. However, the main aim is to raise standards by raising targets to be achieved. In Year 1 the main changes are:
1. Counting, reading and writing numbers to 100 rather than just up to 20. This will prove to be very difficult for many children and will need a great deal of practice.
2. Perhaps even harder is to count in twos, fives and tens from different multiples.
New Statutory Requirements 2014: Number and Place Value
## Perpendicular lines
The term perpendicular was usually introduced to children in the upper primary years and indeed into secondary school, but it is now thought that it should be introduced in year 3. At this stage children just need to be able to recognise two lines which are perpendicular to each other and to make it easier all the pairs of lines shown on the worksheet either cross or touch.
Of course, the key to being able to do this is to recognise a right angle and plenty of practice with recognising right angles must be done before going on to using the term perpendicular. When the lines are vertical and horizontal it is quite easy to spot perpendicular lines but other placements make it more difficult to see just by looking and there is no harm in using the corner of a piece of paper to judge whether this is the case.
Perpendicular lines
## Year 6 resource of the week: factors
Understanding factors and multiples is something many children fail to grasp. Here we have a page on factors and how to find all the factors of a number. The process is fairly straightforward, but can be quite time consuming. A good knowledge of tables and division is also needed. So if children’s knowledge of tables is weak, if they find division difficult and if they have little staying power then it is unlikely that they will enjoy trying to find factors of numbers!
One point that even brighter children take a while to understand is that you only need to continue to divide up to when the number squared is larger than the original number eg to find the factors of 62 you only need to divide 62 by numbers up to 7 because eight eights are 64. It is a good idea to spend some time explaining and showing the logic of this to children.
This can be found in our Year 6 Understanding Number section.
Factors 1
## Roman numeral clock faces (2)
Our Roman numerals clock face worksheet has proved very popular so I thought I would post another similar page, but including quarter to and quarter past the hour times. I have also been asked why the clock face shows IIII for four o’clock rather than IV. In fact most Roman clockfaces do show the four Is and nobody is sure why.
One reason is that when looking at the numerals opposite to each other – all of them are in perfect balance, except for the ‘heavy’ VIII and the ‘light’ IV; optical balance is re-established by printing an also ‘heavy’ IIII.
Another reason which has been given is to do with the old casting process of the numerals; ‘ Since some numerals were cast out of metal, or carved out of wood or bone, you need 20 I’s, 4 V’s, and 4 X’s, even numbers of each, if you use four I’s for “four”. The molds would produce a long centre rod, with 10 I’s, 2 V’s, and 2 X’s on each side.’
A third possibility is that clocks use IIII rather than IV out of respect for the Roman God Jupiter, the king of the Gods, whose name, in Latin, begins IV (the V being the U we now use, the I the J). Very old sun dials seem to use the IIII and early clocks followed suit; it has also been suggested that Wells Cathedral clock, one of the earliest cathedral clocks used the IIII and everyone copied this, and yet another reason is that Louis XIV preferred IIII over IV and ordered all clocks to be made in this way, and it has remained like this ever since.
Finally it has also been suggested that Romans were not great at subtraction so IIII was easier to work out than IV! I have no idea which, if any of these has the best claim to being true but interestingly Big Ben uses the IV convention.
Roman numerals clock 2
## Resource of the Week: Year 5 Number Challenge
I really like this challenge, partly because there is no, one right way to answer it and partly because it really makes children think.
There are 10 digits, from zero to nine to be placed in the 10 boxes in such a way that the targets can be matched as closely as possible. The catch is that each number can only be used once!
Now the obvious way to start is to make 98 the largest even number, but immediately that means that you can not have 97 as the largest odd number. By the time you reach the last target, number closest to 30, you have only two digits left and only two choices!
But what counts as the best possible answers? This is as big a challenge, if not bigger. I have had a class try to make a set of rules to try to be as fair as possible, but it involved a great deal of addition and subtraction. One group made a set of rules that went like this:
1. Find the difference between 98 and the answer given.
2. Find the difference between 99 and the answer given.
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# Addition and Subtraction of Polynomials
## Combining like terms in polynomial expressions
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You are going to build a rectangular garden in your back yard. The garden is 2 m more than 1.5 times as long as it is wide. Write an expression to show the area of the garden.
### Guidance
The word polynomial comes from the Greek word poly meaning “many”. Polynomials are made up of one or more terms and each term must have an exponent that is 0 or a whole number. This means that $3x^2+2x+1$ is a polynomial, but $3x^{0.5}+2x^{-2}+1$ is not a polynomial. Some common polynomials have special names based on how many terms they have:
• A monomial is a polynomial with just one term. Examples of monomials are $3x$ , $2x^2$ and $7$ .
• A binomial is a polynomial with two terms. Examples of binomials are $2x+1$ , $3x^2-5x$ and $x-5$ .
• A trinomial is a polynomial with three terms. An example of a trinomial is $2x^2+3x-4$ .
To add and subtract polynomials you will go through two steps.
1. Use the distributive property to remove parentheses. Remember that when there is no number in front of the parentheses, it is like there is a 1 in front of the parentheses. Pay attention to whether or not the sign in front of the parentheses is $+$ or $-$ , because this will tell you if the number you need to distribute is $+1$ or $-1$ .
2. Combine similar terms. This means, combine the $x^2$ terms with the $x^2$ terms, the $x$ terms with the $x$ terms, etc.
#### Example A
Find the sum: $(3x^2+2x-7)+(5x^2-3x+3)$ .
Solution: First you want to remove the parentheses. Because this is an addition problem, it is like there is a $+1$ in front of each set of parentheses. When you distribute a $+1$ , none of the terms will change.
$1(3x^2+2x-7)+1(5x^2-3x+3)=3x^2+2x-7+5x^2-3x+3$
Next, combine the similar terms. Sometimes it can help to first reorder the expression to put the similar terms next to one another. Remember to keep the signs with the correct terms. For example, in this problem the 7 is negative and the 3x is negative.
$3x^2+2x-7+5x^2-3x+3&=3x^2+5x^2+2x-3x-7+3\\&=8x^2-x-4$
#### Example B
Find the difference: $(5x^2+8x+6)-(4x^2+5x+4)$ .
Solution: First you want to remove the parentheses. Because this is a subtraction problem, it is like there is a $-1$ in front of the second set of parentheses. When you distribute a $-1$ , each term inside that set of parentheses will change its sign.
$1(5x^2+8x+6)-1(4x^2+5x+4)=5x^2+8x+6-4x^2-5x-4$
Next, combine the similar terms. Remember to keep the signs with the correct terms.
$5x^2+8x+6-4x^2-5x-4&=5x^2-4x^2+8x-5x+6-4\\&=x^2+3x+2$
#### Example C
Find the difference: $(3x^3+6x^2-7x+5)-(4x^2+3x-8)$
Solution: First you want to remove the parentheses. Because this is a subtraction problem, it is like there is a $-1$ in front of the second set of parentheses. When you distribute a $-1$ , each term inside that set of parentheses will change its sign..
$1(3x^3+6x^2-7x+5)-1(4x^2+3x-8)=3x^3+6x^2-7x+5-4x^2-3x+8$
Next, combine the similar terms. Remember to keep the signs with the correct terms.
$3x^3+6x^2-7x+5-4x^2-3x+8&=3x^3+6x^2-4x^2-7x-3x+5+8\\&=3x^3+2x^2-10x+13$
#### Concept Problem Revisited
Remember that the area of a rectangle is length times width.
$Area &= l \times w \\Area &= (1.5x + 2) x \\Area &= 1.5x^2 + 2x$
### Vocabulary
Binomial
A binomial has two terms that are added or subtracted from each other. Each of the terms of a binomial is a variable $(x)$ , a product of a number and a variable $(4x)$ , or the product of multiple variables with or without a number $(4x^2y + 3)$ . One of the terms in the binomial can be a number.
Monomial
A monomial can be a number or a variable (like $x$ ) or can be the product of a number and a variable (like $3x$ or $3x^2$ ). A monomial has only one term.
Polynomial
A polynomial , by definition, is also a monomial or the sum of a number of monomials. So $3x^2$ can be considered a polynomial, $2x+3$ can be considered a polynomial, and $2x^2+3x-4$ can be considered a polynomial.
Trinomial
A trinomial has three terms $(4x^2+3x-7)$ . The terms of a trinomial can be a variable $(x)$ , a product of a number and a variable $(3x)$ , or the product of multiple variables with or without a number $(4x^2)$ . One of the terms in the trinomial can be a number $(-7)$ .
Variable
A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is often referred to as the literal coefficient.
### Guided Practice
1. Find the sum: $(2x^2+4x+3) + (x^2-3x-2)$ .
2. Find the difference: $(5x^2-9x+7) - (3x^2-5x+6)$ .
3. Find the sum: $(8x^3+5x^2-4x+2) + (4x^3+7x-5)$ .
1. $(2x^2+4x+3) + (x^2-3x-2)=2x^2+4x+3+x^2-3x-2=3x^2+x+1$
2. $(5x^2-9x+7) - (3x^2-5x+6)=5x^2-9x+7-3x^2+5x-6=2x^2-4x+1$
3. $(8x^3+5x^2-4x+2) + (4x^3+7x-5)=8x^3+5x^2-4x+2+4x^3+7x-5=12x^3+5x^2+3x-3$
### Practice
For each problem, find the sum or difference.
1. $(x^2+4x+5) + (2x^2+3x+7)$
2. $(2r^2+6r+7) - (3r^2+5r+8)$
3. $(3t^2-2t+4) + (2t^2+5t-3)$
4. $(4s^2-2s-3) - (5s^2+7s-6)$
5. $(5y^2+7y-3) + (-2y^2-5y+6)$
6. $(6x^2+36x+13) - (4x^2+13x+33)$
7. $(12a^2+13a+7) + (9a^2+15a+8)$
8. $(9y^2-17y-12) + (5y^2+12y+4)$
9. $(11b^2+7b-12) - (15b^2-19b-21)$
10. $(25x^2+17x-23) - (-14x^3-14x-11)$
11. $(-3y^2+10y-5) - (5y^2+5y+8)$
12. $(-7x^2-5x+11) + (5x^2+4x-9)$
13. $(9a^3-2a^2+7) + (3a^2+8a-4)$
14. $(3x^2-2x+4) - (x^2+x-6)$
15. $(4s^3+4s^2-5s-2) - (-2s^2-5s+6)$
### Vocabulary Language: English
distributive property
distributive property
The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
Polynomial
Polynomial
A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents.
Variable
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
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# Lens
## Focusing or dispersing light emitted from an object can change how we see it in many different ways.
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Lens
Students will learn how light behaves when passing through converging and diverging lenses. Students will also learn how to do ray tracing diagrams and calculate image distances and magnification using the lens' maker's equation.
### Key Equations
; The len's maker's equation
Where f is the focal length of the lens, \begin{align*} d_o \end{align*} is the distance of the object from the lens and \begin{align*} d_i \end{align*} is the distance the image is formed from the lens.
The size of an object’s image is larger (or smaller) than the object itself by its magnification, \begin{align*}M\end{align*}. The level of magnification is proportional to the ratio of \begin{align*}d_i\end{align*} and \begin{align*}d_o\end{align*}. An image that is double the size of the object would have magnification \begin{align*}M = 2\end{align*}.
### Guidance
• For lenses, the distance from the center of the lens to the focus is \begin{align*}f\end{align*}. Focal lengths for foci behind the lens are positive in sign. The distance from the center of the lens to the object in question is \begin{align*}d_0\end{align*}, where distances to the left of the lens are positive in sign. The distance from the center of the lens to the image is \begin{align*}d_i\end{align*}. This number is positive for real images (formed to the right of the lens), and negative for virtual images (formed to the left of the lens).
• Lenses, made from curved pieces of glass, focus or de-focus light as it passes through. Lenses that focus light are called converging lenses, and these are the ones used to make telescopes and cameras. Lenses that de-focus light are called diverging lenses.
• Lenses can be used to make visual representations, called images.
• The focal length, \begin{align*}f\end{align*}, of a lens or mirror is the distance from the surface of the lens to the place where the light is focused. This is called the focal point or focus. For diverging lenses, the focal length is negative.
• For converging lens, one can find the focal point by simply holding a piece of paper near the lens until a distant image is formed. The distance from the paper to the lens is the focal point.
• When light rays converge behind a lens, a real image is formed. Real images are useful in that you can place photographic film at the physical location of the real image, expose the film to the light, and make a two-dimensional representation of the world, a photograph.
• When light rays diverge behind a lens, a virtual image is formed. A virtual image is a manifestation of your brain (it traces the diverging rays backwards and forms an image), like the person you see “behind” a mirror’s surface when you brush your teeth (there's obviously no real light focused behind a mirror!). Since virtual images aren’t actually “anywhere,” you can’t place photographic film anywhere to capture them.
• Real images are upside-down, or inverted. You can make a real image of an object by putting it farther from a mirror or lens than the focal length. Virtual images are typically right-side-up. You can make virtual images by moving the lens closer to the object than the focal length.
In ray tracing problems, you will do a careful ray tracing with a ruler (including the extrapolation of rays for virtual images). It is best if you can use different colors for the three different ray tracings. When sketching diverging rays, you should use dotted lines for the extrapolated lines in front of a lens in order to produce the virtual image. When comparing measured distances and heights to calculated distances and heights, values within \begin{align*}10\end{align*}% are considered “good.” Use the Table (below) as your guide.
Mirror type Ray tracings
Converging lenses
(convex)
Ray #1: Leaves tip, travels parallel to optic axis, refracts and travels through to the focus.
Ray #2: Leaves tip, travels through focus on same side, travels through lens, and exits lens parallel to optic axis on opposite side.
Ray #3: Leaves tip, passes straight through center of lens and exits without bending.
Diverging lenses
(concave)
Ray #1: Leaves tip, travels parallel to optic axis, refracts OUTWARD by lining up with focus on the SAME side as the candle.
Ray #2: Leaves tip, heads toward the focus on the OPPOSITE side, and emerges parallel from the lens.
Ray #3: Leaves tip, passes straight through the center of lens and exits without bending.
#### Example 1
You have a converging lens of focal length 2 units. If you place an object 5 units away from the lens, (a) draw a ray diagram of the situation to estimate where the image will be and (b) list the charactatistics of the image. Finally (c) calculate the position of the image. A diagram of the situation is shown below.
##### Solution
(a): To draw the ray diagram, we'll follow the steps laid out above for converging lenses.
First we draw the a ray that travels parallel to the principle axis and refracts through the focus on the other side (the red ray). Next we draw a ray through the focus on the same side that refracts out parallel (the green ray). Finally we draw the ray that travels straight through the center of the lens without refracting (the blue ray). The result is shown below.
(b): Based on the ray diagram and the initial position of the object, we know that the image is a real, inverted, and smaller than the original object.
(c): To calculate the exact position of the object, we can use the lens maker's equation.
### Simulation
Note: this simulation only shows the effects of a convex lens
### Explore More
1. Consider a converging lens with a focal length equal to three units and an object placed outside the focal point.
1. Carefully trace three rays coming off the top of the object and form the image.
2. Measure \begin{align*}d_o\end{align*} and \begin{align*}d_i\end{align*}.
3. Use the mirror/lens equation to calculate \begin{align*}d_i\end{align*}.
4. Find the percent difference between your measured \begin{align*}d_i\end{align*} and your calculated \begin{align*}d_i\end{align*}.
5. Measure the magnification \begin{align*}M\end{align*} and compare it to the calculated magnification.
2. Consider a converging lens with a focal length equal to three units, but this time with the object placed inside the focal point.
1. Carefully trace three rays coming off the top of the object and form the image.
2. Measure \begin{align*}d_o\end{align*} and \begin{align*}d_i\end{align*}.
3. Use the mirror/lens equation to calculate \begin{align*}d_i\end{align*}.
4. Find the percent difference between your measured \begin{align*}d_i\end{align*} and your calculated \begin{align*}d_i\end{align*}.
5. Measure the magnification \begin{align*}M\end{align*} and compare it to the calculated magnification.
3. Consider a diverging lens with a focal length equal to four units.
1. Carefully trace three rays coming off the top of the object and show where they converge to form the image.
2. Measure \begin{align*}d_o\end{align*} and \begin{align*}d_i\end{align*}.
3. Use the mirror/lens equation to calculate \begin{align*}d_i\end{align*}.
4. Find the percent difference between your measured \begin{align*}d_i\end{align*} and your calculated \begin{align*}d_i\end{align*}.
5. Measure the magnification \begin{align*}M\end{align*} and compare it to the calculated magnification.
4. A piece of transparent goo falls on your paper. You notice that the letters on your page appear smaller than they really are. Is the goo acting as a converging lens or a diverging lens? Explain. Is the image you see real or virtual? Explain.
5. An object is placed \begin{align*}30 \;\mathrm{mm}\end{align*} in front of a lens. An image of the object is located \begin{align*}90 \;\mathrm{mm}\end{align*} behind the lens.
1. Is the lens converging or diverging? Explain your reasoning.
2. What is the focal length of the lens?
6. Little Red Riding Hood (aka \begin{align*}R-\end{align*}Hood) gets to her grandmother’s house only to find the Big Bad Wolf (aka BBW) in her place. \begin{align*}R-\end{align*}Hood notices that BBW is wearing her grandmother’s glasses and it makes the wolf’s eyes look magnified (bigger).
1. Are these glasses for near-sighted or far-sighted people? For full credit, explain your answer thoroughly. You may need to consult some resources online.
2. Create a diagram of how these glasses correct a person’s vision.
1. c. \begin{align*}3\end{align*} units e. \begin{align*}- 2/3\end{align*}
2. c. -6 (so 6 units on left side) e. 3 times bigger
3. c. \begin{align*}-2.54\end{align*} units (2.54 units on the left) e. \begin{align*}M=.36\end{align*}
4. .
5. b. \begin{align*}22.5 \;\mathrm{mm}\end{align*}
6. .
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Show that: $tan\bigg(\frac{1}{2}sin^{-1}\frac{3}{4}\bigg)=\frac{4-\sqrt 7}{3}$
(Note: This question has been split into 2 questions) This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.
Toolbox:
• $sin^{-1}x = \tan^{-1}\large \frac{x}{\sqrt (1-x^2)}$
• $\tan x=\large \frac{2\tan\frac{x}{2}}{1-tan^2\frac{x}{2}}$
Given $tan\bigg(\frac{1}{2}sin^{-1}\frac{3}{4}\bigg)=\frac{4-\sqrt 7}{3}$.
$\textbf{Step 1}$:
Let $\theta = sin^{-1}\frac{3}{4}$
We know that $sin^{-1}x = \tan^{-1}\large \frac{x}{\sqrt (1-x^2)}$
Therefore, $\theta = sin^{-1}\frac{3}{4} = \tan^{-1}\large \frac{\large \frac{3}{4}}{\sqrt (1-(\large \frac{3}{4})^2)}$
$\Rightarrow \theta $$= \tan^{-1}\large \frac{\large \frac{3}{4}}{\sqrt \large\frac{16-9}{16}}$$ = tan^{-1} (\frac{3}{4} \times \frac{\sqrt16}{\sqrt7})$
$\Rightarrow \theta = tan^{-1} \frac{3}{\sqrt7}$
$\Rightarrow tan \theta = \frac{3}{\sqrt7}$
$\textbf{Step 2}$:
We know that $\tan x=\large \frac{2\tan\frac{x}{2}}{1-tan^2\frac{x}{2}}$
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Chapter 3: Graphing
Finding the Distance Between Two Points
The logic used to find the distance between two data points on a graph involves the construction of a right triangle using the two data points and the Pythagorean theorem $(a^2 + b^2 = c^2)$ to find the distance.
To do this for the two data points $(x_1, y_1)$ and $(x_2, y_2)$, the distance between these two points $(d)$ will be found using $\Delta x = x_2 - x_1$ and $\Delta y = y_2 - y_1.$
Using the Pythagorean theorem, this will end up looking like:
$d^2 = \Delta x^2 + \Delta y^2$
or, in expanded form:
$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$
On graph paper, this looks like the following. For this illustration, both $\Delta x$ and $\Delta y$ are 7 units long, making the distance $d^2 = 7^2 + 7^2$ or $d^2 = 98$.
The square root of 98 is approximately 9.899 units long.
Example 3.2.1
Find the distance between the points $(-6,-4)$ and $(6, 5)$.
Start by identifying which are the two data points $(x_1, y_1)$ and $(x_2, y_2)$. Let $(x_1, y_1)$ be $(-6,-4)$ and $(x_2, y_2)$ be $(6, 5)$.
Now:
$\Delta x^2 = (x_2 - x_1)^2$ or $[6 - (-6)]^2$ and $\Delta y^2 = (y_2 - y_1)^2$ or $[5 - (-4)]^2$.
This means that
$d^2 = [6 - (-6)]^2 + [5 - (-4)]^2$
or
$d^2 = [12]^2 + [9]^2$
which reduces to
$d^2 = 144 + 81$
or
$d^2 = 225$
Taking the square root, the result is $d = 15$.
Finding the Midway Between Two Points (Midpoint)
The logic used to find the midpoint between two data points $(x_1, y_1)$ and $(x_2, y_2)$ on a graph involves finding the average values of the $x$ data points $(x_1, x_2)$ and the of the $y$ data points $(y_1, y_2)$. The averages are found by adding both data points together and dividing them by $2$.
In an equation, this looks like:
$x_{\text{mid}}=\dfrac{x_2+x_1}{2}$ and $y_{\text{mid}}=\dfrac{y_2+y_1}{2}$
Example 3.2.2
Find the midpoint between the points $(-2, 3)$ and $(6, 9)$.
We start by adding the two $x$ data points $(x_1 + x_2)$ and then dividing this result by 2.
$x_{\text{mid}} = \dfrac{(-2 + 6)}{2}$
or
$\dfrac{4}{2} = 2$
The midpoint’s $y$-coordinate is found by adding the two $y$ data points $(y_1 + y_2)$ and then dividing this result by 2.
$y_{\text{mid}} = \dfrac{(9 + 3)}{2}$
or
$\dfrac{12}{2} = 6$
The midpoint between the points $(-2, 3)$ and $(6, 9)$ is at the data point $(2, 6)$.
Questions
For questions 1 to 8, find the distance between the points.
1. (−6, −1) and (6, 4)
2. (1, −4) and (5, −1)
3. (−5, −1) and (3, 5)
4. (6, −4) and (12, 4)
5. (−8, −2) and (4, 3)
6. (3, −2) and (7, 1)
7. (−10, −6) and (−2, 0)
8. (8, −2) and (14, 6)
For questions 9 to 16, find the midpoint between the points.
1. (−6, −1) and (6, 5)
2. (1, −4) and (5, −2)
3. (−5, −1) and (3, 5)
4. (6, −4) and (12, 4)
5. (−8, −1) and (6, 7)
6. (1, −6) and (3, −2)
7. (−7, −1) and (3, 9)
8. (2, −2) and (12, 4)
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Examveda
# Rahul is as much younger than Sagar as he is older than Purav. If the sum of the ages of Purav and Sagar is 66 years, and Sagar's age is 48 years, then what is the difference between Rahul and Purav's age ? ( in years)
A. 18
B. 15
C. 16
D. 20
E. Cannot be determined
### Solution(By Examveda Team)
Let the age of Rahul, Sagar and Purav be x, y and z respectively
According to the given information
Age of Sagar - Age of Rahul = Age of Rahul - Age of Purav
\eqalign{ & \Rightarrow y - x = x - z \cr & \Rightarrow 2x = y + z......(i) \cr}
Also y + z = 66 years
From (i) x = 33 years
Also as per equation (i) we have Purav's age + Sagar age = 66 years
by going through option (A) given Purav = 18, and Rahul = 33 years, Sagar = 48 years
Difference between Rahul's and Purav's age = 33 - 18 = 15 years
This Question Belongs to Arithmetic Ability >> Problems On Ages
1. Why you consider option a
2. Seems it has 2 answers.
Rahul=33
If Difference is 18 then
Purav=15
Rahul=15+18=33
Sagar=33+18=51
Purav+Sagar=15+51=66
Again if Difference is 15
Then Purav=18
Rahul=18+15=33
Sagar=33+15=48
Purav+Sagar=18+48=66
|
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# An iron sphere weighs 10N and rests in a V shaped trough whose sides from an angle $+ 6V,{(1/30)^{3/2}}$. The net force exerted by the walls on the sphere in case as shown in figure is:A. 0 NB. 10 NC. $\dfrac{{10}}{{\sqrt 3 }}$ D. 5N
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Hint:- Here in this question we have to resolve the forces into $\cos \theta$ and $\sin \theta$. After resolving the two forces which are exerted by the wall on the ball add the two forces. Put the value of given angle, one would be able to find the net force exerted.
Complete step-by-step solution:-
The forces would become;
$N\cos \theta = - N\cos \theta$;
Now, for the net force add the two forces together,
${F_{Net}} = N\cos \theta + N\cos \theta$ ;
In place of${F_{Net}}$put the force exerted by the ball which is 10 N.
$10 = N\cos \theta + N\cos \theta$
Since the two forces are together add the two forces.
$2N\cos \theta = 10N$;
Take “2” to the RHS and divide by 10N.
$N\cos \theta = \dfrac{{10N}}{2}$;
Here, find the value of N
$N\cos \theta = 5N$;
Solve the equation and put the value of the given angle,
$\cos \theta = \cos 60 = \dfrac{1}{2}$ ;
Put the above value into the equation;
$N = 5 \times 2N$;
The net force comes out to be:
$N = 10N$;
Final Answer: Option “2” is correct. The net force exerted by the walls on the sphere is 10 N.
Note:- Here make sure to make the diagram properly and resolve the forces. Here we have considered only $N\cos \theta$because that is the only force which is acting on the ball; the vertical force $N\sin \theta$ is not acting on the ball so it is not considered.
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# What is the Implicit Function Theorem and how do you prove it?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
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Steve M Share
Mar 8, 2018
The Implicit Function Theorem is a method of using partial derivatives to perform implicit differentiation.
Suppose we cannot find $y$ explicitly as a function of $x$, only implicitly through the equation $F \left(x , y\right) = 0$ which defines $y$ as a function of $x$ and $y$. Therefore we can write $F \left(x , y\right) = 0$ as $F \left(x , y \left(x\right)\right) = 0$. Differentiating both sides of this, using the partial chain rule gives us
(partial F)/(partial x) dx/dx + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))
Example:
If we take a unit circle, with equation:
${x}^{2} + {y}^{2} = 1$
Then differentiating implicitly we have:
$2 z + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$
And if we write this in the form:
$F \left(x , y\right) = 0 \implies F \left(x , y\right) = - {x}^{2} + {y}^{2} - 1$
Then the partial derivatives are:
${F}_{=} x = 2 x$ and ${F}_{=} y = 2 y$
And the Implicit Function Theorem gives
dy/dx = −F_x/F_y
\ \ \ \ \ \ = −((partial F)/(partial x)) / ((partial F)/(partial y))
\ \ \ \ \ \ = −(2x) / (2y)
\ \ \ \ \ \ = −x/y , as before
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NCERT Solutions for Class 6 Math Chapter 4 Basic Geometrical Ideas are provided here with simple step-by-step explanations. These solutions for Basic Geometrical Ideas are extremely popular among class 6 students for Math Basic Geometrical Ideas Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 6 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class 6 Math are prepared by experts and are 100% accurate.
#### Question 1:
Use the figure to name:
(a) Five points
(b) A line
(c) Four rays
(d) Five line segments
#### Answer:
(a) The five points are D, E, O, B, and C.
(b)
(c)
(d)
#### Question 2:
Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given.
#### Question 3:
Use the figure to name:
(a) Line containing point E.
(b) Line passing through A.
(c) Line on which O lies
(d) Two pairs of intersecting lines.
(a)
(b)
(c)
(d) ,
#### Question 4:
How many lines can pass through (a) one given point? (b) Two given points?
#### Answer:
(a) Infinite number of lines can pass through a single point.
(b) Only one line can pass through two given points.
#### Question 5:
Draw a rough figure and label suitably in each of the following cases:
(a) Point P lies on.
(b) and intersect at M.
(c) Line l contains E and F but not D.
(d) and meet at O.
(a)
(b)
(c)
(d)
#### Question 6:
Consider the following figure of line. Say whether following statements are true or false in context of the given figure.
(a) Q, M, O, N, P are points on the line.
(b) M, O, N are points on a line segment.
(c) M and N are end points of line segment.
(d) O and N are end points of line segment.
(e) M is one of the end points of line segment.
(f) M is point on ray.
(g) Ray is different from ray.
(h) Ray is same as ray.
(i) Ray is not opposite to ray.
(j) O is not an initial point of.
(k) N is the initial point of and.
(a) True
(b) True
(c) True
(d) False
(e) False
(f) False
(g) True
(h) False
(i) False
(j) False
(k) True
#### Question 1:
Classify the following curves as (i) Open or (ii) Closed.
(a) Open
(b) Closed
(c) Open
(d) Closed
(e) Closed
#### Question 2:
Draw rough diagrams to illustrate the following:
(a) Open curve (b) Closed curve.
(a) Open curve
(b) Closed curve
#### Question 3:
Draw any polygon and shade its interior.
#### Question 4:
Consider the given figure and answer the questions:
(a) Is it a curve? (b) Is it closed?
(a) Yes
(b) Yes
#### Question 5:
Illustrate, if possible, each one of the following with a rough diagram:
(a) A closed curve that is not a polygon.
(b) An open curve made up entirely of line segments.
(c) A polygon with two sides.
#### Answer:
(a)
(b)
(c) This is not possible as the polygon having the least number of sides is a triangle, which has three sides in it.
#### Question 1:
Name the angles in the given figure.
#### Answer:
∠BAD, ∠ADC, ∠DCB, ∠CBA
#### Question 2:
In the given diagram, name the point (s)
(a) In the interior of ∠DOE
(b) In the exterior of ∠EOF
(c) On ∠EOF
(a) A
(b) C, A, D
(c) B, E, O, F
#### Question 3:
Draw rough diagrams of two angles such that they have
(a) One point in common.
(b) Two points in common.
(c) Three points in common.
(d) Four points in common.
(e) One ray in common.
#### Answer:
(a)
∠COD and ∠AOB have point O in common.
(b)
∠AOB and ∠BOC have points O and B in common.
(c)
∠AOB and ∠BOC have points O, E, B in common.
(d)
∠BOA and ∠COA have points O, E, D, A in common.
(e)
Ray is common between ∠BOC and ∠AOC.
#### Question 1:
Draw a rough sketch of a triangle ABC. Mark a point P in its interior and a point Q in its exterior. Is the point A in its exterior or in its interior?
#### Answer:
Point A lies on the given ΔABC.
#### Question 2:
(a) Identify three triangles in the figure.
(b) Write the names of seven angles.
(c) Write the names of six line segments.
(d) Which two triangles have ∠B as common?
#### Answer:
(a) ΔABC, ΔACD, ΔADB
(b) ∠ABC, ∠ADB, ∠ADC, ∠ACB, ∠BAD, ∠CAD, ∠BAC
(c)
(d) ΔABD and ΔABC
​​
##### Video Solution for Basic Geometrical Ideas (Page: 81 , Q.No.: 2)
NCERT Solution for Class 6 math - Basic Geometrical Ideas 81 , Question 2
#### Question 1:
Draw a rough sketch of a quadrilateral PQRS. Draw its diagonals. Name them. Is the meeting point of the diagonals in the interior or exterior of the quadrilateral?
#### Answer:
Diagonals are PR and QS. They meet at point O which is in the interior of &mnSq1PQRS.
#### Question 2:
Draw a rough sketch of a quadrilateral KLMN. State,
(a) Two pairs of opposite sides,
(b) Two pairs of opposite angles,
(c) Two pairs of adjacent sides,
(d) Two pairs of adjacent angles.
#### Answer:
(a)
(b) ∠KLM and ∠KNM
∠LKN and ∠LMN
(c)
(d) ∠K, ∠L and ∠M, ∠N
∠K, ∠N and ∠L, ∠M
​​
##### Video Solution for Basic Geometrical Ideas (Page: 82 , Q.No.: 2)
NCERT Solution for Class 6 math - Basic Geometrical Ideas 82 , Question 2
#### Question 1:
From the figure, identify:
(a) The centre of circle (e) Two points in the interior
(b) Three radii (f) a point in the exterior
(c) a diameter (g) a sector
(d) a chord (h) a segment
#### Answer:
(a) O
(b)
(c)
(d)
(e) O, P
(f) Q
(g) AOB (shaded region)
(h) DE (shaded region)
​​
##### Video Solution for Basic Geometrical Ideas (Page: 84 , Q.No.: 1)
NCERT Solution for Class 6 math - Basic Geometrical Ideas 84 , Question 1
#### Question 2:
(a) Is every diameter of a circle also a chord?
(b) Is every chord of circle also a diameter?
#### Answer:
(a) Yes. The diameter is the longest possible chord of the circle.
(b) No
#### Question 3:
Draw any circle and mark
(a) Its centre (e) a segment
(b) a radius (f) a point in its interior
(c) a diameter (g) a point in its exterior
(d) a sector (h) an arc
#### Answer:
(a) O
(b)
(c)
(d) COA
(e) DE
(f) O
(g) F
(h)
​​
##### Video Solution for Basic Geometrical Ideas (Page: 84 , Q.No.: 3)
NCERT Solution for Class 6 math - Basic Geometrical Ideas 84 , Question 3
#### Question 4:
Say true or false:
(a) Two diameters of a circle will necessarily intersect.
(b) The centre of a circle is always in its interior.
#### Answer:
(a) True. They will always intersect each other at the centre of the circle.
(b) True
View NCERT Solutions for all chapters of Class 6
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# Intersecting Secants Theorem
When two secant lines intersect each other outside a circle, the products of their segments are equal.
(Note: Each segment is measured from the outside point)
Try this In the figure below, drag the orange dots around to reposition the secant lines. You can see from the calculations that the two products are always the same. (Note: Because the lengths are rounded to one decimal place for clarity, the calculations may come out slightly differently on your calculator.)
This theorem works like this: If you have a point outside a circle and draw two secant lines (PAB, PCD) from it, there is a relationship between the line segments formed. Refer to the figure above. If you multiply the length of PA by the length of PB, you will get the same result as when you do the same thing to the other secant line.
More formally: When two secant lines AB and CD intersect outside the circle at a point P, then
PA.PB = PC.PD
It is important to get the line segments right. The four segments we are talking about here all start at P, and some overlap each other along part of their length; PA overlaps PB, and PC overlaps PD.
## Relationship to Tangent-Secant Theorem
In the figure above, drag point B around the top until it meets point A. The line is now a tangent to the circle, and PA=PB. Since PA=PB, then their product is equal to PA2. So:
PA2 = PC.PD
This is the Tangent-Secant Theorem.
