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# Vector Cross Product
We will prove that:
For any two linearly independent vectors in $$\mathbb{R}^3$$ $$\vec{r}$$ and $$\vec{v}$$, the magnitude of the vector cross product $$\vec{r} \times \vec{v}$$ is equal to the area of the parallelogram with $$\vec{r}$$ and $$\vec{v}$$ as sides.
Additionally, we will show that for any two linearly dependent vectors, the cross product is the null vector and thus the area of the parallelogram described above is zero.
Before we start, it should be mentioned that there are definitely many ways to prove this. This proof is somewhat unconventional.
Proof:
Consider two linearly independent vectors $$\vec{r}$$ and $$\vec{v}$$ in any plane in $$\mathbb{R}^3$$. For the sake of simplicity, let:
$$\vec{r} = r_x\hat{i} + r_y\hat{j}$$
$$\vec{v} = v_x \hat{i} + v_y \hat{j}$$
Our decision to put $$\vec{r}$$ and $$\vec{v}$$ in the plane $$z=0$$ is completely arbitrary. We could have, for instance, just as easily chosen the plane $$x=5$$. This decision is made purely to make the problem more simple.
Define the following:
$$\phi$$ is the angle between $$\vec{r}$$ and $$\vec{v}$$
$$\theta$$ is the angle between $$\vec{v}$$ and either $$\hat{i}$$ or $$\hat{j}$$. (The choice is, again, arbitrary. So that we may be consistent throughout the remainder of this proof, choose $$\hat{i}$$)
The area of a parallelogram is given by: $$A= bh$$, where $$b$$ is the base and $$h$$ is the vertical height.
Now, $$\vec{r}$$ and $$\vec{v}$$ have components that, if expressed in relation to $$\hat{i}$$ and $$\hat{j}$$, are very cumbersome to work with. What we are going to do is construct a linear map $$A$$ such that:
$$A = T(\vec{x}): \mathbb{R}^2 \longrightarrow \mathbb{R}^2$$
And:
$$A\vec{v} = \left[ \begin{array}{c} |\vec{v}|\\ 0 \end{array} \right]$$
In other words, this linear transformation will shift our basis vectors so that one of them is parallel to $$\vec{v}$$. Then we may take the cross product in this new basis, where $$\vec{v}$$ is defined as above and the components of $$\vec{r}$$ are in terms of only one angle.
Now let us construct $$A$$:
$$A\vec{v} = \left[ \begin{array}{c} |\vec{v}|\\ 0 \end{array} \right]$$
$$A\vec{r} = \left[ \begin{array}{c} |\vec{r}|\cos(\phi)\\ |\vec{r}|\sin(\phi) \end{array} \right]$$
Letting $$A =\left[ \begin{array}{cc} a_{11} & a_{12}\\ a_{21} & a_{22}\end{array} \right]$$
$$\Rightarrow$$
$$a_{11}v_x +a_{12}v_y = |\vec{v}|$$
$$a_{21}v_x + a_{22}v_y = 0$$
$$a_{11}\cos(\phi +\theta) + a_{12} \sin(\phi + \theta) = \cos(\phi)$$
$$a_{21} \cos(\phi +\theta) + a_{22} \sin(\phi + \theta) = \sin(\phi)$$
Now, we want to find $$A$$, so we have a linear system of 4 equations in 4 unknowns, which is solvable. I will leave the details of solving this out, as they are somewhat redundant (just use substitution). The elements of $$A$$ are:
$$a_{22} = \frac{\sin(\phi)}{(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))}$$
$$a_{21} = \frac{-\sin(\phi)\tan(\theta)}{(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))}$$
$$a_{12} = \frac{v_x\cos(\phi) - |\vec{v}| \cos(\phi +\theta)}{v_x(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))}$$
$$a_{11} = \frac{ |\vec{v}| \sin(\phi +\theta) - v_x \cos(\phi) \tan(\theta)}{v_x(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))}$$
Now, before we move on, we need to make sure $$\det(A) \neq 0$$, as this would imply that $$T(\vec{x})$$ is not a one-to-one transformation.
Lemma: $$\det(A) = 0$$ if and only if $$\vec{r}$$ and $$\vec{v}$$ are linearly dependent.
Proof:
$$A= \left[ \begin{array}{cc} \frac{ |\vec{v}| \sin(\phi +\theta) - v_x \cos(\phi) \tan(\theta)}{v_x(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))} & \frac{v_x\cos(\phi) - |\vec{v}| \cos(\phi +\theta)}{v_x(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))}\\ \frac{-\sin(\phi)\tan(\theta)}{(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))} & \frac{\sin(\phi)}{(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))}\end{array} \right]$$
Finding $$\det(A)$$, it is apparent that:
$$\det(A) = 0$$ $$\Rightarrow$$ $$0= |\vec{v}|sin(\phi) \sin(\phi+\theta) - |\vec{v}|\cos(\phi +\theta) \sin(\phi) \tan(\theta)$$
$$\Rightarrow$$ $$\tan(\theta) \cos(\phi +\theta) = \sin(\phi + \theta)$$
$$\Rightarrow$$ $$\tan(\theta) =\tan(\phi + \theta)$$ $$\Rightarrow$$ $$\phi = 0$$
$$\phi$$ is the angle between $$\vec{r}$$ and $$\vec{v}$$, hence if $$\phi = 0$$, then $$\exists$$ $$k \in \mathbb{R}$$ such that $$\vec{r} = k \vec{v}$$. This is the definition of linear dependence and we may conclude that $$\det(A) = 0$$ if and only if $$\vec{r}$$ and $$\vec{v}$$ are linearly dependent.
This proves our lemma.
We will return to this case at the end. For now, let us proceed assuming, as we did at the outset, that $$\vec{r}$$ and $$\vec{v}$$ are linearly independent.
Now:
$$T(\vec{r}) = \left[ \begin{array}{c} |\vec{r}|\cos(\phi)\\ |\vec{r}|\sin(\phi) \end{array} \right]$$
$$T(\vec{v}) = \left[ \begin{array}{c} |\vec{v}|\\ 0 \end{array} \right]$$
Now, taking the cross product in this new basis:
$$T(\vec{r}) \times T(\vec{v}) = \frac{-(|\vec{r}||\vec{v}|\sin(\phi))T(\hat{k})}{|T(\hat{k})|}=-(|\vec{r}||\vec{v}|\sin(\phi))\hat{k}$$
$$\Rightarrow$$ $$|T(\vec{r}) \times T(\vec{v})| = |(|\vec{r}||\vec{v}|\sin(\phi))|$$
Now, in our new basis, $$|\vec{v}|$$ is the base of the parallelogram. Then:
$$|\vec{v}| = b$$
$$|\vec{r}|\sin(\phi))|$$ is the vertical component of $$T(\vec{r})$$ in our transformed basis, i.e. the height of the parallelogram. Then:
$$|(|\vec{r}||\vec{v}|\sin(\phi))| = bh =A$$
Thus we have proved the first part of our statement.
Now, return to the case in which $$\vec{r}$$ and $$\vec{v}$$ are linearly dependent. $$T(\vec{x})$$ is no longer valid under this condition, so we cannot use our transformed basis. We can, however, take the cross product in our initial basis:
$$\vec{r} \times \vec{v} = k \vec{v} \times \vec{v} = (kv_xv_y - kv_yv_x)\hat{k} = \vec{0}$$
$$\Rightarrow$$ $$|\vec{r} \times \vec{v} | = 0$$
This proves the second part of our statement.
QED
Image Credit: Wikipedia
One thing that is interesting to note is that the transformation we did in this proof is akin to one many physics students do without even thinking twice about it. Consider a block on an inclined ramp. You wish to find the time it will take for the block to reach the bottom of the ramp. Many students use a "tilted axis" to solve this problem. This is essentially what we did above. However, most physicists will never rigorously justify this. This is an example of the fact that math is defined very strictly by axioms, and most often intuition is not enough to solve a problem. In physics, often times, intuition is sufficient, due to the non-axiomatic structure of the discipline.
Note by Ethan Robinett
3 years, 8 months ago
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3-2 example11
# 3-2 example11 - Student Grady Simonton Course Math119...
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Unformatted text preview: Student: Grady Simonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 2/18/10 Book: Triola: Elementary Statistics, 11e Time: 11:23 ANI Pennies made before 1983 are 97% copper and 3% zinc, whereas pennies made after 1983 are 3% copper and 97% zinc. Listed below are the weights (in grams) of pennies from each of the two time periods. Find the mean and median for each of the two samples, then compare the two sets of results. Before 1983: 3.1032 3.0769 3.1085 3.1281 3.1039 3.1584 After 1983: 2.4848 2.5019 2.4819 2.4999 2.4910 2.4956 The mean of a set of values is the measure of center found by adding the values and dividing the total by the number of values. Mean = 23 w 11 (— number of sample values Start by finding the mean weight of pennies made before 1983. First add the values. Zx 3.1032 +3.0769 +3.1085 +3. 1281 + 3.1039 + 3.1584 18.6?90 grams To find the mean, divide the sum, 18.6790, by the number of observations. In this problem there are 6 observations. 18.6790 mean = — =5 3.1132 grams The median of a data set is the measure of center that is the middle value when the original data values are arranged in order of increasing magnitude. Start by arranging the weights of the pennies before 1983 in ascending order. 3.0769, 3.1032, 3.1039, 3.1035, 3.1281, 3.1534 If the number of values is odd, the median is the number located in the exact middle of the list. If the number of values is even, the median is found by computing the mean of the two middle numbers. Since there are 6 observations, the data set has an even number of values. Thus, the median is the mean of the two middle observations, 3.1039 and 3.1085. 3.1039+ 3.1085 2 3.1062 grams Median The mean weight of the pennies made before 1983 is 3.] 132 grams and the median weight is 3.1062 grams. Use similar methods to find the mean and median weights of the pennies after 1983. Page 1 Student: Grady Silnonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 2/18/10 Book: Triola: Elementary Statistics, 11e Time: 11:23 ANI The mean weight of the pennies made after 1983 is 2.4925 grams. The median weight of the pennies made after 1983 is 2.4933 grams. Compare the means. If the weight has changed by more than 5% conclude that there is a considerable difference between the means. If the weight has changed by less than 5% there is not a considerable difference between the means. Page 2 ...
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# CBSE Class 10th Math 8 – Introduction to Trigonometry MCQs
#### If sin(θ + 24°) = cosθ, where (θ + 24°) is an acute angle, then θ = ____.
Correct! Wrong!
Given sin(θ + 24°) = cosθ ⇒sin(θ + 24) = sin(90°−θ) ⇒θ + 24° = 90°−θ ⇒2θ = 66° ∴ θ = 33°
#### The minimum value of tan2α + cot2α is ____.
Correct! Wrong!
At α π/4 + tan2 α + cot2 α has minimum value. Tan2 π/4 + cot2 π/4 = tan2 450 + cot2 450 = 1 + 1 = 2 Hence, the minimum value of tan2 α + cot2 α is 2.
#### tan 11° tan 31° tan 45° tan 79° tan 59° = ____.
Correct! Wrong!
We have, tan 11° tan 31° tan 45° tan 79° tan 59° = tan 11° tan 31° × 1 × tan(90°−11°) tan(90°−31°) = tan 11° tan 31° × 1 × cot 11° cot 31° (∵ tan(90°−θ) = cotθ) = 1 (∵ tanθ = 1/cotθ) ∴ tan 11° tan 31° tan 45° tan 79° tan 59° = 1.
#### If A and B are acute angles and sin A = cos B, then A + B is ____.
Correct! Wrong!
Given, sin A = cos B ⇒sin A = sin(90°−B) ⇒A = 90°−B ∴ A +B = 90°
#### If cos(81° + θ) = sin (9° + θ), then θ = ____.
Correct! Wrong!