## Relationship to Tangent Theorems
If you move point B around until it overlaps A, the resulting tangent has a length equal to PA2. Similarly, if you drag D around the bottom to point C, the that tangent has a length of PC2. From the this theorem
PA2 = PC2
By taking the square root of each side:
PA = PC
confirming that the two tangents froma point to a circle are always equal.
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Enter both numbers a and b into the calculator to determine the sum, and this calculator can also evaluate any of the variables given the others are known.
## Diamond Problem Formula
The following formula is used to calculate the sum of two numbers in a diamond problem:
Sum = frac{a + b}{2}
Variables:
• Sum is the sum of the two numbers
• a is one of the numbers in the diamond problem
• b is the other number in the diamond problem
To calculate the sum of two numbers in a diamond problem, add the two numbers together and divide the result by 2.
## What is a Diamond Problem?
The Diamond Problem is a term used in computer science, specifically in object-oriented programming, to describe an ambiguity that arises when a class inherits from two or more classes that have a common superclass. This problem is named as such because the class inheritance diagram resembles a diamond shape. The issue arises when a method is invoked and it is unclear which parent class’s method should be accessed, leading to potential confusion and unexpected results. This problem is particularly prevalent in languages that support multiple inheritance, such as C++. Some languages, like Java, avoid the Diamond Problem by not supporting multiple inheritance, while others, like Python, provide explicit ways to control the order of inheritance and thus avoid the ambiguity.
## How to Calculate Diamond Problem?
The following steps outline how to solve a Diamond Problem:
1. First, identify the two numbers that form the top and bottom of the diamond.
2. Next, multiply the two numbers together to find the product.
3. Then, find two numbers that multiply to give the product and add up to the number in the middle of the diamond.
4. Finally, write the two numbers found in the previous step as the left and right sides of the diamond.
Example Problem:
Let’s solve the following Diamond Problem:
Product: 24
Sum: 10
Step 1: Identify the top and bottom numbers.
Top: ?
Bottom: ?
Step 2: Multiply the top and bottom numbers to find the product.
? x ? = 24
Step 3: Find two numbers that multiply to give the product and add up to the sum.
2 x 12 = 24
2 + 12 = 14
Step 4: Write the two numbers as the left and right sides of the diamond.
2 | 12
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877-542-5504 877-542-5504
Probability: The Study of Chance Lesson Plan
Want to Help Fellow Teachers?
Lesson Plan #: AELP-PRB0004
AUTHOR: Shirley LeMoine , Garfield Re-2 School District, Rifle, Co.
Date: 1994
Grade Level(s): 5, 6, 7, 8, 9, 10, 11, 12
Subject(s):
• Mathematics/Probability
OVERVIEW:
The theory of probability is an important branch of mathematics with many practical applications in the physical, medical, biological, and social sciences. An understanding of this theory is essential to understand weather reports, medical findings, political doings and the state lotteries. Students have many misconceptions about probability situations.
PURPOSE:
The purpose of this activity is to begin the process of helping students to learn the basic principles of probability.
OBJECTIVES:
As a result of this activity the student will:
• conduct an experiment
• determine if a game is fair
• collect data (table)
• interpret data ( range, mode, median)
• display data (line graph)
• conduct analysis of game ( tree diagram)
• state and apply the rule (definition) for probability
• RESOURCES AND MATERIALS:
• pencils
• paper
ACTIVITIES AND PROCEDURES:
• introduce activity with a demonstration of game: rock, scissors, paper.
• divide class into pairs (player A and player B) and have them play the game 18 times.
• use overhead graph grid to graph the wins of player A in red (how many A players won one game, two games etc.) Do the same for all B players in a different color.
• Help students determine range, mode and mean for each set of data. Compare the results.
• Do a tree diagram to determine the possible outcomes.
• Answer the following questions to determine if the game is fair.
• How many outcomes does game have ? (9)
• Label each possible outcome on tree diagram as to win for A, B or tie.
• Count wins for A (3)
• Find probability A will win in any round (3/9=1/3) Explain what probability means favorable outcomes/ possible outcomes
• Count wins for B (3)
• Find probability B will win in any round (3/9)
• Is game fair? Do both players have an equal probability of winning in any round? (yes)
• Compare the mathematical model with what happened when the students played the game.
• TYING IT ALL TOGETHER:
• Use this as an introduction to a unit on probability.
• Follow-up with discussion about how probability is used in world.
• Play game again using 3 students. Using the following rules:
• A wins if all 3 hands are same.
• B wins if all 3 hands are different.
• C wins if 2 hands are same.
• There will be 27 outcomes this time. 3 to the third power. 3*3*3=27
May 1994 These lesson plans are the result of the work of the teachers who have attended the Columbia Education Center’s Summer Workshop. CEC is a consortium of teacher from 14 western states dedicated to improving the quality of education in the rural, western, United States, and particularly the quality of math and science Education. CEC uses Big Sky Telegraph as the hub of their telecommunications network that allows the participating teachers to stay in contact with their trainers and peers that they have met at the Workshops.
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## Section7.3Principal Component Analysis
We are sometimes presented with a dataset having many data points that live in a high dimensional space. For instance, we looked at a dataset describing body fat index (BFI) in Activity 6.5.4 where each data point is six-dimensional. Developing an intuitive understanding of the data is hampered by the fact that it cannot be visualized.
This section explores a technique called principal component analysis, which enables us to reduce the dimension of a dataset so that it may be visualized or studied in a way so that interesting features more readily stand out. Our previous work with variance and the orthogonal diagonalization of symmetric matrices provides the key ideas.
### Preview Activity7.3.1.
We will begin by recalling our earlier discussion of variance. Suppose we have a dataset that leads to the covariance matrix
\begin{equation*} C = \begin{bmatrix} 7 \amp -4 \\ -4 \amp 13 \end{bmatrix}. \end{equation*}
1. Suppose that $$\uvec$$ is a unit eigenvector of $$C$$ with eigenvalue $$\lambda\text{.}$$ What is the variance $$V_{\uvec}$$ in the $$\uvec$$ direction?
2. Find an orthogonal diagonalization of $$C\text{.}$$
3. What is the total variance?
4. In which direction is the variance greatest and what is the variance in this direction? If we project the data onto this line, how much variance is lost?
5. In which direction is the variance smallest and how is this direction related to the direction of maximum variance?
Here are some ideas we've seen previously that will be particularly useful for us in this section. Remember that the covariance matrix of a dataset is $$C=\frac 1N AA^T$$ where $$A$$ is the matrix of $$N$$ demeaned data points.
• When $$\uvec$$ is a unit vector, the variance of the demeaned data after projecting onto the line defined by $$\uvec$$ is given by the quadratic form $$V_{\uvec} = \uvec\cdot(C\uvec)\text{.}$$
• In particular, if $$\uvec$$ is a unit eigenvector of $$C$$ with associated eigenvalue $$\lambda\text{,}$$ then $$V_{\uvec} = \lambda\text{.}$$
• Moreover, variance is additive, as we recorded in Proposition 7.1.16: if $$W$$ is a subspace having an orthonormal basis $$\uvec_1,\uvec_2,\ldots,\uvec_n\text{,}$$ then the variance
\begin{equation*} V_W = V_{\uvec_1} + V_{\uvec_2} + \ldots + V_{\uvec_n}\text{.} \end{equation*}
### Subsection7.3.1Principal Component Analysis
Let's begin by looking at an example that illustrates the central theme of this technique.
#### Activity7.3.2.
Suppose that we work with a dataset having 100 five-dimensional data points. The demeaned data matrix $$A$$ is therefore $$5\times100$$ and leads to the covariance matrix $$C=\frac1{100}~AA^T\text{,}$$ which is a $$5\times5$$ matrix. Because $$C$$ is symmetric, the Spectral Theorem tells us it is orthogonally diagonalizable so suppose that $$C = QDQ^T$$ where
\begin{equation*} Q = \begin{bmatrix} \uvec_1 \amp \uvec_2 \amp \uvec_3 \amp \uvec_4 \amp \uvec_5 \end{bmatrix},\hspace{24pt} D = \begin{bmatrix} 13 \amp 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 10 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 2 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{bmatrix}. \end{equation*}
1. What is $$V_{\uvec_2}\text{,}$$ the variance in the $$\uvec_2$$ direction?
2. Find the variance of the data projected onto the line defined by $$\uvec_4\text{.}$$ What does this say about the data?
3. What is the total variance of the data?
4. Consider the 2-dimensional subspace spanned by $$\uvec_1$$ and $$\uvec_2\text{.}$$ If we project the data onto this subspace, what fraction of the total variance is represented by the variance of the projected data?
5. How does this question change if we project onto the 3-dimensional subspace spanned by $$\uvec_1\text{,}$$ $$\uvec_2\text{,}$$ and $$\uvec_3\text{?}$$
6. What does this tell us about the data?
This activity demonstrates how the eigenvalues of the covariance matrix can tell us when data are clustered around, or even wholly contained within, a smaller dimensional subspace. In particular, the original data is 5-dimensional, but we see that it actually lies in a 3-dimensional subspace of $$\real^5\text{.}$$ Later in this section, we'll see how to use this observation to work with the data as if it were three-dimensional, an idea known as dimensional reduction.
The eigenvectors $$\uvec_j$$ of the covariance matrix are called principal components, and we will order them so that their associated eigenvalues decrease. Generally speaking, we hope that the first few principal components retain most of the variance, as the example in the activity demonstrates. In that example, we have the sequence of subspaces
• $$W_1\text{,}$$ the 1-dimensional subspace spanned by $$\uvec_1\text{,}$$ which retains $$13/25 = 52\%$$ of the total variance,
• $$W_2\text{,}$$ the 2-dimensional subspace spanned by $$\uvec_1$$ and $$\uvec_2\text{,}$$ which retains $$23/25 = 92\%$$ of the variance, and
• $$W_3\text{,}$$ the 3-dimensional subspace spanned by $$\uvec_1\text{,}$$ $$\uvec_2\text{,}$$ and $$\uvec_3\text{,}$$ which retains all of the variance.
Notice how we retain more of the total variance as we increase the dimension of the subspace onto which the data are projected. Eventually, projecting the data onto $$W_3$$ retains all the variance, which tells us the data must lie in $$W_3\text{,}$$ a smaller dimensional subspace of $$\real^5\text{.}$$
In fact, these subspaces are the best possible. We know that the first principal component $$\uvec_1$$ is the eigenvector of $$C$$ associated to the largest eigenvalue. This means that the variance is as large as possible in the $$\uvec_1$$ direction. In other words, projecting onto any other line will retain a smaller amount of variance. Similarly, projecting onto any other 2-dimensional subspace besides $$W_2$$ will retain less variance than projecting onto $$W_2\text{.}$$ The principal components have the wonderful ability to pick out the best possible subspaces to retain as much variance as possible.
Of course, this is a contrived example. Typically, the presence of noise in a dataset means that we do not expect all the points to be wholly contained in a smaller dimensional subspace. In fact, the 2-dimensional subspace $$W_2$$ retains $$92\%$$ of the variance. Depending on the situation, we may want to write off the remaining $$8\%$$ of the variance as noise in exchange for the convenience of working with a smaller dimensional subspace. As we'll see later, we will seek a balance using a number of principal components large enough to retain most of the variance but small enough to be easy to work with.
#### Activity7.3.3.
We will work here with a dataset having 100 3-dimensional demeaned data points. Evaluating the following cell will plot those data points and define the demeaned data matrix A whose shape is $$3\times100\text{.}$$
sage.repl.load.load('https://raw.githubusercontent.com/davidaustinm/ula_modules/master/pca_demo.py', globals())
Notice that the data appears to cluster around a plane though it does not seem to be wholly contained within that plane.
1. Use the matrix A to construct the covariance matrix $$C\text{.}$$ Then determine the variance in the direction of $$\uvec=\threevec{1/3}{2/3}{2/3}\text{?}$$
2. Find the eigenvalues of $$C$$ and determine the total variance.
Notice that Sage does not necessarily sort the eigenvalues in decreasing order.
3. Use the right_eigenmatrix() command to find the eigenvectors of $$C\text{.}$$ Remembering that the Sage command B.column(1) retrieves the vector represented by the second column of B, define vectors u1, u2, and u3 representing the three principal components in order of decreasing eigenvalues. How can you check if these vectors are an orthonormal basis for $$\real^3\text{?}$$
4. What fraction of the total variance is retained by projecting the data onto $$W_1\text{,}$$ the subspace spanned by $$\uvec_1\text{?}$$ What fraction of the total variance is retained by projecting onto $$W_2\text{,}$$ the subspace spanned by $$\uvec_1$$ and $$\uvec_2\text{?}$$ What fraction of the total variance do we lose by projecting onto $$W_2\text{?}$$
5. If we project a data point $$\xvec$$ onto $$W_2\text{,}$$ the Projection Formula tells us we obtain
\begin{equation*} \xhat = (\uvec_1\cdot\xvec) \uvec_1 + (\uvec_2\cdot\xvec) \uvec_2. \end{equation*}
Rather than viewing the projected data in $$\real^3\text{,}$$ we will record the coordinates of $$\xhat$$ in the basis defined by $$\uvec_1$$ and $$\uvec_2\text{;}$$ that is, we will record the coordinates
\begin{equation*} \twovec{\uvec_1\cdot\xvec}{\uvec_2\cdot\xvec}. \end{equation*}
Construct the matrix $$Q$$ so that $$Q^T\xvec = \twovec{\uvec_1\cdot\xvec}{\uvec_2\cdot\xvec}\text{.}$$
6. Since each column of $$A$$ represents a data point, the matrix $$Q^TA$$ represents the coordinates of the projected data points. Evaluating the following cell will plot those projected data points.
pca_plot(Q.T*A)
Notice how this plot enables us to view the data as if it were two-dimensional. Why is this plot wider than it is tall?
This example is a more realistic illustration of principal component analysis. The plot of the 3-dimensional data appears to show that the data lies close to a plane, and the principal components will identify this plane. Starting with the $$3\times100$$ matrix of demeaned data $$A\text{,}$$ we construct the covariance matrix $$C=\frac{1}{100} ~AA^T$$ and study its eigenvalues. Notice that the first two principal components account for more than 98% of the variance, which means we can expect the points to lie close to $$W_2\text{,}$$ the two-dimensional subspace spanned by $$\uvec_1$$ and $$\uvec_2\text{.}$$
Since $$W_2$$ is a subspace of $$\real^3\text{,}$$ projecting the data points onto $$W_2$$ gives a list of 100 points in $$\real^3\text{.}$$ In order to visualize them more easily, we instead consider the coordinates of the projections in the basis defined by $$\uvec_1$$ and $$\uvec_2\text{.}$$ For instance, we know that the projection of a data point $$\xvec$$ is
\begin{equation*} \xhat = (\uvec_1\cdot\xvec)\uvec_1 + (\uvec_2\cdot\xvec)\uvec_2, \end{equation*}
which is a three-dimensional vector. Instead, we can record the coordinates $$\twovec{\uvec_1\cdot\xvec}{\uvec_2\cdot\xvec}$$ and plot them in the two-dimensional coordinate plane, as illustrated in Figure 7.3.1.
If we form the matrix $$Q=\begin{bmatrix}\uvec_1 \amp \uvec_2 \end{bmatrix}\text{,}$$ then we have
\begin{equation*} Q^T\xvec = \twovec{\uvec_1\cdot\xvec}{\uvec_2\cdot\xvec}. \end{equation*}
This means that the columns of $$Q^TA$$ represent the coordinates of the projected points, which may now be plotted in the plane.
In this plot, the first coordinate, represented by the horizontal coordinate, represents the projection of a data point onto the line defined by $$\uvec_1$$ while the second coordinate represents the projection onto the line defined by $$\uvec_2\text{.}$$ Since $$\uvec_1$$ is the first principal component, the variance in the $$\uvec_1$$ direction is greater than the variance in the $$\uvec_2$$ direction. For this reason, the plot will be more spread out in the horizontal direction than in the vertical.
### Subsection7.3.2Using Principal Component Analysis
Now that we've explored the ideas behind principal component analysis, we will look at a few examples that illustrate its use.
#### Activity7.3.4.
The next cell will load a dataset describing the average consumption of various food groups for citizens in each of the four nations of the United Kingdom. The units for each entry are grams per person per week.
import pandas as pd
data_mean = vector(df.T.mean())
A = matrix([vector(row) for row in (df.T-df.T.mean()).values]).T
df
We will view this as a dataset consisting of four points in $$\real^{17}\text{.}$$ As such, it is impossible to visualize and studying the numbers themselves doesn't lead to much insight.
In addition to loading the data, evaluating the cell above created a vector data_mean, which is the mean of the four data points, and A, the $$17\times4$$ matrix of demeaned data.
1. What is the average consumption of Beverages across the four nations?
2. Find the covariance matrix $$C$$ and its eigenvalues. Because there are four points in $$\real^{17}$$ whose mean is zero, there are only three nonzero eigenvalues.
3. For what percentage of the total variance does the first principal component account?
4. Find the first principal component $$\uvec_1$$ and project the four demeaned data points onto the line defined by $$\uvec_1\text{.}$$ Plot those points on Figure 7.3.2
5. For what percentage of the total variance do the first two principal components account?
6. Find the coordinates of the demeaned data points projected onto $$W_2\text{,}$$ the two-dimensional subspace of $$\real^{17}$$ spanned by the first two principal components.
Plot these coordinates in Figure 7.3.3.
7. What information do these plots reveal that is not clear from consideration of the original data points?
8. Study the first principal component $$\uvec_1$$ and find the first component of $$\uvec_1\text{,}$$ which corresponds to the dietary category Alcoholic Drinks. (To do this, you may wish to use N(u1, digits=2) for a result that's easier to read.) If a data point lies on the far right side of the plot in Figure 7.3.3, what does it mean about that nation's consumption of Alcoholic Drinks?
This activity demonstrates how principal component analysis enables us to extract information from a dataset that may not be easily obtained otherwise. As in our previous example, we see that the data points lie quite close to a two-dimensional subspace of $$\real^{17}\text{.}$$ In fact, $$W_2\text{,}$$ the subspace spanned by the first two principal components, accounts for more than 96% of the variance. More importantly, when we project the data onto $$W_2\text{,}$$ it becomes apparent that Northern Ireland is fundamentally different from the other three nations.
With some additional thought, we can determine more specific ways in which Northern Ireland is different. On the $$2$$-dimensional plot, Northern Ireland lies far to the right compared to the other three nations. Since the data has been demeaned, the origin $$(0,0)$$ in this plot corresponds to the average of the four nations. The coordinates of the point representing Northern Ireland are about $$(477, 59)\text{,}$$ meaning that the projected data point differs from the mean by about $$477\uvec_1+59\uvec_2\text{.}$$
Let's just focus on the contribution from $$\uvec_1\text{.}$$ We see that the ninth component of $$\uvec_1\text{,}$$ the one that describes Fresh Fruit, is about $$-0.63\text{.}$$ This means that the ninth component of $$477\uvec_1$$ differs from the mean by about $$477(-0.63) = -300$$ grams per person per week. So roughly speaking, people in Northern Ireland are eating about 300 fewer grams of Fresh Fruit than the average across the four nations. This is borne out by looking at the original data, which show that the consumption of Fresh Fruit in Northern Ireland is significantly less than the other nations. Examing the other components of $$\uvec_1$$ shows other ways in which Northern Ireland differs from the other three nations.
#### Activity7.3.5.
In this activity, we'll look at a well-known dataset 1 that describes 150 irises representing three species of iris: Iris setosa, Iris versicolor, and Iris virginica. For each flower, the length and width of its sepal and the length and width of its petal, all in centimeters, are recorded.
Evaluating the following cell will load the dataset, which consists of 150 points in $$\real^4\text{.}$$ In addition, we have a vector data_mean, a four-dimensional vector holding the mean of the data points, and A, the $$4\times150$$ demeaned data matrix.
sage.repl.load.load('https://raw.githubusercontent.com/davidaustinm/ula_modules/master/pca_iris.py', globals())
df.T
Since the data is four-dimensional, we are not able to visualize it. Of course, we could forget about two of the measurements and plot the 150 points represented by their, say, sepal length and sepal width.
sepal_plot()
1. What is the mean sepal width?
2. Find the covariance matrix $$C$$ and its eigenvalues.
3. Find the fraction of variance for which the first two principal components account.
4. Construct the first two principal components $$\uvec_1$$ and $$\uvec_2$$ along with the matrix $$Q$$ whose columns are $$\uvec_1$$ and $$\uvec_2\text{.}$$
5. As we have seen, the columns of the matrix $$Q^TA$$ hold the coordinates of the demeaned data points after projecting onto $$W_2\text{,}$$ the subspace spanned by the first two principal components. Evaluating the following cell shows a plot of these coordinates.
pca_plot(Q.T*A)
Suppose we have a flower whose coordinates in this plane are $$(-2.5, -0.75)\text{.}$$ To what species does this iris most likely belong? Find an estimate of the sepal length, sepal width, petal length, and petal width for this flower.
6. Suppose you have an iris, but you only know that its sepal length is 5.65 cm and its sepal width is 2.75 cm. Knowing only these two measurements, determine the coordinates $$(c_1, c_2)$$ in the plane where this iris lies. To what species does this iris most likely belong? Now estimate the petal length and petal width of this iris.
7. Suppose you find another iris whose sepal width is 3.2 cm and whose petal width is 2.2 cm. Find the coordinates $$(c_1, c_2)$$ of this iris and determine the species to which it most likely belongs. Also, estimate the sepal length and the petal length.
### Subsection7.3.3Summary
This section has explored principal component analysis as a technique to reduce the dimension of a dataset. From the demeaned data matrix $$A\text{,}$$ we form the covariance matrix $$C= \frac1N ~AA^T\text{,}$$ where $$N$$ is the number of data points.
• The eigenvectors $$\uvec_1, \uvec_2, \ldots \uvec_m\text{,}$$ of $$C$$ are called the principal components. We arrange them so that their corresponding eigenvalues are in decreasing order.
• If $$W_n$$ is the subspace spanned by the first $$n$$ principal components, then the variance of the demeaned data projected onto $$W_n$$ is the sum of the first $$n$$ eigenvalues of $$C\text{.}$$ No other $$n$$-dimensional subspace retains more variance when the data is projected onto it.
• If $$Q$$ is the matrix whose columns are the first $$n$$ principal components, then the columns of $$Q^TA$$ hold the coordinates, expressed in the basis $$\uvec_1,\ldots,\uvec_n\text{,}$$ of the data once projected onto $$W_n\text{.}$$
• Our goal is to use a number of principal components that is large enough to retain most of the variance in the dataset but small enough to be manageable.
### Exercises7.3.4Exercises
#### 1.
Suppose that
\begin{equation*} Q = \begin{bmatrix} -1/\sqrt{2} \amp 1/\sqrt{2} \\ 1/\sqrt{2} \amp 1/\sqrt{2} \\ \end{bmatrix}, \hspace{24pt} D_1 = \begin{bmatrix} 75 \amp 0 \\ 0 \amp 74 \end{bmatrix}, \hspace{24pt} D_2 = \begin{bmatrix} 100 \amp 0 \\ 0 \amp 1 \end{bmatrix} \end{equation*}
and that we have two datasets, one whose covariance matrix is $$C_1 = QD_1Q^T$$ and one whose covariance matrix is $$C_2 = QD_2Q^T\text{.}$$ For each dataset, find
1. the total variance.
2. the fraction of variance represented by the first principal component.
3. a verbal description of how the demeaned data points appear when plotted in the plane.
#### 2.
Suppose that a dataset has mean $$\threevec{13}{5}{7}$$ and that its associated covariance matrix is $$C=\begin{bmatrix} 275 \amp -206 \amp 251 \\ -206 \amp 320 \amp -206 \\ 251 \amp -206 \amp 275 \end{bmatrix} \text{.}$$
1. What fraction of the variance is represented by the first two principal components?
2. If $$\threevec{30}{-3}{26}$$ is one of the data points, find the coordinates when the demeaned point is projected into the plane defined by the first two principal components.
3. If a projected data point has coordinates $$\twovec{12}{-25}\text{,}$$ find an estimate for the original data point.
#### 3.
Evaluating the following cell loads a $$2\times100$$ demeaned data matrix A.
sage.repl.load.load('https://raw.githubusercontent.com/davidaustinm/ula_modules/master/pca_ex.py', globals())
1. Find the principal components $$\uvec_1$$ and $$\uvec_2$$ and the variance in the direction of each principal component.
2. What is the total variance?
#### 4.
Determine whether the following statements are true or false and explain your thinking.
1. If the eigenvalues of the covariance matrix are $$\lambda_1\text{,}$$ $$\lambda_2\text{,}$$ and $$\lambda_3\text{,}$$ then $$\lambda_3$$ is the variance of the demeaned data points when projected on the third principal component $$\uvec_3\text{.}$$
2. Principal component analysis always allows us to construct a smaller dimensional representation of a dataset without losing any information.
3. If the eigenvalues of the covariance matrix are 56, 32, and 0, then the demeaned data points all lie on a line in $$\real^3\text{.}$$
#### 5.
In Activity 7.3.5, we looked at a dataset consisting of four measurements of 150 irises. These measurements are sepal length, sepal width, petal length, and petal width.
1. Find the first principal component $$\uvec_1$$ and describe the meaning of its four components. Which component is most significant? What can you say about the relative importance of the four measurements?
2. When the dataset is plotted in the plane defined by $$\uvec_1$$ and $$\uvec_2\text{,}$$ the specimens from the species iris-setosa lie on the left side of the plot. What does this tell us about how iris-setosa differs from the other two species in the four measurements?
3. In general, which species is closest to the “average iris”?
#### 6.
This problem explores a dataset describing 333 penguins. There are three species, Adelie, Chinstrap, and Gentoo, as illustrated on the left of Figure 7.3.9, as well as both male and female penguins in the dataset.
Evaluating the next cell will load and display the data. The meaning of the culmen length and width is contained in the illustration on the right of Figure 7.3.9.
sage.repl.load.load('https://raw.githubusercontent.com/davidaustinm/ula_modules/master/pca_penguins.py', globals())
df.T
This dataset is a bit different from others that we've looked at because the scale of the measurements is significantly different. For instance, the measurements for the body mass are roughly 100 times as large as those for the culmen length. For this reason, we will standardize the data by first demeaning it, as usual, and then rescaling each measurement by the reciprocal of its standard deviation. The result is stored in the $$4\times333$$ matrix A.
1. Find the covariance matrix and its eigenvalues.
2. What fraction of the total variance is explained by the first two principal components?
3. Construct the $$2\times333$$ matrix $$B$$ whose columns are the coordinates of the demeaned data points projected onto the first two principal components. The following cell will create the plot.
pca_plot(B)
4. Examine the components of the first two principal component vectors. How does the body mass of Gentoo penguins compare to that of the other two species?
5. What seems to be generally true about the culmen measurements for a Chinstrap penguin compared to a Adelie?
6. You can plot just the males or females using the following cell.
pca_plot(B, sex='female')
What seems to be generally true about the body mass measurements for a male Gentoo compared to a female Gentoo?
archive.ics.uci.edu
gvsu.edu/s/21D
gvsu.edu/s/21E
gvsu.edu/s/21G
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-1
# ab and ac are tangent to P find ab
ab= 2
ab= 11/2
ab= 1/2
ab=10
in the actual question, AC= 11y
AB= 3y+4
### 2 Answers by Expert Tutors
Rachelle W. | Professional TutorProfessional Tutor
5.0 5.0 (6 lesson ratings) (6)
-1
Two tangents to a circle from the same point outside the circle will always be equal. We set the two expressions equal to each other and solve for y.
11y = 3y + 4
8y = 4
y = 1/2
Once we've found that y = 1/2, we substitute that value back into the original expression for AB and solve.
AB = 3y + 4
AB = 3(1/2) + 4
AB = (3/2) + 4
AB = 11/2
Oops. 8y = 4 translats to y = 1/2
Tamara J. | Math Tutoring - Algebra and Calculus (all levels)Math Tutoring - Algebra and Calculus (al...
4.9 4.9 (51 lesson ratings) (51)
-1
If the two segments are tangent to the same point, then they are congruent (i.e., ab ~ ac).
You are given the following:
ac = 11y and ab = 3y + 4
Since we've already determined that the two segments (ab and ac) and congruent, we can set the expressions that define their lengths equal to one another:
11y = 3y + 4
Now we solve for the unknown variable (y) by first subtracting 3y from both sides of the equation then dividing both sides by the coefficient of y:
11y - 3y = 3y - 3y + 4
8y = 4
8y/8 = 4/8
y = 1/2
To find ab, we plug in the solution for y into the expression defining ab:
ab = 3y + 4
ab = 3(1/2) + 4
= 3/2 + 4
Since one of the terms is a fraction, we need to find a common denominator among the 2 terms so that we can add them. To do so, multiply 4 by 2/2 to get 8/2.
ab = 3/2 + 4(2/2)
= 3/2 + 8/2
= (3+8)/2
ab = 11/2
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# Into Math Grade 3 Module 3 Lesson 3 Answer Key Multiply with 3 and 6
We included HMH Into Math Grade 3 Answer Key PDF Module 3 Lesson 3 Multiply with 3 and 6 to make students experts in learning maths.
## HMH Into Math Grade 3 Module 3 Lesson 3 Answer Key Multiply with 3 and 6
I Can use different strategies to multiply with the factors 3 and 6 and solve equal groups problems.
Your community wants to use wind turbines to bring electricity to homes. Each wind turbine has 3 blades. Choose any number of wind turbines for your community from 5 to 9. How many blades are there?
Show two different ways to find the number of blades.
There are ___ wind turbines with ___ blades on each.
Explanation:
I chose 5 wind turbines
Each wind turbine has 3 blades
There are 5 wind turbines with 3 blades on each
The multiplication equation that model the problem is 5 x 3 =15
Turn and Talk What multiplication equation represents this problem? What if one more wind turbine is added? What multiplication equation represents this problem now?
The multiplication equation that model the problem is 5 x 3 =15
If 1 more is added, the number of blades will be 18
The multiplication equation that model the problem will be 6 x 3 = 18.
Build Understanding
1. There are 4 wind turbines off the coast of a small town. Each wind turbine has 3 blades. How many blades are there on 4 wind turbines?
Count equal groups to find the number of blades. Show your work.
A. How many equal groups are there? _____
B. How many objects are in each group? _____
C. What multiplication equation can you write for the problem? _____
D. How many blades are there on 4 wind turbines? _____
Explanation:
There are 4 wind turbines off the coast of a small town
Each wind turbine has 3 blades
There are 4 equal groups
There are 3 objects in each group
The multiplication equation that solves the problem is 4 x 3 = 12
So, 4 turbines have 12 blades.
Drawing equal groups help to count easily by seeing the picture we can find the product of number of groups and number of objects in each group.
Step It Out
2. A wind farm has rows of wind turbines. The farm has 9 wind turbines in each row. How many wind turbines are there in 6 rows?
6 × 9 =
Use a 5s fact and addition to help.
A. Write a 5s fact with the factor 9. ____
B. Then add 1 more group of 9.
______________
C. Write a multiplication equation to model the problem.
______________
D. How many wind turbines are there in 6 rows?
______________
54 turbines
Explanation:
A wind farm has 6 rows of wind turbines
The farm has 9 wind turbines in each row
Use a 5s fact and addition to help
Write a 5’s fact with the factor 9
5 x 9 = 45
Add 1 more group of 9
45 + 9 = 54
The multiplication equation that model the problem is 6 x 9 = 54
So, there are 54 turbines in 6 rows.
3. Pedro is making 6 small pinwheels for a garden. Each pinwheel will have 8 blades. How many blades will be on 6 pinwheels?
6 × 8 =
Use a 3s fact and doubling to help.
A. Write a 3s fact with the factor 8. ____
B. Then double the product. ____
C. Write a multiplication equation to model the problem.
_______________
D. How many blades will be on 6 pinwheels?
________________
Explanation:
Pedro is making 6 small pinwheels for a garden
Each pinwheel will have 8 blades
Use a 3’s fact and doubling to help
Write a 3’s fact with the factor 8
3 x 8 = 24
Double the product
24 + 24 = 48
So, the multiplication equation that model the problem is 6 x 8 = 48
Therefore, there are 48 blades on 6 pinwheels.
4. Another wind farm has 6 rows of wind turbines. There are 4 wind turbines in each row. How many wind turbines are there in 6 rows?
A. Complete the multiplication table to help solve the problem. Use strategies you know to find the products for the 3s facts and the 6s facts.
B. Circle the product where the row for 6 and the column for 4 meet.
C. How many wind turbines are there in 6 rows?
_______________
24 turbines are in 6 rows
Explanation:
Another wind farm has 6 rows of wind turbines
There are 4 wind turbines in each row
I completed the multiplication table
I doubled the products of 3’s facts to find the products of 6’s facts
I drew a circle around the number 24 as it is the 6’s fact of the factor 4
6 x 4 = 24
So, The are 24 turbines in 6 rows.
Turn and Talk What do you notice about the products in the rows for 3 and 6?
The products in the rows for 6 are the doubles of the products of the rows for 3.
Check Understanding Math Board
Question 1.
Karina is making 6 small pinwheels for the community garden. Each pinwheel will have 7 blades. How many blades will be on 6 pinwheels?
______________
42 blades are on 6 pinwheels
Explanation:
Karina is making 6 small pinwheels for the community garden
Each pinwheel will have 7 blades
The multiplication equation that models the problem is
6 x 7 =
Use a 3’s fact and double the product
3’s fact of 7
3 x 7 = 21
Double the product
21 + 21 = 42
So, 6 x 7 = 42
Therefore, there are 42 blades on 7 pinwheels.
Use a 3s fact and doubles to find the product.
Question 2.
6 × 9 =
6 x 9 = 54
Explanation:
6 x 9
Use a 3’s fact and double the product
3 x 9 = 27
Double the product
27 + 27 = 54
The multiplication equation that model the problem is 6 x 9 = 54.
Use a 5s fact and addition to find the product.
Question 3.
6 × 8 =
6 x 8 = 48
Explanation:
6 x 8 =
Use a 5s fact and addition to help
Write a 5’s fact with the factor 8
5 x 9 = 40
Add 1 more group of 8
40 + 8 = 48
The multiplication equation that model the problem is 6 x 8 = 48.
Question 4.
Use Tools Michael buys 2 packages of hamburger buns. Each package has the number of buns shown. How many hamburger buns are in 2 packages?
• Show the equal groups.
• Write a multiplication equation for the problem.
_____________
There are ___ hamburger buns.
Explanation:
Michael buys 2 packages of hamburger buns
Each package has the 6 number of buns
I drew to show 2 equal groups of 6 objects
The multiplication equation that model the problem is 2 x 6 = 12
So, There are 12 hamburger buns.
Find the product using a 5s fact and addition.
Question 5.
6 × 7 =
6 x 7 = 42
Explanation:
6 x 7 =
Use a 5s fact and addition to help
Write a 5’s fact with the factor 7
5 x 7 = 35
Add 1 more group of 7
35 + 7 = 42
The multiplication equation that model the problem is 6 x 7 = 42.
Question 6.
6 × 6 =
Write a 5s fact with the factor 6.
5 × 6 = 30
Add 1 more group of 6.
30 + 6 = 36
The multiplication equation that model the problem is 6 x 6 = 36.
Find the product using a 3s fact and doubling.
Question 7.
6 × 4 =
6 x 4 = 24
Explanation:
6 x 9
Use a 3’s fact and double the product
3 x 4 = 12
Double the product
12 + 12 = 24
The multiplication equation that model the problem is 6 x 4 = 24.
Question 8.
6 × 6 =
6 x 6 = 36
Explanation:
6 x 6
Use a 3’s fact and double the product
3 x 6 = 18
Double the product
18 + 18 = 36
The multiplication equation that model the problem is 6 x 6 = 36.
Question 9.
Use Tools Diane has 4 rolls of fabric. She can make 3 costumes with each roll. How many costumes can Diane make? Show the equal groups. _____
Explanation:
Diane has 4 rolls of fabric. She can make 3 costumes with each roll
I drew 4 equal groups of 3 objects each
The multiplication equation is
4 x 3 =
Use a 2’s fact with factor 3 and double the product
2 x 3 = 6
6 + 6 = 12
So, 4 x 3 = 12
Therefore, Diane can make 12 costumes.
Question 10.
Use Structure Devon plants 6 rows of tomato plants in a garden. There are 4 plants in each row. How many tomato plants does Devon plant?
Given,
Devon plants 6 rows of tomato plants in a garden. There are 4 plants in each row.
6 × 4 = 24
Thus Devon plant 24 tomato plants.
Use the multiplication table to find the product.
Question 11.
3 × 9 = ___
27
Explanation:
3 x 9 = 27
The product of 3 and 9 is 27.
Question 12.
6 × 9 = ___
6 x 9 = 54
Explanation:
Use 3’s fact with 9
3 x 9 = 27
Double it
27 + 27 = 54
So, 6 x 9 = 54.
Question 13.
3 × 7 = ___
21
Explanation:
3 x 7 = 21
The product of 3 and 7 is 21.
Question 14.
6 × 7 = ___
6 x 7 = 42
Explanation:
Use 3’s fact with 7
3 x 7 = 21
Double it
21 + 21 = 42
So, 6 x 7 = 42.
Find the product.
Question 15.
3 × 8 = ___
24
Explanation:
3 x 8 = 24
The product of 3 and 8 is 24.
Question 16.
___ = 6 × 6
36
Explanation:
6 x 6
Use a 3’s fact and double the product
3 x 6 = 18
Double the product
18 + 18 = 36
So, 6 x 6 = 36.
Question 17.
8 × 6 = ___
48
Explanation:
8 x 6 =
Use a 3’s fact and double the product
8 x 3 = 24
Double the product
24 + 24 = 48
So, 8 x 6 = 48.
Question 18.
Explanation:
Any number multiplied by 10 will have 0 in ones place and the number itself in tens place
So, 3 x 10 = 30.
Question 19.
Explanation:
5 x 6
Use 3’s fact for 5 and double the product
5 x 3 = 15
15 + 15 = 30
So, 5 x 6 = 30.
Question 20.
Explanation:
4 x 3 =
Use a 2’s fact with 3 and double the product
2 x 3 = 6
6 + 6 = 12
So, 4 x 3 = 12.
Question 21.
Explanation:
6 x 7 =
Use a 3’s fact with 7 and double the product
3 x 7 = 21
21 + 21 = 42
So, 6 x 7 = 42.
Question 22.
Explanation:
9 x 6 =
Use a 3’s fact with 9 and double the product
9 x 3 = 27
27 + 27 = 54
So, 9 x 6 = 54.
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# Question: What Is 1 Through 50 Added Together?
## What is the sum of all odd numbers from 1 to 100?
Sum of odd consecutive integers from 1 to 100 = (Sum of all consecutive integers from 1 to 100) – (Sum of even consecutive integers from 1 to 100).
Sum of odds = (100 x 101/2) – [ (50 x 51)] = 5050 – 2550 = 2500..
## What is the sum of 1 to 50 numbers?
1275 is a sum of number series from 1 to 50 by applying the values of input parameters in the formula.
## How many 1’s are there from 1 to 100?
They are 10, 11,12, 13, 14, 15, 16, 17, 18, 19, 21, 31,41, 51, 61, 71, 81, 91. There are 18 not counting 1 and 100, because as per the wording of the question ‘between’ means 1 and 100 are not to be included. So, there are 21 ones from 1 to 100.
## What is Gauss formula?
Gauss’s method forms a general formula for the sum of the first n integers, namely that 1+2+3+\ldots +n=\frac{1}{2}n(n+1) One way of presenting Gauss’ method is to write out the sum twice, the second time reversing it as shown. If we add both rows we get the sum of 1 to n, but twice.
## What is the sum of 48?
What three consecutive integers have a sum of 48? Which means that the first number is 15, the second number is 15 + 1 and the third number is 15 + 2. Therefore, three consecutive integers that add up to 48 are 15, 16, and 17. We know our answer is correct because 15 + 16 + 17 equals 48 as displayed above.
## What 4 numbers make 100?
Using the numerals 1,7,7,7 and 7 (a “1” and four “7”s) create the number 100.
## How do you find the sum of integers from 1 to 50?
Solution: To find for the sum of arithmetic sequence of integers 1-50. d = 1 , solving for d – second term minus first term. = 1,275 is the sum of integers from 1-50.
## What is 1 200 added together?
The sum of the integers from 1 to n is given by n(n+1)/2. The sum of the integers from 1 to 200 = 200×201/2 = 20100.
## What is the sum of 1 to 40?
Any sum like this is easy. take the lowest and highest… 1 +40 = 41, then the next highest and the next lowest. 2 + 39 = 41.
## What is the sum of all odd numbers from 1 to 50?
Answer. =625. Therefore the sum of all odd numbers between 0 and 50 is 625.
## What is the sum of 51 to 100?
The number series 51, 52, 53, 54, . . . . , 99, 100. Therefore, 3775 is the sum of positive integers between 51 and 50.
## What is the sum of 1 to 365?
66795 is a sum of number series from 1 to 365 by applying the values of input parameters in the formula.
## What is 1 100 added up?
The sum of the numbers 1-100 would be equal to the number of pairs (50) multiplied by the sum of each pair (101), or 50 x 101 = 5,050.
## What is the sum of 1 to 75?
2850 is a sum of number series from 1 to 75 by applying the values of input parameters in the formula.
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# Algebraic Expressions Lesson 3 Episode 5 (Teachers)
### Reflecting
No captions Captions
Stop the video above first if it is playing.
The students work to explain the relationship between the two algebraic expressions they have written.
### Episode Supports
Students’ Conceptual Challenges
In this episode, ET and Haleemah are asked to use part of each of their two equations, 3c + 2c + 4c = T and (3 + 2 + 4)c = T, to create a new equation. They are not sure what to do [2:37]. Eventually ET writes the equation 9c = T, which was not formed using the other two equations [3:05]. The teacher helps them by selecting 3c + 2c + 4c from one of their equations and 9c from ET’s new equation. She then asks Haleemah and ET whether or not these two expressions are equal. ET thinks they are equal, but his explanation relies only on symbolic reasoning (namely that you add the 3 + 2 + 4 to get 9, and then you put the c back in next to the 9 at the end). Later in the episode, they check the equation 3c + 2c + 4c = 9c by letting c = 5. In Episode 6, Haleemah and ET make sense of the meaning of each part of the equation in the game app context.
Focus Questions
For use in a classroom, pause the video and ask these questions:
1. [Pause the video at 2:11] Ask your students to consider the task in front of Haleemah and ET—how might they combine parts of the equations (3 + 2 + 4)c = T and 3c + 2c + 4c = T to form a new equation that is also true.
2. [Pause the video at 3:41] ET has written a third equation, 9c = T. Ask your students if they can think of any new true equations they could make given this new equation.
3. [Pause the video at 4:44] If you posed Focus Question #2 and your students did not discuss this new equation (3c + 2c + 4c = 9c), ask them if they think this equation is true. Encourage them to explain their reasoning to each other, and then ask for volunteers to explain their reasoning to the class.
Supporting Dialogue
1. After watching the video, ask your students what they think it means for two equations to be equal. Highlight student ideas about equivalence. Perhaps students say that one side of one equation can be substituted for one side of the other equation and the other equation stays true. Perhaps they mention that both equations work the same way. Encourage them to explain their thinking.
2. If your students have created their own equations that look different from the ones in the video, you can ask them to display them on whiteboards or through a collaborative app (e.g., see www.Padlet.com). Then ask pairs of students to pick two or three equations and debate whether or not the equations are equivalent. Ask for volunteers to share their partner’s thinking in their own words.
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# NCERT solutions for class 9 Maths chapter 13 Surface Areas and Volumes (Exercise – 13.3)
## EX 13.2 QUESTION 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution:
Radius (r) = $\frac { 10.5 }{ 2 }$ cm = 5.25 cm
and slant height (l) =10 cm
Curved surface area of the cone = πrl
$\frac { 22 }{ 7 }$ x $\frac { 10.5 }{ 2 }$ x 10cm2
= 11 x 15 x 1 cm2 = 165cm2
## Ex 13.3 Question 2.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Radius (r) = $\frac { 24 }{ 2 }$ m = 12 m
and slant height (l) = 21 m
∴ Total surface area of a cone = πr(r +l)
$\frac { 22 }{ 7 }$ * 12 * (12 + 21 ) = = $\frac { 22 }{ 7 }$ * 12 * 33 = 1244.57 m2
## Ex 13.3 Question 3.
Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find
(i) radius of the base and
(ii) total surface area of the cone.
Solution:
Curved surface area = πrl = 308 cm2
Slant height (l) = 14 cm
(i) Let the radius of the base of cone = r cm
= 308 ⇒ $\frac { 22 }{ 7 }$ x r x 14 = 308
r = $\frac { 308\times 7 }{ 22\times 14 }$ = 7cm
Hence, the radius of the cone is 7 cm
(ii) Total surface area of the cone = πr(r + l)
$\frac { 22 }{ 7 }$ * 7 * (7 + 14) = 22 * 21 = 462 cm2
## Ex 13.3 Question 4.
A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ₹70.
Solution:
Height tent (h) = 10 m
Radius of the cone(r) = 24 m
(i) The slant height, l m
l2 = r2 + h2l2 = 242 + 102 = 576 + 100= 676
l =$\sqrt { 676 }$ m = 26m
(ii) Curved surface area of the cone = πrl
= $\frac { 22 }{ 7 }$ * 24 * 26 m2
Cost of 1m2 canvas =₹70
∴ The cost of $\frac { 22 }{ 7 }$ * 24 * 26 m2 canvas = 70 * $\frac { 22 }{ 7 }$ * 24 * 26 = ₹137280
## Ex 13.3 Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π = 3.14)
Solution:
Radius of cone (r) = 6 m
Height(h) = 8m
∴ Slant height (l) = l2 = r2 + h2
l2 = 62 + 82 = 36 + 64 = 100
= $\sqrt { 100 }$m = 10 m
Area of tarpaulin to make the tent = πrl
= 3.14 x 6 x 10 = 1884 m2
Let the length of 3 m wide tarpaulin = L m
Therefore, the area of tarpaulin required = 3 * L
According to question,
3 * L = 188.4 ⇒ L = $\frac { 188.4 }{ 3 }$ = 62.80 m
Extra tarpaulin for stitching margins and wastage = 20cm = 0.2m
Thus, total length of tarpaulin = 62.8 m + 0.2 m = 63 m
## Ex 13.3 Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m2 .
Solution:
Radius of conical tomb (r) = $\frac { 14 }{ 2 }$ m = 7 m
Slant height (l) = 25 m
∴ Curved surface area of conical tomb = πrl
$\frac { 22 }{ 7 }$ x 7 x 25 m2 = 550 m2
Cost of white-washing for 100 m2 area = Rs. 210
∴ Cost of white-washing for 550 m2 area
= Rs. $\frac { 210 }{ 100 }$ x 550 = Rs. 1155
## Ex 13.3 Question 7.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Solution:
Radius of right circular cone (r) = 7 cm and height (h) = 24 cm
Slant height (l)l m
l2 = r2 + h2= l2 = 72 + 242 = 49 + 576 = 625.
l = $\sqrt { 625 }$ cm = 25 cm
∴Area of sheet required to make 1 cap = πrl = $\frac { 22}{ 7 }$ x 7 x 25 cm2 = 550 cm2
∴ Area of sheet required to make 10 caps = 10 x 550 cm2
= 5500 cm2
## Ex 13.3 Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take $\sqrt{104}$ = 1.02)
Solution:
Radius of cone (r) = 40/2 = 20 cm and height (h) = 1 cm
Slant height (l)l m
l2 = r2 + h2= l2 = (0.2)2 + 12 = 0.04+ 1 = 1.04
l = $\sqrt{104}$ = 1.02m
Curved surface area of cone = πrl = 3.14 * 7 * 25 cm2 = 6.4056 m2
∴ Curved surface area of 50 cone s = 50 x 6.4056 = 32.028 m2
Cost of painting at rate of ₹12 per m2 = ₹12 * 32.028 = Rs. 384.34 (approx.)
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# SSAT Upper Level Math : Lines
## Example Questions
### Example Question #11 : Lines
Find the equation of a line that has a slope of and passes through the points .
Explanation:
In finding the equation of the line given its slope and a point through which it passes, we can use the slope-intercept form of the equation of a line:
, where is the slope of the line and is its -intercept.
Since the problem gives us the slope of the line , we just need to use the point that is given to us to find the -intercept. Plug in known values for and taken from the given point into the equation to find the -intercept:
Multiply:
Subtract from each side of the equation:
Now that you've solved for , you can plug the given slope and the -intercept into the slope-intercept form of the equation of a line to figure out the answer:
### Example Question #12 : Lines
Find the equation of the line that has a slope of and passes through the point .
Explanation:
In finding the equation of the line given its slope and a point through which it passes, we can use the slope-intercept form of the equation of a line:
, where is the slope of the line and is its -intercept.
Since the problem gives us the slope of the line , we just need to use the point that is given to us to find the -intercept. Plug in known values for and taken from the given point into the equation and solve for to find the -intercept:
Multiply:
Subtract from each side of the equation:
Now, we can write the final equation by plugging in the given slope and the -intercept :
### Example Question #13 : Lines
Find the equation of the line that has a slope of and passes through the point .
Explanation:
In finding the equation of the line given its slope and a point through which it passes, we can use the slope-intercept form of the equation of a line:
, where is the slope of the line and is its -intercept.
Since the problem gives us the slope of the line , we just need to use the point that is given to us to find the -intercept. Plug in known values for and taken from the given point into the equation and solve for to find the -intercept:
Multiply:
Add to each side of the equation:
Now, we can write the final equation by plugging in the given slope and the -intercept :
### Example Question #14 : Lines
Find the equation of a line that has a slope of and passes through the points .
Explanation:
In finding the equation of the line given its slope and a point through which it passes, we can use the slope-intercept form of the equation of a line:
, where is the slope of the line and is its -intercept.
Since the problem gives us the slope of the line , we just need to use the point that is given to us to find the -intercept. Plug in known values for and taken from the given point into the equation and solve for to find the -intercept:
Multiply:
Subtract from both sides of the equation:
Now, we can write the final equation by plugging in the given slope and the -intercept :
### Example Question #15 : Lines
Find the equation of the line that has a slope of and passes through the point .
Explanation:
The question gives us both the slope and the -intercept of the line, allowing us to write the following equation by inserting those values into the slope-intercept form of the equation of a line, :
Alternatively, if you did not realize that the problem gives you the -intercept, you could solve it by using the slope-intercept form of the equation of a line. Since the problem gives us the slope of the line , we would just need to use the point that is given to us to find the -intercept. We could plug in the known values for and taken from the given point into the equation and solve for to find the -intercept:
Multiplying leaves us with:
We could then substitute in the given slope and the -intercept into the equation to arrive at the correct answer:
### Example Question #16 : Lines
Find the equation of a line that has a slope of and passes through the point .
Explanation:
The question gives us both the slope and the -intercept of the line. Remember that represents the slope, and represents the -intercept to write the following equation:
Alternatively, if you did not realize that the problem gives you the -intercept, you could solve it by using the slope-intercept form of the equation of a line:
, where is the slope of the line and is its -intercept.
Since the problem gives us the slope of the line , we would just need to use the point that is given to us to find the -intercept. We could plug in the known values for and taken from the given point into the equation and solve for to find the -intercept:
Multiplying leaves us with:
.
We could then substitute in the given slope and the -intercept into the equation to arrive at the correct answer:
### Example Question #17 : Lines
Find the equation of the line that passes through and .
Explanation:
First, notice that our -intercept for this line is ; we can tell this because one of the points, , is on the -axis since it has a value of for
Now, we need to find the slope of the line. We can do that by using the slope equation:
We can substitute in the values of the provided points—, and —and then solve for the slope of the line that connects them:
Now, put the two pieces of information together and substitute them into the equation to solve the problem:
### Example Question #18 : Lines
Find the equation of the line that passes through the points and .
Explanation:
First, notice that our -intercept for this line is ; we can tell this because one of the points, , is on the -axis since it has a value of for
Now, we need to find the slope of the line. We can do that by using the slope equation:
We can substitute in the values of the provided points—, and —and then solve for the slope of the line that connects them:
Now, put the two pieces of information together and substitute them into the equation to solve the problem:
### Example Question #19 : Lines
Find the equation of the line that passes through the points .
Explanation:
First, notice that our -intercept for this line is ; we can tell this because one of the points, , is on the -axis since it has a value of for
Now, we need to find the slope of the line. We can do that by using the slope equation:
We can substitute in the values of the provided points—, and —and then solve for the slope of the line that connects them:
Now, put the two pieces of information together and substitute them into the equation to solve the problem:
### Example Question #20 : Lines
Find the equation of the line that passes through the points and .
Explanation:
First, we need to find the slope of the line. We can do that by using the slope equation:
We can substitute in the values of the provided points—, and —and then solve for the slope of the line that connects them:
Next, plug one of the points' coordinates and the slope to the equation and solve for to find the -intercept. For this example, let's use the point :
Multiply:
Change from a whole number to a mixed number with in the denominator, just like in the fraction :
Subtract from each side of the equation:
Finally, put the slope and the -intercept into the equation to arrive at the correct answer:
|
## Basic College Mathematics (9th Edition)
Published by Pearson
# Chapter 4 - Decimals - 4.4 Multiplying Decimal Numbers - 4.4 Exercises - Page 294: 56
#### Answer
(a) $\$175.35$(b)$\$5.45$
#### Work Step by Step
(a) We know that you order one of each type of shirt for yourself. From the table, the sum is: $14.75+16.75+18.95+21.95=\$72.40$In addition, you order three XXL shirts at$\$21.95$ each plus \2$for the size. Adding this to the previous total, we get:$72.40+3*21.95+3*2=72.40+65.85+6=\$144.25$ We calculate the shipping based on this total, which comes out to $\$11.95$plus$\$4.25$ for the addition address. This brings the total to: $144.25+11.95+4.25=\$160.45$Finally, we need to include the cost of the monograms (2 at$\$4.95$ each) and the gift box ($\$5$):$160.45+2*4.95+5=\$175.35$ (b) We see from part (a) that the cost of the shirts for yourself (with monograms) comes out to: $14.75+16.75+18.95+21.95+4.95*2=\$82.30$And the cost of the shirts for dad (with gift box) comes out to:$21.95*3+3*2+5=\$76.85$ Thus the difference is: $82.30-76.85=\$5.45\$
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# Visualizing the General Case of Pythagorean Theorem
## Tapping It Up a Notch: Pythagorean Theorem – Part 2
In our last post, we used an inquiry/discovery approach to help students visualize how we can find the hypotenuse of a right-angle triangle when given the lengths of the two legs.
In this post, we will now introduce the General Case for Pythagorean Theorem in an attempt to use the same visual model to derive the formula for Pythagorean Theorem.
## Using Pythagorean Theorem to Find the Length of the Hypotenuse
### Visual Representation of Any Right Triangle [General Case]
In this video, we now introduce the concept for any right-angle triangle. Using the video from the previous post is probably a good idea as a starting point.
## Summary of Pythagorean Theorem Video
### Starting With The General Case Visually
We now start with variables representing any side lengths for a right-angle triangle.
### Connect Squaring Side-Lengths to Area
Once again, we show that squaring side lengths on the right-angle triangle will yield an area since we are multiplying length by width (or side by side, in this case):
### Segment the Area of the Shortest Leg into Pieces
In order to preserve the visualization of the sum of the squares of both legs being equivalent to the square of the length of the hypotenuse, we will need to chop up one of the areas into pieces:
### Show The Sum of the Squares of the Leg Lengths Algebraically
As we did in the 3, 4, 5 case in the previous post, we now must use algebra to show what the sum of the squares of the leg lengths look like. This can help students start to see what the formula must be.
### Finding the Length of the Hypotenuse
In the previous case, finding the side length was pretty easy for students to do without necessarily consciously thinking about opposite operations. In this case, we must square-root the resulting area to find the length of the hypotenuse.
### Introduce c^2 as a Variable to Replace a^2 + b^2
Using substitution, we will replace a^2 + b^2 with a new variable, c^2.
After students are comfortable with the algebra behind Pythagorean Theorem, you may choose to discuss this more deeply. Having students understand that we are actually substituting c with the square-root of a^2 + b^2 could be beneficial as they get closer to advanced functions and calculus.
### Deriving the Formula for Pythagorean Theorem
Students may now be more comfortable understanding and using the algebraic representation of the Pythagorean Theorem Formula:
Another post is forthcoming to help students better connect the visual representation of Pythagorean Theorem and the algebraic representation.
What do you think? How can we improve the introduction of this very important mathematical concept? Leave a comment below!
## Other Related Pythagorean Theorem Posts:
[postshortcode the_query=”post_type=post&p=14140″]
[postshortcode the_query=”post_type=post&p=14185″]
[threeactshortcode the_query=”post_type=realworldmath&p=3760″]
## WANT TO LEARN HOW TO TEACH THROUGH TASK?
Download our Complete Guide to successfully implementing our Make Math Moments 3-Part Framework in your math class!
## Share With Your Learning Community:
I’m Kyle Pearce and I am a former high school math teacher. I’m now the K-12 Mathematics Consultant with the Greater Essex County District School Board, where I uncover creative ways to spark curiosity and fuel sense making in mathematics. Read more.
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# How do you solve x/(x-1)>2?
Feb 18, 2017
$1 < x < 2$ or in interval form: $\left(1 , 2\right)$
#### Explanation:
Put the inequality in $> 0$ form:
$\frac{x}{x - 1} - 2 > 0$
Find the common denominator:
$\frac{x}{x - 1} + \frac{- 2 \left(x - 1\right)}{x - 1} > 0$;
Remember: $- 2 \left(x - 1\right) = - 2 x + 2$
Combine under the same denominator:
$\frac{x - 2 x + 2}{x - 1} = \frac{- x + 2}{x - 1} > 0$
Key Points: $- x + 2 > 0$ and $x - 1 > 0$
Simplify: $- x > - 2$ and $x > 1$
When you divide by -1, the inequality changes direction:
(-x/-1) < (-2/-1); x < 2
Therefore: $1 < x < 2$ or in interval form: $\left(1 , 2\right)$
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# How do you find area with length and width?
## How do you find area with length and width?
Multiply the length times the width to find the area. State your answer in square units. For instance, if a room is 8 feet by 10 feet, multiply 8 times 10 to get an area of 80 square feet.
## How do we calculate area?
Area is calculated by multiplying the length of a shape by its width. In this case, we could work out the area of this rectangle even if it wasn’t on squared paper, just by working out 5cm x 5cm = 25cm² (the shape is not drawn to scale).
How do you find the area of a house shape?
The simplest (and most commonly used) area calculations are for squares and rectangles. To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area.
How do I calculate sqm?
Multiply the length and width together. Once both measurements are converted into metres, multiply them together to get the measurement of the area in square metres.
### How fo you find the area of a rectangle?
To find the area of a rectangle, we multiply the length of the rectangle by the width of the rectangle.
### How do you find the length of an area?
To find Length or Breadth when Area of a Rectangle is given
1. When we need to find length of a rectangle we need to divide area by breadth.
2. Length of a rectangle = Area ÷ breadth.
3. ℓ = A ÷ b.
4. Similarly, when we need to find breadth of a rectangle we need to divide area by length.
5. Breadth of a rectangle = Area ÷ length.
How do you find the area of combined shapes?
To calculate the area of a composite shape you must divide the shape into rectangles, triangles or other shapes you can find the area of and then add the areas back together. You may have to calculate missing lengths before finding the area of some of the shapes.
How do you calculate plot area in square feet?
Multiply the length by the width and you’ll have the square feet. Here’s a basic formula you can follow: Length (in feet) x width (in feet) = area in sq. ft.
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CurveSketching.mws
Applications of Differentiation: Summary of Curve Sketching
Vocabulary
domain, intercept, symmetry, asymptote, critical number, concavity, inflection point
Objective
Sketch accurate graphs of functions, identifying important properties which have been discussed in the course up to this point.
Lecture Outline
Thus far we have discussed many properties of functions. We list some of the most important of these properties below.
Properties of Functions
• 1. Domain
• 2. Intercepts (x-intercept(s) and y-intercept)
• 3. Symmetry (even, odd, and periodic)
• 4. Asymptotes (vertical, horizontal, oblique)
• 5. Intervals of Increase or Decrease
• 6. Local Extrema
• 7. Concavity and inflection point(s)
Properties one through four in the list above can often be determined without the use of calculus techniques while properties five through seven require consideration of the first and second derivatives of the function.
Given any function, we can methodically determine each of the properties and then sketch a graph of a function satisfying those properties. The sketch obtained in this way is often a reasonably accurate representation of the graph of the initial function.
Example:
We methodically determine each of the properties in the list.
> f:=x->x^2/(x^2-9);
The domain of f consists of all real numbers x such that the denominator is not equal to zero.
> solve(x^2-9,x);
Hence the domain is all real numbers excluding 3 and -3.
The y-intercept is the point (0,f(0)) provided that 0 is in the domain of f.
> f(0);
We see that (0,0) is the y-intercept.
The x-intercepts occur wherever the numerator of f is equal to 0.
> solve(x^2,x);
Hence there is only one x-intercept and it has x-coordinate 0. The x-intercept itself is the point (0,0).
Now we consider the symmetry of f.
> f(-x);
Note that f(-x)=f(x). We conclude that f is an even function (i.e., the graph of f is symmetric with respect to the y-axis).
Since f is a rational function in reduced form (i.e., the numerator and denominator share no common factors), the vertical asymptotes of f are obtained by finding where the denominator is equal to 0. We have already determined that the denominator is 0 when x=3 or x=-3 when we considered the domain of f. Hence the graph of f has two vertical asymptotes, the lines x=3 and x=-3.
The horizontal asymptote is found by determining
> limit(x^2/(x^2-9),x=infinity);
We thus have that the line y=1 is the horizontal asymptote of the graph of f. Recall that a rational function has at most one horizontal asymptote, so we do not need to consider .
The intervals of increase and decrease may be determined using the increasing/decreasing test. We compute the first derivative of f and determine where it is positive or negative.
> fprime:=x->D(f)(x);
> solve(fprime(x)>0,x);
> solve(fprime(x)<0,x);
We conclude that f is increasing on the intervals ( ,-3) and (-3,0) while f is decreasing on the intervals (0,3) and (3, ).
The critical numbers of f occur where the first derivative is equal to 0 or is undefined. Since f is a rational function, the only critical numbers where f can have local extrema are those for which the first derivative is equal to 0 (recall that these are also called stationary numbers). We solve for those critical numbers.
> solve(fprime(x),x);
There is only one stationary number, x=0. We can classify this critical number as a local extremum using the first derivative test. The first derivative changes sign from positive to negative across 0, so f has a local maximum at x=0.
The concavity of f is determined by using the concavity test.
> fdoubleprime:=x->D(fprime)(x);
> solve(fdoubleprime(x)>0,x);
> solve(fdoubleprime(x)<0,x);
We conclude that the graph of f is concave upward on the intervals ( ,-3) and (3, ) while the graph of f is concave downward on the interval (-3,3).
Inflection points occur at points on the graph where concavity changes direction. In this case, it is apparent that concavity changes across x=-3 and x=3, but these are not in the domain of f, so there are no inflection points.
Let us verify the information about the properties of f we have obtained above by comparison with a Maple plot of the graph of f.
> plot(f(x),x=-5..5,y=-20..20);
Exercise:
Use the same process as above to determine the important properties of the graph of g, make a sketch of a function having those properties, and then compare this sketch with a Maple plot of the graph of g.
>
>
>
>
>
|
×
# Coefficient of Variation Calculator
## How to use Coefficient of Variation Calculator?
• Input the comma-separated values.
• Select the data type “sample” or “population”
• Click on the calculate button.
## Other Calculators
The coefficient of variation calculates the CV of a sample or a population data set and gives its step-by-step solution. Firstly, it calculates the average value (mean) and the standard deviation of the entered data, and also provides the steps of the calculation.
## What is the coefficient of variation?
The coefficient of variation (CV) is defined as the ratio of the standard deviation to the mean. It is a statistical measure of the dispersion of data points in a data series around the mean.
## Formulas of CV
The general formula for the coefficient of variation is as follows:
CV = Standard deviation/mean
For sample
CV = s/x̄
• s = sample standard deviation
• x̄ = sample mean
For population
CV = σ/µ
• σ = population standard deviation
• µ = population mean
## Steps to calculate the coefficient of variation.
To calculate the coefficient of variation of a sample or a population data manually, you just need to follow the below steps:
1. Check whether the data is sample data or population data.
2. Calculate the total terms
3. Calculate the mean or average value.
4. Calculate the standard deviation of the particular data, this process is quite lengthy.
5. Divide the SD by the average value to get the final result.
In the following examples, the procedure of calculating the coefficient of variation is completely described with steps.
Example
Calculate the CV of sample data 9, 3, 5, 8, 2, 64, 75, 23, 10, 13
Solution:
Step 1: Calculate the total terms
Total terms = n = 10
Step 2: Calculate the sample mean.
Mean = (9 + 3 + 5 + 8 + 2 + 64 + 75 + 23 + 10 + 13) / 10
Mean = 212 / 10
Mean = 21.2
Step 3: Calculate the standard deviation.
S = [∑(xi – x̄)/n-1]
1. Calculating the deviation scores
Deviations = 9 – 21.2, 3 – 21.2, 5 – 21.2, 8 – 21.2, 2 – 21.2, 64 – 21.2, 75 – 21.2, 23 – 21.2, 10 – 21.2, 13 – 21.2
Deviations = -12.2, -18.2, -16.2, -13.2, -19.2, 42.8, 53.8, 1.8, -11.2, -8.2
1. Calculating the squared deviations
Squared deviations = (-12.2)2, (-18.2)2, (-16.2)2, (-13.2)2, (-19.2)2, (42.8)2, (53.8)2, (1.8)2, (-11.2)2, (-8.2)2
Squared deviations = 148.84, 331.24, 262.44, 174.24, 368.64, 1831.84, 2894.44, 3.24, 125.44, 67.24
1. Calculating the sum of squares
Sum of squares = 148.84 + 331.24 + 262.44 + 174.24 + 368.64 + 1831.84 + 2894.44 + 3.24 + 125.44 + 67.24
Sum of squares = 6207.6
Putting all values in the formula
S = √(6207.6/10-1)
S = √(6207.6/9)
S = √(689.733)
S = 26.26
Step 5: Calculating the final result
CV = s/x̄
CV = (26.26/21.2)
CV = 1.238
|
# Using The Constant Multiple Rule To Find The Derivative Of A Fractional Function Video
Using the constant multiple rule to find the derivative of a fractional function video involves algebraic fractions, constant multiple rule, derivatives, fractions, number sense, numbers.
# Using The Constant Multiple Rule To Find The Derivative Of A Fractional Function Video Tutorial
algebraic fractions video, constant multiple rule video, derivatives video, fractions video, number sense video, numbers video.
# Using The Constant Multiple Rule To Find The Derivative Of A Fractional Function
This math video tutorial gives a step by step explanation to a math problem on "Using The Constant Multiple Rule To Find The Derivative Of A Fractional Function".
# Derivatives
In calculus, the derivative is a measurement of how a function changes when the values of its inputs change. Loosely speaking, a derivative can be thought of as how much a quantity is changing at some given point. For example, the derivative of the position or distance of a car at some point in time is the instantaneous velocity, or instantaneous speed (respectively), at which that car is traveling (conversely the integral of the velocity is the car's position).
A closely related notion is the differential of a function.
The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. For a real-valued function of a single real variable, the derivative at a point equals the slope of the tangent line to the graph of the function at that point.
The process of finding a derivative is called differentiation. The fundamental theorem of calculus states that differentiation is the reverse process to integration.
# Fractions
In mathematics, a fraction is a concept of a proportional relation between an object part and the object whole. Each fraction consists of a denominator (bottom) and a numerator (top), representing (respectively) the number of equal parts that an object is divided into, and the number of those parts indicated for the particular fraction.
|
Visualizing the chain rule and product rule | Essence of calculus, chapter 4
In the last videos I talked about the derivatives of simple functions, things like powers of
x, sin(x), and exponentials, the goal being to have a clear picture or intuition to hold
in your mind that explains where these formulas come from.
Most functions you use to model the world involve mixing, combining and tweaking these
these simple functions in some way; so our goal now is to understand how to take derivatives
of more complicated combinations; where again, I want you to have a clear picture in mind
for each rule.
This really boils down into three basic ways to combine functions together: Adding them,
multiplying them, and putting one inside the other; also known as composing them.
Sure, you could say subtracting them, but that’s really just multiplying the second
by -1, then adding.
Likewise, dividing functions is really just the same as plugging one into the function
1/x, then multiplying.
Most functions you come across just involve layering on these three types of combinations,
with no bound on how monstrous things can become.
But as long as you know how derivatives play with those three types of combinations, you
can always just take it step by step and peal through the layers.
So, the question is, if you know the derivatives of two functions, what is the derivative of
their sum, of their product, and of the function compositions between them?