We have, cos(81° + θ) = sin (9° + θ) ⇒cos(81° + θ) = cos(90° − 9° − θ) ⇒81° + θ = 81° − θ ⇒2θ = 0 ∴ θ = 0
#### Choose the following correct option.
Correct! Wrong!
sin(900 – θ) =cos θ
#### The value of sec4A − tan4A is ____.
Correct! Wrong!
sec4A − tan4A = (sec2 A)2 – (tan2 A)2 = (sec2 A - tan2 A)( sec2 A + tan2 A) = (1) ( sec2 A + tan2 A) (∵sec2 A - tan2 A = 1) = sec2 A + tan2 A ∵ sec4A − tan4A = sec2 A + tan2 A
#### The value of sin4 θ – cos4 θ =____.
Correct! Wrong!
sin2θ – cos2 θ = (sin2θ)2 – (cos2 θ)2 = (sin2θ + cos2 θ)( sin2θ – cos2 θ) = (1)( sin2θ – cos2 θ)( ∵ sin2θ + cos2 θ = 1) = sin2θ – cos2 θ Hence, the value of sin4 θ – cos4 θ is sin2θ – cos2 θ.
#### tan 25° tan 45° tan 65° = ____.
Correct! Wrong!
We have, tan 25° tan 45° tan 65° = tan 25° × 1 × tan (90−25)° = tan 25° cot 25°(∵ tan190° − θ) = cotθ) = tan 25° × 1 tan 25° = 1 ∴ tan 25° tan 45° tan 65° = 1
#### sin 270/cos 630= ____.
Correct! Wrong!
We have, sin 270/cos 630 = sin(900 – 630)/cos 630 = cos 630/cos 630 (∵ sin(90°−θ) = cosθ) = 1 sin 270/cos 630= 1
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# Class10 Maths Exercise 5.1 Q(2) Q(3) Q(4)
## Class10 Maths Exercise 5.1 – Q(2) Q(3) Q(4)
Class10 Maths Exercise 5.1 – In this Question, I have explained Arithmetic Progression Concepts. Some series are AP or not is explained in the video and Written solution of Chapter 5 Exercise 5.1 Q(2) .
CLASS 10 MATHEMATICS
CH-5 ARITHMETIC PROGRESSION (AP)
EX-5.1
Q2. Write first four terms of the A.P. when the first term a and the common difference are given as follows:
(i) a = 10, d = 10
a = 10, d = 10
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
a1 = a = 10
a2 = a1+d = 10+10 = 20
a3 = a2+d = 20+10 = 30
a4 = a3+d = 30+10 = 40
a5 = a4+d = 40+10 = 50
THUS,
Therefore, the A.P. series will be 10, 20, 30, 40, 50 …
And First four terms of this A.P. will be 10, 20, 30, and 40.
(ii) a = – 2, d = 0
Let the AP be a1, a2, a3, a4, a5 …
a1 = a = -2
a2 = a1+d = – 2+0 = – 2
a3 = a2+d = – 2+0 = – 2
a4 = a3+d = – 2+0 = – 2
Therefore, the A.P. series will be – 2, – 2, – 2, – 2 …
THUS,
First four terms of this A.P. will be – 2, – 2, – 2 and – 2.
(iii) a = 4, d = – 3
Let the AP be a1, a2, a3, a4, a5 …
a1 = a = 4
a2 = a1+d = 4-3 = 1
a3 = a2+d = 1-3 = – 2
a4 = a3+d = -2-3 = – 5
Therefore, the A.P. series will be 4, 1, – 2 – 5 …
THUS,
first four terms of this A.P. will be 4, 1, – 2 and – 5.
(iv) a = – 1, d = 1/2
Let the AP be a1, a2, a3, a4, a5 …
a2 = a1+d = -1+1/2 = -1/2
a3 = a2+d = -1/2+1/2 = 0
a4 = a3+d = 0+1/2 = 1/2
Therefore, the A.P. series will be-1, -1/2, 0, 1/2
THUS, First four terms of this A.P. will be -1, -1/2, 0 and 1/2.
(v) a = – 1.25, d = – 0.25
:
Let the AP be a1, a2, a3, a4, a5 …
a1 = a = – 1.25
a2 = a1 + d = – 1.25-0.25 = – 1.50
a3 = a2 + d = – 1.50-0.25 = – 1.75
a4 = a3 + d = – 1.75-0.25 = – 2.00
Therefore, the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 ……..
THUS, first four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.
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CLASS 10 MATHEMATICS
CH-5 ARITHMETIC PROGRESSION (AP)
EX-5.1
Q3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, – 1, – 3 …
Given series;
3, 1, – 1, – 3 …
First term, a = 3
Common difference, d = a2-a1 or
an – an-1
therefore,
d=1 – 3 = -2
THUS, d = -2
(ii) Given series, – 5, – 1, 3, 7 …
First term, a = -5
Common difference, d = a2-a1
therefore,
d= ( – 1)-( – 5) = – 1+5 = 4
THUS, d=4
(iii) Given series, 1/3, 5/3, 9/3, 13/3 ….
First term, a = 1/3
Common difference, d = a2-a1
therefore,
d= 5/3 – 1/3 = 4/3
THUS,
d=4/3
(iv) Given series, 0.6, 1.7, 2.8, 3.9 …
First term, a = 0.6
Common difference, d = a2-a1
therefore,
d=1.7 – 0.6=1.1
THUS,
d=1.1
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CLASS 10 MATHEMATICS
CH-5 ARITHMETIC PROGRESSION (AP)
EX-5.1
Q4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) 2, 4, 8, 16 …
given,
2, 4, 8, 16 …
Here, the common difference is;
a2 – a1 = 4 – 2 = 2
a3 – a2 = 8 – 4 = 4
a4 – a3 = 16 – 8 = 8
Since, an+1 – an or the common difference is not the same every time.
Therefore, the given series are not forming an A.P.
(ii) 2, 5/2, 3, 7/2 ….
Given, 2, 5/2, 3, 7/2 ….
Here,
a2 – a1 = 5/2-2 = 1/2
a3 – a2 = 3-5/2 = 1/2
a4 – a3 = 7/2-3 = 1/2
Since, an+1 – an or the common difference is same every time.
Therefore, d = 1/2 and the given series are in A.P.
The next three terms are;
a5 = 7/2+1/2 = 4
a6 = 4 +1/2 = 9/2
a7 = 9/2 +1/2 = 5
(iii) -1.2, -3.2, -5.2, -7.2 …
Given
, -1.2, – 3.2, -5.2, -7.2 …
Here,
a2 – a1 = (-3.2)-(-1.2) = -2
a3 – a2 = (-5.2)-(-3.2) = -2
a4 – a3 = (-7.2)-(-5.2) = -2
Since, an+1 – an or common difference is same every time.
Therefore, d = -2 and the given series are in A.P.
Hence, next three terms are;
a5 = – 7.2-2 = -9.2
a6 = – 9.2-2 = – 11.2
a7 = – 11.2-2 = – 13.2
(iv) -10, – 6, – 2, 2 …
given,
-10,-6,-2,2
Here, the terms and their difference are;
a2 – a1 = (-6)-(-10) = 4
a3 – a2 = (-2)-(-6) = 4
a4 – a3 = (2 -(-2) = 4
Since, an+1 – an or the common difference is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Hence, next three terms are;
a5 = 2+4 = 6
a6 = 6+4 = 10
a7 = 10+4 = 14
(v) 3, 3+√2, 3+2√2, 3+3√2
GIVEN,
3, 3+√2, 3+2√2, 3+3√2
Here,
a2 – a1 = 3+√2-3 = √2
a3 – a2 = (3+2√2)-(3+√2) = √2
a4 – a3 = (3+3√2) – (3+2√2) = √2
Since, an+1 – an or the common difference is same every time.
Therefore, d = √2 and the given series forms a A.P.
Hence, next three terms are;
a5 = (3+√2) +√2 = 3+4√2
a6 = (3+4√2)+√2 = 3+5√2
a7 = (3+5√2)+√2 = 3+6√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
GIVEN,
3, 3+√2, 3+2√2, 3+3√2
Here,
a2 – a1 = 0.22-0.2 = 0.02
a3 – a2 = 0.222-0.22 = 0.002
a4 – a3 = 0.2222-0.222 = 0.0002
Since, an+1 – an or the common difference is not same every time.
Therefore, and the given series doesn’t forms a A.P.
(vii) 0, -4, -8, -12 …
GIVEN,0, -4, -8, -12 …
Here,
a2 – a1 = (-4)-0 = -4
a3 – a2 = (-8)-(-4) = -4
a4 – a3 = (-12)-(-8) = -4
Since, an+1 – an or the common difference is same every time.
Therefore, d = -4 and the given series forms a A.P.
Hence, next three terms are;
a5 = -12-4 = -16
a6 = -16-4 = -20
a7 = -20-4 = -24
(viii) -1/2, -1/2, -1/2, -1/2 ….
GIVEN, -1/2, -1/2, -1/2, -1/2 ….
Here,
a2 – a1 = (-1/2) – (-1/2) = 0
a3 – a2 = (-1/2) – (-1/2) = 0
a4 – a3 = (-1/2) – (-1/2) = 0
Since, an+1 – an or the common difference is same every time.
Therefore, d = 0 and the given series forms a A.P.
Hence, next three terms are;
a5 = (-1/2)-0 = -1/2
a6 = (-1/2)-0 = -1/2
a7 = (-1/2)-0 = -1/2
(ix) 1, 3, 9, 27 …
GIVEN,1, 3, 9, 27 …
Here,
a2 – a1 = 3-1 = 2
a3 – a2 = 9-3 = 6
a4 – a3 = 27-9 = 18
Since, an+1 – an or the common difference is not same every time.
Therefore, and the given series doesn’t form a A.P.
(x) a, 2a, 3a, 4a …
GIVEN,a, 2a, 3a, 4a …
Here,
a2 – a1 = 2aa
a3 – a2 = 3a-2a = a
a4 – a3 = 4a-3a = a
Since, an+1 – an or the common difference is same every time.
Therefore, d = a and the given series forms a A.P.
Hence, next three terms are;
a5 = 4a+a = 5a
a6 = 5a+a = 6a
a7 = 6a+a = 7a
(xi) aa2, a3, a4 …
GIVEN,aa2, a3, a4 …
Here,
a2 – a1 = a2–a = a(a-1)
a3 – a2 = a3 – a2 = a2(a-1)
a4 – a3 = a4 – a3 = a3(a-1)
Since, an+1 – an or the common difference is not same every time.
Therefore, the given series doesn’t forms a A.P.
(xii) √2, √8, √18, √32 …
GIVEN,√2, √8, √18, √32 …
Here,
a2 – a1 = √8-√2 = 2√2-√2 = √2
a3 – a2 = √18-√8 = 3√2-2√2 = √2
a4 – a3 = 4√2-3√2 = √2
Since, an+1 – an or the common difference is same every time.
Therefore, d = √2 and the given series forms a A.P.
Hence, next three terms are;
a5 = √32+√2 = 4√2+√2 = 5√2 = √50
a6 = 5√2+√2 = 6√2 = √72
a7 = 6√2+√2 = 7√2 = √98
(xiii) √3, √6, √9, √12 …
GIVEN,√3, √6, √9, √12 …
Here,
a2 – a1 = √6-√3 = √3×√2-√3 = √3(√2-1)
a3 – a2 = √9-√6 = 3-√6 = √3(√3-√2)
a4 – a3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)
Since, an+1 – an or the common difference is not same every time.
Therefore, the given series doesn’t form a A.P.
(xiv) 12, 32, 52, 72 …
GIVEN,12, 32, 52, 72 …
Here,
a2 − a1 = 32−12 = 20
a3 − a2 = 52−32 = 20
a4 − a3 = 72−52= 20
Since, an – an-1 or the common difference is same every time.
Therefore, the given series forms a A.P.
The next three terms are
a5=a4+d=92
a6=a5+d=112
a7=a6+d=132
12,32,52,72,92, 112 ,132 are in AP
(xv) 12, 52, 72, 73 …
Given, 12, 52,72,73….
Here,
a2 − a1 = 52−12 = 40
a3 − a2 = 72−52= 20
a4 − a3 = 73−72 = 1
Since, anan-1 or the common difference isn’t same every time.