The sum rule is the easiest, if somewhat tounge-twisting to say out loud: The derivative of a sum of
two functions is the sum of their derivatives.
But it’s worth warming up with this example by really thinking through what it means to
take a derivative of a sum of two functions, since the derivative patterns for products
and function composition won’t be so straight forward, and will require this kind of deeper
thinking.
The function f(x) = sin(x) + x2 is a function where, for every input, you add together the
values of sin(x) and x2 at that point.
For example, at x = 0.5, the height of the sine graph is given by this bar, the height
of the x2 parabola is given by this bar, and their sum is the length you get by stacking
them together.
For the derivative, you ask what happens as you nudge the input slightly, maybe increasing
it to 0.5+dx.
The difference in the value of f between these two values is what we call df.
Well, pictured like this, I think you’ll agree that the total change in height is whatever
the change to the sine graph is, what we might call d(sin(x)), plus whatever the change to
x2 is, d(x2).
We know the derivative of sine is cosine, and what that means is that this little change
d(sin(x)) would be about cos(x)dx.
It’s proportional to the size of dx, with a proportionality constant equal to cosine
of whatever input we started at.
Similarly, because the derivative of x2 is 2x, the change in the height of the x2 graph
So, df/dx, the ratio of the tiny change to the sum function to the tiny change in x that
caused it, is indeed cos(x)+2x, the sum of the derivatives of its parts.
But like I said, things are a bit different for products.
Let’s think through why, in terms of tiny nudges.
In this case, I don’t think graphs are our best bet for visualizing things.
Pretty commonly in math, all levels of math really, if you’re dealing with a product
of two things, it helps to try to understand it as some form of area.
In this case, you might try to configure some mental setup of a box whose side-lengths are
sin(x) and x2.
What would that mean?
Well, since these are functions, you might think of these sides as adjustable; dependent
on the value of x, which you might think of as a number that you can freely adjust.
So, just getting the feel for this, focus on that top side, whose changes as the function
sin(x).
As you change the value of x up from 0, it increases in up to a length of 1 as sin(x)
moves towards its peak.
After that, it starts decreasing as sin(x) comes down from 1.
And likewise, that height changes as x2.
So f(x), defined as this product, is the area of this box.
For the derivative, think about how a tiny change to x by dx influences this area; that
resulting change in area is df.
That nudge to x causes the width to change by some small d(sin(x)), and the height to
change by some d(x2).
This gives us three little snippets of new area: A thin rectangle on the bottom, whose
area is its width, sin(x), times its thin height, d(x2); there’s a thin rectangle
on the right, whose area is its height, x2, times its thin width, d(sin(x)).
And there’s also bit in the corner.
But we can ignore it, since its area will ultimately be proportional to dx2, which becomes
negligible as dx goes to 0.
This is very similar to what I showed last video, with the x2 diagram.
Just like then, keep in mind that I’m using somewhat beefy changes to draw things, so
we can see them, but in principle think of dx as very very small, meaning d(x2) and d(sin(x))
are also very very small.
Applying what we know about the derivative of sine and x2, that tiny change d(x2) is
2x*dx, and that tiny change d(sin(x)) is cos(x)dx.
Dividing out by that dx, the derivative df/dx is sin(x) by the derivative of x2, plus x2
by the derivative of sine.
This line of reasoning works for any two functions.
A common mnemonic for the product rule is to say in your head “left d right, right
d left”.
In this example, sin(x)*x2, “left d right” means you take the left function, in this
case sin(x), times the derivative of the right, x2, which gives 2x.
Then you add “right d left”: the right function, x2, times the derivative of the
left, cos(x).
Out of context, this feels like kind of a strange rule, but when you think of this adjustable
box you can actually see how those terms represent slivers of area.
“Left d right” is the area of this bottom rectangle, and “right d left” is the area
of this rectangle on the right.
By the way, I should mention that if you multiply by a constant, say 2*sin(x), things end up
much simpler.
The derivative is just that same constant times the derivative of the function, in this
case 2*cos(x).
I’ll leave it to you to pause and ponder to verify that this makes sense.
Aside from addition and multiplication, the other common way to combine functions that
comes up all the time is function composition.
For example, let’s say we take the function x2, and shove it on inside sin(x) to get a
new function, sin(x2).
What’s the derivative of this new function?
Here I’ll choose yet another way to visualize things, just to emphasize that in creative
math, we have lots of options.
I’ll put up three number lines.
The top one will hold the value of x, the second one will represent the value of x2,
and that third line will hold the value of sin(x2).
That is, the function x2 gets you from line 1 to line 2, and the function sine gets you
from line 2 to line 3.
As I shift that value of x, maybe up to the value 3, then value on the second shifts to
whatever x2 is, in this case 9.
And that bottom value, being the sin(x2), will go over to whatever the sin(9) is.
So for the derivative, let’s again think of nudging that x-value by some little dx,
and I always think it’s helpful to think of x starting as some actual number, maybe
1.5.
The resulting nudge to this second value, the change to x2 caused by such a dx, is what
we might call d(x2).
You can expand this as 2x*dx, which for our specific input that length would be 2*(1.5)*dx,
but it helps to keep it written as d(x2) for now.
In fact let me go one step further and give a new name to x2, maybe h, so this nudge d(x2)
is just dh.
Now think of that third value, which is pegged at sin(h).
It’s change d(sin(h)); the tiny change caused by the nudge dh.
By the way, the fact that it’s moving left while the dh bump is to the right just means
that this change d(sin(h)) is some negative number.
Because we know the derivative of sine, we can expand d(sin(h)) as cos(h)*dh; that’s
what it means for the derivative of sine to be cosine.
Unfolding things, replacing h with x2 again, that bottom nudge is cos(x2)d(x2).
And we could unfold further, noting that d(x2) is 2x*dx.
And it’s always good to remind yourself of what this all actually means.
In this case where we started at x = 1.5 up top, this means that the size of that nudge
on the third line is about cos(1.52)*2(1.5)*(the size of dx); proportional to the size of dx,
where the derivative here gives us that proportionality constant.
Notice what we have here, we have the derivative of the outside function, still taking in the
unaltered inside function, and we multiply it by the derivative of the inside function.
Again, there’s nothing special about sin(x) and x2.
If you have two functions g(x) and h(x), the derivative of their composition function g(h(x))
is the derivative of g, evaluated at h(x), times the derivative of h.
This is what we call the “chain rule”.
Notice, for the derivative of g, I’m writing it as dg/dh instead of dg/dx.
On the symbolic level, this serves as a reminder that you still plug in the inner function
to this derivative.
But it’s also an important reflection of what this derivative of the outer function
actually represents.
Remember, in our three-lines setup, when we took the derivative of sine on the bottom,
we expanded the size of the nudge d(sin) as cos(h)*dh.
This was because we didn’t immediately know how the size of that bottom nudge depended
on x, that’s kind of the whole thing we’re trying to figure out, but we could take the
derivative with respect to the intermediate variable h.
That is, figure out how to express the size of that nudge as multiple of dh.
Then it unfolded by figuring out what dh was.
So in this chain rule expression, we’re saying look at the ratio between a tiny change
in g, the final output, and a tiny change in h that caused it, h being the value that
we’re plugging into g.
Then multiply that by the tiny change in h divided by the tiny change in x that caused
it.
The dh’s cancel to give the ratio between a tiny change in the final output, and the
tiny change to the input that, through a certain chain of events, brought it about.
That cancellation of dh is more than just a notational trick, it’s a genuine reflection
of the tiny nudges that underpin calculus.
So those are the three basic tools in your belt to handle derivatives of functions that
combine many smaller things: The sum rule, the product rule and the chain rule.
I should say, there’s a big difference between knowing what the chain rule and product rules
are, and being fluent with applying them in even the most hairy of situations.
I said this at the start of the series, but it’s worth repeating: Watching videos, any
videos, about these mechanics of calculus will never substitute for practicing them
yourself, and building the muscles to do these computations yourself.
I wish I could offer to do that for you, but I’m afraid the ball is in your court, my
friend, to seek out practice.
What I can offer, and what I hope I have offered, is to show you where these rules come from,
to show that they’re not just something to be memorized and hammered away; but instead
are natural patterns that you too could have discovered by just patiently thinking through
what a derivative means.
Thank you to everyone who supported this series, and once more I’d like to say a special
thanks to Brilliant.org.
For those of you who want to go flex those problem solving muscles, Brilliant offers
a platform aimed at training you to think like a mathematician.
I don’t know about you, but I’ve always found it all too easy to fall into the habit
of just reading math or watching lectures without taking the time to do some real problem-solving
in between, even though that’s always the part where I learn the most.
Brilliant is a great place to get that practice, and if you visit brilliant.org/3b1b, or more
simply follow the link on the screen and in the description, it lets them know you came
from this channel.
Their calculus material is a nice complement to this series, but some of my other favorites
are their probability and complex algebra sequences.
|
Q:
# How do you find the area of a sector of a circle?
A:
The equation for the area of a circle is pi times the square of the radius or A = πr2. Calculating the area of a sector involves figuring out what fraction of the circle's area the sector covers.
## Keep Learning
Credit: Matteo Colombo Moment Getty Images
1. Understand A = πr2
The radius of a circle is defined as a line segment drawn from the circle's center to any point on its edge. Pi is the ratio of a circle's circumference to its diameter. Pi is an irrational number, meaning it is a decimal that does not terminate or repeat, but for everyday purposes, it can be rounded to 3.14.
2. Apply the circle area formula
Suppose a circle has a radius of 10 centimeters. To find its area, square 10 to get 100, and then multiply by 3.14. The circle's area is 314 square centimeters.
3. Figure out units of arc
Recall that the edge of a circle spans 360 degrees. A right angle drawn from the circle's center and intersecting its edge covers 90 degrees of arc. The area of this sector would be 90/360 or 1/4 the area of the whole circle. Other angles work the same way when drawn at the circle's center. This is the remaining step necessary to calculate the area of a sector.
4. Try a sample problem
Go back to the circle in step 2. Suppose a 60-degree angle is drawn at the circle's center. The area of the circle is 314, and 314 divided by 6 is 52.33. Therefore, a sector with a central angle of 60 degrees inscribed a circle with a radius of 10 centimeters has an area of 52.33 square centimeters.
Sources:
## Related Questions
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# How much work does it take to raise a 6 kg weight 28 m ?
Aug 10, 2017
$W = 168 \text{J}$
#### Explanation:
The work done by a constant force is given by the dot product:
$W = \vec{F} \cdot \vec{d}$
Giving the familiar equation:
$\implies \textcolor{\mathrm{da} r k b l u e}{W = F \mathrm{dc} o s \left(\theta\right)}$
where $F$ is the force applied to the object, $d$ is the displacement of the object, and $\theta$ is the angle between the force and displacement vectors
In raising the object, there is only the force which is being applied to cause the lift and the force of gravity acting against it.
Therefore, we can write:
$\sum F = {F}_{a} - {F}_{G} = m a$
We'll assume dynamic equilibrium (no acceleration but motion), giving:
$F a = {F}_{G}$
$\therefore$ The applied force must be equal to the force of gravity (opposite in direction).
The force of gravity acting on a stationary object at or near the surface of the earth is given by:
${F}_{G} = m g$
Which gives a new equation for work:
$\textcolor{\in \mathrm{di} g o}{{W}_{\text{lift}} = m g \cos \left(\theta\right)}$
We are given the following information:
• $\mapsto m = 6 \text{kg}$
• $\mapsto d = 28 \text{m}$ (vertically)
• $\mapsto \theta = {0}^{o}$ (implied)
Therefore:
$W = \left(6 \text{kg")(28"m}\right)$
$= 168 \text{J}$
$- - - -$
Note that $\theta = 0$ because the applied force and displacement occur in the same direction: upward. Therefore the angle between the vectors is ${0}^{o}$.
|
## MAT 115 Course Prep
This collection of problems is meant to provide practice for a range of skills needed for MAT 115. Read and attempt each problem first; if you aren’t sure how to start a problem, explore the resources on the right to refresh your memory and try again. A drop-down button is found beneath each problem for you to compare both your logic and final answers. Keep track of the skills you aren’t comfortable with, and reacquaint yourself with them so you’re fully prepared for the topics you’ll grapple with soon. Get help in the Math Lab. Return to the Course Prep page.
Problem 1: Using the functions defined below, simplify each expression. If possible, find exact values.
$\boldsymbol{f(x)=x^2} \qquad \boldsymbol{g(x) = -x^2} \qquad \boldsymbol{h(x)=1-2x}$ Part a. $f(-1)$ Click here to show solution. $f(-1) = (-1)^2 = 1.$ Video: Worked Example: Evaluating Functions from Equation Part b. $g(-1)$ Click here to show solution. $g(-1) = -(-1)^2 = -(1) = -1.$ Part c. $h(-1)$ Click here to show solution. $h(-1) = 1-2(-1) = 1+2 = 3.$ Part d. $f(x)-g(x)$ Click here to show solution. $f(x)-g(x) = (x^2) - (-x^2) = x^2+x^2=2x^2.$ Since no value of $x$ is given, an exact value is not possible as an answer. Part e. $g(x)+f(x)$ Click here to show solution. $g(x)+f(x) = (-x^2)+x^2 = 0.$ Part f. $\frac{f(x)}{g(x)}$ Click here to show solution. $\frac{f(x)}{g(x)} = \frac{x^2}{-x^2} = -1,$ since the top and bottom share a common factor of $x^2$ which cancels (assuming $x \neq 0$). Part g. $f(x)g(x)$ Click here to show solution. $f(x)g(x)=(x^2)(-x^2)=-x^4$. Part h. $g(x)h(x)$ Click here to show solution. $g(x)h(x) = (-x^2)(1-2x) = -x^2+2x^3,$ by the distributive property.
Problem 2: Solve the following equations for x, and verify your solution.
Part a. $100 = 0.2x$ Click here to show solution. Divide both sides by 0.2: $500=x$. To verify, plug this value of $x$ back into the original question and make sure that the statement is true. In this case, the choice of $x = 500$ gives us $100 = 0.2(500) \Rightarrow 100 = 100$. Since this is true, we know our solution is correct. To give you an idea of what it would look like if we somehow got the incorrect answer, pretend we solved the original equation and got $x = 80$. Then, in the verification step, we would have $100 = 0.2(80) \Rightarrow 100 = 16$. This is obviously false, and so $x= 80$ is not a valid solution to the equation. Video: Solving Linear Equations I Write-up: One Step Equations Part b. $-80=\frac{4}{3}x$ Click here to show solution. Divide both sides by $\frac{4}{3}$. This has the effect of multiplying each side by $\frac{3}{4}$: $\frac{-240}{4} = x$, which reduces to show that $x=-60$. We verify by plugging this value back into the original question: $-80 = \frac{4}{3}(-60) \Rightarrow -80 = \frac{-240}{3} \Rightarrow -80 = -80$. This is true, and so we’ve verified our answer. Part c. $44 = 8-6x$ Click here to show solution. Begin by subtracting 8 from each side, yielding $36 = -6x$. Division by -6 gives $x=-6$. Verify the solution by plugging this value of $x$ back into the original equation: $44 = 8-6(-6) \Rightarrow 44 = 8-(-36) \Rightarrow 44 = 8+36 \Rightarrow 44=44$. This is true, and so we’ve verified our answer. Video: Solving Linear Equations II Write-up: Two Step Equations Part d. $\frac{x^2-6x}{x}=3$ Click here to show solution. On the left-hand side, a factor of $x$ is common to each term in the numerator, and cancels with the $x$ found in the denominator assuming $x \neq 0$: $\frac{x^2-6x}{x} = \frac{x(x-6)}{x} = x-6$. Simplification gives $x-6=3$, and adding 6 to both sides gives $x=9$. Verify the solution by substitution: $\frac{(9)^2-6(9)}{9} = 3 \Rightarrow \frac{81-54}{9} = 3 \Rightarrow \frac{27}{9} = 3 \Rightarrow 3 = 3$. This is true, and so we’ve verified our answer. Part e. $x^5=243$ Click here to show solution. To isolate the $x$, raise both sides to the $\frac{1}{5}$ power. This gives $x=243^{\frac{1}{5}} = \sqrt[5]{243} = 3$. We plug in this value of $x$ to verify the answer: $(3)^5 = 243 \Rightarrow 3\cdot3\cdot3\cdot3\cdot3=243 \Rightarrow (9)\cdot3\cdot3\cdot3 = 243 \Rightarrow (27)\cdot3\cdot3 = 243 \Rightarrow \dots \Rightarrow 243 = 243$. This is true, and so we’ve verified our answer. Video: Solving Equations with Rational Exponents Part f. $x^9=27.8$ Click here to show solution. In a similar fashion as the previous problem, raise both sides to the $\frac{1}{9}$ power, giving $x=27.8^{\frac{1}{9}} = \sqrt[9]{27.8}$. Using a calculator, we find that this is approximately 1.447. To verify our answer, we can compute: $(1.447)^9 \approx 27.811 \approx 27.8$. This shows that x = 1.447 is a correct approximation of the exact solution, which is x = $\sqrt[9]{27.8}$. Part g. $xy=k$ Click here to show solution. In earlier questions, $x$ was the only variable present, and “solving the equation” ended with a numerical solution. In cases like these, we must recall that “solving for $x$” is synonymous with isolating $x$ on one side of the equation. In this equation, divide both sides by $y$: $\frac{xy}{y} = \frac{k}{y} \Rightarrow x = \frac{k}{y}$ due to the cancellation on the left-hand side of the equation. We can still verify this solution via substitution for $x$ in the original equation: $\frac{k}{y} y = k \Rightarrow \frac{ky}{y} = k \Rightarrow k = k$ where the final step is possible due to the cancellation of $y$‘s on the left hand side. This statement is true, and so we’ve verified our answer. Part h. $y=xk$ Click here to show solution. In a similar fashion as the previous problem, we simply aim to isolate $x$ on one side of the equation. Divide by $k$ on both sides: $\frac{y}{k} = \frac{xk}{k} \Rightarrow \frac{y}{k} = x$. Verify via substitution: $y = \frac{y}{k}k \Rightarrow y = \frac{yk}{k} \Rightarrow y = y$, where the final step is possible due to the cancellation of $k$‘s on the right hand side. This statement is true, and so we’ve verified our answer.
Problem 3: Expand the following expressions.
Part a. $(a+b)^2$ Click here to show solution. This can be rewritten as $(a+b)(a+b)$, which can be FOILed (recall this tells you the order in which to multiply terms – First, Outer, Inner, Last). This gives $(a)(a)+(a)(b)+(b)(a)+(b)(b) = a^2+2ab+b^2$. Video: Multiplying Binomials Write-up: FOIL Method Part b. $(a-b)^2$ Click here to show solution. Following the same logic as the last problem, the multiplication gives $(a)(a)+(a)(-b)+(-b)(a)+(-b)(-b) = a^2-2ab+b^2$.
Problem 4: Find the slope of the line passing through the following points.
Part a. (0,5) and (4,19) Click here to show solution. Slope (commonly denoted by the symbol $m$) can be found via the familiar phrase “rise over run”. More accurately, we are concerned with the change in rise $(y_2-y_1)$ over the change in run $(x_2-x_1)$. In this case, this comes out to $m = \frac{19-5}{4-0} = \frac{14}{4} = \frac{7}{2}$. Worked Example: Slope from Two Points Write-up: Slope Formula Part b. (2,11) and (3,-3) Click here to show solution. Following the same logic as the previous problem, we have $m = \frac{-3-11}{3-2} = \frac{-14}{1} = -14$. Part c. $(x_1,y_1)$ and $(x_2,y_2)$ Click here to show solution. This represents the general case: $m = \frac{y_2-y_1}{x_2-x_1}$.
Problem 5: Find the equation of the line passing through the following points.
Part a. (0,5) and (4,19) Click here to show solution. There are two lanes of thought here, corresponding to the two different generic forms of a line: slope-intercept form $[y = mx+b]$ and point-slope form $[y-y_i = m(x-x_i)]$. In slope-intercept form, the slope and y-intercept of the line need to be known (unsurprisingly). We calculated the slope between these points above: $m = \frac{7}{2}$. Separately, when $x=0$, we know that $y=5$ since $(0,5)$ is given as a point on this line. Therefore, the y-intercept, commonly denoted with the symbol $b$, is 5. The equation of the line is therefore $y=\frac{7}{2}x+5$. In point-slope form, the slope and any one point on the line must be known (unsurprisingly). Choose $(x_i,y_i) = (0,5)$, for example: then $y-5=\frac{7}{2}(x-0),$ which reduces to $y-5=\frac{7}{2}x$. Adding 5 to both sides gives the same answer as before: $y=\frac{7}{2}x+5$. Point-slope form is generally more helpful, as needing any point on the line is a less strict requirement than knowing the y-intercept. The slope is needed in either case. Video: Point-Slope and Slope-Intercept Equations Write-up: Slope-Intercept Form Write-up: Point-Slope Form Part b. (2,11) and (3,-3) Click here to show solution. We don’t know the y-intercept here, so use point-slope form with either of the points provided. For the purpose of this example, we’ll use $(2,11)$; recall from the earlier question that the slope between these points is -14. The equation of the line is $y-11=-14(x-2) = -14x+28$. Adding 11 to both sides gives $y=-14x+39$. (Note that from this manipulation, we can actually determine the y-intercept to be 39, as the final simplified answer is in slope-intercept form with $m=-14, b=39$. This isn’t obvious from the information given initially.) Part c. $(x_1,y_1)$ and $(x_2,y_2)$ Click here to show solution. Use point-slope form again. For slope, recall that $m = \frac{y_2-y_1}{x_2-x_1}$, and note that either point can be plugged into the point-slope formula. Both $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$ and $y-y_2=\frac{y_2-y_1}{x_2-x_1}(x-x_2)$ are correct.
Problem 6: Your annual salary is $40,000 for the year 2018. Use the information below to calculate your salary in 2019, 2020, and 2021. Part a. Assume your salary increases by$1500 per year.
Year 2018 2019 2020 2021 Salary $40,000 ??? ??? ??? Click here to show solution. Take the salary for each year and add 1500 to get the salary for the next year. Year 2018 2019 2020 2021 Salary$40,000 $41,500$43,000 $44,500 Part b. Assume your salary increases by 3% per year. Year 2018 2019 2020 2021 Salary$40,000 ??? ??? ???
Let’s start with the calculation for 2019. An increase of 3% over the salary in 2018 means that $40,000+\frac{3}{100}(40,000)$ is earned in 2019. Factor 40,000 out of the left hand side to find $40,000(1+\frac{3}{100})=40,000(1+0.03) = (1.03)(40,000) = 41,200$. This logic can be applied recursively for the other future years:
Year 2018 2019 2020 2021 Salary $40,000 (1.03)(40,000) =$41,200 (1.03)(41200) = $42,436 (1.03)(42436) =$43,709.08
Problem 7: Use the following graph to answer the questions below.
Need Help ? MathPlanet – Coordinates & Ordered Pairs Part a. Does the point (6, 5) fall on the graph of the line ? Click here to show solution. A coordinate point tells you the x-coordinate first, followed by the y-coordinate. We can see that the point (6, 5) has an x-coordinate of 6. We can see on the x-axis, between 0 and 5, that this region is divided into 5 equal parts, marked by the light grey lines. This means that each square cell on the grid is one unit wide. To find 6 on the x-axis, we just need to go one unit to the right of the 5. Then, we follow the light grey vertical grid-line up to the red line, and we have found the point on the line with x-coordinate 6. Finally, we follow the light grey horizontal grid-line from this point to the left, until we reach the y-axis. This horizontal grid-line intersects the y-axis right at the number 5, so we know that 5 is the y-coordinate that corresponds to the x-coordinate of 6. Thus, we have determined that (6, 5) does in fact fall on the graph of the line.
Problem 8: Rewrite each expression in the form: xa (this is called base-exponent form).
Part a. $\dfrac{1}{x^{2}}$ Click here to show solution. We can move a factor from the denominator to the numerator as long as we change the sign of the exponent: $\dfrac{1}{x^{2}} = \dfrac{x^{-2}}{1}$ If it seems like the 1 in the numerator disappeared, remember that multiplying by one won’t change the value of $x^{-2}$. Dividing by 1 also won’t change the value of $x^{-2}$, so we can just write: $\dfrac{1}{x^{2}} = \boxed{x^{-2}}$ If we want to show all the steps, we could write: $\dfrac{1}{x^{2}} = \dfrac{1}{1 \cdot x^{2}} = \dfrac{1 \cdot x^{-2}}{1} =\dfrac{x^{-2}}{1} = x^{-2}$ Need Help ? BrownMath – Negative and Fractional Exponents MathPlanet – Exponent Rules Part b. $\sqrt{x}$ Click here to show solution. We can rewrite radical expressions using fractional exponents. For a square root, we use the fraction 1/2: $\sqrt{x} = \boxed{x^{\frac{1}{2}}}$ If this seems strange, think of it this way: 20 = 1 and 21 = 2 so we would expect that $2^{\frac{1}{2}}$ would produce a number between 1 and 2, and $\sqrt{2}$ is between 1 and 2. Part c. $\left(\dfrac{1}{x^{3}}\right)^{4}$ Click here to show solution. We can distribute an exponent over division, which means the 4 becomes the exponent of the 1 in the numerator and the x^3 in the denominator: $\left(\dfrac{1}{x^{3}}\right)^{4} = \dfrac{(1)^{4}}{\left(x^{3}\right)^{4}}$ Now 1 multiplied by itself 4 times is still 1, but in the denominator we have to combine the exponents. Recall that successive exponents can be combined by multiplication: $\left(\dfrac{1}{x^{3}}\right)^{4} = \dfrac{1}{x^{12}}$ Finally, we can rewrite this by moving the x into the numerator, and changing its exponent from positive to negative: $\left(\dfrac{1}{x^{3}}\right)^{4} = \boxed{x^{-12}}$ We could also have written out 4 copies of the original fraction: $\left(\dfrac{1}{x^{3}}\right)^{4} = \dfrac{1}{x^{3}} \cdot \dfrac{1}{x^{3}} \cdot \dfrac{1}{x^{3}} \cdot \dfrac{1}{x^{3}}$ This could then be expanded as: $\left(\dfrac{1}{x^{3}}\right)^{4} = \dfrac{1}{x\cdot x\cdot x} \cdot \dfrac{1}{x\cdot x\cdot x} \cdot \dfrac{1}{x\cdot x\cdot x} \cdot \dfrac{1}{x\cdot x\cdot x}$ Part d. $\sqrt[3]{x^{2}}$ Click here to show solution. Recall that the 3 is called the index of the root, and indicates that this is not a square root but a cube root. Following our earlier rule, we can rewrite this root using a 1/3 exponent: $\sqrt[3]{x^{2}} = \left(x^{2}\right)^{\frac{1}{3}}$ Exponent rules allow us to multiply successive exponents, so we calculate 2 times 1/3 to get 2/3 as the exponent: $\sqrt[3]{x^{2}} = \boxed{x^{\frac{2}{3}}}.$
Problem 9: Expand the following expressions. Remember to use FOIL.
F.O.I.L. stands for First, Outer, Inner, Last. When two binomials are being multiplied together, we can use FOIL to remember all the products that appear in the expansion. For example: $(a + b)(c + d) = ac + ad + bc + bd$ $\boxed{ac}$ is the product that results from multiplying the First terms together. $\boxed{ad}$ is the product of the Outer terms. $\boxed{bc}$ is the product of the Inner terms. $\boxed{bd}$ is the product of the Last terms. Need Help ? MathPlanet – Multiplication of Polynomials Part a. $(x+4)(x+6)$ Click here to show solution. Following FOIL will give us the following: $x \cdot x + x \cdot 6 + 4 \cdot x + 4 \cdot 6$ We can take each product and simplify it as: $x^{2} + 6x + 4x + 24$ Now we can combine 6x and 4x to make 10x: $\boxed{x^{2} + 10x + 24}.$ Part b. $(x+8)(x-3)$ Click here to show solution. Following FOIL will give us the following: $x \cdot x + x \cdot -3 + 8 \cdot x + 8 \cdot -3$ We can take each product and simplify it as: $x^{2} - 3x + 8x - 24$ Now we can combine -3x and 8x to make 5x: $\boxed{x^{2} + 5x - 24}.$ Part c. $(-x+2)(x-7)$ Click here to show solution. Following FOIL will give us the following: $-x \cdot x + -x \cdot -7 + 2 \cdot x + 2 \cdot -7$ We can take each product and simplify it as: $-x^{2} + 7x + 2x - 14$ Notice how -x times -7 gave us a positive result, because the negatives cancel each other. Now we can combine 7x and 2x to make 9x: $\boxed{x^{2} + 9x - 14}.$ Part d. $(-x-1)(-x+5)$ Click here to show solution. Following FOIL will give us the following: $-x \cdot -x + -x \cdot 5 + -1 \cdot -x + -1 \cdot 5$ We can take each product and simplify it as: $x^{2} - 5x + 1x - 5$ Notice how the first term is positive, because it is the product of two negatives. This is also true of the third term, 1x. Also, instead of writing 1x, we usually just write x, and the coefficient of 1 is implied Now we can combine -5x and x to make -4x: $\boxed{x^{2} - 4x - 5}.$
Problem 10: Rewrite each expression in factored form.
Part a. $x^{2}+8x+12$ Click here to show solution. List the factors of the constant term, 12: $\begin{array}{c|c} & \\\hline 1 & 12 \\2 & 6 \\3 & 4 \\-1 & -12 \\-2 & -6 \\-3 & -4 \end{array}$ Notice that we include pairs where both numbers are negative. We want to choose the pair that adds up to the coefficient of x, which is 8, so we pick 2 and 6. $x^{2}+8x+12 = \boxed{(x+2)(x+6)}.$ Need Help ? MathPlanet – Factoring Quadratics Part b. $x^{2}+9x+8$ Click here to show solution. List the factors of the constant term, 8: $\begin{array}{c|c} & \\\hline 1 & 8 \\2 & 4 \\-1 & -8 \\-2 & -4 \end{array}$ Notice that we include pairs where both numbers are negative. We want to choose the pair that adds up to the coefficient of x, which is 9, so we pick 1 and 8. $x^{2}+9x+8 = \boxed{(x+1)(x+8)}.$ Part c. $x^{2}+2x-15$ Click here to show solution. List the factors of the constant term, -15: $\begin{array}{c|c} & \\\hline 1 & -15 \\3 & -5 \\-1 & 15 \\-3 & 5 \end{array}$ Notice that we include each pair twice, once with the smaller number negative and once with the larger number negative. We want to choose the pair that adds up to the coefficient of x, which is 2, so we pick -3 and 5, because adding them together gives us -3 + 5 = 2. $x^{2}+2x-15 = \boxed{(x-3)(x+5)}.$ Part d. $x^{2}-x-42$ Click here to show solution. List the factors of the constant term, -42: $\begin{array}{c|c} & \\\hline 1 & -42 \\2 & -21 \\3 & -14 \\6 & -7 \\-1 & 42 \\-2 & 21 \\-3 & 14 \\-6 & 7 \end{array}$ Notice that we include each pair twice, once with the smaller number negative and once with the larger number negative. We want to choose the pair that adds up to the coefficient of x, which is -1, so we pick 6 and -7, because when we add them together we get 6 + -7 = -1. $x^{2}-x-42 = \boxed{(x+6)(x-7)}.$ Part e. $x^{2}-7x+12$ Click here to show solution. List the factors of the constant term, 12: $\begin{array}{c|c} & \\\hline 1 & 12 \\2 & 6 \\3 & 4 \\-1 & -12 \\-2 & -6 \\-3 & -4 \end{array}$ Notice that we include pairs where both numbers are negative. We want to choose the pair that adds up to the coefficient of x, which is -7, so we need to pick two negative numbers. We pick -3 and -4 because adding them gives us -3 + -4 = -7 $x^{2}-7x+12 = \boxed{(x-3)(x-4)}.$ Part f. $x^{2}+3x-40$ Click here to show solution. List the factors of the constant term, -40: $\begin{array}{c|c} & \\\hline 1 & -40 \\2 & -20 \\4 & -10 \\5 & -8 \\-1 & 40 \\-2 & 20 \\-4 & 10 \\-5 & 8 \end{array}$ Notice that we include each pair twice, once with the smaller number negative and once with the larger number negative. We want to choose the pair that adds up to the coefficient of x, which is 3, so we pick -5 and 8, because when we add them together we get -5 + 8 = 3. $x^{2}+3x-40 = \boxed{(x-5)(x+8)}.$
Problem 11: Solve the linear inequality.
Part a. $4x + 5 < 7x - 13$ Click here to show solution. Notice that there are terms containing x on both sides of the inequality. First, we can subtract 4x from both sides so that we can combine these like terms: $5 < 3x - 13$ Now we can move all the constants to the other side, which means we must add 13 to both sides: $18 < 3x$ Finally, we can divide both sides by 3 to isolate the variable: $6 < x$ This says 6 is less than x, which is another way of saying that x is greater than 6. Thus, we have: $\boxed{x > 6}.$ Need Help ? Video: KhanAcademy – Solving Inequalities
Problem 12: Express the following using interval notation.
Part a. $7 < x < 13$ Click here to show solution. The statement literally says that 7 is less than x, which is less than 13. In other words, x is greater than 7 and x is less than 13. This indicates a single interval, from 7 to 13. Because the original statement uses $\boxed{<}$ and not $\boxed{\leq}$ we know that 7 and 13 are NOT included in the interval. We use “soft brackets” (parentheses) to indicate that the interval includes all numbers between 7 and 13, but does NOT include the endpoints: $\boxed{( 7, 13 )}.$ Need Help ? Video: KhanAcademy – Interval Notation Text: Lumen – Interval Notation Part b. $-16 \leq x \leq 10$ Click here to show solution. The statement literally says that -16 is less than or equal to x, which is less than or equal to 10. In other words, x not greater than 10, and is not less than -16. This indicates a single interval with endpoints at -16 and 10. Because the original statement uses $\boxed{\leq}$ and not $\boxed{<}$ we know that -16 and 10 ARE included in the interval. We use “hard brackets” [square braces] to indicate that the interval includes all numbers between 7 and 13 and ALSO includes the endpoints: $\boxed{[ -16, 10 ]}.$ Part c. $-8 \leq x < -4$ Click here to show solution. Compared to the previous two problems, we simply need to use a hard bracket to start our interval, because the -8 is NOT included, and a soft bracket to end the interval, because the -4 IS included. $\boxed{[ -8, -4 )}.$ Part d. $x > 56$ Click here to show solution. The statement says x is greater than 56. This excludes the possibility that x = 56, so we will use a soft bracket. All the numbers greater than 56 together form a sort of interval with a left endpoint at 56, but no right endpoint. Since the interval starts at 56 and never ends, we can think of the right endpoint as being “at” infinity. Infinity is not a number, so we do not want to include it, which means we must use a soft bracket here as well. Using infinity as our right endpoint means that all numbers between 56 and infinity are included, even if both endpoints are not. $\boxed{( 56, \infty )}.$ Part e. $x \leq -42$ Click here to show solution. The statement says that x is less than or equal to -42. This includes the possibility that x = -42, so we must use a hard bracket. Since all negative numbers less than -42 are included in this interval, we need to start our interval at negative infinity (but not include it). -42 is actually where our interval ends. In other words, negative infinity is the left endpoint of this interval, and -42 is the right endpoint. $\boxed{( -\infty, -42 ]}.$
Part a. $x^2 - 8x + 5 = 0$ Click here to show solution. Recall the quadratic formula: $x = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ In the quadratic equation given above, we can see that a = 1, b = -8, and c = 5. Notice that, even thought there is no visible coefficient in front of the x2 term, we always assume a coefficient of 1, which is why a = 1. Also, the minus sign in front of the 8 must be included in the b value, which is why we have b = -8. $x = \dfrac{-(-8) \pm \sqrt{(-8)^{2}-4(1)(5)}}{2(1)}$ Notice how we placed parentheses around every place where we plugged in a value. Now we want to simplify -(-8) = 8 and (-8)2 = 64 and 4(1)(5) = 20 and 2(1) = 2. $x = \dfrac{8 \pm \sqrt{64-20}}{2}$ Since 64 – 20 = 44, and 44 = 4 times 11, we can write: $x = \dfrac{8 \pm \sqrt{4 \cdot 11}}{2}$ 4 is a square number, so we can pull it out from under the square root as a 2: $x = \dfrac{8 \pm 2\sqrt{11}}{2}$ Now, we can divide by 2. Recall that each term in the numerator must be divided by 2: $x = 4 \pm \sqrt{11}$ Finally, recall that the symbol $\boxed{\pm}$ means that we actually have two separate solutions for x: $\boxed{4 + 2\sqrt{11} \quad \text{or} \quad 4 - 2\sqrt{11}}.$ Need Help ? Video: KhanAcademy – Quadratic Formula
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## Friday, January 15, 2010
### Addition and the Part-Whole Circle
The other day Little Bean and I worked on addition up to 10 using the Part-Whole circle and tally marks. This Right Start lesson called for Little Bean to visualize the tally sticks moving from circle to circle so that he could easily solve problems up to 10 in his head (without the abacus).
Here is one of our problems: an addend goes in each smaller circle and then Bean needs to find the sum. 4 + 3 is not hard if he uses the abacus, but to do it in his mind he must visualize changing the problem into something he can understand more easily. Remember how Right Start teaches the child to think in terms of "5 and"? (5 and 2 is 7, 5 and 3 is 8, etc). This comes in super handy for doing mental math.
Here is where Bean's mind went to right off the bat: 4 + 3 is the same as 5 + 2 and 5 and 2 is 7! Therefore, 4 + 3 = 7. He decided to take away one of the tally sticks from the right circle and add it to the one on the left, giving him 5 and 2 as his parts. From there, it was a matter of recalling what 5 and 2 is from the very beginnings of Right Start Level A. I was SO impressed that Little Bean could do this in his head by thinking in terms of "5 and". We did many problems like this last week, and he loved them all. It was a fun challenge for him to answer my question, "What could we change in this problem so that we can solve it without counting?" He figured it out every time.
Just in case Little Bean wasn't getting it, there was the option of using the abacus. This doesn't require such a high level of critical thinking, but it is another way to come up with the answer without counting (counting is extremely inefficient, not to mention inaccurate; Right Start does not encourage counting at all). To do 4 + 3 on the abacus, you first push over 4 beads, then leaving a finger width gap, push over 3 beads. To see the answer, you just push the two addends together. You do not need to count the beads because the child has been trained to know that there are 5 blue beads and 5 yellow beads on each rod. So when you push them together, you'll immediately see 5 blue beads and 2 yellow beads have been entered and you have your answer because the child memorizes all the "5 and" facts in the very beginning. Easy peasy! Little Bean does this sometimes when solving math problems on worksheets or doing story problems.
Here is the completed problem 4 +3 = 7. See how easy it is to see the sum without counting? Even if you don't purchase Right Start for your math curriculum, I highly recommend getting an AL abacus from them!
Raising a Happy Child said...
I am interested to see how math is being taught here in US. I completely don't remember ever seeing those strange circles when I learned math, but we did a lot of manipulatives (so, yes, counting with props). It sounds though that the system works well for Little Bean.
Mom with the best job in the world! said...
Wow! That is impressive! I was horrible at math (doing it in my head) still am. I don't want to pass that on to my kids. I will look into Right Start and the abacus. Did you get it through Right Start? Where did you order it from?
I am beginning homeschooling this fall with my son who will be 5 in December. Can you send me the website?
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# Common Core: High School - Algebra : Solve System of Equations Using Substitution: CCSS.Math.Content.HSA-REI.C.5
## Example Questions
← Previous 1
### Example Question #1 : Solve System Of Equations Using Substitution: Ccss.Math.Content.Hsa Rei.C.5
Explanation:
The first step to do is solve for a variable. Let's solve for first.
Now we subtract from each side.
Now we divide by
Now since we found , we can plug this into the other equation and find .
Now we solve for .
Now we divide by .
Now round to the nearest hundredth.
Now since we have a value for , we plug this into the first equation and find .
Now we subtract from each side.
Now we divide by
Now round to the nearest hundredth.
### Example Question #2 : Solve System Of Equations Using Substitution: Ccss.Math.Content.Hsa Rei.C.5
Explanation:
The first step to do is solve for a variable. Let's solve for first.
Now we subtract from each side.
Now we divide by
Now since we found , we can plug this into the other equation and find .
Now we solve for .
Now we divide by
Now round to the nearest hundredth.
Now since we have a value for , we plug this into the first equation and find .
Now we add to each side.
Now we divide by
Now round to the nearest hundredth.
### Example Question #3 : Solve System Of Equations Using Substitution: Ccss.Math.Content.Hsa Rei.C.5
Explanation:
The first step to do is solve for a variable. Let's solve for first.
Now we subtract from each side.
Now we divide by
Now since we found , we can plug this into the other equation and find .
Now we solve for .
Now we divide by .
Now round to the nearest hundredth.
Now since we have a value for , we plug this into the first equation and find .
Now we add to each side.
Now we divide by
Now round to the nearest hundredth.
### Example Question #4 : Solve System Of Equations Using Substitution: Ccss.Math.Content.Hsa Rei.C.5
Explanation:
The first step to do is solve for a variable. Let's solve for first.
Now we subtract from each side.
Now we divide by
Now since we found , we can plug this into the other equation and find .
Now we solve for .
Now we divide by .
Now round to the nearest hundredth.
Now since we have a value for , we plug this into the first equation and find .
Now we subtract from each side.
Now we divide by
Now round to the nearest hundredth.
### Example Question #5 : Solve System Of Equations Using Substitution: Ccss.Math.Content.Hsa Rei.C.5
Explanation:
The first step to do is solve for a variable. Let's solve for first.
Now we subtract from each side.
Now we divide by
Now since we found , we can plug this into the other equation and find .
Now we solve for .
Now we divide by .
Now round to the nearest hundredth.
Now since we have a value for , we plug this into the first equation and find .
Now we add on each side.
Now we divide by
Now round to the nearest hundredth.
### Example Question #6 : Solve System Of Equations Using Substitution: Ccss.Math.Content.Hsa Rei.C.5
Explanation:
The first step to do is solve for a variable. Let's solve for first.
Now we subtract from each side.
Now we divide by
Now since we found , we can plug this into the other equation and find .
Now we solve for .
Now we divide by .
Now round to the nearest hundredth.
Now since we have a value for , we plug this into the first equation and find .
Now we add -17.8349082823791 to each side.
Now we divide by
Now round to the nearest hundredth.
### Example Question #7 : Solve System Of Equations Using Substitution: Ccss.Math.Content.Hsa Rei.C.5
Explanation:
The first step to do is solve for a variable. Let's solve for first.
Now we subtract from each side.
Now we divide by
Now since we found , we can plug this into the other equation and find .
Now we solve for .
Now we divide by .
Now round to the nearest hundredth.
Now since we have a value for , we plug this into the first equation and find .
Now we add to each side.
Now we divide by
Now round to the nearest hundredth.
### Example Question #8 : Solve System Of Equations Using Substitution: Ccss.Math.Content.Hsa Rei.C.5
Explanation:
The first step to do is solve for a variable. Let's solve for first.
Now we subtract from each side.
Now we divide by
Now since we found , we can plug this into the other equation and find .
Now we solve for .
Now we divide by .
Now round to the nearest hundredth.
Now since we have a value for , we plug this into the first equation and find .
Now we subtract from each side.
Now we divide by
Now round to the nearest hundredth.
### Example Question #9 : Solve System Of Equations Using Substitution: Ccss.Math.Content.Hsa Rei.C.5
Explanation:
The first step to do is solve for a variable. Let's solve for first.
Now we subtract from each side.
Now we divide by
Now since we found , we can plug this into the other equation and find .
Now we solve for .
Now we divide by .
Now round to the nearest hundredth.
Now since we have a value for , we plug this into the first equation and find .
Now we add to each side.
Now we divide by
Now round to the nearest hundredth.
### Example Question #10 : Solve System Of Equations Using Substitution: Ccss.Math.Content.Hsa Rei.C.5
Explanation:
The first step to do is solve for a variable. Let's solve for first.
Now we subtract from each side.
Now we divide by
Now since we found , we can plug this into the other equation and find .
Now we solve for .
Now we divide by .
Now round to the nearest hundredth.
Now since we have a value for , we plug this into the first equation and find .
Now we subtract from each side.
Now we divide by
Now round to the nearest hundredth.
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# ACT Math : How to find a missing side with tangent
## Example Questions
### Example Question #1 : How To Find A Missing Side With Tangent
For the triangles in the figure given, which of the following is closest to the length of line NO?
10
9
7
6
8
9
Explanation:
First, solve for side MN. Tan(30°) = MN/16√3, so MN = tan(30°)(16√3) = 16. Triangle LMN and MNO are similar as they're both 30-60-90 triangles, so we can set up the proportion LM/MN = MN/NO or 16√3/16 = 16/x. Solving for x, we get 9.24, so the closest whole number is 9.
### Example Question #1 : How To Find A Missing Side With Tangent
Josh is at the state fair when he decides to take a helicopter ride. He looks down at about a 35 ° angle of depression and sees his house. If the helicopter was about 250 ft above the ground, how far does the helicopter have to travel to be directly above his house?
sin 35 ° = 0.57 cos 35 ° = 0.82 tan 35 ° = 0.70
142.50 ft
438.96 ft
304.88 ft
357.14 ft
205.00 ft
357.14 ft
Explanation:
The angle of depression is the angle formed by a horizontal line and the line of sight looking down from the horizontal.
This is a right triangle trig problem. The vertical distance is 250 ft and the horizontal distance is unknown. The angle of depression is 35°. We have an angle and two legs, so we use tan Θ = opposite ÷ adjacent. This gives an equation of tan 35° = 250/d where d is the unknown distance to be directly over the house.
### Example Question #1 : How To Find A Missing Side With Tangent
Consider the triangle where . Find to the nearest decimal place.
Note: The triangle is not necessarily to scale
None of the other answers
Explanation:
To solve this equation, it is best to remember the mnemonic SOHCAHTOA which translates to Sin = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, and Tangent = Opposite / Adjacent. Looking at the problem statement, we are given an angle and the side opposite of the angle, and we are looking for the side adjacent to the angle. Therefore, we will be using the TOA part of the mnemonic. Inserting the values given in the problem statement, we can write . Rearranging, we get . Therefore
### Example Question #1 : How To Find A Missing Side With Tangent
A piece of wire is tethered to a building at a angle. How far back is this wire from the bottom of said building? Round to the nearest inch.
Explanation:
Begin by drawing out this scenario using a little right triangle:
Note importantly: We are looking for as the the distance to the bottom of the building. Now, this is not very hard at all! We know that the tangent of an angle is equal to the ratio of the side adjacent to that angle to the opposite side of the triangle. Thus, for our triangle, we know:
Using your calculator, solve for :
This is . Now, take the decimal portion in order to find the number of inches involved.
Thus, rounded, your answer is feet and inches.
### Example Question #14 : Trigonometry
What is the value of in the right triangle above? Round to the nearest hundredth.
Explanation:
Recall that the tangent of an angle is the ratio of the opposite side to the adjacent side of that triangle. Thus, for this triangle, we can say:
Solving for , we get:
or
### Example Question #15 : Trigonometry
In the right triangle shown above, let , and . What is the value of Reduce all fractions.
Explanation:
First we need to find the value of . Use the mnemonic SOH-CAH-TOA which stands for:
.
Now we see at point we are looking for the opposite and adjacent sides, which are and respectively.
Thus we get that
and plugging in our values and reducing yields:
### Example Question #16 : Trigonometry
In a given right triangle , leg and . Using the definition of , find the length of leg . Round all calculations to the nearest hundredth.
Explanation:
In right triangles, SOHCAHTOA tells us that , and we know that and hypotenuse . Therefore, a simple substitution and some algebra gives us our answer.
Use a calculator or reference to approximate cosine.
Isolate the variable term.
Thus, .
### Example Question #17 : Trigonometry
In a given right triangle , leg and . Using the definition of , find the length of leg . Round all calculations to the nearest tenth.
Explanation:
In right triangles, SOHCAHTOA tells us that , and we know that and hypotenuse . Therefore, a simple substitution and some algebra gives us our answer.
Use a calculator or reference to approximate cosine.
Isolate the variable term.
Thus, .
### Example Question #18 : Trigonometry
In a given right triangle , leg and . Using the definition of , find the length of leg . Round all calculations to the nearest tenth.
Explanation:
In right triangles, SOHCAHTOA tells us that , and we know that and leg. Therefore, a simple substitution and some algebra gives us our answer.
Use a calculator or reference to approximate cosine.
Isolate the variable term.
Thus, .
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Example 4: Simplify the improper fraction, $$\frac{24}{9}$$. Your IP: 173.212.240.46 For example, if your improper fraction was 10/4, you would start by dividing 10 by 4 to get 2 with a remainder of 2. • In other words, what's the biggest number that goes into '2' (the numerator) and '4', the denominator? For example, simplify the fraction 24/ 108. Examples of improper fractions are 16/3, 81/9, 525/71. There is some extra work, however, if you are asked to rewrite the simplified improper fraction into a mixed number. As long as the denominator is less than 124 you have an improper fraction, and you can use the methods presented here to solve. https://www.mathsisfun.com/improper-fractions.html, http://www.coolmath.com/prealgebra/01-fractions/fractions-07-improper-01, http://downloads.bbc.co.uk/skillswise/maths/ma17frac/factsheet/ma17frac-l1-f-improper-fractions-and-simplifying.pdf, http://www.coolmath.com/prealgebra/01-fractions/fractions-07-improper-02, consider supporting our work with a contribution to wikiHow. Step 1: Make It a Fraction ( for Mixed Numbers Only ) So, what is a mixed number? There are two methods to simplify fractions. X % of people told us that this article helped them. If I multiply both terms of 2/3 by 4, what do I get? Then, turn the remainder into a fraction by placing it over the denominator of the original fraction. Let's examine the fraction 2/4. wikiHow is where trusted research and expert knowledge come together. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. Label the end '/12'. In this example, we will multiply two very simple proper fractions and apply the 3 steps above. In the first method, try to divide (Only the whole number) both the numerator and denominator of the fraction by 2, 3, 5, 7, etc, until we cannot go any further. Mixed fractions: A fraction is called a mixed fraction, which is represented by a whole number and a fraction. The first step is to determine the largest number that evenly divides the numerator and the denominator (also called the Greatest Common Factor of these numbers). Include your email address to get a message when this question is answered. This method works pretty well if the numerator and denominators of your fractions are relatively small. What are improper fractions and mixed numbers or mixed fractions, How to convert between improper fractions and mixed numbers, examples and step by step solutions. Learn more... Fractions are numbers that represent parts of whole numbers. Step 4: Add or subtract the numerators and keep the denominator the same. Methods to Simplify Fraction General Steps. One way to reduce fractions is to find the greatest common factor of the numerator and the denominator. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. by example, 1/2 *2/3. If there is no remainder, then there is no fraction - the result is simply a whole number. We’ll get into mixed numbers a little later. You cannot convert an improper fraction into a proper fraction. That's as far as the fraction will reduce, because there is no whole number that will divide evenly into both the numerator and the denominator of this last fraction. Convert an Improper Fraction to a Mixed Number. You can simplify fractions by dividing the numerator and denominator by the same number. An improper fraction can actually be converted into a mixed number and vice versa. To learn how to simplify an improper fraction by using a model, keep reading! Mixed fraction to Improper Fraction. The quotient will be the mixed number. Simplifying (or reducing) fractions means to make the fraction as simple as possible. $$\frac{18}{24}$$ Step 3: Simplify the fraction if needed. Step 1. ... You can follow the below steps to simplify the fraction: Step1: Make a list of factors of both numerator and denominator. If you really can’t stand to see another ad again, then please consider supporting our work with a contribution to wikiHow. If there are no mixed fractions, proceed to Step 2 right away. Let’s do a few more examples. Step 1 – Multiply the numerators Step 2 – Multiply the denominators Step 3 – Simply the fraction to it’s the simplest form (You have to follow this step, only when we have common factors for the numerator and denominator.Otherwise keep it as it is. ) For example how would i simplify this question ? Continue until you can’t go any further. Step 1: Roman, Rae and Trig enter Dr S. Cape’s escape room laboratory. Step1: Convert both mixed numbers into improper fractions. This article has been viewed 68,593 times. (You can also arrive at the same result by dividing the original numerator and denominator by 6, which is 2 x 3.) The steps on how to convert an improper fraction into a mixed number will be discussed below. For example, if your denominator is 4, then divide each circle you draw into 4 equal pieces, or quarters. Last Updated: October 1, 2020 If the whole number part of the mixed number is zero, you can leave it off the simplified version. Please help us continue to provide you with our trusted how-to guides and videos for free by whitelisting wikiHow on your ad blocker. There, they solve a number of clues involving graduated cylinders and a periodic table that both require simplifying fractions. Divide both numerator and denominator by the largest factor that both numbers have in common. Simplifying fractions means dividing the top and bottom by the same number so that the top and bottom become as small or as simple as possible. It is simply a number that has a whole part, and a part of a whole, such as one and one seventh, three and three fourths, etc. All you have to do is provide the input value i.e. Study the fraction part, and simplify it, if possible by finding and dividing out the common factors of the numerator and denominator. Simplifying fractions using their GCF is more efficient than dividing the numerator and denominator of the fraction by prime numbers until their only shared factor is 1. To do this conversion on a mixed fraction, simply multiply the denominator by the whole number, and then add the numerator. How to Convert an Improper Fraction to a Mixed Number. If necessary, simplify the final fraction to get your answer. 7 × 2 =14. If a fraction has a numerator greater than its denominator it is termed an “improper fraction” and can be simplified as a mixed number (a number that combines a whole number and a fraction). Calculator to add, subtract, multiply, divide, convert and simplify fractions or mixed numbers with step by step explanation. What if I have a big numerator, like 124? How do we simplify mixed numbers? Performance & security by Cloudflare, Please complete the security check to access. Then subtract 18 from 41, and subtract 5/9 from 9/9. If you prefer, consider converting a40 from 1 scale to the other. Example 1 - Multiplying proper fractions. Sometimes you may try to identify the GCF mentally, and although you may find a common factor, it may not be the greatest common factor. To simplify an improper fraction, start by turning it into a mixed number by dividing the numerator by the denominator. Research source Divide the numerator and denominator by 3. By using our site, you agree to our. To use this method of simplification, use the following steps: Well, simplify mixed fractions calculator is 100% free tool that simplifies the given mixed fractions number within a blink of eyes, just follow the given step to attain instant outcomes: Inputs: All you need to enter the values of mixed fraction into designated boxes of this calculator The difference is 23 4/9. Do you want to know how to solve simplifying fractions? Follow steps 1 through 4 as explained above: Step 1 Step 2. Like simplifying it into a regular fraction? OK, let us do a non-mixed first and then add the steps for mixed. Some improper fractions can also represent whole numbers, such as, All tip submissions are carefully reviewed before being published, This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. Step 2. For example, if you are given the problem “3 1/2 / 1 3/4,” you would first convert the two mixed fractions: We know ads can be annoying, but they’re what allow us to make all of wikiHow available for free. First. How to Simplify Fractions Step by Step. • If you multiply both numerator and denominator by 4, you get 8/12 or 2/3. Divide 3 into both the numerator and denominator to simplify the fraction to its lowest terms. Thanks to all authors for creating a page that has been read 68,593 times. We simplify mixed numbers in three steps: Step 1) Convert mixed number to an improper fraction. Free Simplify Fractions Year 6 Fractions Learning Video Clip. Convert mixed numbers to improper fractions; Simplify and cancel; Multiply the numerators and denominators; Convert any improper fraction answer back to a mixed number How to simplify fractions using their GCF. This gives you your improper fraction. In other words, you're multiplying 2/3 by 4/4, which is multiplying by 1. What if I want to simplify it without making it a mixed number? The remainder will be interpreted as the fraction part of your mixed number. The steps below show an example of how to do this: Mixed fraction includes an entire number and a fraction. Cloudflare Ray ID: 6005c088cd1fdfbb Step 3) Convert the simplified improper fraction back to a mixed number. The denominator will not divide evenly into the numerator. fraction in the input field and click on the enter button to get the output displayed in no time. Solution: 24 / 108 ÷ 2 = 12 / 56 When 12 / 56 ÷ 2 = 6 / 27 6 / 27 ÷ 3 = 2 / 9 To convert a mixed numbers back to improper fraction form, multiply the whole number by the denominator and add the product to the numerator. Draw a bar and split it into 12 equal segments: the denominator of the fraction. Step 2: Add the numerator of the Fraction to the result in step 1. i.e Add 1+ 14 Step 1: Convert to improper fractions Check for any mixed fractions and change them to improper fractions first. Then, place the remainder over the denominator of the original fraction to get 2/4, and simplify to get 1/2, making the answer 2 1/2. First, multiply the whole number by the denominator of the fractional part. If necessary, simplify the final fraction to get your answer. Simplifying Fractions Simplifying Fractions – Example 1: Simplify . That's ok.... You can still continue to simplify the fraction even if you didn't identify the GCF correctly the first time. Let's check it out. You're in the right place! The result is the numerator of the improper fraction. Take this final figure as your new numerator and place it over the original denominator. Exercise 3: Subtract the fractions, . We have walked through the four steps to multiply fractions. The division method would be easiest, since drawing a model with 124 pieces would take some time. The next step is to rewrite the fraction for a ratio. Welcome to Simplifying Fractions Step by Step with Mr. J! Step 5: If the answer is an improper form, reduce the fraction into a mixed number. so it is helpful to know how to create them. Then, turn the remainder into a fraction by placing it over the denominator of the original fraction. Method 1. {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/5\/5f\/Simplify-an-Improper-Fraction-Step-1-Version-3.jpg\/v4-460px-Simplify-an-Improper-Fraction-Step-1-Version-3.jpg","bigUrl":"\/images\/thumb\/5\/5f\/Simplify-an-Improper-Fraction-Step-1-Version-3.jpg\/aid4531059-v4-728px-Simplify-an-Improper-Fraction-Step-1-Version-3.jpg","smallWidth":460,"smallHeight":345,"bigWidth":"728","bigHeight":"546","licensing":"
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Expanding two brackets
1. 1. Objective Be able to expand brackets and simplify the expressions that arise
2. 2. Question (Level 6) • Expand and simplify the following expression (2a + 3)(4a + 2)
3. 3. Step 1 (2a + 3)(4a + 2) 2a +3 4a +2 Multiply the individuals components together
4. 4. 2a +3 4a 8a2 +2
5. 5. 2a +3 4a 8a2 +12a +2
6. 6. 2a +3 4a 8a2 +12a +2 +4a
7. 7. 2a +3 4a 8a2 +12a +2 +4a +6
8. 8. Step 2 • Now collect the like terms 2a +3 4a 8a2 +12a +2 +4a +6 16a
9. 9. Step 3 So we have 2a +3 4a 8a2 +12a +2 +4a +6 8a2 + 16a + 6
10. 10. This will always work! You may have to deal with negatives too, but don’t worry it’s just as easy.
11. 11. Dealing with negatives Expand and simplify the following expression (3b + 2)(4b – 6)
12. 12. Step 1 (3b + 2)(4b – 6) 3b +2 4b -6 Multiply the individuals components together
13. 13. 3b +2 4b 12b2 -6
14. 14. 3b +2 4b 12b2 +8b -6
15. 15. 3b +2 4b 12b2 +8b -6 -18b
16. 16. 3b +2 4b 12b2 +8b -6 -18b -12
17. 17. Step 2 • Now collect the like terms 2a +3 4a 12b2 +8b +2 -18b -12 -10b
18. 18. Step 3 So we have 2a +3 4a 12b2 +8b +2 -18b -12 12b2 –10b – 12
19. 19. That’s it! Just be careful when you are multiplying negative numbers and take your time when collecting like terms.
20. 20. Let’s try some questions – You only have five minutes! • Expand and simplify: 1) (2a + 7)(6a + 2) 2) (4b + 5)(3b +9) 3) (5c + 4)(8c – 2) 4) (7d + 5)(3d – 4) 5) (9y –2)(3y + 6) 6) (4p – 3)(2p – 6) 7) (8m – 2)(7m – 4)
21. 21. Time is up! Let’s check some answers
22. 22. Did you get them correct? 1) (2a + 7)(6a + 2) 2) (4b + 5)(3b +4) 3) (5c + 4)(8c – 2) 4) (7d + 5)(3d – 4) 5) (9y –2)(3y + 4) 6) (4p – 3)(2p – 6) 7) (8m – 2)(7m – 4) 12a2 + 46a +14 12b2 + 31b +20 40c2 +22c - 8 21d2 – 13d -20 27y2 +30y - 8 8p2 –30p + 18 56m2 – 46m +8
23. 23. THINK! Did you meet the objective of the lesson? Be able to expand brackets and simplify the expressions that arise
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# Wild About Math!Making Math fun and accessible
11Nov/07271
## Impress your friends with mental Math tricks
See Math tricks on video at the Wild About Math! mathcasts page.
_____________________
Being able to perform arithmetic quickly and mentally can greatly boost your self-esteem, especially if you don't consider yourself to be very good at Math. And, getting comfortable with arithmetic might just motivate you to dive deeper into other things mathematical.
This article presents nine ideas that will hopefully get you to look at arithmetic as a game, one in which you can see patterns among numbers and pick then apply the right trick to quickly doing the calculation.
Don't be discouraged if the tricks seem difficult at first. Learn one trick at a time. Read the description, explanation, and examples several times for each technique you're learning. Then make up some of your own examples and practice the technique.
As you learn and practice the tricks make sure you check your results by doing multiplication the way you're used to, until the tricks start to become second nature. Checking your results is critically important: the last thing you want to do is learn the tricks incorrectly.
1. Multiplying by 9, or 99, or 999
Multiplying by 9 is really multiplying by 10-1.
So, 9x9 is just 9x(10-1) which is 9x10-9 which is 90-9 or 81.
Let's try a harder example: 46x9 = 46x10-46 = 460-46 = 414.
One more example: 68x9 = 680-68 = 612.
To multiply by 99, you multiply by 100-1.
So, 46x99 = 46x(100-1) = 4600-46 = 4554.
Multiplying by 999 is similar to multiplying by 9 and by 99.
38x999 = 38x(1000-1) = 38000-38 = 37962.
2. Multiplying by 11
To multiply a number by 11 you add pairs of numbers next to each other, except for the numbers on the edges.
Let me illustrate:
To multiply 436 by 11 go from right to left.
First write down the 6 then add 6 to its neighbor on the left, 3, to get 9.
Write down 9 to the left of 6.
Then add 4 to 3 to get 7. Write down 7.
Then, write down the leftmost digit, 4.
So, 436x11 = is 4796.
Let's do another example: 3254x11.
The answer comes from these sums and edge numbers: (3)(3+2)(2+5)(5+4)(4) = 35794.
One more example, this one involving carrying: 4657x11.
Write down the sums and edge numbers: (4)(4+6)(6+5)(5+7)(7).
Going from right to left we write down 7.
Then we notice that 5+7=12.
So we write down 2 and carry the 1.
6+5 = 11, plus the 1 we carried = 12.
So, we write down the 2 and carry the 1.
4+6 = 10, plus the 1 we carried = 11.
So, we write down the 1 and carry the 1.
To the leftmost digit, 4, we add the 1 we carried.
So, 4657x11 = 51227 .
3. Multiplying by 5, 25, or 125
Multiplying by 5 is just multiplying by 10 and then dividing by 2. Note: To multiply by 10 just add a 0 to the end of the number.
12x5 = (12x10)/2 = 120/2 = 60.
Another example: 64x5 = 640/2 = 320.
And, 4286x5 = 42860/2 = 21430.
To multiply by 25 you multiply by 100 (just add two 0's to the end of the number) then divide by 4, since 100 = 25x4. Note: to divide by 4 your can just divide by 2 twice, since 2x2 = 4.
64x25 = 6400/4 = 3200/2 = 1600.
58x25 = 5800/4 = 2900/2 = 1450.
To multiply by 125, you multipy by 1000 then divide by 8 since 8x125 = 1000. Notice that 8 = 2x2x2. So, to divide by 1000 add three 0's to the number and divide by 2 three times.
32x125 = 32000/8 = 16000/4 = 8000/2 = 4000.
48x125 = 48000/8 = 24000/4 = 12000/2 = 6000.
4. Multiplying together two numbers that differ by a small even number
This trick only works if you've memorized or can quickly calculate the squares of numbers. If you're able to memorize some squares and use the tricks described later for some kinds of numbers you'll be able to quickly multiply together many pairs of numbers that differ by 2, or 4, or 6.
Let's say you want to calculate 12x14.
When two numbers differ by two their product is always the square of the number in between them minus 1.
12x14 = (13x13)-1 = 168.
16x18 = (17x17)-1 = 288.
99x101 = (100x100)-1 = 10000-1 = 9999
If two numbers differ by 4 then their product is the square of the number in the middle (the average of the two numbers) minus 4.
11x15 = (13x13)-4 = 169-4 = 165.
13x17 = (15x15)-4 = 225-4 = 221.
If the two numbers differ by 6 then their product is the square of their average minus 9.
12x18 = (15x15)-9 = 216.
17x23 = (20x20)-9 = 391.
5. Squaring 2-digit numbers that end in 5
If a number ends in 5 then its square always ends in 25. To get the rest of the product take the left digit and multiply it by one more than itself.
35x35 ends in 25. We get the rest of the product by multiplying 3 by one more than 3. So, 3x4 = 12 and that's the rest of the product. Thus, 35x35 = 1225.
To calculate 65x65, notice that 6x7 = 42 and write down 4225 as the answer.
85x85: Calculate 8x9 = 72 and write down 7225.
6. Multiplying together 2-digit numbers where the first digits are the same and the last digits sum to 10
Let's say you want to multiply 42 by 48. You notice that the first digit is 4 in both cases. You also notice that the other digits, 2 and 8, sum to 10. You can then use this trick: multiply the first digit by one more than itself to get the first part of the answer and multiply the last digits together to get the second (right) part of the answer.
An illustration is in order:
To calculate 42x48: Multiply 4 by 4+1. So, 4x5 = 20. Write down 20.
Multiply together the last digits: 2x8 = 16. Write down 16.
The product of 42 and 48 is thus 2016.
Notice that for this particular example you could also have noticed that 42 and 48 differ by 6 and have applied technique number 4.
Another example: 64x66. 6x7 = 42. 4x6 = 24. The product is 4224.
A final example: 86x84. 8x9 = 72. 6x4 = 24. The product is 7224
7. Squaring other 2-digit numbers
Let's say you want to square 58. Square each digit and write a partial answer. 5x5 = 25. 8x8 = 64. Write down 2564 to start. Then, multiply the two digits of the number you're squaring together, 5x8=40.
Double this product: 40x2=80, then add a 0 to it, getting 800.
Add 800 to 2564 to get 3364.
This is pretty complicated so let's do more examples.
32x32. The first part of the answer comes from squaring 3 and 2.
3x3=9. 2x2 = 4. Write down 0904. Notice the extra zeros. It's important that every square in the partial product have two digits.
Multiply the digits, 2 and 3, together and double the whole thing. 2x3x2 = 12.
Add a zero to get 120. Add 120 to the partial product, 0904, and we get 1024.
56x56. The partial product comes from 5x5 and 6x6. Write down 2536.
5x6x2 = 60. Add a zero to get 600.
56x56 = 2536+600 = 3136.
One more example: 67x67. Write down 3649 as the partial product.
6x7x2 = 42x2 = 84. Add a zero to get 840.
67x67=3649+840 = 4489.
8. Multiplying by doubling and halving
There are cases when you're multiplying two numbers together and one of the numbers is even. In this case you can divide that number by two and multiply the other number by 2. You can do this over and over until you get to multiplication this is easy for you to do.
Let's say you want to multiply 14 by 16. You can do this:
14x16 = 28x8 = 56x4 = 112x2 = 224.
Another example: 12x15 = 6x30 = 6x3 with a 0 at the end so it's 180.
48x17 = 24x34 = 12x68 = 6x136 = 3x272 = 816. (Being able to calculate that 3x27 = 81 in your head is very helpful for this problem.)
9. Multiplying by a power of 2
To multiply a number by 2, 4, 8, 16, 32, or some other power of 2 just keep doubling the product as many times as necessary. If you want to multiply by 16 then double the number 4 times since 16 = 2x2x2x2.
15x16: 15x2 = 30. 30x2 = 60. 60x2 = 120. 120x2 = 240.
23x8: 23x2 = 46. 46x2 = 92. 92x2 = 184.
54x8: 54x2 = 108. 108x2 = 216. 216x2 = 432.
Practice these tricks and you'll get good at solving many different kinds of arithmetic problems in your head, or at least quickly on paper. Half the fun is identifying which trick to use. Sometimes more than one trick will apply and you'll get to choose which one is easiest for a particular problem.
Multiplication can be a great sport! Enjoy.
See Math tricks on video at the Wild About Math! mathcasts page.
Check out these related articles:
1. Hey there!
Just wanted to stop by and say thanks for a great post and read – math doesn’t have to be boring!
Have a nice day,
Alex
2. Alex,
You’re quite welcome. Glad you liked it.