Therefore, the series isn’t AP
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## 0 thoughts on “Class10 Maths Exercise 5.1 Q(2) Q(3) Q(4)”
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гидра официальный сайт, конечно же, реализует анонимность в сети, но тем не менее, этой защищенности не хватает и заниматься с проектом с простого браузера нельзя. При открытии ресурса через обыкновенный для вас браузер интернет-провайдер проследит все проекты, на которые вы заходили, и столь подозрительная активность заинтересует правоохранительные службы. Потому требуется подумать о особой безопасности.
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hydra onion ссылка доступна в тор Браузер, тор браузер это открытое и свободное ПО для реализации второго уровня луковой маршрутизации. Это цепочка прокси-серверов последовательно соединенных друг с другом в длинную цепь online соединений, позволяющая устанавливать секретное скрытное сетевое соединение. Можно рассматривать как секретную сетку виртуальных туннелей (VPN), оказывающая трафик данных в зашифрованном варианте. Известность приобрел как инструментарий для “свободного” online-серфинга, в частности просмотра блокированных вебсайтов таких как Гидра и аналогичных ресурсов из нелегального интернета (Darknet). Применяя тор браузер Вы остаетесь анонимными только лишь до тех пор пока не станете сохранять собственные индивидуальные данные, не надо забывать о собственной безопасности, по этой причине мы рекомендуем Вам не хранить пароли и иную информацию, применяя какую злодеи смогут Вам причинить вред, чистите кэш, куки и удаляйте историю посещений.
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Взять ссылку на гидру и безопасно покупать можно на нашем проекте. В онлайне очень часто можно наткнуться на мошенников и утерять собственные личные денежные средства. По этой причине для Вашей защиты мы подготовили данный интернет-портал где Вы стабильно сможете получить доступ к магазину торговой платформы ссылка на гидру в тор. Для выполнения закупок на торговой площадке гидра наш сайт ежедневно посещает большое количество пользователей, для принятия действующей работоспособной ссылки, надо надавить на кнопку раскрыть и надежно совершить закупку, а если Вы первый раз зашли на интернет-сайт до покупки товара нужно зарегистрироваться и пополнить баланс. Ваша защищенность наша главная цель, какую мы с гордостью осуществляем.
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Теперь в подробностях обсудим, как заниматься с платформой, так как здесь есть ряд особенностей, которые следует принимать к сведению. Поэтому пошагово разберем момент работы с проектом, приобретение товаров и их продажу. Независимо от того, зачем вы зашли на hydra onion, интернет-сайт потребует регистрации для выполнения действий.
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hydra зеркало это сайт из online-сети тор, основанный для безымянного и защищенного серфинга в запрещенном онлайне, в сети тор насчитается около миллиона веб-сайтов различной тематики как правило это интернет-магазины и форумы, обнаруживаются и ресурсы с довольно подозрительной активностью, касаться их мы здесь мы не будем (если они Для вас занятны сможете пользоваться особой поисковой машиной DuckDuckGo, она по умолчанию интегрирована в TOR браузер), портал гидра онион доступен по актуальному зеркалу выставленному на страничках этого сайта. Чтобы перейти на гидру в зоне онион Вам понадобится тор браузер так как домены в зоне onion специально созданы для анонимной сети TOR, надавите на кнопку “скопировать” (она расположена перед этим) далее вставьте ссылку в строку адреса в тор браузере и перейдите на портал магазина-online гидра, по окончании совершения закупки не забудьте подчистить интернет-браузер, благополучных закупок.
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hydra onion относительно объёмистый, в основном, это одна из наиболее известных платформ в государствах СНГ. Поэтому, если вам требуются какие-либо запрещенные категории изделий, то вы точно подберете их тут.И значительное количество других товаров, какие относятся к этим обобщенным категориям. Кроме того, Гидра и портал платформы регулярно развиваются, web магазинов делается все больше, перечень товаров увеличивается, потому, если здесь чего-либо не существовало прежде, может появится сейчас.
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Пользоваться торговой площадкой гидра сайт трудно. Специально для Вас мы приготовили все потенциальные способы облегчения этой проблемы. Разработанная нами постоянно работающая гидра ссылка поможет легко и быстро раскрыть вебсайт в обыкновенных интернет-браузерах, в том числе Google Chrome, Opera, Яндекс.Браузер и т.д. Для захода необходимо лишь кликнуть на кнопку ОТКРЫТЬ и воспользоваться сервисами трейдерской платформы Hydra. Имея цель защищенности абонента от обмана и предупреждения захода по фишинговому адресу, мы показали гиперссылку на трейдерскую платформу Гидра, с опцией ее тиражирования (путем нажатия кнопки СКОПИРОВАТЬ) и использования в анонимном Tor браузере. Наша задача упростить пользователям вход к гидре и тем самым позволить площадке развиваться и преуспевать, мы за беспрепятственный online без политических запретов.
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Как войти на hydra? Данным вопросом задаются все пользователи гидры, каждый день необходимо отыскивать рабочее зеркало гидры т.к. каждый день рабочие зеркала блокируются правительством и входа к ресурсу не существует, применять VPN непросто и недешево, тор на английском языке, что также не всем подойдет. Преднамеренно для наибольшего упрощения данной задачи мы разработали наш сайт. Для открытия hydra Вам надо перейти по действующему работающему зеркалу показанному выше или скопировать ссылку для тор браузера какая также показана на страничках нашего сайта и раскрыть ее в тор браузере, после этого пройти регистрацию, дополнить баланс и радоваться приобретениям. Не забывайте помогать развитию ресурса делитесь нашим ресурсом с приятелями и знакомыми.
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Получить ссылку на гидру и безопасно покупать возможно на нашем проекте. В интернете зачастую возможно натолкнуться на жуликов и потерять собственные личные средства. По этой причине для Вашей защиты мы спроектировали данный интернет-портал где Вы всегда можете иметь вход к магазину трейдерской платформы гидра зеркало. Для совершения закупок на торговой платформе гидра наш сайт каждый день посещает масса клиентов, для принятия действующей работоспособной гиперссылки, надо надавить на кнопочку открыть и безопасно совершить покупку, а если Вы впервые зашли на сайт перед приобретением товара требуется зарегистрироваться и дополнить баланс. Ваша собственная безопасность наша основная цель, которую мы с гордостью выполняем.
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Каким способом зайти на hydra? Данным моментом задаются все пользователи гидры, ежедневно требуется искать работающее зеркало гидры т.к. изо дня в день рабочие зеркала банятся властью и входа к online-ресурсу не существует, использовать VPN сложно и дорого, тор на британском языке, что тоже далеко не всем подходит. Специально для максимального упрощения этой задачи мы спроектировали наш сайт. Для раскрытия hydraruzxpnew4af Вам нужно зайти по действующему рабочему зеркалу указанному выше либо скопировать ссылку для тор браузера какая точно также показана на нашем сайте и раскрыть ее в тор браузере, после чего зарегистрироваться, пополнить счет и восторгаться покупкам. Не забывайте при этом помогать совершенствованию портала обмениваетесь нашим ресурсом с приятелями и родственниками.
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гидра зеркало довольно большой, в основном, это одна из особо известных площадок в государствах СНГ. Поэтому, если вам требуются какие-либо запрещенные категории изделий, то вы гарантированно найдете их здесь.И большое число других изделий, какие относятся к таким обобщенным группам. Кроме того, Гидра и сайт площадки непрерывно прогрессируют, магазинов он-лайн становится все больше, выбор изделий подрастает, поэтому, если тут чего-то не существовало прежде, сможет обнаружиться сегодня.
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гидра официальный сайт довольно большой, в целом, это одна из особенно известных площадок в странах СНГ. Поэтому, если вам необходимы определенные не разрешенные группы изделий, то вы точно отыщите их тут.И большое число иных изделий, какие имеют отношение к этим обобщенным категориям. Помимо этого, Гидра и сайт платформы регулярно прогрессируют, магазинов online становится все больше и больше, ассортиментный выбор товаров растет, поэтому, если здесь чего-либо не было вчера, сможет обнаружиться сейчас.
13. hydraruzxpnew4af будет доступна в тор Браузер, тор браузер это открытое и свободное программное обеспечение для выполнения 2-го уровня так называемой луковой маршрутизации. Это сегодня цепочка прокси-серверов последовательно соединенных друг с другом в длинную цепочку online соединений, позволяющая устанавливать секретное скрытное соединение в сети. Рассматривается как анонимная сетку виртуальных туннелей (VPN), предоставляющая трафик данных в зашифрованном виде. Свою известность приобрел как инструментарий для “свободного” интернет-серфинга, в частности просмотра заблокированных сайтов например Гидра и аналогичных порталов из теневого интернета (Darknet). Применяя тор браузер Вы останетесь неизвестными только до тех пор пока не начнете сохранять свои индивидуальные данные, не надо забывать о собственной безопасности, поэтому мы советуем Вам не сохранять пароли на вход и прочую информацию, используя какую злодеи смогут Вам навредить, чистите кэш, куки и удаляйте историю.
14. Как войти на гидру? Названным вопросом озадачены все пользователи гидры, ежедневно требуется отыскивать рабочее зеркало гидры т.к. постоянно зеркала банятся властью и входа к ресурсу нет, применять VPN непросто и недешево, тор на британском языке, что тоже далеко не всем подходит. Преднамеренно для максимального упрощения данной задачи мы создали сайт. Для раскрытия hydra onion обход блокировки Вам надо зайти по актуальному рабочему зеркалу указанному перед этим либо скопировать гиперссылку для тор браузера которая также указана на страницах нашего ресурса и раскрыть ее в тор браузере, после чего пройти регистрацию, дополнить счет и наслаждаться приобретениям. Не забывайте при этом содействовать развитию ресурса делитесь нашим ресурсом с товарищами и знакомыми.
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hydra ссылка довольно крупный, в основном, это одна из наиболее популярных платформ в государствах СНГ. Потому, если для вас требуются определенные нелегальные категории изделий, то вы наверняка разыщете их здесь.И значительное число других товаров, которые имеют касательство к таким обобщенным группам. Кроме того, Гидра и сайт платформы непрерывно раскручиваются, он-лайн магазинов делается все больше, перечень изделий подрастает, поэтому, если тут чего-либо не было вчера, может обнаружиться теперь.
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Получить ссылку на гидру и безопасно сделать покупку можно на страницах нашего вебсайта. В глобальной сети интернет очень часто можно натолкнуться на жуликов и потерять свои личные деньги. Именно поэтому для Вашей защиты мы создали данный интернет-портал где Вы постоянно можете получить вход к online-магазину трейдерской площадки hydraruzxpnew4af. Для выполнения закупок на трейдерской площадке гидра наш портал каждый день посещает большое количество пользователей, для получения действующей работоспособной гиперссылки, нужно нажать на кнопочку раскрыть и безопасно совершить закупку, а если Вы в первый раз зашли на портал до приобретения изделия требуется зарегистрироваться и пополнить баланс. Ваша безопасность наша важнейшая задача, которую мы с достоинством осуществляем.
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Воспользоваться торговой площадкой hydraruzxpnew4af трудно. Для Вас мы приготовили все вероятные способы упрощения этой проблемы. Созданная нами непрерывно работающая hydra ссылка позволит свободно и быстро раскрыть интернет-сайт в обыкновенных интернет-браузерах, в том числе Google Chrome, Яндекс.Браузер, Opera и т.д. Для перехода необходимо лишь нажать на кнопочку ОТКРЫТЬ и воспользоваться услугами трейдерской платформы Hydra. С целью защищенности пользователя от подлога и предотвращения перехода по фишинговому интернет-адресу, мы указали гиперссылку на трейдерскую площадку Гидра, с опцией ее копирования (посредством щелчка кнопки СКОПИРОВАТЬ) и использования в защищенном Tor браузере. Наша роль упростить абонентам доступ к гидре и таким образом сделать возможным платформе развертываться и преуспевать, мы за свободный online без политических ограничений.
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hydraruzxpnew4af это трейдерская площадка различных изделий некоторой тематики. Интернет-сайт действует с 2015 г. и на сегодняшний день деятельно развивается. Основная валюта – криптовалюта Bitcoin. Дополнительно для покупки данной денежной единицы на сайте работают штатные обменники. Закупить или обменять Биткоин сможете с помощью раздела “Баланс” в кабинете пользователя. Гидра предлагает два варианта получения товаров: первый – это клад (тайник, магнит, закладки, прикоп); другой – транспортировка по стране (почта, транспортные фирмы, курьерские службы). Громадное число проверенных магазинов он-лайн результативно осуществляют свои реализации на протяжении ряда лет. На сайте имеется система отзывов, посредством которой Вы сможете убедиться в честности торговца. Торговая площадка Hydra приспособлена под всякие устройства. В связи с блокированием гиперссылки Hydra систематично ведутся обновления рабочих зеркал для обхождения блокировки. Вслед за новейшими зеркалами возникают и “фейки” торговой платформы Гидра. Обычно фейк идентичен официальному веб-сайту гидра, однако зайти в личный кабинет не выйдет, т.к. это фейк и его цель накопление логинов и паролей. Всегда проверяйте ссылка на гидру по которой Вы переходите, а лучше всего применяйте действующие ссылки на гидру представленные на страничках этого сайта и Ваши данные не угодят во владение мошенников.
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hydra ссылка доступна в тор Браузер, тор браузер это открытое и свободное ПО для реализации второго поколения луковой маршрутизации. Это цепочка прокси-серверов последовательно связанных между собой в длинную цепочку online соединений, позволяющая настраивать секретное не оставляющее следов соединение в сети. Рассматривается как анонимная сеть условных туннелей (VPN), оказывающая передачу данных в зашифрованном варианте. Свою популярность приобрел как инструментарий для “свободного” интернет-серфинга, к примеру просмотра блокированных сайтов например Гидра и подобных интернет-ресурсов из нелегального онлайна (Darknet). Применяя тор браузер Вы будете анонимными только лишь до тех пор пока не начнете сохранять собственные индивидуальные данные, не нужно забывать о своей защищенности, по этой причине мы советуем Вам не сохранять пароли и иную информацию, используя какую злодеи сумеют Вам навредить, очищайте кэш, куки и стирайте историю.
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hydra onion, разумеется, гарантирует секретность в сети, но тем не менее, этой защищенности не хватает и заниматься с проектом с простого браузера нереально. При входе на сайт используя привычный для вас браузер интернет-провайдер отследит все проекты, на которые вы заходили, и столь сомнительная интенсивность заинтересует структуры правопорядка. Вследствие этого надо задуматься о дополнительной защищенности.
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Как зайти на гидру? Таким моментом задаются все участники гидры, каждый день приходится отыскивать рабочее зеркало гидры т.к. каждый день зеркала банятся правительством и доступа к online-ресурсу не существует, использовать VPN сложно и недешево, тор на английском языке, что также далеко не всем подходит. Преднамеренно для максимального упрощения этой задачи мы спроектировали сайт. Для раскрытия гидра сайт анонимных покупок Вам нужно перейти по действующему работающему зеркалу показанному перед этим или скопировать гиперссылку для тор браузера какая также показана на страничках нашего сайта и открыть ее в тор браузере, после чего зарегистрироваться, пополнить счет и наслаждаться покупкам. Не забывайте при этом содействовать развитию портала обмениваетесь нашим интернет-ресурсом с друзьями и родственниками.
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гидра ссылка доступна в тор Браузер, тор браузер это открытое и свободное ПО для реализации второго уровня так называемой луковой маршрутизации. Это цепочка прокси-серверов в последовательности связанных между собой в длинную цепь интернет соединений, позволяющая настраивать анонимное бесследное соединение в сети. Рассматривается как секретная сеть виртуальных туннелей (VPN), предоставляющая передачу данных в закодированном варианте. Популярность приобрел как инструмент для “свободного” интернет-серфинга, к примеру просмотра блокированных вебсайтов таких как Гидра и подобных ресурсов из теневого интернета (Darknet). Применяя тор браузер Вы будете анонимными только лишь до того времени пока не станете хранить собственные индивидуальные данные, не надо забывать о собственной защищенности, поэтому мы рекомендуем Вам не сохранять пароли и прочую информацию, применяя которую злоумышленники сумеют Вам причинить вред, чистите кэш, куки и удаляйте историю.
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гидра ссылка, естественно, обеспечивает секретность в глобальной интернет-сети, и все же, этой защищенности не хватает и заниматься с платформой с обычного браузера невозможно. При открытии ресурса используя обычный для вас браузер online-провайдер проследит все разделы, на какие вы заходили, и настолько подозрительная активность заинтересует органы правопорядка. Потому требуется поразмышлять о особой защищенности.
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hydra ссылка это наиболее широкая трейдерская площадка не разрешенных изделий в Рф и государствах СНГ. Тут можно купить такие товары как разнообразные эйфоретики, диссоциативы, всевозможные аптечные средства, экстази, марихуана, психоделические препараты, энетеогены, опиаты, хим реактивы и конструкторы, разные стимуляторы, кроме того возможно анонимно обналичить Ваши биткоины и заказать всевозможные виды документов разных государств. На трейдерской площадке гидра происходит большое количество заказов каждый день, множество тысяч удовлетворенных заказчиков и позитивных мнений. Наш вэб-портал помогает всем заказчикам иметь надежный вход к трейдерской платформе hydra и ее изделиям и услугам. Гидра онион открыта в тор браузере, ссылку на актуальное рабочее зеркало Вы можете скопировать выше, достаточно надавить на кнопку СКОПИРОВАТЬ.
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Воспользоваться трейдерской площадкой гидра ссылка сложно. Намеренно для Вас мы подготовили все вероятные варианты облегчения этой задачи. Разработанная нами неизменно рабочая hydra ссылка поможет легко и быстро открыть портал в классических интернет-браузерах, таких как Opera, Яндекс.Браузер, Google Chrome и т.д. Для захода стоит только нажать на кнопочку ОТКРЫТЬ и пользоваться услугами торговой площадки Hydra. Имея цель защищенности абонента от обмана и предотвращения перехода по фишинговому адресу, мы показали ссылку на торговую площадку Гидра, с опцией ее копирования (посредством щелчка кнопки СКОПИРОВАТЬ) и применения в защищенном Tor браузере. Наша задача упростить абонентам доступ к гидре и тем самым позволить площадке развертываться и процветать, мы за свободный интернет без общественно-политических запретов.
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рабочее зеркало гидры это трейдерская площадка всевозможных товаров определенной направленности. Сайт функционирует с 2015 года и сегодня активно раскручивается. Основная валюта – криптовалюта Bitcoin. Специально для закупки данной денежной единицы на проекте работают штатные обменники. Купить либо обменять Bitcoin возможно посредством раздела “Баланс” в кабинете пользователя. Гидра может предложить два метода получения товаров: главный – это клад (закладки, прикоп, тайник, магнит); второй – транспортировка по всей России (курьерские службы, почта, транспортные компании). Громадное число испытанных online магазинов удачно осуществляют свои реализации несколько лет подряд. На сайте существует система ответов, с помощью которой Вы можете удостовериться в честности торговца. Площадка торговли Гидра приспособлена под всякие устройства. В связи с блокировкой ссылки Hydra регулярно выполняются обновления сайтов-зеркал для обхождения блокировки. Прямо за новыми зеркалами появляются и “фейки” трейдерской площадки Hydra. В основном фейк аналогичен официальному сайту гидра, однако войти в личный кабинет не выйдет, т.к. это фейк и его задача сбор логинов и паролей. Всегда контролируйте ссылка на гидру по которой Вы переходите, а надежнее используйте действующие ссылки на hydra выставленные на нашем ресурсе и Ваши данные не попадут во владение мошенников.
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hydraruzxpnew4af это трейдерская платформа разнообразных товаров определенной тематики. Ресурс действует с 2015 года и сегодня динамично развертывается. Главная валюта – криптовалюта Bitcoin. Специально для приобретения этой валюты на ресурсе функционируют штатные обменники. Купить или обменять Bitcoin можно с помощью раздела “Баланс” в личном кабинете. Гидра предлагает два вида получения изделий: первый – это клад (закладки, магнит, тайник, прикоп); следующий – транспортировка по всей России (почта, курьерские службы, транспортные фирмы). Огромное количество испытанных он-лайн магазинов успешно осуществляют свои продажи на протяжении ряда лет. На сайте существует система отзывов, посредством какой Вы можете удостовериться в честности продавца. Площадка торговли Hydra адаптирована под разные девайсы. В связи с блокированием ссылки Hydra периодически выполняются ревизии зеркал для обхождения блокирования. Прямо за новыми зеркалами возникают и “фейки” торговой площадки Hydra. В основном фейк идентичен официальному ресурсу hydra, однако зайти в личный кабинет не выйдет, т.к. это фейк и его задача сбор логинов и паролей. Всякий раз контролируйте ссылка на гидру по которой Вы заходите, а надежнее используйте действующие ссылки на hydra представленные на нашем веб-сайте и Ваши данные не попадут в руки жуликов.
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гидра официальный сайт это трейдерская площадка разнообразных изделий некоторой тематики. Сайт действует с 2015 года и на сегодняшний день энергично развивается. Главная денежная единица – криптовалюта Bitcoin. Специально для покупки данной валюты на нашем сайте работают штатные обменники. Купить или поменять Bitcoin сможете посредством раздела “Баланс” в кабинете пользователя. Гидра может предложить два вида получения товаров: главный – это клад (прикоп, тайник, закладки, магнит); второй – доставка по всей России (курьерские службы, транспортные фирмы, почта). Большое число проверенных магазинов online с успехом осуществляют свои продажи на протяжении ряда лет. На сайте существует система ответов, посредством которой Вы сможете удостовериться в добросовестности торговца. Площадка торговли Гидра приспособлена под всякие устройства. В связи с блокированием гиперссылки Hydra периодично проводятся ревизии сайтов-зеркал для обхода блокировки. Вслед за новыми зеркалами появляются и “фейки” трейдерской площадки Hydra. Обычно фейк идентичен главному веб-сайту гидра, однако зайти в кабинет пользователя не удастся, т.к. это фейк и его задание сбор логинов и паролей. Каждый раз проверяйте ссылка на гидру по какой Вы переходите, а лучше используйте действующие гиперссылки на гидру представленные на нашем вебсайте и Ваши данные не угодят во владение жуликов.
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# CLASS-11RELATION & FUNCTION - ABSOLUTE VALUE FUNCTION
ABSOLUTE VALUE FUNCTION -
Example.1) The absolute value function y = f(x) =│x│
Ans.) The domain consists of all real numbers, but the range is made up of all non-negative real numbers. The graph is shown in below figure -
When x ≥ 0, y =│x│is equivalent to y = x, the equation of the straight line through the origin with slope 1. When x < 0, y =│x│is equivalent to y = -x, the equation of the straight line through the origin with slope -1.
Example.2) Draw the graph of the function f(x) = │x - 1│
Ans.) The domain is the set of all real numbers. The range is the set of all non-negative real numbers.
The graph is the graph of y =│x│ shifted one unit to the right. (Ans.)
Example.3) Draw the graph of the function f(x) = - x│x│.
Ans.) If x > 0, then y = f(x) = - x . x = - x²,
So, y = - x², for x > 0 ..........(i)
If, x = 0, then y = f(x) = 0,
So, y = f(x) = 0, for x = 0 ............(ii)
If, x < 0, then y = f(x) = (- x) (- x) = x²
so, y = f(x) = x² for x < 0 .................(iii)
From (i), as x ⟶ ∞, y = - ∞ for x > 0.
From (iii), as x ⟶ - ∞, y ⟶ ∞ for x < 0,
From (ii), y = f(x) = 0
Example.4) Draw the graph of the following functions
│x│
(i) f(x) = ------, (ii) f(x) = │x│+│x - 1│
x
Ans.) (i) The domain is the set of all non-zero real numbers. The graph is the right half of the line y = 1 for x > 0, plus the left half of the line y = - 1 for x < 0
The range is {1, -1}. The graph is not defined for x = 0.