Sol
3. Hey Sol i don’t expect to be using them to impress anybody, but some of those techniques will come in very handy. Thanks a lot. I’ll give it a Stumble so others can learn them too.
4. Karen,
Thanks for the kind words and the stumbling.
Sol
5. WOW!!!
I wish I knew this while I was in school!!!
That is neat…
6. Very cool – I plan to use this as often as possible. By the way there is a minor typo…
“So, 46×99 = 4600x(100-1) = 4600-46 = 4554.” should read “So, 46×99 = 46x(100-1) = 4600-46 = 4554.”
that’s ok though – this page is so cool i think i’ll let this one slide.
7. Alan,
I’m glad you like the page. I’m not seeing the error, though. Your correction looks to me the same as what I wrote. Please elaborate.
Thanks.
8. mental math tricks helps to perform arithmetic calculations quickly
Mental math
9. Multiplying by 11 is easier when adjusting the rule for multiplying by 9. Just think of 11 as (10+1)
So 436 × 11 = 4360 + 436 = 4796…that’s the simpler way of explaining why the digits add up to each other like you wrote:
4360
+ 436
—–
4796
10. Excellent Stumble. Thumb Up.
@IB – yes, your way of showing why this “adding the pair” approach for multiplying by 11 is right on.
12. You are saying 46×99 = 4600x(100-1) when it should be 46 instead of 4600.
13. Fred, Alan:
Thanks for the correction. Now I see it!
Article is now fixed.
14. These are just retarded, any idiot could think them up…
15. omg the trick for multiplying squares is awesome.
-Rob
16. Rob,
17. Even primary school students know these simple tricks in China…..
18. Shao Han,
Can you recommend any books in English where I could learn about what Math Chinese students learn?
19. Hi, I has got a problem with multipling:
If i multiply some numbers in that metod some things go wrong:
E.G. (agree):
87 * 81
88*80 + 7 = 7047
38*20
38*20 + 0 = 760
E.G.(doesnt agree):
56 * 17
63*10 + 42 =’ 672
Real: 952
75 * 88
93*70 + 40 =’ 6550
Real:6000
85 * 26
91*20+30 =’ 1850
Real: 2210
Please tell me what mistake was doing!
20. Hi BlueS,
I assume you’re trying to use technique #4 in the article. That technique only works when the two numbers differ by a small even amount and when you can easily calculate the square of the number in the middle of the two numbers (i.e. the average).
In your example of 56×17 I see what you’re trying to do but it’s different than this trick.
Let a=56
Let b=17
You want to calculate a*b, right?
I see that you added 7 to 56 and subtracted 7 from 17 so that you could multiply by 10. That’s a good idea.
So, you were computing (a+7)x(b-7).
(a+7)x(b-7) =
(axb)-(7xa)+(7xb)-49 =
(axb)-7x(a-b)-49
So, (axb) = (a+7)x(b-7) + 7x(a-b)+49
Or, 56×17 = 63×10 + 7x(39)+49 = 630 + 273 + 49
= 952
This approach is not easy for these two numbers.
What you could do with what you’ve noticed is to say that 56×17 = 56x(10+7) = 56×10 + 56×7
= 560+392 = 952.
Does this help?
21. everyone should already know this in my opinion. it’s basic basic math.
22. I agree that everyone should know things like how to multiply by 9 or 11. However, the method used to achieve the answer may be quite different. I was taught multiplication and agree that it is basic math, however I was never taught “tricks” such as this; basically easy ways to remember how to multiply certain numbers. I am horrible with math so ordinarily I cannot do multiplication in my head. However, with these tips, I may get better at it.
23. @Alex: Knowing these tricks is largely about having a relationship with numbers. I’m glad you have it but not everyone does.
@Amanda: Do report back on how these techniques help you if they do. The Vedic Math approach allows people to do multiplication without knowing more than up to 5×5 in their multiplication tables. I’ll post some Vedic Math techniques in the future.
24. good tricks mate. you make it easier. nice one.
dont listen to them < >.
25. Your annotation for the multiplying by 9’s is wrong. You have:
9×9 is just 9x(10-1) which is 9×10-9 which is 90-9 or 81
If you follow the acronym PEMDAS you would do what is in the parenthesis first and then multiply which would give you 9×9. You should have stated you need to use the distributive method. Which would mean it would read (9×10)-(9×1)= 90-9= 81.
26. This is a great system developed a long time ago by Jakow Trachtenberg whilst in a Nazi camp. More info here; http://en.wikipedia.org/wiki/Trachtenberg_system
27. there is an easier way to multiply by 9.
this is the way I learned when i was in school.
this works all the way up to 9 x 9, but if you’re in elementary school, it can come in handy.
take the number you are multiplying 9 by, and subtract one. then figure out what number plus that number equals nine.
put the first and second answer beside each other and you get tour answer.
it’s simpler than it sounds…..
example:
9 x 7 = ?
7 – 1 = 6
6 + 3 = 9
9 x 3 = ?
3 – 1 = 2
2 + 7 = 9
this was the easiest way for me.
28. Well let me put the technique #4. Generically it uses the fact that (a+b)*(a-b) = a^2 – b^2. Here b is half of the difference between the numbers, hence a is the average. In the example where difference is 6, the minus 9 comes coz (6/2)^2 = 9. Similary it can be done for large differences also, but it depends how comfortable is one is with squaring numbers.
29. I love this!
There is one problem with technique #6 though:
Take 81*89 where the first digits are equal, and where the last digits’ sum is 10.
By applying your method the answer should be: (8*9)(9*1) which gives 729, when the correct answer is 7209. This happens in all cases where the product of the last digits is less than 10 (ie with 9 and 1), so a quick fix would be to always make sure you have two digits in the product, and if not then add a zero in front.
30. Oh!
I just dicovered that the problem I mentioned above also applies to technique #7!
If the square of the second number is less than 10, you also have to add a zero before it in order to get the right answer:
Wrong:
92^2
=> 180 =>360
814
360
1174
Right:
92^2
=> 180 => 360
8104
360
8464
31. This is great for “impressing” the somewhat math smart girls. I’m sure that the math wizzes already know this.
Bookmarked
===
http://www.BecomingAPUA.com – V is the #1 Pick Up Artist
===
32. I invited #8 a long time ago… at least, I thought I did. LOL
33. very nice math trick.. wish i had known when i was in school.. if you ask me, i would suggest to put them in elementary school’s curriculum
34. Great work, very useful.
35. A fantastic work! I didn’t just like it, I loved it. I have so much respect and appreciation for people like you who spend their time doing something productive on the net.
Thanks man!
36. finally :)!!
37. have never been good with numbers, but these things do help quite a bit. how come we were never taught this at school???
38. Anyone who has taken CAT exam in India know all these techniques and more.
39. :%s/math/maths/g
40. Great post! Good tricks for life.
41. amazing tricks!
42. heres a real trick
(a+b)^2 = 2a^2 + 2ab + b^2, so
86^2 = (80+6)^2 = 80^2 + 2*80*6 + 6^2
so 80^2 is easy 8^2+10^=6400
2*80*6=2*10*6*8=2*480=960
6^2 = 36
=7396
works for any number 0-99 pretty easily
43. @Varun: True.. very true!
44. @ EPIc
my little brother when he was 7 taught me (age 21) the easiest way to multiply by 9 (only works up to 9×9):
Hold your ten fingers out in front of you. Now let’s say we multiply 9×3.
Starting from your left pinkie, count three fingers (end up at your left middle finger). Fold it down. Now read your fingers. 2 (fold) 7. 27.
Try 9×6. Count from left pinkie, end up at right thumb. Fold down. Read… five fingers, fold, four fingers. 54.
EASY!
And to all of you from China, India, etc, it’s great that your primary schools taught math tricks. But many (most?) in America do not, and your comments are not constructive, almost hurtful. I go to MIT yet I still can’t do basic arithmitic in my head. Don’t belittle people because they want to learn- it’s never too late to learn!
45. @ naomi
You’re not quite right about that finger trick- I use it too, and it can be easily adapted to 9×10-9×20 (it works higher than this too but takes some playing around, and not all numbers work out perfectly- I’ll let you try that!)
ie 9×13
Do the same thing for 9×3 as naomi says. Except now your leftmost pinkie is 100. Read left to right- 1 finger (100), 1 finger (10), fold, seven fingers. Answer = 117.
9×16 therefore would be left pinkie (100), four, fold, four. 144.
46. Boy this was cool one. Simply splendid.After reading this my maths seem to have improved
47. Hi My wife and I would like to thank you all for this web site. Hours of pleasure and all
48. BlueS, Sol –
For 56×17, the 2x rule seems easier here. 2x2x2x7 = 56, so do 7×17, then x2 three times.
7×17 = 119 (7×10 + 7×7)
x2 = 238
x2 = 476
x2 = 952
Great article, Sol. Lots of fun.
49. Regarding #2. A similar trick works for multiplications with 111, 1111, … you just need to make the “pipeline” longer.
Example:
24253 * 111 = (2) (2+4) (2+4+2) (4+2+5) (2+5+3) (5+3) (3) = (2) (6) (8) (11) (10) (8) (3) = 2692083
It works for 101, 1001 too, where you need to “skip” a position or two when adding. Example:
24253 * 101 = (2) (4) (2+2) (4+5) (2+3) (5) (3) = 2449553
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## 统计代写|假设检验代写hypothesis testing代考|Computing the P-value for a Z-test
A Z-test for testing the mean and proportion values for different types of hypotheses using the critical value procedure was studied. The steps for computing the $P$-value for a Z-test including one-sided and two-sided tests will be illustrated step by step, with examples using the general procedure for hypothesis testing.
Example 6.1: Compute the $P$-value to the left of a $\mathbf{Z}$ value: Compute the $P$-value to the left of a negative $Z$ value $(Z=-1.25)$ (left-tailed test). Use a significance level of $0.01(\alpha=0.01)$
Computing the $P$-value for a left-tailed Z-test can be achieved employing the general procedure for testing a hypothesis.
Step 1: Specify the null and alternative hypotheses
The two hypotheses (the null and alternative) for a left-tailed test can be written as presented in Eq. (6.1):
$$H_0: \mu \geq c \text { vs } H_1: \mu<c$$
The hypothesis in Eq. (6.1) represents a one-tailed test because the alternative hypothesis is $H_1: \mu<c$ (left-tailed), where $c$ is a given value.
Step 2: Select the signilicance level $(\alpha)$ for the study
The level of significance is chosen to be $0.01$.
Step 3: Use the sample information to calculate the test statistic value
The test statistic value of $Z$ is given to be $Z=-1.25$, otherwise we have to calculate it using a formula.
Step 4: Calculate the $P$-value and identify the critical and noncritical regions for the study
The $P$-value for the Z-test can be computed easily by employing the standard normal table (Table A in the Appendix) to calculate the probability of $Z(-1.25)$ or less, which represents the required $P$-value. The area to the left of $-1.25$ is $0.1056$ $(1-0.8944=01056)$ which represents the $P$-value. The shaded area in Fig. $6.1$ represents the required $P$-value.
## 统计代写|假设检验代写hypothesis testing代考|Testing one sample mean when the variance is known: P-value
We have studied hypothesis testing for one sample mean when the sample size is large using the critical value (traditional) procedure. The $P$-value procedure for one sample mean using a Z-test will be illustrated employing the same examples presented in Chapter 2, Z-test for one-sample mean, and solved using the critical value procedure. The three situations of hypothesis testing are covered and the $P$-value are calculated.
Example 6.4: The concentration of cadmium of surface water: Example $2.5$ is reproduced “A professor at an environmental section wanted to verify the claim that the mean concentration of cadmium (Cd) of surface water in Juru River is $1.4(\mathrm{mg} / \mathrm{L})$. He selected 35 samples and tested for cadmium concentration. The collected data showed that the mean concentration of cadmium is $1.6$ and the standard deviation of the population is $0.4$. A significance level of $\alpha=0.01$ is chosen to test the claim. Assume that the population is normally distributed.”
The five steps for conducting hypothesis testing employing the $P$-value procedure can be used to test the hypothesis regarding the mean concentration of cadmium of surface water in Juru River. The results of the $P$-value procedure will be compared with the critical value (traditional) procedure.
Step 1: Specify the null and alternative hypotheses
The two hypotheses regarding the mean concentration of cadmium of surface water in Juru River are presented in Eq. (6.4).
$$H_0: \mu=1.4 \text { vs } H_1: \mu \neq 1.4$$
Step 2: Select the significance level ( $\alpha$ ) for the study
The level of significance is chosen to be $0.01$. The $\mathrm{Z}$ critical values for a twotailed test with $\alpha=0.01$ are $\pm 2.58$.
The two procedures will be used to solve this problem; namely the critical value and $P$-value procedures.
Step 3: Use the sample information to calculate the test statistic value
The test statistic value for the Z-test is used to make a decision regarding the mean concentration of cadmium of surface water in Juru River. The test statistic value using the Z-test formula was calculated to be $2.96$.
# 假设检验代考
## 统计代写|假设检验代写hypothesis testing代考|计算z检验的p值
$$H_0: \mu \geq c \text { vs } H_1: \mu<c$$
Eq.(6.1)中的假设代表一个单尾检验，因为备选假设是$H_1: \mu<c$(左尾)，其中$c$是一个给定值。
$Z$的检验统计值被给定为$Z=-1.25$，否则我们必须使用公式计算
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# How do you calculate the derivative of [int(sqrt(1+(dy/dx)^2)/60 dx)]?
Aug 13, 2015
Notice how the integral is the opposite of a derivative. The first part of the Fundamental Theorem of Calculus states that, and the term "antiderivative" essentially describes the basic function of an integral. Therefore, the derivative of an integral is the integrand. In other words:
$\frac{d}{\mathrm{dx}} \left\{\int \textcolor{g r e e n}{\left[f \left(x\right) + C\right]} \mathrm{dx}\right\} = \frac{d}{\mathrm{dx}} \left[\textcolor{g r e e n}{g \left(x\right)}\right] = \frac{\mathrm{dg}}{\mathrm{dx}} = \textcolor{b l u e}{f \left(x\right) + C}$
where $g \left(x\right) = \int f \left(x\right) \mathrm{dx} + C x$ (and thus $\frac{\mathrm{dg}}{\mathrm{dx}} = f \left(x\right) + C$). $C$ may be zero as well as nonzero.
$\int \left\{\frac{d}{\mathrm{dx}} \left[f \left(x\right) + C\right]\right\} \mathrm{dx} = \int \left[\frac{\mathrm{df}}{\mathrm{dx}} + 0\right] \mathrm{dx} = \textcolor{b l u e}{f \left(x\right) + C}$
(yes, the indefinite integral of zero is a constant)
For this, we would be using the upper case, differentiating an integral.
$\frac{1}{60} \int \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$
which, by the way, is the equation for 1/60th times an arc length over an interval $\left[a , b\right]$.
If we were to try to evaluate this integral manually, even with something like $f \left(x\right) = {x}^{2}$, it would be very difficult. Thankfully, all we need to write out is:
$\frac{d}{\mathrm{dx}} \left[\frac{1}{60} \int \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}\right]$
$= \frac{1}{60} \cdot \frac{d}{\mathrm{dx}} \left[\int \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}\right]$
$= \textcolor{b l u e}{\frac{1}{60} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}}}$
Here, $C = 0$, so that's what we get.
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# Number System: Cyclicity of Remainders
Views:57489
Number system is the most fundamental topic in mathematics. Questions from this topic are seen in almost every entrance exam. It is a vast topic which mainly includes concept of HCF and LCM, concept of unit digit, concept of factors, concept of cyclicity, concept of factorials, etc.
Cyclicity of remainders is an important concept which can be used to solve questions based on remainders. This concept utilizes the fact that remainders repeat themselves after a certain interval when divided by a number.
First of all, we know that Remainder = 0 to d – 1; where d= number by which the divisor is divided.
If we divide an by d, the remainder can be any value from 0 to d-1. If we keep on increasing the value of n, the remainders are cyclical in nature. The pattern of the remainders would repeat. Let us understand the concept of repetition with the help of an example.
Example 1: 4ˆ1 divided by 9, leaves a remainder of 4.
4ˆ2 divided by 9, leaves a remainder of 7. {Rem(16/9) = 7}
4ˆ3 divided by 9, leaves a remainder of 1. {Rem (64/9) = 1}
4ˆ4 divided by 9, leaves a remainder of 4. {Rem (256/9) = 4}
4ˆ5 divided by 9, leaves a remainder of 4. {Rem (1024/9) = 7}
4ˆ6 divided by 9, leaves a remainder of 4. {Rem (4096/9) = 1}
4ˆ(3k+1) leaves a remainder of 4
4ˆ(3k+2) leaves a remainder of 7
4ˆ3k leaves a remainder of 1
As you can see above, the remainder when 4n is divided by 9 is cyclical in nature. The remainders obtained are 4, 7, 1, 4, 7, 1, 4, 7, 1 and so on. They will always follow the same pattern.
###### Concept
an when divided by d, will always give remainders which will have a pattern and will move in cycles of r such that r is less than or equal to d.
Let us apply this concept to some examples to getter a better understanding of it.
###### Solved Examples
Example 1: What will be the remainder when 41^43 is divided by 9?
Solution:
Step 1: We know that 41ˆ1 when divided by 9 gives a remainder= 5.
41ˆ2 when divided by 9 gives a remainder= 7.
41ˆ3 when divided by 9 gives a remainder= 8.
41ˆ4 when divided by 9 gives a remainder= 0.
41ˆ5 when divided by 9 gives a remainder=2.
41ˆ6 when divided by 9 gives a remainder= 1.
So, the cycle/pattern is 5,7,8,0, 2, and 1.
Step 2: The cyclicity is 6.
Step 3: 43 when divided by 6 is 1
Step 4: The answer is the 1st value in the cyclic pattern i.e. 5
Alternatively, we can solve this question using the divisibility of 9.
Example 2: Find out the remainder when 32ˆ32ˆ32 is divided by 7.
Solution: Step 1: We know that 32ˆ1 when divided by 7 gives a remainder= 4
32ˆ2 when divided by 7 gives a remainder= 2.
32ˆ3 when divided by 7 gives a remainder= 1.
32ˆ4 when divided by 7 gives a remainder= 4.
32ˆ5 when divided by 7 gives a remainder= 2.
32ˆ6 when divided by 7 gives a remainder= 1.
So, the cycle/pattern is 4, 2, 1.
Step 2: The cyclicity is 3.
Step 3: 32ˆ32 when divided by 3 is 1
Step 4: The answer is the 1st value in the cyclic pattern of 4, 2, and 1 i.e. 4.
Check your understanding of the topic Number systems by taking this test now.
###### Key Learning:
• The concept of cyclicity is all about identifying the last digit of the number.
• Remainders have a repetitive pattern.
###### Questions for Practice
Question 1: Find the remainder when 10ˆ20ˆ30 is divided by 3.
Question 2: Find the remainder when 47ˆ45 is divided by 4.
Question 3: Find the remainder when 52ˆ28 is divided by 6.
1. 1
2. 3
3. 4
For doubts, post your comments below and our experts will provide you with the solutions.
Views:57489
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# Substitution. How to substitute numbers into expressions.
If you are given an expression and you are asked to substitute certain values into the expression, then you need to change the variables (letters) in the expression to their given values and work out the answer. Substitution is also referred to as plugging. It’s always a good idea to work down the page and not across the page. On the first line of working out substitute your values into the expression and then do any working directly underneath.
Example
Substitute the values x = 5 and y = 8 into the following expressions:
a) x + 6
b) x + y
c) 3x + 9
d) 2x + 6y
e) 7x – y
f) 3x² + 2y
g) x - 2y
a) x + 6
= 5 + 6 (first substitute your value of x into the expression)
= 11
b) x + y
= 5 + 8 (this time substitute both values for x and y into the expression)
= 13
c) 3x + 9
= 3 × 5 + 8 (this time substitute x into the expression and remember that 3x means 3 multiplied by x).
= 23
d) 2x + 6y
= 2 × 5 + 6 × 8 (this time substitute x and y into the expression)
= 10 + 48 (work out the multiplications first then the addition)
= 58
e) 7x – y
= 7 × 5 – 8 (again both values of x and y need to be plugged into the expression)
= 35 – 8 (work out the multiplication before the subtraction)
= 27
f) 3x² + 2y
3 × 5² + 2 × 8 (again plug in the values of x and y)
3 × 25 + 16 (first square the 5)
75 + 16 (next do the multiplication)
= 91 (finally carry out the addition)
g) x - 2y
= 5 - 2 × 8 (sub in x = 5 and y = 8)
= 5 -16 (work out the multlication before the subtraction)
= -11 (make sure you include the negative sign)
Example 2
If z = 2ab + c², work out the value of z when a = 4 b = 10 and c = 9.
All you need to do here is substitute the values of a, b and c to work out the value of z.
z = 2ab + c²
z = 2 × 4 × 10 + 9² (2ab means 2 × a × b)
z = 80 + 81
z = 161
So to recap, substitute means change the letters in the expression or formula to their numerical values. Also set your
## More by this Author
Joan Whetzel 5 years ago
Great Explanation.
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CHAPTER 1. Mathematics Mixed Numbers A mixed number is a combination of a whole number and a fraction. Addition of Mixed Numbers To add mixed numbers, add the whole numbers together. Then add the fractions together by finding a common denominator. The final step is to add the sum of the whole numbers to the sum of the fractions for the final result. Example: The cargo area behind the rear seat of a small airplane can handle solids that are 43/4 feet long. If the rear seats are removed, then 21/3 feet is added to the cargo area. What is the total length of the cargo area when the rear seats are removed? Subtraction of Mixed Numbers To subtract mixed numbers, find a common denominator for the fractions. Subtract the fractions from each other (it may be necessary to borrow from the larger whole number when subtracting the fractions). Subtract the whole numbers from each other. The final step is to combine the final whole number with the final fraction. Example: What is the length of the grip of the bolt shown in Figure 1-3? The overall length of the bolt is 31/2 inches, the shank length is 31/8 inches, and the threaded portion is 15/16 inches long. To find the grip, subtract the length of the threaded portion from the length of the shank. To subtract, start with the fractions. Borrowing will be necessary because 5/16 is larger than 1/8 (or 2/16). From the whole number 3, borrow 1, which is actually 16/16. After borrowing, the first mixed number will now be Therefore, the grip length of the bolt is 113/16 inches. (Note: The value for the overall length of the bolt was given in the example, but it was not needed to solve the problem. This type of information is sometimes referred to as a “distracter" because it distracts from the information needed to solve the problem.) The Decimal Number System The Origin and Definition The number system that we use every day is called the decimal system. The prefix in the word decimal is a Latin root for the word “ten." The decimal system probably had its origin in the fact that we have ten fingers (or digits). The decimal system has ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. It is a base 10 system and has been in use for over 5,000 years. A decimal is a number with a decimal point. For example, 0.515, .10, and 462.625 are all decimal numbers. Like whole numbers, decimal numbers also have place value. The place values are based on powers of 10, as shown in Figure 1-4.
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## College Algebra (6th Edition)
$\{ (1,2,3) \}$
Plan (see p.533): 1. Eliminate one variable and arrive at a system of two equations in two variables. 2. Solve the system of two equations in two variables. 3. Back-substitute the solutions of (2) to find the eliminated varable in (1) ------------ (1.) Perform elimination by addition method (eliminate $x$) by $2\times Eq.1-3\times Eq.2 \Rightarrow$ $(6-6)x+(4+15)y+(-6-6)z=-4+6$ $4\times Eq.1-3\times Eq3 \Rightarrow$ $(12-12)x+(8+9)y+(-12-12)z=-8-30$ $\left[\begin{array}{ll} 19y-12z=2 & I\\ 17y-24z=-38 & II \end{array}\right.$ ..., (2.) ...$II-2\times I$ eliminates z: $(17-38)y=-38-4$ $-21y=-42$ $y=2$ Back-substitute into$19y-12z=2$ $19(2)-12z=2$ $-12z=2-38$ $z=3$ (3.) Back-substitute into one of the initial equations: $3x+2y-3z=-2$ $3x+2(2)-3(3)=-2$ $3x-5=-2$ $3x=3$ $x=1$ Solution set: : $\{ (1,2,3) \}$
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## 3 Must Know Math Concepts for Algebra
Algebra is the language of using letters to represent numbers. Do you know what are the important Algebra Math concepts in PSLE Math?
You will also be using Algebra often in Secondary school and later years. So primary 6 is a good chance to build up your understanding of Algebra.
Here are 3 must know math concepts for Primary 6 algebra. Also check out the 3 mistakes which you must avoid for Algebra
## 1. Simplification
Algebraic simplification is the first must know math concepts.
When you are given numbers in the question, you can solve the question for a final numerical answer.
For instance, “John has 3 sweets. He buys 2 more sweets. How many sweets does John has?”
The answer is “3 + 2 = 5”
But in Algebra, you are given letters to represent unknown numbers. So you can’t solve the question for a final numerical answer.
But you can simplify and give a final algebraic answer.
For example, “John has m sweets. He buys 2 more sweets. How many sweets does John has?”
The answer is “m + 2”. And that is the final answer.
Do note that you cannot add m to 2 to become 3m as they are considered “unlike” terms. This is a very common mistake.
## 2. Substitution
After you have known how to simplify algebraic expressions, the next math concept will be substitution.
As I mentioned earlier, Algebra is the language of using letters to represent unknown numbers. Sometimes, the question will give you the value of the letter and require you to give a numerical answer.
In this case, you need to substitute (or replace) the letter with the number given and work out the final answer.
When you do substitution, keep in mind the order of operations.
Example: If a = 2, find the value of a + 2 x 3?
The order of operations states that multiplication comes before addition. So you do “2 x 3” first which is 6. Then you substitute the letter a by 2, and add 2 to 6. The final answer is 8.
Now take a look at another example.
If a = 2, find the value of (a + 2) x 3?
The order of operations states that bracket comes before multiplication. So you replace the letter a by 2 and do the addition in the bracket first. Then you multiply the answer by 3.
So answer is “4 x 3 = 12”.
Notice that the final answer is different.
## 3. Solving Problem Sums
To solve problem sums using Algebra, you must first understand the question clearly, followed by learning how to express the question in algebraic form.
For instance, “Subtract p from 20” can be expressed as “20 – p”.
Keep in mind on when to put brackets and always put units at the end.
For example, “John has \$p. He receives \$20. The final answer is \$(p + 20). Don’t forget the \$ and bracket!
After you are clear with this, you can extend the same skill to using “Units and Parts” which is a very powerful algebra math concept to solve challenging problem sums.
For examples of using “Units and Parts”, you can enrol in my online course, “8 Must Know Methods of Solving Math Problem Sums“. You will learn how to use “units and parts” as well as 7 other important methods to solve many problem sums.
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# Introduction to Finding Pythagoras
## Presentation on theme: "Introduction to Finding Pythagoras"— Presentation transcript:
Introduction to Finding Pythagoras
Graphing with Coordinates on the Cartesian Plane
Ancient Math Geek, Rene Descartes
took a number line and ran it left to right and called it the x-axis. The smaller numbers are on the left of the number line and the higher numbers are to the right of the line.
Ancient Math Geek, Rene Descartes
then took a number line and ran it up and down and called it the y-axis. The smaller numbers are down the line and the higher numbers are up on the line. 4 3 2 1 -1 -2 -3 -4
By the way, how many points are on my number line?
There are an infinite number of points on any number line. Between each whole number are countless fractions and decimals. Also each line continues on to the right or up to infinity ∞ and to the left or down to negative infinity -∞ the line never stops.
The Cartesian Plane: x & y -axis
x-axis Remember too, the x-axis is for the INDEPENDENT variable & the y-axis is for the DEPENDENT variable.
The Cartesian Plane: origin
You must walk before you can fly.
Coordinates Remember, when you give coordinates - or ordered pairs- you must give the x-coordinate and then the y-coordinate. You write them in parentheses with a comma in the middle. You must walk before you can fly. ( , ) (-5,7 ) (x,y)
The Cartesian Plane: Quadrants
y-axis Quadrant II (-x, +y) Quadrant I (+x, +y) x-axis Quadrant III (-x, -y) Quadrant IV (+x, -y)
Let’s Zoom in to the origin
Let’s Zoom in to the origin. There are countless fractions and decimals between the whole numbers! 3 2 1 0 -1 -2 -3
Write the coordinates for each point.
(-3,4) A ____ B ____ C ____ D ____ E ____ F ____ G ____ H ____ (0,6) (6,-3) (-1,1) (-4, 0) (-7, -5) (2,2) (0,-5)
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# 3.1 solve linear systems by graphing note: definitely need graph paper for your notes today
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3.1 Solve Linear Systems by Graphing Note: Definitely need graph paper for your notes today
Post on 18-Jan-2016
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TRANSCRIPT
3.1 Solve Linear Systems by Graphing
Note: Definitely need graph paper for your notes today
System of Two Linear Equations
• Also called a linear system
• Simply two linear equations
• Solution of the system: an ordered pair that satisfies both equations
EXAMPLE 1 Solve a system graphically
Graph the linear system and estimate the solution. Then check the solution algebraically.
4x + y = 8
2x – 3y = 18
Equation 1
Equation 2
SOLUTION
Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (3, – 4). You can check this algebraically as follows.
EXAMPLE 1 Solve a system graphically
Equation 1 Equation 2
4x + y = 8
4(3) + (– 4) 8=?
=?12 – 4 8
8 = 8
2x – 3y = 18
=?2(3) – 3( – 4) 18
=?6 + 12 18
18 = 18
The solution is (3, – 4).
SOLUTION
GUIDED PRACTICE for Example 1
Graph the linear system and estimate the solution. Then check the solution algebraically.1. 3x + 2y = – 4
x + 3y = 13x + 2y = – 4x + 3y = 1
Equation 1Equation 2
Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (–2, 1). You can check this algebraically as follows.
GUIDED PRACTICE for Example 1
Equation 1 Equation 23x + 2y = –4
=?–6 + 2 –4
x + 3y = 1
–2 + 3 1=?
1 = 1
The solution is (–2, 1).
=?3(–2) + 2(1) –4
–4 = –4
(–2 ) + 3( 1) 1=?
SOLUTION
GUIDED PRACTICE for Example 1
Graph the linear system and estimate the solution. Then check the solution algebraically.
Equation 1Equation 2
2. 4x – 5y = – 102x – 7y = 4
2x – 7y = 44x – 5y = – 10
Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (–5, –2). You can check this algebraically as follows.
GUIDED PRACTICE for Example 1
Equation 1 Equation 2 2x –7y = 4
–10 + 14 4=?
4 = 4
The solution is (–5, –2).
–10 = –10
4x – 5y = –10=?4(–5) – 5(–2 ) –10 2(–5) – 7(–2 ) 4=?
=?–20 + 10 –10
SOLUTION
GUIDED PRACTICE for Example 1
Graph the linear system and estimate the solution. Then check the solution algebraically.
Equation 1Equation 2
3. 8x – y = 83x + 2y = – 16
8x – y = 83x + 2y = – 16
Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (0, –8). You can check this algebraically as follows.
GUIDED PRACTICE for Example 1
Equation 1 Equation 2
=? 0 + 8 8 0 – 16 –16=?
–16 = –16
The solution is (0, –8).
8(0) – (–8 ) 8=?
8 = 8
8x – y = 8 3x + 2y = –163(0) + 2(–8) –16=?
How to do it with a graphing calculator
Classifying Systems page 154
• Consistent: at least one solution
• Inconsistent: no solution (lines never intersect)
Types of Consistent Systems
• Independent- exactly one solution (graphs have one intersection point)
• Dependent- infinitely many solutions (two graphs are the same line)
Keep in Mind…
• If it’s inconsistent- parallel lines, neither independent or dependent
• If it’s consistent- either one intersection point or two equations that represent the same line- classify as either independent or dependent
EXAMPLE 2 Solve a system with many solutions
Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent.
4x – 3y = 8
8x – 6y = 16
Equation 1
Equation 2
SOLUTION
The graphs of the equations are the same line. So, each point on the line is a solution, and the system has infinitely many solutions. Therefore, the system is consistent and dependent.
If it is an inconsistent system (no solution)
• The lines are parallel– The lines have the same _______.
EXAMPLE 3 Solve a system with no solution
Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent.
2x + y = 4
2x + y = 1
Equation 1
Equation 2
SOLUTION
The graphs of the equations are two parallel lines.Because the two lines have no point of intersection, the system has no solution. Therefore, the system is inconsistent.
Guided Practice 4 – 6
EXAMPLE 4 Standardized Test Practice
SOLUTION
Equation 1 (Option A)
y = 1 x + 30
Equation 2 (Option B)
EXAMPLE 4 Standardized Test Practice
y = x2.5
To solve the system, graph the equations y = x + 30 and y = 2.5x, as shown at the right.
EXAMPLE 4 Standardized Test Practice
Notice that you need to graph the equations only in the first quadrant because only nonnegative values of x and y make sense in this situation.
The lines appear to intersect at about the point (20, 50). You can check this algebraically as follows.
Equation 1 checks.
Equation 2 checks.
50 = 20 + 30
50 = 2.5(20)
ANSWERThe total costs are equal after 20 rides.
Extra Word Problem Example
• A soccer league offers two options for membership plans. Option A incluces an initial fee of \$40 and costs \$5 for each game played. Option B costs \$10 for each game played. After how many games will the total cost of the two options be the same?
Review Questions
• How do you solve a linear system by graphing?
• How can you tell just by looking at the system that the same graph is shown twice?
Homework
• 1, 2, 6-11, 15, 16, 20 – 25, 28-31, 35 – 39
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Fractions Part 4: Compare Fractions
by C. Elkins, OK Math and Reading Lady
Starting in 3rd grade, students start using words and symbols to read and write fractions (Oklahoma Academic Standards OAS 3.N.3.1), construct fractions (3.N.3.2), compose and decompose them (3.N.3.3), and order and compare them using models and number lines (3.N.3.4). Fourth and Fifth graders continue to refine these skills. In this post, I will address different ways to compare fractions (keeping in mind the concrete-pictorial-abstract progression) by comparing numerators, comparing denominators, comparing to half, and utilizing knowledge of unit fractions. Students should have extensive experience utilizing models such as fraction strips, fraction circles, pattern blocks, number lines, pictures, and drawings to help build the concepts of fractional parts before being asked to put a <, >, or = sign between two fractions. See the end for a FREE comparing fractions guide.