(ii) Case.1) x ≥ 1, then f(x) = x + x - 1 = 2x - 1.
Case.2) 0 ≤ x < 1, then f(x) = x - (x - 1) = 1.
Case.3) x < 0, then f(x) = - x - (x - 1) = - 2x + 1.
So, the graph consists of a horizontal line segment and two half-lines.
Example.5) Draw the graph of the function y =│x - 2│+│x - 3│
Ans.)
y = │x - 2│+│x - 3│
Here, we consider separate intervals.
When x > 3, (x - 2), (x - 3) both will be positive,
therefore,
y = x - 2 + x - 3
=> y = 2x - 5
Some points of the graph are -
When, x = 3, y = │x - 2│+│x - 3│
= │3 - 2│+│3 - 3│ = 1
When, x = 2, y = │x - 2│+│x - 3│
= │2 - 2│+│2 - 3│ = 1
and if 2 < x < 3 then (x - 2) will be positive and (x - 3) will be negative, thus,
y = (x - 2) - (x - 3) = 1
so, for 2 ≤ x ≤ 3, y = 1 (a line parallel to the x-axis)
Now, if x < 2, in this case both (x - 2) and (x - 3) will be negative and, therefore, y = - (x - 2) - (x - 3) = 5 - 2x.
Some points on the line y = 5 - 2x for x < 2 are -
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# Our Blog
Strategies, Sample Questions, and Random Ramblings.
# GMAT Statistics
by ejkiv
July 19, 2015
GMAT Statistics. It is probably one of the most hated classes in the entire business school experience. Luckily, you don’t have to worry about a lot of the complexities for quite some time, however this is a section that is getting more play in exams lately, in fact you are likely to see 2 or 3 questions on this topic come exam day. Average is the one concept that is covered, and we have already covered 2 sections related to this subject. Here are a few more:
Range: The difference between the largest value of a set and the smallest value of a set.
The range of the set {2,3,5,7,10} is 8, which comes from 10-2.
Median: The middle number of a set. If there are an even number of terms, it is the average of the two middle most terms in the set.
The median of the set {2,3,5,7,10} is 5 because it is the middle term. The median of the set {2,3,5,7,7,10} is 6 because the average of the two middle terms, 5 and 7, is 6.
A special type of set is one that is evenly spaced. {1,2,3,4,5} or {10, 20, 30} or {150, 300, 450, 600} or {11,13,15,17} are all evenly spaced sets. We discussed this earlier, but there is still a small amount of information to add with our new terminology: The median will always equal the mean. Even better than this is the fact that the mean and median can be calculated from taking the average of the first and the last term if you can find the median by traditional methods. This will become a great shortcut on longer problems.
Mode: The number that occurs most in a set
The mode of the set {2,3,5,7,7,10} is 7. There can be more than one mode. In the set {2,2,3,3,7,10} both 2 and 3 are modes of this set.
The final piece of statistics that has been creeping up more and more on the GMAT is standard deviation. You will not have to calculate standard deviation (you can wait until your 1st year stats class for that) but you should know how it is calculated and what exactly it stands for.
Standard Deviation: A measure of how spread out numbers are. This can also be describe as variability.
Variance: The square of standard deviation.
I threw variance in there because it is so closely related to standard deviation. Let’s look at an example calculation with a few sets just to understand it a bit better;
A = {3, 4, 5, 6, 7}
B = {10, 10, 20, 30, 30}
The first thing that you will need to know is the mean of each set; in this case, the mean of A is 5 and B is 20. You can probably see that the numbers in A are closer to the mean than in set B, and this is likely how you will have to evaluate things on the GMAT; however, we will still go through the calculation.
To calculate standard deviation you need “the average of the sum of the squared differences from the mean.” It sounds like a mouth full, but in practice it isn’t so bad. Start with the squaring the differences from the mean for each set:
The sum of the differences in Set A is 10 and in Set B is 400. The average difference is equal to the variance, or 2 for Set A and 80 for Set B. The standard deviation is simply the square root of variance or sqrt(2) for set A and sqrt(80) for set B. Here is what the entire calculation looks like for set A, just so you can see.
And because variance is just the square of standard deviation it is 2. Again, you will not have to calculate this for the exam, but it is worth remembering how to do if for no other reason than you will see it again. In actuality, knowing it will prepare you better for any question you may face.
The next question that you will face is related to the number of standard deviations from the mean. If a set had an average of 50 and a standard deviation of 5 then each standard deviation is spaced 5 above or below the mean of 50. Thus, one standard deviation above the mean is 55 and two standard deviations above is 60. Similarly, 1 standard deviation below the mean is 45 and two standard deviations below is 40.
Within one standard deviation: 50 ± 5 or between 45 and 55
Within two standard deviation: 50 ± 5(2) or between 40 and 60
Within three standard deviation: 50 ± 5(3) or between 35 and 65
Within four standard deviation: 50 ± 5(4) or between 30 and 70
Thus, if you were to be asked for a number that is between 1 and 2 standard deviations from the mean, both 42 and 58 would be acceptable answers.
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# Learning How to Solve the Equation For a Moving Object
Linear algebra is a branch of mathematics which deals with linear equations like:
Let’s look at some of these equations and how they can be solved using the concepts of linear algebra. We will also look at the various properties of linear equations and what it means to linear algebra.
The first equation, ‘r’, is an important one because it tells us how much force is exerted on a mass by a push stroke. This is known as the power of gravity or the gravitational force. This is a fairly simplistic form of linear algebra, but it does tell us how much force is exerted.
The second equation, ‘s’, is known as the sine wave. It tells us that this acceleration is equal to the sine wave frequency. If we apply this equation to the first equation we get:
By using this formula we can calculate the value of r for any point on a line. This value is then compared to the value of S. This gives us the acceleration, a which is measured in terms of the velocity of the object.
The third equation, ‘m’ is known as the mirror image of r. It tells us the magnitude of the acceleration of the object.
Finally, the fourth equation, ‘x’, is equal to the slope of the line or the curve of the velocity. As long as the speed of the object is constant then the slope of the line is always constant and this gives us an expression for the acceleration.
These four equations are the basics of linear algebra. In particular you should be able to do all of these problems with ease if you have a basic knowledge of this subject.
‘m’ is known as the mover of the mover. When you change the direction of the mass, i.e., if you move the object from right to left then the slope of the line will be higher. It is more of an accelerating force. You can also apply this rule to accelerate the object when you drop it.
‘s’ is the slope of the line for the opposite direction. This is actually what we would do if we were to accelerate the object at the same velocity.
‘x’ is equal to the value of the velocity multiplied by the angle of attack. For example, if you drop the object in front of the curve at an angle of 45 degrees it will travel in a horizontal direction.
A more complex example is to use the formula ‘x’ x = cos(x) where x is the angle of attack. As long as we know that the equation is correct then it will give a good approximation to the object’s velocity.
Using the above formula we can now solve the equation, ‘y = m’. We can now solve for y by knowing the equation’s’. To do this we need to take the square root of the equation.
So, using the above formula and using the formula we can solve for m, we can then solve for m using the formula, ‘y’ and then solve for the slope of the line. Using the first two, we can then find the acceleration, s. Therefore, by doing this we can find the value of m.
Now, you need to think of some ways that you can combine this information with others so that you get the value of m for a number of different types of objects. For example, if we want to find the value of m for a cylinder then we can multiply the above equation by the velocity, Vm and then find out the acceleration of the object under the effect of gravity.
However, what if you are only interested in the acceleration of an object in free space? The easiest way to do this is to add the equations for the velocity and for the acceleration together and then find out their sum. {or mat value. The sum is the total weight of the object over the time interval, which includes the speed of light. The length of the time interval is the time being used will depend on whether you are interested in a straight line or a curved line.
However, don’t feel that linear algebraic equations are too difficult. If you understand the basics of this subject you can easily learn more complex equations and can even do your calculations on the computer. Some of the online sources that you can access will offer a variety of formulas that you can try out, depending on your level.
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## Convergents of e
Published on Friday, 12th March 2004, 06:00 pm; Solved by 26779;
Difficulty rating: 15%
### Problem 65
The square root of 2 can be written as an infinite continued fraction.
$\sqrt{2} = 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + ...}}}}$
The infinite continued fraction can be written, $\sqrt{2} = [1; (2)]$, $(2)$ indicates that 2 repeats ad infinitum. In a similar way, $\sqrt{23} = [4; (1, 3, 1, 8)]$.
It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for $\sqrt{2}$.
$1 + \dfrac{1}{2} = \dfrac{3}{2}\\ 1 + \dfrac{1}{2 + \dfrac{1}{2}} = \dfrac{7}{5}\\ 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}} = \dfrac{17}{12}\\ 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}}} = \dfrac{41}{29}$
Hence the sequence of the first ten convergents for $\sqrt{2}$ are:
$1, \dfrac{3}{2}, \dfrac{7}{5}, \dfrac{17}{12}, \dfrac{41}{29}, \dfrac{99}{70}, \dfrac{239}{169}, \dfrac{577}{408}, \dfrac{1393}{985}, \dfrac{3363}{2378}, ...$
What is most surprising is that the important mathematical constant,
$e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, ... , 1, 2k, 1, ...]$.
The first ten terms in the sequence of convergents for e are:
$2, 3, \dfrac{8}{3}, \dfrac{11}{4}, \dfrac{19}{7}, \dfrac{87}{32}, \dfrac{106}{39}, \dfrac{193}{71}, \dfrac{1264}{465}, \dfrac{1457}{536}, ...$
The sum of digits in the numerator of the 10th convergent is $1 + 4 + 5 + 7 = 17$.
Find the sum of digits in the numerator of the 100th convergent of the continued fraction for $e$.
|
# what is the difference between tangent and slope of tangent?
I need your help, my question is what is the difference between tangent and slope of tangent ?
A clear example would be appreciated.
Thank you.
The tangent to a curve at a point is a straight line just touching the curve at that point; the slope of the tangent is the gradient of that straight line.
Here's a picture to help.
The green line is the tangent line to the point $$(1,1)$$. It is a geometric object.
The slope at $$(1,1)=2$$ (if you want to know why, it's to do with differentiating, or finding the gradient of the tangent at a point). This is a number.
So, to conclude: the tangent's a line; the slope is a number.
Qualitatively, your question is equivalent to the following:
"What's the difference between a hill and the steepness of a hill?".
To answer your second question (in the comments): $$\tan(\theta)=\frac{\textrm{change in y}}{\textrm{change in x}}=\frac{\Delta y}{\Delta x}=\rm{gradient}$$ (we can see this from the right-angled triangle, if we use some basic trigonometry).
• does that mean that when we compute the tan(theta) we get the slop of the tangent. Jun 14, 2014 at 21:49
• What is $\theta$? Jun 14, 2014 at 21:50
• theta is the angle Jun 14, 2014 at 21:53
• If $\theta$ is the angle between the tangent and the horizontal, then, yes: $\tan(\theta)=\rm{slope}$. Jun 14, 2014 at 21:53
• Picture made with $\LaTeX$ of 1950. Jun 14, 2014 at 21:55
The English word 'Tangent' also means the tangent line (at a point of a circle, say), and its slope is the numeric value $\tan(\alpha)$ where $\alpha$ is the angle of the tangent line and the right wing and (any parallel to) the $x$-axis.
|
Triangles Exercise 6.4 – class 10
1. Let ∆ABC∼ ∆PQR and their areas be, respectively, 64 cm² and 121 cm² . If EF = 15.4 cm, find BC.
Solution:
2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Solution:
Since AB || CD,
∴ ∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles)
In ∆AOB and ∆COD, ∠AOB = ∠COD (Vertically opposite angles)
∠OAB = ∠OCD (Alternate interior angles)
∠OBA = ∠ODC (Alternate interior angles)
∴ ∆AOB ∼ ∆COD (By AAA similarity criterion)
3. In the following figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, such that, [area(∆ABC)]/[area(∆DBC)] = AO/DO
Solution:
Let us draw two perpendiculars AP and DM on line BC.