In my opinion, determining if (or how) two fractions are equivalent is also a very important step when comparing fractions. However, regarding the OAS, students are not asked to represent or rename equivalent fractions until 4th grade (4.N.2.1). I will address equivalent fractions in the next post – just know that sometimes this skill goes hand in hand with comparing fractions. AND keep in mind that most of the standards for fractions through 4th grade stipulate “using concrete and pictorial models, fraction strips, number lines.” Students in 4th grade should not be expected to do abstract paper-pencil steps to simplify or “reduce” fractions to simplest terms, nor cross multiply to compare, etc. They need hands-on experience to more fully understand the concepts about fractions that are so difficult to grasp abstractly. Then in 5th grade students should have enough visual pictures in their head to solve operational problems with fractions. OK, that’s my soapbox. Don’t make it harder than it should be.
Materials to use: pattern blocks, fraction strips, fraction circles, cubes, tiles, two-color counters, Cuisenaire rods, number lines, paper plates, graham crackers, candy (m and m’s, skittles, etc.)
Ways to Compare (when using same size wholes – you can’t compare 3/4 of a donut with 1/2 of a birthday cake):
• Using unit fractions: If the fraction is a unit fraction, it has a 1 as a numerator. This should form the first type of comparison: 1/2 > 1/3 and 1/5 < 1/4 and 1/6 > 1/10, etc. This type of comparison is critical to fractional understanding.
• Same denominator: When the denominators are the same, then compare the numerators. 2/4 > 1/4.
• Same numerator: When the numerators are the same, compare the denominators. For example: When comparing 2/5 with 2/10, since fifths are larger parts than tenths, 2/5 will be larger than 2/10. This is hard for some students to think about, because the smaller the number designated for the denominator, the larger the part (when comparing the same size whole).
• Unit fractions one away from the whole: These are fractions in which there is one unit to be added to make it a whole (1). The numerators of these fractions will be one less than the denominator. 11/12 is 1/12 away from the whole (1). 7/8 is 1/8 away from the whole (1). Example: To compare 3/4 with 5/6, use manipulatives or a number line to see that 3/4 is 1/4 away from 1, while 5/6 is 1/6 away from 1. Since 1/4 is a bigger part than 1/6, then 3/4 < 5/6.
• Less than half? More than half? Learn all of the fractions that equal half. While this might sound simple, students often have misconceptions that 1/2 is the only way to describe half, or that a 5 must be in the fraction to be half (because 5 is the midpoint when used on a number line for rounding). I ask students to recall their addition facts dealing with doubles from 2nd grade. Since 2 + 2 = 4, two is half of 4, and 2/4 = 1/2. Repeat that with other forms of 1/2. Students should learn that finding half of an even-numbered denominator should be figured quickly (7/14, 9/18, 25/50, 50/100, etc.). Then use knowledge of half to determine if a fraction is less than half or more than half. Since 7/14 = 1/2, then I know that 6/14 < 1/2 and 9/14 > 1/2.
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CATEGORIES
# Montessori Monday: The Fraction Skittles
The Montessori Math materials are distinct for their ability to make concrete otherwise abstract concepts and, by doing so, to allow children to master mathematical concepts and relationships earlier than you may remember having learned them in school. Take the fraction materials, for example. While children use fractions all the time ("Can I have just half the cookie?") most traditional math sequences don't introduce them until Elementary school. In Montessori classrooms, though, they are presented first sensorially in early childhood, then in more complicated lessons as the child masters the basic relationships of parts comprising wholes. The Montessori Skittles introduce wholes, halves, thirds and quarters. In the Early Childhood classroom, these are presented in equivalences, with the child able to compare halves and quarters, for example, and to make combinations of fractions that create one complete whole. In later Elementary, these same materials will be used when children learn to divide fractions by fractions. First, the child is introduced to the whole skittle, a large figure they can move with their whole hand. Then, the child is introduced to the halves, which can be held together with one hand to explore with the other. Afterward, the child explores quarters, which are conceptually easier because they are halves of halves, and finally to thirds, the most difficult of these initial fractions. At each step, the child can explore how these fractions can combine to create wholes. These simple exercises can be presented to children even in their first year in the Montessori Early Childhood classroom, as young as 2.5 years old. Although the concept of fractions is more complicated than that of whole numbers, when presenting them through materials the child can hold in their hand, move about and compare, children are able to master the language and equivalences. Early games to manage the nomenclature, to compare different fractions to each other, and to compile whole figures quickly expand to manipulating fractions in other mathematical operations. It may be earlier than you'd expect, but it is presented in a way that reflects what we know about how children learn and, thus, is just right for the children in our classrooms.
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# Using the Remainder Theorem to Test for Divisibility
Title: Using the Remainder Theorem to Test for Divisibility
Subtitle: A Comprehensive Guide for Polynomials
Introduction
The Remainder Theorem is a useful tool for testing divisibility of polynomials. It is a simple yet powerful theorem that can help you determine whether a given polynomial is divisible by another polynomial. This guide will explain what the Remainder Theorem is and how to use it to test for divisibility. It will also provide examples and a FAQ section to help you get the most out of the theorem.
Body
The Remainder Theorem states that if a polynomial f(x) is divided by another polynomial g(x), then the remainder is equal to f(a) where a is a number. This means that if you divide a polynomial f(x) by a polynomial g(x) and the remainder is 0, then f(x) is divisible by g(x). This theorem can be used to test for divisibility of polynomials.
Let’s look at an example. Suppose we have the polynomial f(x) = x^4 + 2x^3 + 3x^2 + 4x + 5 and we want to test if it is divisible by x + 2. We can use the Remainder Theorem to do this. We can divide f(x) by x + 2 and calculate the remainder. If the remainder is 0, then f(x) is divisible by x + 2.
To calculate the remainder, we first need to divide f(x) by x + 2. We can do this using long division. The result of the division is:
x^3 + x^2 + 2x + 5
The remainder is 5, so f(x) is not divisible by x + 2.
Examples
Let’s look at another example. Suppose we have the polynomial f(x) = x^5 + x^4 + x^2 + 3x + 1 and we want to test if it is divisible by x + 2. We can divide f(x) by x + 2 and calculate the remainder. The result of the division is:
x^4 + x^3 + x + 1
The remainder is 1, so f(x) is not divisible by x + 2.
We can also use the Remainder Theorem to test for divisibility of polynomials of higher degree. For example, suppose we have the polynomial f(x) = x^7 + x^6 + x^4 + 3x^2 + 2x + 5 and we want to test if it is divisible by x + 2. We can divide f(x) by x + 2 and calculate the remainder. The result of the division is:
x^6 + x^5 + x^3 + 3x + 4
The remainder is 4, so f(x) is not divisible by x + 2.
FAQ Section
Q: What is the Remainder Theorem?
A: The Remainder Theorem states that if a polynomial f(x) is divided by another polynomial g(x), then the remainder is equal to f(a) where a is a number. This means that if you divide a polynomial f(x) by a polynomial g(x) and the remainder is 0, then f(x) is divisible by g(x).
Q: How can I use the Remainder Theorem to test for divisibility of polynomials?
A: To test for divisibility of polynomials using the Remainder Theorem, you need to divide the polynomial f(x) by the polynomial g(x) and calculate the remainder. If the remainder is 0, then f(x) is divisible by g(x).
Q: What if the remainder is not 0?
A: If the remainder is not 0, then f(x) is not divisible by g(x).
Summary
The Remainder Theorem is a useful tool for testing divisibility of polynomials. It states that if a polynomial f(x) is divided by another polynomial g(x), then the remainder is equal to f(a) where a is a number. This means that if you divide a polynomial f(x) by a polynomial g(x) and the remainder is 0, then f(x) is divisible by g(x). This theorem can be used to test for divisibility of polynomials of any degree.
Conclusion
The Remainder Theorem is a powerful tool for testing divisibility of polynomials. It is a simple yet powerful theorem that can help you determine whether a given polynomial is divisible by another polynomial. This guide has provided an explanation of the Remainder Theorem and how to use it to test for divisibility, as well as examples and a FAQ section to help you get the most out of the theorem.
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# How do you foil (x-8)(x-4)?
Aug 10, 2015
Sum (product of First terms, product of Outside terms, product of Inside terms, and product of Last terms) to get
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 12 x + 32$
#### Explanation:
Evaluating the product of two binomials using FOIL:
Sum
$\textcolor{w h i t e}{\text{XXXX}}$product of First terms
$\textcolor{w h i t e}{\text{XXXX}}$product of Outside terms
$\textcolor{w h i t e}{\text{XXXX}}$product of Inside terms
$\textcolor{w h i t e}{\text{XXXX}}$product of Last terms
For $\left(x - 8\right) \left(x - 4\right)$
$\textcolor{w h i t e}{\text{XXXX}}$First terms: $x , x$$\textcolor{w h i t e}{\text{XXXXXXX}}$Product $= {x}^{2}$
$\textcolor{w h i t e}{\text{XXXX}}$Outside terms: $x , - 4$$\textcolor{w h i t e}{\text{XXX}}$Product $= - 4 x$
$\textcolor{w h i t e}{\text{XXXX}}$Inside terms: $- 8 , x$$\textcolor{w h i t e}{\text{XXXXX}}$Product $= - 8 \times$
$\textcolor{w h i t e}{\text{XXXX}}$Last terms: $- 8 , - 4$$\textcolor{w h i t e}{\text{XXXXX}}$Product $= + 32$
Sum
$\textcolor{w h i t e}{\text{XXXX}}$$= {x}^{2} - 4 x - 8 x + 32$
Combining terms with same exponent of $x$
$\textcolor{w h i t e}{\text{XXXX}}$$= {x}^{2} - 12 x + 32$
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## University Calculus: Early Transcendentals (3rd Edition)
Published by Pearson
# Chapter 8 - Section 8.4 - Integration of Rational Functions by Partial Fractions - Exercises - Page 445: 7
#### Answer
$$\frac{t^2+8}{t^2-5t+6}=1-\frac{12}{t-2}+\frac{17}{t-3}$$
#### Work Step by Step
$$\frac{t^2+8}{t^2-5t+6}$$ Since the degree of the numerator is not lower than that of the denominator, we need to divide $t^2+8$ by $t^2-5t+6$ to get the remainder first. Dividing them gives us $$\frac{t^2+8}{t^2-5t+6}=1+\frac{5t+2}{t^2-5t+6}=1+\frac{5t+2}{(t-2)(t-3)}$$ Now we use partial fraction method for the remainder: $$\frac{5t+2}{(t-2)(t-3)}=\frac{A}{t-2}+\frac{B}{t-3}$$ To find $A$ and $B$, we clear fractions and get $$5t+2=A(t-3)+B(t-2)$$ $$5t+2=At-3A+Bt-2B$$ $$5t+2=t(A+B)+(-3A-2B)$$ Equating coefficients of corresponding powers of $t$, we get - $A+B=5$ - $-3A-2B=2$ So $A=-12$ and $B=17$ Therefore, $$\frac{t^2+8}{t^2-5t+6}=1+\frac{5t+2}{t^2-5t+6}=1-\frac{12}{t-2}+\frac{17}{t-3}$$
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Chi-Square Test for Goodness of Fit
This lesson explains how to conduct a chi-square goodness of fit test.
When to Use a Chi-Square Goodness of Fit Test
We use a chi-square goodness of fit test when we want to formally test whether or not a categorical variable follows a hypothesized distribution.
Checking Conditions
Before we can conduct a chi-square goodness of fit test, we first need to make sure the following conditions are met to ensure that our test will be valid:
• Random: A random sample or random experiment should be used to collect the data.
• Categorical: The variable we are studying should be categorical.
• Size: The expected number of observations at each level of the variable should be at least 5.
If these conditions are met, we can then conduct the test. The following example show how to conduct a chi-square goodness of fit test.
Example: Chi-Square Goodness of Fit Test
An owner of a shop claims that 30% of all his weekend customers visit on Friday, 50% on Saturday, and 20% on Sunday. An independent researcher visits the shop on a random weekend and finds that 91 customers visit on Friday, 104 visit on Saturday, and 65 visit on Sunday.
Is this data consistent with the shop owner’s claim? Use a 0.05 level of significance.
Step 1. State the hypotheses.
The null hypothesis (H0): The shop owner’s claim is correct: 30% of customers visit on Friday, 50% on Saturday, and 20% on Sunday.
The alternative hypothesis: (Ha): At least one of the proportions in the null hypothesis is not correct.
Step 2. Determine a significance level to use.
The problem tells us that we are to use a .05 level of significance.
Step 3. Find the test statistic.
The test statistic is X2 = Σ [ (Oi – Ei)2 / Ei ]
Where Σ is just a fancy symbol that means “sum”, Oi is the observed frequency at level i of the variable, and Ei is the expected frequency at level i of the variable.
There were 260 customers who visited the shop on this particular weekend (91 on Friday + 104 on Saturday + 65 on Sunday).
According to the shop owner, we should expect 30% * 260 = 78 of the total customers to visit on Friday. The observed number of people who visited on Friday was 91. So for Friday we have:
(O – E)2 / E = (91 – 78)2 / 78 = 2.167
According to the shop owner, we should expect 50% * 260 = 130 of the total customers to visit on Saturday. The observed number of people who visited on Saturday was 104. So for Saturday we have:
(O – E)2 / E = (104 – 130)2 / 130 = 5.2
According to the shop owner, we should expect 20% * 260 = 52 of the total customers to visit on Sunday. The observed number of people who visited on Sunday was 65. So for Sunday we have:
(O – E)2 / E = (65 – 52)2 / 52 = 3.25
To find the test statistic, we simply sum up these numbers: 2.167 + 5.2 + 3.25 = 10.617
Use the Chi-Square Calculator with a degrees of freedom = k-1 (k is the number of levels of the variable) = 3-1 = 2, Chi-square critical value = 10.617, and click “Calculate p-value” to find that the p-value = .99505. Then 1 – .99505 = .00495.
Step 4. Reject or fail to reject the null hypothesis.
Since the p-value (.00495) is less than our significance level of .05, we reject the null hypothesis.
Step 5. Interpret the results.
Since we rejected the null hypothesis, we have sufficient evidence to say the true distribution of customers who come in to this shop on weekends is not equal to 30% on Friday, 50% on Saturday, and 20% on Sunday.
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## MP Board Class 8th Maths Solutions Chapter 14 Factorization Ex 14.4
Find and correct the errors in the following mathematical statements.
Question 1.
4(x – 5) = 4x – 5
Solution:
The given statement is incorrect.
The correct statement is 4(x – 5) = 4x – 20.
Question 2.
x(3x + 2) = 3x2 + 2
The given statement is incorrect.
The correct statement is x(3x + 2) = 3x2 + 2x.
Question 3.
2x + 3y = 5xy
Solution:
The given statement is incorrect.
The correct statement is 2x + 3y = 2x + 3y.
Question 4.
x + 2x + 3x = 5x
Solution:
The given statement is incorrect.
The correct statement is x + 2x + 3x = 6x.
Question 5.
5y + 2y + y – 7y = 0
Solution:
The given statement is incorrect.
The correct statement is 5y + 2y + y – 7y = y.
Question 6.
3x + 2x = 5x2
The given statement is incorrect.
The correct statement is 3x + 2x = 5x.
Question 7.
(2x)2 + 4(2x) + 7 = 2x2 + 8x + 7
EH The given statement is incorrect.
The correct statement is (2x)2 + 4(2x) + 7
= 4x2 + 8x + 7.
Question 8.
(2x)2 + 5x = 4x + 5x = 9x
EH The given statement is incorrect.
The correct statement is (2x)2 + 5x = 4x2 + 5x.
Question 9.
(3x + 2)2 = 3x2 + 6x + 4
Solution:
The given statement is incorrect.
The correct statement is (3x + 2)2
= (3x)2 + 2(3x)(2) + 22 = 9x2 + 12x + 4.
Question 10.
Substituting x = -3 in
(a) x2 + 5x + 4
gives (-3)2 + 5 (-3) + 4 = 9 + 2 + 4 = 15
(b) x2 – 5x + 4
gives (-3)2 – 5(-3) + 4 = 9 – 15 + 4 = -2
(c) x2 + 5x gives (-3)2 + 5 (-3) = -9 -15 = – 24
Solution:
(a) Incorrect statement
Substituting x = -3 in x2 + 5x + 4 gives
(-3)2 + 5(-3) + 4 = 9 – 15 + 4 = -2
(b) Incorrect statement
Substituting x = -3 in x2 – 5x + 4 gives
(-3)2 – 5(-3) + 4 = 9 + 15 + 4 = 28
(c) Incorrect statement
Substituting x = -3 in x2 + 5x gives
(-3)2 + 5(-3) = 9 – 15 = – 6
Question 11.
(y – 3)2 = y2 – 9
Solution:
The given statement is incorrect.
The correct statement is (y – 3)2 = y2 – 6y + 9
Question 12.
(z + 5)2 = z2 + 25
Solution:
The given statement is incorrect.
The correct statement is (z + 5)2 = z2 + 10z + 25
Question 13.
(2a + 3b)(a – b) = 2a2 – 3b2
Solution:
The given statement is incorrect.
The correct statement is (2a + 3b)(a – b)
= 2a(a – b) + 3b(a – b)
= 2a2 – 2ab + 3ab – 3b2
= 2a2 + ab – 3b2
Question 14.
(a + 4)(a + 2) = a2 + 8
Solution:
The given statement is incorrect.
The correct statement is (a + 4) (a + 2)
= a(a + 2) + 4 (a + 2)
= a2 + 2a + 4a + 8 = a2 + 6a + 8
Question 15.
(a – 4)(a – 2) = a2 – 8
Solution:
The given statement is incorrect.
The correct statement is (a – 4) (a – 2)
= a(a – 2) – 4(a – 2) = a2 – 2a – 4a + 8
= a2 – 6a + 8
Question 16.
$$\frac{3 x^{2}}{3 x^{2}}$$ = o
Solution:
The given statement is incorrect.
The correct statement is
$$\frac{3 x^{2}}{3 x^{2}}$$ = 1
Question 17.
$$\frac{3 x^{2}+1}{3 x^{2}}$$ = 1 + 1 = 2
Solution:
The given statement is incorrect.
The correct statement is
$$\frac{3 x^{2}+1}{3 x^{2}}=\frac{3 x^{2}}{3 x^{2}}+\frac{1}{3 x^{2}}=1+\frac{1}{3 x^{2}}$$
Question 18.
$$\frac{3 x}{3 x+2}=\frac{1}{2}$$
Solution:
The given statement is incorrect.
The correct statement is
$$\frac{3 x}{3 x+2}=\frac{3 x}{3 x+2}$$
Question 19.
$$\frac{3}{4 x+3}=\frac{1}{4 x}$$
Solution:
The given statement is incorrect.
The correct statement is $$\frac{3}{4 x+3}=\frac{3}{4 x+3}$$
Question 20.
$$\frac{4 x+5}{4 x}$$ = 5
Solution:
The given statement is incorrect.
The correct statement is
$$\frac{4 x+5}{4 x}=\frac{4 x}{4 x}+\frac{5}{4 x}=1+\frac{5}{4 x}$$
Question 21.
$$\frac{7 x+5}{5}$$ = 7x
Solution:
The given statement is incorrect.
The correct statement is
$$\frac{7 x+5}{5}=\frac{7 x}{5}+\frac{5}{5}=\frac{7 x}{5}+1$$
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# How to explain fractions to 3rd Graders: Worksheets and tips
• Spark your 3rd grader’s interest in fractions with these super fun fractions worksheets and tips designed for Grade 3 tutors. Given that fractions can be tricky and confusing for some students, we have resolved to help explain to teachers and parents how to explain fractions to 3rd graders in a way that makes sense.
Fractions are one of the most important and fun topics in math for 3rd graders. They help us understand how to divide things into equal parts, how to compare different parts of a whole, and how to solve problems involving fractions.
Before introducing fractions to your students, ensure they have a solid understanding of a fraction. A fraction is a way of representing a part of a whole or a part of a group.
For example, if you have a pizza and cut it into four equal slices, each slice is one-fourth of the pizza. You can write this as 1/4. The number on the top of the fraction is called the numerator, which tells you how many parts you have. The number on the bottom of the fraction is called the denominator, which tells you how many parts the whole is divided into. So 1/4 means one part out of four.
• ### Why fractions are fun and important
Fractions are fun and important for 3rd graders to learn for many reasons. First, they are not only useful for math, but they are also fun and relevant to everyday life. You can use fractions to measure ingredients for cooking, tell time, divide money, make art, play music, and much more.
Fractions also help you develop critical thinking and problem-solving skills to help you in other subjects and the future. Learning fractions can be fun and rewarding by using engaging activities and games that make fractions come alive.
• ### What Are Fractions? - A Simple Definition and Examples
To help your students understand fractions, you can use simple definitions and examples they can relate to. Here are some examples of how you can explain fractions to your 3rd graders:
• A fraction is a part of a whole. For example, if you break a chocolate bar into two pieces, each piece is half of the chocolate bar. You can write this as 1/2.
• A fraction is a part of a group. For example, if you have six apples and three of them are red, three out of six apples are red. You can write this as 3/6.
• A fraction can also be written as a division. For example, 1/2 means one divided by two. You can use a division symbol or a horizontal line to write fractions.
• A fraction can have different names depending on how you write or say it. For example, 1/2 can be called one-half, two-fourths, or four-eighths. These are called equivalent fractions because they represent the same amount.
• ### How to use visual aids to teach fractions
Mathskills4kids.com has designed one of the best ways to teach or explain fractions to your 3rd graders using visual aids that show them what fractions look like and how they work. Visual aids can help your students see the relationship between the numerator and the denominator, compare different fractions, and find equivalent fractions.
Some of these common visual aids for teaching fractions are:
• Fraction circles: These are circles that are divided into equal parts with different colors or patterns. You can use fraction circles to show your students how to make halves, thirds, fourths, sixths, eighths, etc.
You can also use them to show how to compare fractions with the same denominator or find equivalent fractions by matching different circles.
• Fraction bars: These are rectangles that are divided into equal parts with different colors or patterns. You can use fraction bars to show your students how to make halves, thirds, fourths, sixths, eighths, etc.
You can also use them to show how to compare fractions with the same numerator or find equivalent fractions by lining up different bars.
• Fraction models: These are objects or pictures representing fractions in real-life situations. You can use fraction models to show your students how to make fractions from pizzas, cakes, cookies, candy bars, etc.
You can also use them to show how to solve word problems involving fractions by drawing or using manipulatives.
• Number lines: These lines have numbers marked at equal intervals. You can use number lines to show your students how to locate fractions on a line using unit fractions (fractions with 1 as the numerator) as reference points.
You can also use them to show how to compare fractions with different denominators or find equivalent fractions using equivalent unit fractions.
• ### Understanding fractions: Some examples of Mthskills4kids’ Grade 3 fractions worksheets
To help your students practice and understand fractions, visit Mathskills4kids.com, an amazing website with thrilling worksheets, including different exercises and challenges.
These Worksheets can help your students review what they have learned, apply their knowledge to new situations, and check their understanding and progress.
Here are some examples of worksheets and tips illustrating how to explain fractions to 3rd Graders:
• How to make halves, thirds, fourths, sixths, and eighths: This worksheet will help your students learn how to divide shapes into equal parts and name the fractions. It will also help them recognize that different shapes can have the same fraction name. For example, a circle and a square can both be divided into four equal parts, and each part is one-fourth.
• Understanding fraction bars: This worksheet will help your students learn how to use fraction bars to represent fractions. It will also help them compare fractions with the same numerator or denominator and find equivalent fractions by lining up different bars.
• Understand fractions as area models: This worksheet will help your students learn how to use area models to represent fractions. It will also help them compare fractions with the same numerator or denominator and find equivalent fractions by shading or coloring different model parts.
• Matching unit fractions to models: This worksheet will help your students learn how to identify and match unit fractions (fractions with 1 as the numerator) to models. It will also help them understand that unit fractions are the building blocks of all other fractions and can be used to make other fractions by adding or multiplying them.
• Identifying and matching unit fractions on number lines: This worksheet will help your students locate and match unit fractions on number lines. It will also help them understand that number lines can be used to show fractions as points or distances on a line and that they can be used to compare fractions with different denominators.
• Graphing and comparing fractions on number lines: This worksheet will help your students learn how to graph and compare fractions on number lines. It will also help them understand that number lines can be used to show equivalent fractions by using equivalent unit fractions as reference points used to order fractions from least to greatest or vice versa.
• Fractions of a whole modeling word problems: This worksheet will help your students learn how to solve word problems involving fractions of a whole by using models or pictures. It will also help them understand how to write fractions as part of a whole or as a division and how to use keywords like of, out of, or from to identify fractions.
• ### Tips for making fractions fun and engaging: Games, songs, and activities
To make learning fractions more fun and engaging for your 3rd graders, you can use games, songs, and activities that involve fractions in creative and interactive ways. Games, songs, and activities can help your students practice and reinforce fraction concepts, develop their math skills, and enjoy learning math.
Here are some examples of games, songs, and activities that you can use to teach fractions to your 3rd graders:
• Fraction bingo: This is a game where you give your students bingo cards with different fractions and call out fraction names or show fraction models. Your students will mark the matching fraction on their cards and shout "bingo" when they have a full row, column, or diagonal. You can use fraction circles, bars, models, or number lines as the calling cards.
• Fraction memory: This is a game where you give your students cards with different fractions on them and ask them to find pairs of equivalent fractions by flipping two cards at a time. You can use fraction circles, bars, models, or number lines as cards.
• Fraction war: This game can be played with fraction cards. You can use the same cards from the bingo game or make your own. The game is similar to the regular war card game but with fractions. Each player gets a stack of cards and turns over one card at a time. The player with the larger fraction wins both cards.
If the fractions are equal, the players have a war and turn over two more cards each. The player with the larger fraction wins all the cards. The game ends when one player has all the cards or when the time is up.
• Fraction pizza: This is an activity where you give your students paper plates and ask them to make pizzas with different toppings and cut them into equal slices. You then ask them to write the fraction of each topping on their pizzas and compare them with their classmates.
You can also ask them to solve word problems involving fraction pizzas using models or pictures.
• Fraction Puzzles: This is a hands-on activity that helps students practice identifying and matching equivalent fractions. You can use free printable puzzles or make your own by cutting up fraction strips or circles. The students have to find and put together the pieces that show the same fraction in different ways.
• Fraction Flip Books: This creative activity helps students review fraction concepts and vocabulary. You can use free printable templates or make your own by folding and cutting paper.
The students will write and illustrate different fraction terms and examples on each flip book page, such as numerator, denominator, equivalent fractions, mixed numbers, improper fractions, etc.
### Bonus: Find more resources to reinforce fraction understanding in Third Grade here!
If you're looking for more ways to reinforce fraction understanding in third grade, you're in luck! Many online resources offer fun and engaging worksheets with tips on how to explain fractions to 3rd graders.
Here are some of the best ones we've found:
• Math Playground: This website has a variety of games and puzzles that let kids explore fractions in different ways. They can play with fraction bars, fraction circles, fraction tiles, fraction models, and more.
They can also compare fractions, add fractions, subtract fractions, and solve fraction word problems. You can find all the fraction games here: https://www.mathplayground.com/index_fractions.html.
• com: This website has a vast collection of worksheets, games, and activities that cover all the fraction topics for third grade. They have worksheets on identifying fractions, comparing fractions, equivalent fractions, fractions on a number line, fractions of a set, and more.
They also have games that make learning fractions fun, such as Fraction Bingo, Fraction War, Fraction Pizza Party, and more. You can find all the fraction resources here: https://www.education.com/resources/third-grade/fractions/
• Khan Academy: This website has a comprehensive curriculum that teaches fractions from the basics to the advanced levels. They have videos, exercises, quizzes, and tests covering all the third-grade fraction skills.
They also have interactive tools that let kids manipulate fractions and see how they work. You can find all the fraction lessons here: https://www.khanacademy.org/math/cc-third-grade-math/imp-fractions
• Math Blaster: This website has a fun and futuristic theme that makes math exciting for kids. They have games and missions that challenge kids to use their fraction skills in different scenarios.
They can blast asteroids with fractions, race with fractions, save aliens with fractions, and more. You can find all the fraction games here: https://www.mathblaster.com/parents/math-activities/view-all-math-activities/fraction-puzzles
Thank you for sharing the links of MathSkills4Kids.com with your loved ones. Your choice is greatly appreciated.
### Conclusion: Keep Practicing and Have Fun
Fractions are an important and fun part of math that 3rd graders need to learn and master. By using worksheets, visual aids, games, songs, and activities, you can help your students understand what fractions are, how to compare and order them, how to find equivalent fractions, and how to apply them to real-world situations.
We hope you enjoyed this fun fractions article from Mathskills4kids.com and found some useful resources, worksheets, and tips for explaining fractions to 3rd graders!
•
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Projection Transformations
# Projection Operators
Definition: For any vector $\vec{x} \in \mathbb{R}^2$, a projection operator $T: \mathbb{R}^2 \to \mathbb{R}^2$ projects every vector $\vec{x}$ onto some axis. For any vector $\vec{x} \in \mathbb{R}^3$, a projection operator projects every vector $\vec{x}$ onto some plane.
## Projection Transformations in 2-Space
Let $\vec{x} \in \mathbb{R}^2$ such that $\vec{x} = (x, y)$. Recall that we can imagine a projection in $\mathbb{R}^2$ of a vector to be a "shadow" that the vector casts onto another vector, or in this case an axis. For example, consider the transformation that maps $\vec{x}$ onto to $x$-axis as illustrated:
We note that the x-coordinate of our vector stays the same while the y-coordinate becomes a zero. Thus, the following equations define the image under our transformation:
(1)
\begin{align} w_1 = x + 0y \\ w_2 = 0x + 0y \end{align}
Thus, we obtain that our standard matrix is $A = \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}$ and in $w = Ax$ form:
(2)
\begin{align} \quad \begin{bmatrix} w_1\\ w_2 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} \end{align}
Of course we could always project $\vec{x}$ onto the $y$-axis like the following diagram illustrates:
In this case, we note that the x-coordinate of our vector becomes zero while the y-coordinate stays the same, and the following equations define our image:
(3)
\begin{align} w_1 = 0x + 0y \\ w_2 = 0x + y \end{align}
Thus our standard matrix is $A = \begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}$.
## Projection Transformations in 3-Space
Let $\vec{x} \in \mathbb{R}^3$. We can orthogonally project $\vec{x}$ onto either the $xy$, $xz$ or $yz$ planes by mapping exactly one coordinate to zero.
In the case above, suppose that we map $\vec{x}$ onto the $xy$-plane. It thus follows that the x and y coordinates stay the same while our z-coordinate becomes zero, resulting in the following equations defining our image:
(4)
\begin{align} w_1 = x + 0y + 0z \\ w_2 = 0x + y + 0z \\ w_3 = 0x + 0y + 0z \end{align}
Hence our standard matrix for this transformation is $A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}$. The following table describes other possible orthogonal projections to other planes:
Operator Equations Defining the Image Standard Matrix
Orthogonal projection onto the $xz$-plane $w_1 = x + 0y + 0z \\ w_2 = 0x + 0y + 0z \\ w_3 = 0x + 0y + z$ $\begin{bmatrix}1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix}$
Orthogonal projection onto the $yz$-plane $w_1 = 0x + 0y + 0z \\ w_2 = 0x + y + 0z \\ w_3 = 0x + 0y + z$ $\begin{bmatrix}0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$
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We've updated our
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# Representing Inequalities on a Number Line and with Interval Notation
### Learning Outcomes
• Represent inequalities on a number line
• Represent inequalities using interval notation
• Describe solutions to inequalities
## Represent Inequalities on a Number Line
First, let us define some important terminology. An inequality is a mathematical statement that compares two expressions using the ideas of greater than or less than. Special symbols are used in these statements. When you read an inequality, read it from left to right—just like reading text on a page. In algebra, inequalities are used to describe large sets of solutions. Sometimes there are an infinite amount of numbers that will satisfy an inequality, so rather than try to list off an infinite amount of numbers, we have developed some ways to describe very large lists in succinct ways. The first way you are probably familiar with—the basic inequality. For example:
• ${x}\lt{9}$ indicates the list of numbers that are less than $9$. Would you rather write ${x}\lt{9}$ or try to list all the possible numbers that are less than $9$? (Most people would answer with the former.)
• $-5\le{t}$ indicates all the numbers that are greater than or equal to $-5$.
Note how placing the variable on the left or right of the inequality sign can change whether you are looking for greater than or less than. For example:
• $x\lt5$ means all the real numbers that are less than 5, whereas;
• $5\lt{x}$ means that 5 is less than x, or we could rewrite this with the x on the left: $x\gt{5}$. Note how the inequality is still pointing the same direction relative to x. This statement represents all the real numbers that are greater than 5, which is easier to interpret than 5 is less than x.
The second way is with a graph using a number line: And the third way is with an interval. We will explore the second and third ways in depth in this section. Again, the three ways to write solutions to inequalities are:
• an inequality
• a graph
• an interval
### Inequality Signs
The box below shows the symbol, meaning, and an example for each inequality sign. Sometimes, it is easy to get tangled up in inequalities; just remember to read them from left to right.
Symbol Words Example
$\neq$ not equal to ${2}\neq{8}$, 2 is not equal to 8
$\gt$ greater than ${5}\gt{1}$, 5 is greater than 1
$\lt$ less than ${2}\lt{11}$, 2 is less than 11
$\geq$ greater than or equal to ${4}\geq{ 4}$, 4 is greater than or equal to 4
$\leq$ less than or equal to ${7}\leq{9}$, 7 is less than or equal to 9
The inequality $x>y$ can also be written as ${y}<{x}$. The sides of any inequality can be switched, as long as the inequality symbol between them is also reversed.
## Graphing an Inequality
Inequalities can also be graphed on a number line. Below are three examples of inequalities and their graphs. Graphs are a very helpful way to visualize information, especially when that information represents an infinite list of numbers! $x\leq -4$. This translates to all the real numbers on a number line that are less than or equal to $4$. ${x}\geq{-3}$. This translates to all the real numbers on the number line that are greater than or equal to -3. Each of these graphs begins with a circle—either an open or closed (shaded) circle. This point is often called the end point of the solution. A closed, or shaded, circle is used to represent the inequalities greater than or equal to $\displaystyle \left(\geq\right)$ or less than or equal to $\displaystyle \left(\leq\right)$. The point is part of the solution. An open circle is used for greater than (>) or less than (<). The point is not part of the solution. The graph then extends endlessly in one direction. This is shown by a line with an arrow at the end. For example, notice that for the graph of $\displaystyle x\geq -3$ shown above, the end point is $−3$, represented with a closed circle since the inequality is greater than or equal to $−3$. The blue line is drawn to the right on the number line because the values in this area are greater than $−3$. The arrow at the end indicates that the solutions continue infinitely.
### Example
Graph the inequality $x\ge 4$
Answer: We can use a number line as shown. Because the values for $x$ include $4$, we place a solid dot on the number line at $4$. Then we draw a line that begins at $x=4$ and, as indicated by the arrowhead, continues to positive infinity, which illustrates that the solution set includes all real numbers greater than or equal to $4$.