We know that area of a triangle = 1/2 x height x base
In ∆APO and ∆DMO,
∠APO = ∠DMO (Each = 90°)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ∆APO ∼ ∆DMO (By AA similarity criterion)
4. If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Let us assume two similar triangles as ∆ABC ∼ ∆PQR.
5. D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Find the ratio of the area of ∆DEF and ∆ABC.
Solution:
D and E are the mid-points of ∆ABC.
6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:
Let us assume two similar triangles as ∆ABC ∼ ∆PQR.
Let AD and PS be the medians of these triangles.
∆ABC ∼ ∆PQR
∴ AB/PQ = AC/PR = BC/QR ————(1)
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ————(2)
Since AD and PS are medians,
BD = DC = BC/2 and QS = SR = QR/2
Equation (1) becomes
AB/PQ = BD/QS = BC/QR —————(3)
In ∆ABD and ∆PQS,
∠B = ∠Q [Using equation (2)]
and AB/PQ = BD/QS [using (3)]
∴ ∆ABD ∼ ∆PQS (SAS similarity criterion) Therefore, it can be said that
7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution:
Let ABCD be a square of side a.
Therefore, its diagonal = √2a
Two desired equilateral triangles are formed as ∆ABE and ∆DBF.
Side of an equilateral triangle, ∆ABE, described on one of its sides = a
Side of an equilateral triangle, ∆DBF, described on one of its diagonals√2a
We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.
8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
Solution:
We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other.
Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.
Let side of ∆ABC = x
Therefore, side of ΔBDE = x/2
Hence, the correct answer is (C).
9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio (A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81
Solution:
If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles. It is given that the sides are in the ratio 4:9.
Therefore, ratio between areas of these triangles = [4/9]² = 16/81
Hence, the correct answer is (D).
|
# How to Divide Polynomials Using the Box Method
By Bon Crowder
Dividing polynomials using the box method is actually a really great way to save yourself a lot of time. Divide polynomials using the box method with help from an experienced math professional in this free video clip.
### Transcript
How to divide polynomials using the box method? My name is Bon Crowder and I'm with MathFour.com. And we're going to divide these two polynomials using the box method. So, here I have my boxes laid out and if you'll notice, they're color coded. Because what I'm going to do, is this, the 12-X cubed goes here. And what I'm going to do is, have this piece along the side. And I'm going to treat this kind of like a regular multiplication table. So, here's the answer in a sense, and I need three-X times what, gives me 12-X cubed. Well, three-X times four-X squared, gives me 12-X cubed. Then I look at this and go, well, gee, four-X squared times negative two gives me negative eight-X squared. But this 23-X squared needs to be these two put together. So, negative eight-X squared plus what, gives me negative 23-X squared, negative 15-X squared. So, then I look and I go, gee, three-X times something gives me 15-X squared. Three-X times negative five-X gives me 15-X squared. And then, I look to the next, again it's just like a multiplication chart, down and across. Negative five-X times negative two is ten-X. But here we go again, we need the 10-X to combine with this other blue box to give me 13-X. So, that's three-X. Then I need to know, well, gee, three-X times what, gives me three-X. No, this three-X times what, gives me this three-X? They're the same. So, I need a one there, because one times three-X is three-X. And then, I multiply one times negative two and I get negative two. Is that the same there, sure is. So, which means my answer to the whole problem is this piece across the top. Four-X squared minus five-X plus one. And that's my answer. And that's how you divide polynomials using the box method. I'm Bon Crowder with MathFour.com, have fun with it.
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```Chapter 3
Functions and
Equations
1
3.2
Equations and
Problem Solving
♦
equations
♦ Use factoring, the square root property,
completing the square, and the quadratic
♦ Understand the discriminant
♦ Solve problems involving quadratic equations
2
A quadratic equation in one variable is an
equation that can be written in the form
ax2 + bx + c
where a, b, and c are constants with a ≠ 0.
3
Quadratic equations can have no real
solutions, one real solution, or two real
solutions.
The are four basic symbolic strategies in
which quadratic equations can be solved.
• Factoring
• Square root property
• Completing the square
4
Factoring
Factoring is a common technique used to
solve equations. It is based on the zeroproduct property, which states that if
ab = 0, then a = 0 or b = 0 or both. It is
important to remember that this property
works only for 0. For example, if ab = 1,
then this equation does not imply that
either a = 1 or b = 1. For example, a = 1/2
and b = 2 also satisfies ab = 1 and neither
a nor b is 1.
5
equations with factoring
Solve the quadratic equation 12t2 = t + 1. Check
Solution
2
12t t 1
12t t 1 0
2
3t 14t 1 0
3t 1 0 or 4t 1 0
1
t
or
3
1
t
4
6
equations with factoring
Solution continued
Check:
2
1
1
12 1
3
3
2
1
1
12 1
4
4
4 4
3 3
3 3
4 4
7
The Square Root Property
Let k be a nonnegative number. Then the
solutions to the equation
x2 = k
are given by
x k.
8
Example: Using the square root
property
If a metal ball is dropped 100 feet from a
water tower, its height h in feet above the
ground after t seconds is given by
h(t) = 100 – 16t2. Determine how long it takes
the ball to hit the ground.
Solution
The ball strikes the ground when the equation
100 – 16t2 = 0 is satisfied.
9
Example: Using the square root
property
Solution continued 100 16t 2 0
100 16t
2
100
t
16
2
100
10
t
16
4
The ball strikes the ground after 10/4, or 2.5,
seconds.
10
Completing the Square
Another technique that can be used to
solve a quadratic equation is completing
the square. If a quadratic equation is
written in the form x2 + kx =d, where k and
d are constants, then the equation can be
solved using
2
2
k
k
x kx x .
2
2
2
11
Example: Completing the Square
Solve 2x2 – 8x = 7.
Solution
Divide each side by 2:
7
2
x 4x
2
7
2
x 4x 4 4
2
2
15
x2
2
15
x2
2
15
x 2
2
12
Symbolic, Numerical and Graphical
Solutions
symbolically, numerically, and graphically.
The following example illustrates each
technique for the equation x(x – 2) = 3.
13
Symbolic Solution
14
Numerical Solution
15
Graphical Solution
16
The solutions to the quadratic equation
ax2 + bx + c = 0, where a ≠ 0, are given by
b b 4ac
x
.
2a
2
17
formula
Solve the equation 3x2 – 6x + 2 = 0.
Solution
Let a = 3, b = 6, and c = 2.
b b 2 4ac
x
2a
x
6
6 4 3 2
2 3
2
6 12
1
1
x
1
4 3 1
3
6
6
3
18
The Discriminant
If the quadratic equation ax2 + bx + c = 0 is
solved graphically, the parabola
y = ax2 + bx + c can intersect the x-axis zero,
one, or two times. Each x-intercept is a real
b 2 4ac 0
b 2 4ac 0
b 2 4ac 0
19
To determine the number of real solutions to
ax2 + bx + c = 0 with a ≠ 0, evaluate the
discriminant b2 – 4ac.
1. If b2 – 4ac > 0, there are two real
solutions.
2. If b2 – 4ac = 0, there is one real solution.
3. If b2 – 4ac < 0, there are no real
solutions.
20
Example: Using the discriminant
Use the discriminant to find the number of
solutions to 4x2 – 12x + 9 = 0. Then solve the
equation by using the quadratic formula.
Solution
Let a = 4, b = –12, and c = 9
b2 – 4ac = (–12)2 – 4(4)(9) = 0
Discriminant is 0, there is one solution.
21
Example: Using the discriminant
Solution continued
b b 4ac
x
2a
2
x
12 0
8
3
x
2
The only solution is 3/2.
22
Example: Using the discriminant
Solution continued
The graph suggests there is only one
intercept 3/2.
23
Problem Solving
Many types of applications involve
problems, we use the steps for “Solving
Application Problems” from Section 2.2.
24
Example: Solving a construction
problem
A box is being constructed by cutting
2-inch squares from the corners of a rectangular
piece of cardboard that is 6 inches longer than it is
wide. If the box is to have a volume of 224 cubic
inches, find the dimensions of the piece of
cardboard.
25
Example: Solving a construction
problem
Solution
Step 1: Let x be the width and x + 6 be the
length.
Step 2: Draw a picture.
26
Example: Solving a construction
problem
Since the height times the width times the length
must equal the volume, or 224 cubic inches, the
following can be written
2(x – 4)(x + 2) = 224
Step 3: Write the quadratic equation in
the form ax2 + bx + c = 0 and factor.
x 2x 8 112
2
x 2 2x 120 0
x 12x 10 0
x 12 or x 10
27
Example: Solving a construction
problem
The dimensions can not be negative, so the
width is 12 inches and the length is 6
inches more, or 18 inches.
Step 4:
After the 2-inch-square corners are cut out,
the dimensions of the bottom of the box are
12 – 4 = 8 inches by 18 – 4 = 14 inches.
The volume of the box is then 2•8•14 = 224
cubic inches, which checks.
28
```
|
LESSON
# Inverse Kinematics for a 2-Joint Robot Arm Using Geometry
#### Transcript
We saw this simple two-link robot in the previous lecture about forward kinematics. The tooltip pose of this robot is described simply by two numbers, the coordinates x and y with respect to the world coordinate frame. So, the problem here is that given x and y, we want to determine the joined angles, Q1 and Q2. The solution that we’re going to follow in this particular section is a geometric one. We’re going to start with a simple piece of construction. We’re going to overlay the red triangle on top of our robot. We know that the end point coordinate is x, y, so the vertical height of the triangle is y, the horizontal width is x. And, using Pythagoras theorem, we can write r squared equals x squared plus y squared. So far, so easy.
Now, we’re going to look at this triangle highlighted here in red and we want to determine the angle alpha. In order to do that, we need to use the cosine rule. And, if you’re a little rusty on the cosine rule, here is a bit of a refresher. We have an arbitrary triangle. We don’t have to have any right angles in it and we’re going to label the length of this edge as A and the angle opposite that edge, we’re going to label as little a. And, we do the same for this edge and this angle, and this edge and this angle. So, all together, the sides are labelled capitals A, B and C, and the angles are labelled little a, little b, and little c. So, the cosine rule is simply this relationship here. It’s a bit like Pythagoras’ theorem except for this extra term on the end with the cos a in it.
Now, let’s apply the cosine rule to the particular triangle we looked at a moment ago. It’s pretty straightforward to write down this particular relationship. We can isolate the term cos alpha which gives us the angle alpha that we’re interested in. And, it’s defined in terms of the constant link lengths, A1 and A2 and the position of the end effector, x and y. We can write this simple relationship between the angles alpha and Q2. And, we know from the shape of the cosine function that cos of Q2 must be equal to negative of cos alpha. This time, let’s just write an expression for the cosine of the joined angle Q2. Now, we’re going to draw yet another red triangle and we’re going apply some simple trigonometry here. If we know Q2, then we know this length and this length of the red triangle. We can write this relationship for the sine of the joined angle Q2.
Now, we can consider this bigger triangle whose angle is beta and this side length of the triangle is given here in blue. And, the length of the other side of the triangle is this. So, now we can write an expression for the angle beta in terms of these parameters here. Going back to the red triangle that we drew earlier, we can establish a relationship between Q1 and the angle beta. Introduce yet another angle, this one gamma and we can write a relationship between the angle gamma and the tooltip coordinates x and y. Now, we can write a simple relationship between the angles that we’ve constructed, gamma and beta and the joined angle we’re interested in which is Q1. And, the total relationship looks something like this.
Quite a complex relationship, it gives us the angle of joined one, that’s Q1 in terms of the end effector coordinates y and x, and a bunch of constants, a1 and a2, and it’s also a function of the second joint angle, Q2. So, let’s summarize what it is that we have derived here. We have an expression for the cosine of Q2 and we have an expression for Q1. Now, the cosine function is symmetrical about 0. So, if we know the value of the cosine of Q2, then there are two possible solutions, a positive angle and a negative angle. We’re going to explicitly choose the positive angle, which means that I can write this expression here. And now, we have what we call the inverse kinematic solution for this two-link robot. We have an expression for the two joined angles, Q1 and Q2 in terms of the end effector pose x and y, and a bunch of constants.
You notice that the two equations are not independent. The equation for Q1, in fact, depends on the solution for Q2. In this case, Q2 is negative and we’re going to write the solution for Q2 with a negative sign in front of the inverse cosine. Now, we need to solve for Q1, so we’re going to introduce this particular red triangle, the angle beta that we solved previously, and the angle gamma which is defined in terms of y and x. Now, we write a slightly different relationship between Q1, gamma and beta, different to what we had before. There’s a change of sign involved. Then, we can substitute all that previous equation and come up with this expression for Q1. Again, there is a change of sign here. Previously, this was a negative sign.
And, here in summary form is the solution for the inverse kinematics of our two-link robot when it is in this particular configuration, where Q2 is negative. Let’s compare the two solutions, the case where Q2 is positive and the case where Q2 is negative.
#### Code
There is no code in this lesson.
We revisit the simple 2-link planar robot and determine the inverse kinematic function using simple geometry and trigonometry.
### Professor Peter Corke
Professor of Robotic Vision at QUT and Director of the Australian Centre for Robotic Vision (ACRV). Peter is also a Fellow of the IEEE, a senior Fellow of the Higher Education Academy, and on the editorial board of several robotics research journals.
### Skill level
This content assumes an understanding of high school-level mathematics, e.g. trigonometry, algebra, calculus, physics (optics) and some knowledge/experience of programming (any language).
Average
## Discussion
1. cesar says:
Maybe adding a video on the analitical solution for 3d robot would be helpful.
1. Peter Corke says:
Thanks for the suggestion. You can find that solution in a lot of standard textbooks such as Siciliano et al.
2. Prasanna says:
1. What if I give x and y in other quadrants?
( I did it in MATLAB, theta 2 is becoming complex for negative x value)
2. Is there any solution (inverse kinematics) for 3 link planar robot?
1. Peter Corke says:
Q1. Other quadrants are fine. I cannot see how you can get a complex results because the equation for q1 uses x^2 which will always be positive. q1 is computed using atan2 which cannot give a complex result.
Q2. There are multiple solutions, not sure how many. You can frame the inverse kinematics problem as solving for just the end-effector position (x,y) or for end-effector pose in the plane (x,y,theta). I suspect in the first case there will be an infinite number of solutions.
3. zahid says:
I have confusion.
How
q2 = Pi-alpha
become
cos(q2)=-cos(alpha)
1. CG-Omri says:
By definition q2 = Pi -alpha, and we know that: cos(Pi-x)=-cos(x).
Applying that rule gives cos(q2)=cos(Pi-alpha)=-cos(alpha).
1. TheOracle says:
Where is the Pi from though?
2. ghofran966 says:
take cos for both sides of equation then use the difference identity for right hand side of the equation:
cos(a−b)=cos(a)cos(b)+sin(a)sin(b)
4. Rob says:
Hi,
I think there is an error in the q1 solution in the second part (elbow up)
I found q1 = atan(y/x)-atan(a2*sin(q2)/(a1+a2*cos(q2))) (same expression as in the first part “elbow down”)
It is geometricaly logical because your β is a negative angle at this point of the calculation (because q2 is negative)
Please tell me if I’m wrong
Thanks
5. TheOracle says:
Those diagrams look beautiful. What software did you draw them in??
6. bharath says:
Hi Prof Corke,
In Q2, the end effector pose includes the angle i.e. (x,y, theta) as well. So I think position can be placed in the place of the pose. Because if we consider it as a pose then for a given pose only unique joint values are possible.
right me if i am wrong
thanks
|
# Diagonal of a Rectangle – Properties, Formula, Examples, FAQs
Home » Math Vocabulary » Diagonal of a Rectangle – Properties, Formula, Examples, FAQs
## What Are Diagonals of a Rectangle?
Diagonals of a rectangle are line segments connecting opposite vertices of a rectangle. A rectangle is a quadrilateral in which opposite sides are equal and all angles measure $90^\circ$.
A diagonal is a line segment connecting two non-adjacent vertices of a polygon. The word diagonal comes from the ancient Greek word “diagonios,” (Latin diagonalis) which means “from angle to angle.”
Both Euclid and Strabo used it to describe a line that connects two vertices of a cuboid or a rhombus; later, it became known in Latin as diagonus (slanting line).
### Diagonals of a Rectangle Definition
A line segment that connects any two opposite vertices of a rectangle is called diagonal of a rectangle.
Consider ▭ABCD. A rectangle with diagonals AC and BD as shown below.
Here, $AC = BD$.
## Properties of Diagonals of a Rectangle
1. The diagonals of a rectangle are congruent.
$\left[AC = BD\right]$
1. Each diagonal divides a rectangle into two congruent right-angled triangles.
1. The diagonals of a rectangle are not perpendicular to each other.
1. If two diagonals of a rectangle bisect each other at $90^\circ$, it is called a square.
1. When two diagonals intersect, they form one obtuse angle and one acute angle. The opposite central angles are equal.
1. Adjacent angles formed by the diagonals are supplementary.
1. Diagonals of a rectangle form the hypotenuse of the right triangle.
## Diagonal of a Rectangle Formula
The diagonal of a rectangle formula helps in finding the length of the diagonal when the length and width of the rectangle are known.
A diagonal divides a rectangle into two right triangles, each of which has a hypotenuse and sides that are equal to the sides of the rectangle. The diagonal is that hypotenuse.
So, how to find the diagonal of a rectangle? What is the diagonal measurement of a rectangle?
The diagonal length of a rectangle is calculated using the formula, $d = \sqrt{(w^2 + l^2)}$
where, $d = diagonal$
$l = length$
$w = width$
This formula can be used to find the measurement of the length of a diagonal of a rectangle.
## Diagonal of a Rectangle Formula Derivation
The diagonal of a rectangle formula is derived using the Pythagoras’ Theorem. Split the rectangle into two right triangles formed by a diagonal.
Applying Pythagoras’ Theorem in the right triangle PSR, we have
$d^2 = b^2 + l^2$
Here, $d =$ diagonal, $l =$ length, $b =$ breadth
Taking square root on both sides,
$d = \sqrt{(b^2 + l^2)}$
• $l =$ length of the rectangle
• $b =$ breadth of the rectangle
Simply substitute the values of the length and breadth in the formula to get the answer.
Example:
In the above rectangle,
Length of diagonal $= d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ units
## Angles Made by Diagonals of a Rectangle
As discussed above, if the diagonals of the polygon bisect each other, it is called a square.
Diagonals of a rectangle are equal in length, bisect each other. They do not meet at a right angle in the center. The diagonals of a rectangle do not bisect the interior angles of the rectangle.
Example: Take a look at the image given below.
In the image given above, AC and BD are the diagonals.
Angles formed by diagonal AC and BD are $\angle AED,\; \angle AEB,\; \angle BEC$, and $\angle DEC$.
These angles do not meet at right angles, but the adjacent angles are supplementary, which means the adjacent angle-pairs add up to $180^\circ$.
• $\angle AED + \angle AEB = 180^\circ$
• $\angle AEB + \angle BEC = 180^\circ$
• $\angle BEC + \angle DEC = 180^\circ$
• $\angle AED + \angle DEC = 180^\circ$
## Facts about Diagonals of a Rectangle
• The two diagonals of a rectangle divide the rectangle into four triangles.
• Each diagonal of a rectangle divides it into two congruent right triangles.
• The intersection of the diagonals is the circumcenter. That is, you can draw a circle with that at the center to pass through the four corners.
Thus, diagonal $= 2r$ …where r is the radius of the circumcircle
• We can calculate the length of the diagonal of the rectangle using the formula:
$d = \sqrt{(b^2 + l^2)}$
• The rectangle is called a square if its diagonals bisect each other at right angles.
• As the diagonals are generally slanted, a slanted symbol “ / ” is used to denote diagonals.
## Conclusion
In this article, we have learned about diagonals of the rectangle, its properties, formulas, and facts. Now let’s solve some examples based on the formulas learned above.
## Solved Examples for Diagonals of a Rectangle
1. Identify the length, width and diagonal in the given rectangle.
Solution:
Length $\rightarrow$ AD and BC
Width $\rightarrow$ AB and CD
Diagonals $\rightarrow$ AC and BD
2. Find the length of each diagonal of a rectangle of length 15 units and width 8 units.
Solution:
Length of the rectangle, $l = 15$ units.
Breadth of the rectangle, $b = 8$ units.
Length of the diagonal $= d = \sqrt{(b^2 + l^2)}$
$d = \sqrt{(8^2 + 15^2)}$
$d = \sqrt{(64 + 225)}$
$d = \sqrt{289}$
$d = 17$ units.
The length of the diagonal is 17 units.
3. A rectangular park is 30 ft long and 16 ft wide. Determine the diagonal of the rectangular park.
Solution:
Length of the rectangular park $= 30$ ft
Breadth of the rectangular park $= 15$ ft
Length of the diagonal $d = \sqrt{(b^2 + l^2)}$
$= sqrt{(16^2 + 30^2)}$
$= \sqrt{(256+ 900)}$
$= \sqrt{1156}$
$= 34$ ft
The diagonal of the rectangular park is 34 ft.
4. Find the length of the diagonal of the given rectangle.
Solution:
Length $= 12$ in
Breadth $= 5$ in
Diagonal $= ?$
Using the Pythagorean theorem, we get:
$l^2 + b^2 = d^2$
$12^2 + 5^2 = d^2$
$d^2 = 144 + 25$
$d^2 = 169$
Taking the square root
$d = 13$ in
5. Find the length of the rectangle if the breadth of the rectangle is 3 inches, and the diagonal of the rectangle is 5 inches.
Solution:
Breadth of the rectangle $= 3$ in
Diagonal of the rectangle (hypotenuse) $= 5$ in
Here we will apply Pythagoras’ Theorem.
$(Hypotenuse)^2 = (Length)^2 + (Breadth)^2$
$5^2 = l^2 + 3^2$
$l^2 = 5^2 \;-\; 3^2$
$l^2 = 25 \;-\; 9$
$l = \sqrt{16}$
$l = 4$ in
Length of the rectangle is 4 inches.
## Practice Problems on Diagonals of a Rectangle
1
### A line segment that connects any two opposite vertices of a rectangle is said to be its _________.
diagonal
length
None of the above
CorrectIncorrect
A line segment that connects any two opposite vertices of a rectangle is said to be its diagonal.
2
### The diagonals of a rectangle are ______.
perpendicular
equal
not equal
None of the above.
CorrectIncorrect
The diagonals of a rectangle are equal in length.
3
### Which of the following is NOT TRUE ?
Each diagonal divides a rectangle into 2 congruent right triangles.
Diagonals of a rectangle bisect each other.
The length of the diagonals of a rectangle is equal.
The diagonals of a rectangle are perpendicular to each other.
CorrectIncorrect
Correct answer is: The diagonals of a rectangle are perpendicular to each other.
The diagonals of a rectangle are not perpendicular to each other. When two diagonals bisect each other at $90^\circ$, it is called a square.
4
### The length of diagonals of the rectangle is calculated using formula _________.
$d = \sqrt{2(b^2 + l^2)}$
$d = \sqrt{(b^2\;-\;l^2)}$
$d = \sqrt{(b^2+l^2)}$
$d = \sqrt{(b^2 +2\;l^2)}$
CorrectIncorrect
Correct answer is: $d = \sqrt{(b^2+l^2)}$
The length of diagonals of the rectangle is calculated using formula: $d = \sqrt{(b^2 + l^2)}$
where, $d =$ diagonal of rectangle
$l =$ length of the rectangle
$b =$ breadth of the rectangle
5
### Adjacent angles formed by the diagonals of a rectangle are ________.
supplementary
congruent
complementary
straight
CorrectIncorrect
Adjacent angles formed by the diagonals are supplementary.
A rectangle is a quadrilateral where all the angles are right angles. A rectangle is also a parallelogram where the opposite sides are equal.
Consider this rectangle ABCD.
In the above rectangle, consider triangles ABC and DCB.