This video shows an example of how to draw the graph of an inequality. https://youtu.be/-kiAeGbSe5c
### Try It
[ohm_question]26182[/ohm_question]
### Example
Write an inequality describing all the real numbers on the number line that are less than $2$. Then draw the corresponding graph.
Answer: We need to start from the left and work right, so we start from negative infinity and end at $2$. We will not include either because infinity is not a number, and the inequality does not include $2$. Inequality: $x\lt2$ To draw the graph, place an open dot on the number line first, and then draw a line extending to the left. Draw an arrow at the leftmost point of the line to indicate that it continues for infinity.
## Represent Inequalities Using Interval Notation
Another commonly used, and arguably the most concise, method for describing inequalities and solutions to inequalities is called interval notation. With this convention, sets are built with parentheses or brackets, each having a distinct meaning. The solutions to $x\geq 4$ are represented as $\left[4,\infty \right)$. This method is widely used and will be present in other math courses you may take. The main concept to remember is that parentheses represent solutions greater than or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be "equaled." A few examples of an interval, or a set of numbers in which a solution falls, are $\left[-2,6\right)$, or all numbers between $-2$ and $6$, including $-2$, but not including $6$; $\left(-1,0\right)$, all real numbers between, but not including $-1$ and $0$; and $\left(-\infty,1\right]$, all real numbers less than and including $1$. The table below outlines the possibilities. Remember to read inequalities from left to right, just like text. The table below describes all the possible inequalities that can occur and how to write them using interval notation, where a and b are real numbers.
Inequality Words Interval Notation
${a}\lt{x}\lt{ b}$ all real numbers between a and b, not including a and b $\left(a,b\right)$
${x}\gt{a}$ All real numbers greater than a, but not including a $\left(a,\infty \right)$
${x}\lt{b}$ All real numbers less than b, but not including b $\left(-\infty ,b\right)$
${x}\ge{a}$ All real numbers greater than a, including a $\left[a,\infty \right)$
${x}\le{b}$ All real numbers less than b, including b $\left(-\infty ,b\right]$
${a}\le{x}\lt{ b}$ All real numbers between a and b, including a $\left[a,b\right)$
${a}\lt{x}\le{ b}$ All real numbers between a and b, including b $\left(a,b\right]$
${a}\le{x}\le{ b}$ All real numbers between a and b, including a and b $\left[a,b\right]$
${x}\lt{a}\text{ or }{x}\gt{ b}$ All real numbers less than a or greater than b $\left(-\infty ,a\right)\cup \left(b,\infty \right)$
All real numbers All real numbers $\left(-\infty ,\infty \right)$
### Example
Describe the inequality $x\ge 4$ using interval notation
Answer: The solutions to $x\ge 4$ are represented as $\left[4,\infty \right)$. Note the use of a bracket on the left because 4 is included in the solution set.
### Try It
[ohm_question]72501[/ohm_question]
### Example
Use interval notation to indicate all real numbers greater than or equal to $-2$.
Answer: Use a bracket on the left of $-2$ and parentheses after infinity: $\left[-2,\infty \right)$. The bracket indicates that $-2$ is included in the set with all real numbers greater than $-2$ to infinity.
### Think About It
In the previous examples you were given an inequality or a description of one with words, and asked to draw the corresponding graph and write the interval. In this example, you are given an interval and asked to write the inequality and draw the graph. Given $\left(-\infty,10\right)$, write the associated inequality and draw the graph. In the box below, write down whether you think it will be easier to draw the graph first or write the inequality first. [practice-area rows="1"][/practice-area]
Answer: We will draw the graph first. The interval reads "all real numbers less than 10," so we will start by placing an open dot on 10 and drawing a line to the left with an arrow indicating the solution continues to negative infinity. To write the inequality, we will use < since the parentheses indicate that 10 is not included. $x<10$
In the following video, you will see examples of how to write inequalities in the three ways presented here: as an inequality, in interval notation, and with a graph. https://youtu.be/X0xrHKgbDT0
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By going through these CBSE Class 12 Maths Notes Chapter 11 Three Dimensional Geometry, students can recall all the concepts quickly.
## Three Dimensional Geometry Notes Class 12 Maths Chapter 11
1. Direction cosines of a line:
Let AB be a line in space.
Through O, draw a line OP parallel to AB. Let OP makes angles α, β, and γ with OX, OY, and OZ respectively.
Cosines of the angles α, β, and γ,
i.e., cos α, cos β, and cos γ are known as the direction cosines of line AB.
Let l = cos α, m = cos β, n = cos γ
⇒ l, m, n are the direction cosines of the line AB Let us consider the ray BA. OQ is drawn parallel to BA. Now, OQ makes angles n – α, n – β, and n – γ with coordinates axes OX, OY, and OZ respectively.
∴ Direction cosines of BA are cos(π – α), cos (π – β) and cos (π – γ),
i.e., – cos α, – cos β, – cos γ or -l, -m, -n.
→ Relation between l, m, and n
AB is any ray having direction cosines l, m, n. Now OP is drawn parallel to AB, where P is the point (x, y, z). Let PM be drawn perpendicular to OY.
In ΔOPM OP = r (say). PM⊥OY.
∴ ∠PMO = 90°
Also, ∠POM = β.
∴ $$\frac{\mathrm{OM}}{\mathrm{OP}}$$ = β
∴ $$\frac{y}{r}$$ = m,
∴ y = rm
Similarly, x = rl and z = rn.
Now, OP2 = x2 + y2 + z2
r2 = (rl)2 + (rm)2 + (rn)2
or
r2 = r2(l2 + m2 + n2)
⇒ l2 + m2 + n2 = 1.
→ Direction Ratios of a line
Definition: The numbers which are proportional to the direction cosines of a line are known as direction ratios of the line.
Let Z, m, n be the direction cosines of a line.
Multiplying each by r, we ger rl, rm, rn the direction ratios.
Let rl = a,rm = b and rn – c.
r2l2 + r2m2 + r2n2 = a2 + b2 + c2
∴ r2(l2 + m2 + n2) = a2 + b2 + c2
∵ l2 + m2 + n2 = 1
∴ l = $$\frac{a}{r}=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}$$
Similarly, m = $$\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}$$ and n = $$\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$$
Thus, if a, b and c are the direction ratios of a line, the direction cosines are $$\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}$$, $$\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}$$ and $$\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$$.
→ Direction cosines of the line passing through the points P(x1, y1, z1) and Q(x2, y2, z2)
Direction ratios of the line PQ are x2 – x1, y2 – y1, z2 – z1
∴ PQ = $$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$$
∴Direction cosines of PQ are $$\frac{x_{2}-x_{1}}{\mathrm{PQ}}$$, $$\frac{y_{2}-y_{1}}{\mathrm{PQ}}$$, $$\frac{z_{2}-z_{1}}{P Q}$$
2. Angle between the two lines
1. Let l1, m1, n1, and l2, m2, n2 be the direction cosines of the lines OP and OQ.
The angle θ between these lines is given by
cos θ = l1l2 + m1m2 + n1n2
and sin θ = $$\sqrt{\left(m_{1} n_{2}-m_{2} n_{1}\right)^{2}+\left(n_{1} l_{2}-n_{2} l_{1}\right)^{2}+\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2}}$$
2. If a1, b1, c1 and a2, b2, c2 are the direction ratios of two lines, then
3. Two lines are perpendicular to each other,
if θ = $$\frac{π}{2}$$ ⇒ cos θ = cos $$\frac{π}{2}$$ = 0
⇒ l1l2 + m1m2 + n1n2 = 0
or
a1a2 + b1b2 + c1c2 = 0
4. When the lines are parallel
θ = 0 ⇒ sin θ = sin 0 = 0.
STRAIGHT LINE
3. Equation of a line through a given point:
(a) Let the line passes through $$\vec{a}$$ and is parallel to vector $$\vec{b}$$
Then, the equation of the line is
$$\vec{r}$$ – $$\vec{a}$$ = λ$$\vec{b}$$
or
$$\vec{r}$$ = $$\vec{a}$$ + λ$$\vec{b}$$
(b) Let the point A be (x1, y1, z1) and a, b, c are the direction ratios of the line. The equation of the line is
$$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$$
If l, m, n are the direction cosines, then equation of the line is
$$\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}$$
4. Equation of the line passing through two points:
(a) Let $$\vec{a}_{1}$$, $$\vec{a}_{2}$$ be the position vectors of two points P and Q respectively.
⇒ $$\vec{b}$$ = $$\vec{a}_{1}$$ – $$\vec{a}_{2}$$
∴ Equation of PQ is
$$\vec{r}$$ = $$\vec{a}$$ + λ($$\vec{a}_{2}$$ – $$\vec{a}_{1}$$)
(b) Direction ratios of the line passing through P(x1, y1, z1) and Q(x2, y2, z2) are x2 – x1, y2 – y1 and z2 – z1
∴ Equation of the line PQ is
$$\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$$
5. Angle between two straight lines:
(a) Let the two lines be
$$\vec{r}$$ = $$\vec{a}_{1}$$ + λ$$\vec{b}_{1}$$
$$\vec{r}$$ = $$\vec{a}_{2}$$ + λ$$\vec{b}_{2}$$
If θ be the angle between them, then
cos θ = $$\frac{\vec{b}_{1} \vec{b}_{2}}{\left|\vec{b}_{1}\right|\left|\vec{b}_{2}\right|}$$
(b) Let the lines be
$$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$$
and $$\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$$
i.e., a1, b1, c1 and a2, b2, c2 are the direction ratios of the lines.
∴ cos θ = $$\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$$
If l1, m1, n1 and l2, m2, n2 are the direction cosines, then
cos θ = l1l2 + m1m2 +n1n2
6. Shortest Distance:
(a)Let $$\vec{r}$$ = $$\vec{a}_{1}$$ + λ$$\vec{b}_{1}$$ and
$$\vec{r}$$ = $$\vec{a}_{2}$$ + λ$$\vec{b}_{2}$$ be the two non-intersecting lines.
The shortest distance between the given lines = $$\left|\frac{\left(\vec{b}_{1} \times \vec{b}_{2}\right) \cdot\left(\vec{a}_{2}-\vec{a}_{1}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|$$
(b) Let the lines be
If the lines are intersecting, then lines are coplanar.
⇒ S.D. = 0
⇒ ($$\vec{b}_{1} \times \vec{b}_{2}$$) – ($$\vec{a}_{1} \times \vec{a}_{2}$$) = 0
or
$$\left|\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|$$ = 0
7. Distance between parallel lines:
Let the parallel lines be
$$\vec{r}$$ = $$\vec{a}_{1}$$ + λ$$\vec{b}_{1}$$ and
$$\vec{r}$$ = $$\vec{a}_{2}$$ + μ$$\vec{b}_{2}$$
Shortest distance d between these lines is given by,
d = $$\left|\frac{\vec{b} \times\left(\vec{a}_{1}-\vec{a}_{2}\right)}{|\vec{b}|}\right|$$
PLANES
8. Different forms of equations of a plane:
1. Normal Form.
(a) Let the plane ABC be at a distance d from the origin. ON is the normal to the plane in the directon n̂. Equation of the plane is $$\vec{r}$$. n̂ = d.
(b) If l, m, n are the direction cosines of the normal to the plane which is at distance d from the origin. The equation of the plane is lx + my + nz = d.
However, general form of the equation of a plane are $$\vec{r}$$ .$$\overrightarrow{\mathrm{N}}$$ = D and Ax + By + Cz + D = 0.
2. (a) Let the plane passes through a point A and let it perpendicular to the vector $$\overrightarrow{\mathrm{N}}$$
∴ Equation of the plane
($$\vec{r}$$ – $$\vec{a}$$).$$\overrightarrow{\mathrm{N}}$$ = 0.
(b) If a plane passes through (x1, y1, z1) and perpendicular to the line with direction ratios a, b, c the equation of the plane is
a(x – x1) + b(y – y1)+c(z – z1) =0
3. Equation of the plane passing through three points.
(a) Let the three points be A, B and C whose position vectors be, $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$
The equation of the plane is
$$(\vec{r}-\vec{a}) \cdot[(\vec{b}-\vec{a}) \times(\vec{c}-\vec{a})]$$ = 0
(b) Let the three points through which the plane is passing be A(x1, y1, z1), B(x2, y2, z2), and C(x3, y3, z3).
Then, equation of the plane is
$$\left|\begin{array}{llll} x & y & z & 1 \\ x_{1} & y_{1} & z_{1} & 1 \\ x_{2} & y_{2} & z_{2} & 1 \\ x_{3} & y_{3} & z_{3} & 1 \end{array}\right|$$ = 0
or
$$\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|$$ = 0
4. Intercepts form of the equation of the plane
Let the plane make the intercepts a, b, and c on coordinate axes OX, OY, and OZ respectively.
Then, the equation of the plane is
$$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}$$ = 1
9. Plane passing through the intersection of the two planes:
Let the equation of two planes be
$$\vec{r}$$. $$\vec{n}_{1}$$ = d1 and
$$\vec{r}$$ .$$\vec{n}_{2}$$ = d2
The equation of the plane passing through the line of intersection of the given planes is
($$\vec{r}$$. $$\vec{n}_{1}$$ – d1) + λ($$\vec{r}$$ .$$\vec{n}_{2}$$ – d2) = 0
or
$$\vec{r}$$ .($$\vec{n}_{1}$$ + λ$$\vec{n}_{2}$$) = (d1 + λd2)
(b) Equation of the plane passing through the line of intersection of the planes
a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is (a1x + b1 y + c1z + d1 + λ(a2x + b2y + c2z + d2) = 0
or
(a1 + λa2)x + (b1 + λb2)y + (c1 + λc2)z + d1 + λd2 = 0,
where λ is determined according to the given condition.
10. Coplanarity of two lines:
(a) The two lines
$$\vec{r}$$ = $$\vec{a}_{1}$$ + λ$$\vec{b}_{1}$$ and
$$\vec{r}$$ = $$\vec{a}_{2}$$ + λ$$\vec{b}_{2}$$ intersect each other, if $$\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{2} \times \vec{b}_{1}\right)$$ = 0
11. The angle between the planes:
Definition : The angle between the two planes is the angle between their normals. The angle 0 between the planes
$$\vec{r}$$ = $$\vec{n}_{1}$$ = d1 and $$\vec{r}$$ = $$\vec{n}_{1}$$ = d2 is given by cos θ = $$\frac{\left|\vec{n}_{1} \cdot \vec{n}_{2}\right|}{\left|\vec{n}_{1}\right|\left|\vec{n}_{2}\right|}$$
(b) If the planes are
a1x + b1y + C1z + d1 = 0
a2x + b2y + c2z + d2 = 0, then the angle θ between these planes is given by
cos θ = $$\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$$
(c) The planes are perpendicular to each other, if θ = $$\frac{π}{2}$$
∴ cos $$\frac{π}{2}$$ = 0 ⇒ $$\vec{n}_{1}$$.$$\vec{n}_{2}$$ = 0
or
a1a2 + b1b2 + c1c2 = 0
(d) The planes a1x + b1y +c1z + d1 = 0
and a2x + b2y + c2z + d2 = O
are parallel, if $$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$.
(e) Equation of the plane parallel to
$$\vec{r} \cdot \vec{n}$$ = d is $$\vec{r} \cdot \vec{n}$$ = λ
Plane parallel to
ax + by + cz + d = 0 is ax + by + cz + λ = 0.
12. Distance of a point from th plane:
(a) Let the plane be $$\vec{r} \cdot \vec{n}$$ = d.
∴ Perpendicular distance of the point $$\vec{a}$$ from the plane
=|d – $$\vec{a} \cdot \hat{n}$$|
(b) Let the equation of the plane
be Ax + By + Cz + D= O.
The distance of the point (x1, y1, z1) from this plane
= $$\left|\frac{A x_{1}+B y_{1}+C z_{1}+D}{\sqrt{A^{2}+B^{2}+C^{2}}}\right|$$
13. Angle between a line and a plane:
Definition: The angle between a line and plane is said to be the complement of the angle between the line and the normal to the plane.
(a) Let the line and plane be
$$\vec{r}$$ = $$\vec{a}$$ + λ$$\vec{b}$$ and $$\vec{r} \cdot \vec{n}$$ = d.
If θ be the angle between the plane and the line, then
sin θ = $$\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}$$
(b) ¡et the line and the plane be $$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$$ and Ax + By + Cz + D = O.
The angle Φ between them is given by
sin Φ = $$\frac{a \cdot \mathrm{A}+b \cdot \mathrm{B}+c \cdot \mathrm{C}}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+\mathrm{C}^{2}}}$$.
1. IMPORTANT RESULT
l2 + m2 + n2 = 1, where < l, m, n > are direction-cosines of a st. line.
2. DIRECTION-RATIOS
The direction-ratios of the line joining of the points (x1, y1, z1) and (x2, y2, z2) are :
<x2-x1, y2-y1 ,z2-z1 >.
3. (i) Angle between two lines. The angle between two lines having direction-cosines
< l1 m1 n1 > and < l2, m2, n2 > is given by :
cos θ = |l1l2+ m1m2 + n1n2|.
(ii) The lines are:
(a) perpendicular l1l2+ m1m2 + n1n2 = 0.
(b) parallel iff l1 = l2, m1 = m2, + n1 = n
4. SHORTEST DISTANCE
The shortest distance between two lines :
$$\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \text { and } \vec{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}} \text { is }\left|\frac{\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|$$
5. EQUATIONS OF PLANES
(i) Equation of a plane, which is at a distance ‘p’ from the origin and perpendicular to the unit
vector $$\hat{n}$$ is $$\vec{r} \cdot \hat{n}=p$$
(ii) General Form. The general equation of first degree i.e.ax + by + cz + d- 0 represents a plane,
(iii) One-point Form. The equation of a plane through (*,, y,, z,) and having <a,b,c> as direction-
ratios of the normal is a (x – x1) + b (y – y1) + c (z – z1) = 0.
(iv) Three-point Form. The equation of the plane through (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is:
$$\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ x-x_{2} & y-y_{2} & z-z_{2} \\ x-x_{3} & y-y_{3} & z-z_{3} \end{array}\right|=0$$
(v) Intercept Form. Equation of the plane, which cuts off intercepts a, b, c on the axes, is :
$$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}$$ = 1
6. ANGLE BETWEEN TW O PLANES
The angle between the planes :
a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given by :
cos θ = $$\frac{\left|a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$$
7. BISECTING PLANES
The equations of the planes bisecting the planes :
a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are:
$$\frac{a_{1} x+b_{1} y+c_{1} z+d_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}=\pm \frac{a_{2} x+b_{2} y+c_{2} z+d_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$$
|
Lesson Objectives
• Learn how to determine if three points are collinear using determinants
## How to Determine if Three Points are Collinear Using Determinants
In the last lesson, we learned how to determine the area of a triangle using determinants. Now, we will see another application of this formula. If our formula yields a result of zero, the three points given are collinear. This means they lie on the same line.
### Test for Collinear Points
$$\left| \begin{array}{ccc}x_{1}&y_{1}&1\\ x_{2}& y_{2}& 1\\ x_{3}& y_{3}& 1\end{array}\right|=0$$ Let's look at an example.
Example #1: Determine if the points given are collinear. $$(4, 0)$$ $$(9, 3)$$ $$(-1, -3)$$ Let's label our points and then plug into the formula: $$\text{Point 1}: (4, 0)$$ $$\text{Point 2}: (9, 3)$$ $$\text{Point 3}: (-1, -3)$$ $$\left| \begin{array}{ccc}4&0&1\\9&3&1\\-1&-3& 1\end{array}\right|=0$$ Since our formula gives us a result of zero, we know these three points are collinear.
#### Skills Check:
Example #1
Determine if collinear. $$(-2, -2), (1, -1), (7, 5)$$
A
Yes
B
No
Example #2
Determine if collinear. $$(3, 0), (0, -12), (1, -8)$$
A
Yes
B
No
Example #3
Determine if collinear. $$(15, 3), (2, -1), (-3, 5)$$
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# 0.51111111111111111111111111111111111111111111111111111111111111111111 as a fraction
## 0.51111111111111111111111111111111111111111111111111111111111111111111 as a fraction - solution and the full explanation with calculations.
Below you can find the full step by step solution for you problem. We hope it will be very helpful for you and it will help you to understand the solving process.
If it's not what You are looking for, type in into the box below your number and see the solution.
## What is 0.51111111111111111111111111111111111111111111111111111111111111111111 as a fraction?
To write 0.51111111111111111111111111111111111111111111111111111111111111111111 as a fraction you have to write 0.51111111111111111111111111111111111111111111111111111111111111111111 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.
0.51111111111111111111111111111111111111111111111111111111111111111111 = 0.51111111111111111111111111111111111111111111111111111111111111111111/1 = 5.1111111111111/10 = 51.111111111111/100 = 511.11111111111/1000 = 5111.1111111111/10000 = 51111.111111111/100000 = 511111.11111111/1000000 = 5111111.1111111/10000000 = 51111111.111111/100000000 = 511111111.11111/1000000000 = 5111111111.1111/10000000000 = 51111111111.111/100000000000 = 511111111111.11/1000000000000 = 5111111111111.1/10000000000000 = 51111111111111/100000000000000
And finally we have:
0.51111111111111111111111111111111111111111111111111111111111111111111 as a fraction equals 51111111111111/100000000000000
|
Intermediate Algebra (12th Edition)
$2q^2\sqrt[3]{5q}$
$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $6q^2\sqrt[3]{5q}-2q\sqrt[3]{40q^4} ,$ simplify first each term by expressing the radicand as a factor that is a perfect power of the index. Then, extract the root. Finally, combine the like radicals. $\bf{\text{Solution Details:}}$ Expressing the radicand as an expression that contains a factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} 6q^2\sqrt[3]{5q}-2q\sqrt[3]{8q^3\cdot5q} \\\\= 6q^2\sqrt[3]{5q}-2q\sqrt[3]{(2q)^3\cdot5q} .\end{array} Extracting the roots of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} 6q^2\sqrt[3]{5q}-2q(2q)\sqrt[3]{5q} \\\\= 6q^2\sqrt[3]{5q}-4q^2\sqrt[3]{5q} .\end{array} By combining like radicals, the expression above is equivalent to \begin{array}{l}\require{cancel} (6q^2-4q^2)\sqrt[3]{5q} \\\\= 2q^2\sqrt[3]{5q} .\end{array}
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# Algorithmic Analysis
In the previous article on our introduction to algorithms, we spoke about algorithmic analysis. And in this article we will go in depth.
Writing An Algorithm
Before we can analyze an algorithm, we need to write it first. So how do we write it?
``` ```Algorithm Swap(a,b):
temp = a
a = b
b = temp
return```
```
Above is a simple algorithm, I have borrowed some Python syntax. I said in the previous articles algorithms are syntax independent, which I still stand by.
I just borrowed the Python syntax so we can see where the name of the algorithm starts. The commands within the algorithm and where the algorithm ends.
We do not have variable declarations as they are not needed, syntax to start a function, etc.
Analyzing An Algorithm
Now that we now how to write an algorithm, we can move on to analyzing an algorithm.
We need to consider the following for algorithms;
• Time
• Memory
• Data Consumption(How much data your algorithm consumes)
• Power Consumption
We will only be focusing on time and space analysis for our algorithmic article series. These two are measured in Big O notation
Time Taken By An Algorithm
How long will an algorithm take to complete a specific task.
Say we want an algorithm to find the shortest path between A.B.C.D and return the answer to the user.
We will write this algorithm and after that check how much time it takes to complete this task. We do not want algorithms that take long to complete a task.
A simple direct statement in an algorithm takes one unit of time.
Example:
temp = a, this is one unit of time.
a = b, this is also another unit of time.
The above statements are very direct, straight to the point and not calling any other algorithm, hence they are not nested statements.
Let’s analyze the time analyses of our above function:
``` ```Algorithm swap(a,b):
temp = a
a = b
b = temp
return```
```
We count all of our statements, temp = a, a = b and b = temp. Each of these statements take 1 unit of time.
temp = a (1)
a = b (1)
b = temp (1)
So our f(n) is 3, which is a constant value. Any integer number is considered as a constant value.
We can already see the pattern that all assignment statements take one unit of time.
We can represent constant values with O(1), all the integer numbers are O(1).
Lets analyze the space complexity of this algorithm:
To analyze space complexity we look at the variables, in our case we have a,b, and temp.
All in all we have used three variables. Our space complexity is O(3).
We can calculate time complexity in two ways, first one being frequency count and the other asymptotic notations.
Frequency Count
Frequency count specifies the number of times a statement is to be executed, so how many times the statement within our algorithm is executed.
I generally follow this four rules basis:
• For comments and declarations the step count is 0, as comments are not executed, and declarations are not needed while writing the algorithm.
• For the return and assignment statement the step count is 1, assignment statement is simply assigning a left hand value to a right hand value. While the return statement is just returning a value.
• Only consider higher order exponents. E.g.; 3n² +4n. We will only consider 3n² as it has the higher order exponent.
• Ignore the constant value multipliers. We are left with 3n², our constant multiplier is 3, we ignore it and are left with n². Our time complexity is O(n²).
Lets get our feet wet with some examples;
``` ```Algorithm Sum(A,n):
s = 0
for i in A:
s = s + A[i]
return s```
```
The algorithm above is an array of a certain size, let us say the array has 4 elements within it. The elements within the array are 9,6,10,5,6
A is our array and n is the number of elements. The algorithm above is finding the sum of all the elements in the array. We want to find the time taken by the algorithm, we will do this by finding the frequency count method, and our four rules.
Breaking down the algorithm;
Initially i is 0, as we have a for loop we know our algorithm is repeating a certain number of times, and this number of times is n.
Lets count with it;
i = 0
i = 1
i = 2
i = 3
i = 4
i = 5
The loop stops and returns the value at s, this is when the algorithm is done looping. The logic of this goes as follows; i = 0 and 0 < n, i = 1 and 1<n, i = 2 and 2 < n, i = 3 and 3 < n, i = 4 and 4 < n, i = 5 and 5 > n. The loop will stop at i = 5 as 5 is greater than n.
The condition is checked 5 times all in all. Four times the condition was true i < n and the fifth time false, i > n. Our loop is executing for n + 1 times. How would we count the For loop, as it executes n + 1 times then we count it as n+1 steps.
Summarizing;
``` ```Algorithm Sum(A,n): 0
s = 0 -1
for i in range(A): n+1
s = s + A n(1)
return s 1```
```
The first line of our algorithm;
Algorithm Sum(A,n): this does not count for anything as it is a declaration and the step count is 0.
The second line of our algorithm;
s = 0; this is given a value of 1 in the step count frequency method as it is an assignment statement.
The third line in our algorithm;
for i in range A: the statement will tell our interpreter that loop inside the array until you complete the range of the array. This means we will loop n + 1 times.
The fourth line in our algorithm;
s = s + A[i]; this an assignment statement as we are assigning s to s + A[i], and because this statement is within a loop, we will assign n as its value.
The 5th line in our algorithm;
return s; as it is a return statement then the step count is 1
Tally up our values;
1+n+1+n+1 = 2n + 3
2n + 3
We only consider the higher order exponents, this means we will only be left with 2n.
Then we will ignore will constant multiplier, and we will be left with n, hence we be left with O(n).
Moving on to space complexity of the algorithm. We only count the variables in our algorithm.
We have s, A ,n, i;
A = n ( A is our array which has n elements, hence it is n)
s = 1 ( s only occupies one element)
n =1 ( n only occupies one element)
i = 1 ( i only occupies one element)
Tally count;
n+1+1+1+1 = n + 4, we only consider the higher order exponents and are left with n.
O(n).
Example 2:
``` ```Algorithm Add(A,B,n):
for i in range(A):
for j in range(A):
C = A + B
return```
```
With our example two we have two nested loops, the outer For loop and the inner for loop. Additionally we have a metrices or what is known as a two dimensional array.
Which are size 3 * 3, also expressed as n * n, hence we can say our above algorithm finds the sum of two metrices.
``` ```Algorithm ADD(A,B,n): 0
for i in range(A): n+1
for j in range(A): n(n+1)
c = A + B n(n)
return 1```
```
Starting from the first statement, we ignore Algorithm ADD(A,B,n):
Second line we have our outer for loop and our step count is n+1.
Third line we have our inner for loop, which has n as it is within a loop, in this case the outer loop. As it is a loop itself we also assign n+1 to it. The second loop will have n as it is within a loop and n+1 as it is a loop itself(it loops n+1 times even though it is within a loop).
The fourth line we have c = A + B, it is within two for loops, for the first loop we will assign n and the second loop another n.
The last line we have< return 1, it returns after the For loop.
Tally: n + 1 +n² +1 + n² + 1
2n² n +3
Only consider higher order exponents, we will be left with 2n².
2n², we ignore the multiplier constant.
We are left with n², hence we will have O(n²).
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# Rounding Numbers
Purpose
In this unit students will discuss why it is important to round numbers sensibly. They will practice choosing and rounding to a sensible level of accuracy for different contexts.
Achievement Objectives
NA5-6: Know and apply standard form, significant figures, rounding, and decimal place value.
Specific Learning Outcomes
• Round whole numbers sensibly in context.
• Round decimal numbers sensibly.
Activity
#### Session 1
1. Rounding numbers for newspaper headlines. The government of Outer Australis reports spending \$33,883,641.31 in the 2002-2003 financial year. Discuss how to put this number into a newspaper headline. A sensible answer here is "Government spends \$34 million last year".
While many students will round the 8 hundred thousand ‘up’ because it is over ‘5’ it is desirable to present this as a number line and see which end the number is closer to.
Discuss why this line has 33,500,000 in the middle and why 33,883,641.31 is roughly where the arrow is.
1. Student Exercises.
Round these numbers suitably for use in newspaper headlines
• Quality Stores make a profit of \$3,493,631.
• Prime Minister paid \$251,419.91 last year.
• Scientist estimates there are 56,409,100 possums in New Zealand.
• Cost of producing cheese drops to 101.8 cents per kilogram due to improved efficiency at the cheese factory.
• A milk factory reports it bought 27,309,604 litres of milk from farmers last year.
1. 7 plates are to be sold for \$75. The price of 1 plate is 75 ÷ 7 = \$10.71428571. Discuss how to round to get these different answers:
• The supermarket charges to the nearest cent so one plate is priced at \$10.71.
• The petrol station does not use cents only multiples of 5 cents. So it charges \$10.70
Number lines are very useful.
Supermarket:
Petrol Station:
1. Student Exercise.
Complete the table.
Cost per Item Items Total Cost Number Calculator Supermarket Petrol Station Plates \$75.00 7 10.71428571 \$10.71 \$10.70 Tins of Soup \$49.40 13 Cola cans \$104.10 97 Frozen Peas \$71.11 21 Fly Spray \$811.00 206 Chocolate bars \$111.00 73
#### Session 2
1. Jerry measures the width of the school football field and finds it is 70.4m wide. Discuss why this means the true width is between 70.35m and 70.45m. Melissa measures the length of the field and finds it is 101.8m long. Discuss why it is between 101.75m and 101.85m long. Jerry and Melissa now collaborate to find the area by using the calculator. Unfortunately 70.4 x 101.8 = 7166.75 is not a sensible answer. Discuss why this is overly accurate given the possible errors in the data.
Discuss why 70.35 x 101.75 < True Area < 70.45 x 101.85 that is 7158.1125 m2< True Area < 7175.3325m2. Rounding the lower and upper areas to 3 significant figures is a sensible answer but rounding to 2 decimal places (7166.75) is not. In practice an answer like 7170m2 is good enough for practical problems like how much grass seed is needed to resow the field.
2. Student Exercise.
Complete the table for area of rectangles.
Length Width Area Lower Measured Upper Lower Measured Upper Lowest Rounded Highest 101.75 101.8 101.85 70.35 70.4 70.45 7158.1125 7170 7175.3325 111.7 61.3 88.7 9.81 202 11.62 181.7 161.7 9.31 6.34 1.61 0.86
1. Discuss why, when multiplying numbers from measurements the answer is never more accurate that the lowest number of significant figures in the factors.
2. The area of a rectangle is 100.4m and has one side of 20.4m. So the other side is
100.4 ÷ 20.4 = 4.92156862m.
Discuss why the exact answer is between 100.35 ÷ 20.45 and 100.45 ÷ 20.35 that is 4.907m and 4.936m. Discuss why 4.9m is a sensible answer.
#### Session 3
How many significant figures are there in a number?
1. At the Olympics the running track is 400m long. If this number has 1 significant figure the length of the track is between 350 and 450 which is ridiculous.
Discuss and fill in the table
Length Number of Sig. Figs. Least possible length Largest possible length Sensible 400m 1 350 450 No 400m 2 395 405 No 400m 3 399.5 400m 4 399.95 400m 5
So a 400m running track is laid out to an accuracy of 5 significant figures or perhaps 6.
1. Student Exercises.
• Julie says she is about 50 metres from home. How many significant figures are in this number?
• A building plan shows the length of a room is 6700mm long. How many significant figures?
• The Lotto people predict a first division prize of \$1,400,000. How many significant figures?
• The census shows New Zealand population is 4 270 000. How many significant figures?
1. A supermarket works out its prices to be to the nearest cent. Suppose a kilogram of chicken costs \$14.91. Discuss why it is more sensible to say this number has been rounded to 2 decimal places rather than 4 significant figures.
2. If trigonometry has been studied, discuss calculated lengths need to be rounded sensibly. For example;
Discuss why a sensible answer for x is between 2.6 and 2.8. So 2.7 is reasonable.
1. Student Exercises.
#### Session 4
Rounding decimals by using number lines.
1. To round 16.469 to 1 decimal place discuss why the choices are either 16.4 or 16.5 and so a suitable number line is
Discuss why the middle is 16.45 and fill in ten divisions and add the arrow for 16.469.
Discuss why 16.469 is rounded to 16.5 and not 16.4.
1. Student Exercises.
Students who have a limited understanding of decimal place value often find dividing a scale into 10 and labelling each division difficult. It is worth doing nevertheless to assist their understanding. Draw number lines divided into ten labelled parts to solve these problems.
13.89143 (1 decimal place) 707.6341 (2 decimal places) 3.04192 (3 sig. fig.) 80915.81 (nearest 10) 0.0473816 (3 decimal places) 4.004916 (4 sig. fig)
1. This a true story. Mr. Brown rounded 14.486 to the nearest whole number by rounding 14.486 to 14.49 by the "over 5" rule. Then he rounded 14.49 to 14.5 by the same rule. Then he rounded 14.5 to 15 by the rule. Unfortunately this is wrong. Discuss why. This highlights the danger of the rule versus realising that the number is rounded to the closer of the left hand end or right hand end on a number line.
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