$\angle ABC = \angle DCB = 90$ [Angles of rectangle]
$BC = BC$ (common side)
$AB = DC$ (Opposite sides of a parallelogram are equal)
Hence, $ABC\congDCB$
$AC = DB$ [CPCTC]
Hence, the diagonals of a rectangle are equal.
The diagonals of a rectangle always bisect each other. However, they do not meet at 90 degrees except if the rectangle is a square.
Area of a rectangle is given by the product of length and breadth.Area of rectangle $= l\timesb$
Let us understand why the area of the rectangle is $length \times breadth$.
Consider ▭ ABCD. We can see that the diagonal AC divides the rectangle into two congruent triangles, $\Delta ABC$ and $\Delta ADC$.
Area of Rectangle $ABCD =$ Area of Triangle $ABC +$ Area of Triangle ADC
$= 2\times$ Area of Triangle ABC
$= 2 \times (\frac{1}{2} \times Base \times Height)$
$= AB \times BC$
$= Length \times Breadth$
The area of the rectangle is the product of length and breadth.
A rectangle has two diagonals. Each diagonal divides a rectangle into two right triangles. The two diagonals of a rectangle are always congruent.
No! The two diagonals of a rectangle do not intersect at right angles.
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## Boats and Streams Formula
All the students who are searching the Boats and Streams Formula they have finally reached at right place because in this article mostly all
the types of questions are covered by the team members of the portal for all those participants who are ready to become master in the Boats and Streams chapter. We all know that quantitative aptitude section possess Boats and Streams questions definitely. Applicants who don’t know the basics of the problem solving they face problem in this topic. Here we are providing you Boats and stream formula and problem solving shortcut tricks under one roof. These tricks and solved questions will be useful in your preparation of competitive government exams.
In those exams where Boats and streams topic covers it is said that at least 2-3 questions will also be there in the Aptitude section. These types of questions are generally asked in most of the aptitude tests for companies like Infosys, TCS, HCL, all government job exams etc. Advantage to ask this question on boats and streams is to search the capacity of the mind to the appliers who are willing to appoint in a special organization. There are only two fundamental concepts following in such questions and students can solve several questions with these concepts. Hey guys learn all the given Shortcut Tips & Tricks to Solving Problem from the below portion of the article. All the best!!
### Boats and Streams Formula Shortcut Tips to Solving Problem
1) Given a boat travels downstream with speed d km/hr and it travels with speed u km/hr upstream. Find the speed of stream and speed of boat in still water?
Solution: Let speed of boat in still water is b km/hr and speed of stream is s km/hr.
Then b + s = d and b – s = u
Solving the 2 equations we get,
b = (d + u)/2
s = (d – u)/2
2) A man can row a boat, certain distance downstream in td hours and returns the same distance upstream in tu hours. If the speed of stream is s km/h, then the speed of boat in still water is?
Solution: We know distance = speed * time
Let the speed of boat be b km/hr
Case downstream:
d = (b + s) * td
Case upstream:
d = (b - s) * tu
=> (b + s) / (b - s) = tu / td
b = [(tu + td) / (tu - td)] * s
3) A man can row in still water at b km/h. In a stream flowing at s km/h, if it takes him t hours to row to a place and come back, then the distance between two places, d is given by
Downstream: Let the time taken to go downstream be td
d = (b + s) * td
Upstream: Let the time taken to go upstream be tu
d = (b - s) * tu
td + tu = t
[d / (b + s)] + [d / (b - s)] = t
So, d = t * [(b2 - s2) / 2b]
OR
Short trick: d = [t * (Speed to go downstream) * (Speed to go upstream)]/[2 * Speed of boat or man in still water]
4) A man can row in still water at b km/h. In a stream flowing at s km/h, if it takes t hours more in upstream than to go downstream for the same distance, then the distance d is given by
Solution:
Time taken to go upstream = t + Time taken to go downstream
(d / (b - s)) = t + (d / (b + s))
=> d [ 2s / (b2 - s2 ] = t
So, d = t * [(b2 - s2) / 2s]
OR
Short trick: d = [t * (Speed to go downstream) * (Speed to go upstream)] / [2 * Speed of still water]
If speed of boat or swimmer is x km/h and the speed of stream is y km/h then,
Speed of boat downstream = (x + y) km/h
Important Points of Boats and Streams Formula: When speed of boat is given then it means speed in the still water, unless it is stated otherwise.
Some Basic Formulas-
Speed of boat in still water is
= ½ (Downstream Speed + Upstream Speed)
Speed of stream is
= ½ (Downstream Speed – Upstream Speed)
Boats and streams formulas: Short tricks with formula for boats and streams is given as you expected,
Downstream / Upstream:
Direction along the stream is called Downstream.
Direction against the stream is called upstream.
If the speed in the still water is x km / hr and the speed of the stream is y km / hr then,
Speed downstream = (x + y) km / hr.
Speed upstream = (x – y) km / hr.
If the speed downstream is u km / hr and the speed upstream is v km / hr then,
Speed in still water = (1 / 2) (a + b) km / hr.
Rate of stream = (1 / 2) (a – b) km / hr.
Some Important Solved Questions
Types of Questions asked in Previous Exam by SSC:
Type 1: When the distance covered by boat in downstream is same as the distance covered by boat upstream. The speed of boat in still water is x and speed of stream is y then ratio of time taken in going upstream and downstream is,
Short Trick:
Time taken in upstreamTime taken in Downstream = (x+y)/(x-y)
1) A man can row a boat @ 9 km ph in still water. He takes double the time to move upstream than to move the downstream – the same distance. Find the speed of the stream?
According to Question and formula given above:
Let the downward time = 1 hour and so the upward time = 2 hours.
1/9+s = 2/9-s (Since distance is the same)
18 + 2 s = 9 – s (By cross multiplication)
18 – 9 = s + 2 s
9 = 3 s
Hence s or Speed of stream = 9/3 = 3 km ph Answer.
OR
Simply
b + s = 2(b –s)
b + s = 2b – 2s
s + 2s = 2b –b
OR
b = 3s or 9 = 3s (b = 9 is given) = 3 km ph Answer
2) A boat runs at 20 km ph along the stream and 10 km ph against the stream. Find the ratio of speed of the boat in still water to that of the speed of the stream?
ATQ (According to Question) and formula given above:
Speed of Boat = ½ (20 + 10) = 15 km ph.
Speed of Stream = ½ (20 – 10) = 5 km ph.
Ratio: 15:5 = 3:1 Answer.
3) Find the speed of the stream when a boat takes 5 hours to travel 60 kms downstream at a rate of 10 kms per hour in still water?
According to Question and formula given above:
Speed b + s = 60/5 = 12 km ph
Speed b = 10 km ph
So speed is = 12-10 = 2 km ph Answer.
4) If a man rows 6 km downstream in 3 hours and 2 km upstream in 2 hours then how long will he take to cover 9 kms in stationary (still) water?
According to Question and formula given above:
Speed of Boat in still waters = ½ (6/3 + 2/2) = ½ (2 + 1) = 1.5 km ph
Time taken for 9 kms = 9/1.5 = 6 hours Answer
5) A boat covers a certain distance in one hour downstream with the speed of 10 km ph in still water and the speed of current is 4 km ph. Then find out the distance travelled?
According to Question and formula given above:
Distance = Speed x Time = 1 x (10+4) = 14 kms.
Reminder: Dear candidates to get the complete details of other such formulas stay tuned with ejobhub.
Take a Look on Below Table
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# 4.13: Convert Customary Units of Measurement in Real-World Situations
Difficulty Level: At Grade Created by: CK-12
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Practice Applications of Customary Unit Conversions
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Have you ever tried to measure accurately using different liquid units of measurement? Take a look at this dilemma.
Evan is making a recipe for fruit punch that uses 3 cups of pineapple juice. If he makes 5 batches of the recipe, how many quarts of pineapple juice will he need?
Pay attention to this Concept and you will learn how to apply customary units of measurement in real-world situations.
### Guidance
When we measure in the United States, we often use the customary system of measurement. The customary system is made up of units such as inches, feet, cups, gallons and pounds.
Now let's look at conversions within the customary system of measurement.
Customary Units of Measurement
Take a few minutes to copy all of these units of measurement down in your notebook.
Now let’s look at how we convert among customary units of measurement.
While you may be able to complete some of the mathematics in your head, it may make more sense to use a proportion. Because there is a relationship between different units of measure, you can use proportions to help you convert between customary units of measurement.
Then you can apply these conversions to real-world problems.
Take a look at this situation.
The distance from John’s house to Mike’s house on a map is 4.5 inches. The scale of the map is 1.5 inches = 2 miles. What is the actual distance from John’s house to Mike’s house in feet?
First, find the actual distance in miles. Then convert miles to feet.
Write a proportion to find the actual distance between the two houses.
1.5 inches2 miles=4.5 inchesx miles\begin{align*}\frac{1.5 \ inches}{2 \ miles} = \frac{4.5 \ inches}{x \ miles}\end{align*}
Now cross-multiply and solve for x\begin{align*}x\end{align*}.
(1.5)x1.5xx=4.5(2)=9=6
So the two houses are 6 miles apart. Now convert miles to feet.
1 mile5280 feet=6 milesx feet\begin{align*}\frac{1 \ mile}{5280 \ feet} = \frac{6 \ miles}{x \ feet}\end{align*}
Now cross multiply to solve for x\begin{align*}x\end{align*}.
(1)xx=6(5280)=31,680
The two houses are 31.680 feet apart.
Here is another one.
Jeff ran his weekly long run of 13 miles in 2 hours. If his rate is 9.23 per mile, how long would it take Jeff to run 5280 feet?
To figure this out, you can use a proportion, but it might make more sense to think in terms of customary units of measurement. Here Jeff runs 9.23 per mile.
How many feet are in one mile?
Yes, there are 5280 feet in one mile.
Therefore, Jeff runs 1 mile in 9.23.
Convert each customary unit of measurement.
#### Example A
Karin has a recipe that calls for 3 gallons of cider. How many quarts will she need?
Solution: 12\begin{align*}12\end{align*} quarts
#### Example B
Josea threw the ball 12 feet. How many inches did he throw the ball?
Solution: 144\begin{align*}144\end{align*} inches
#### Example C
Carl drank 3 pints of lemonade. How many ounces did he drink?
Solution: 48\begin{align*}48\end{align*} ounces
Now let's go back to the dilemma from the beginning of the Concept.
First find the total number of cups he needs.
If there are 3 cups in one batch, and he is making 5 batches, then he will need 3×5=15 cups\begin{align*}3 \times 5 = 15 \ cups\end{align*}.
Set up a proportion.
The conversion factor is the number of cups in a quart: 4 cups1 quart\begin{align*}\frac{4 \ cups}{1 \ quart}\end{align*}.
Now write the second ratio, making sure it follows the form of the first ratio.
4 cups1 quart=15 cupsx quarts\begin{align*}\frac{4 \ cups}{1 \ quart} = \frac{15 \ cups}{x \ quarts}\end{align*}
Next cross-multiply and solve for x\begin{align*}x\end{align*}.
(4)x4xx=1(15)=15=154=334
He will need 334\begin{align*}3 \frac{3}{4}\end{align*} quarts of pineapple juice.
### Vocabulary
Customary System
the system of measurement that includes inches, feet, miles, pounds, tons, cups, quarts, gallons, etc.
### Guided Practice
Here is one for you to try on your own.
A scale model of a building has a height of 3.5 feet. The scale of the model is 112 inch=10 feet\begin{align*}1 \frac{1}{2} \ inch = 10 \ feet\end{align*}. What is the actual height of the building?
Solution
The scale is in inches, but the scale model height is given in feet. First convert the scale height to inches. Then find the height of the building.
1 foot12 inches=3.5 feetx inches
Now cross multiply to solve for x\begin{align*}x\end{align*}.
(1)xx=3.5(12)=42
So the height of the scale model is 42 inches. Now find the height of the actual building.
1.5 inch10 feet=42 inchesx feet
Now cross multiply and solve for x\begin{align*}x\end{align*}.
(1.5)x1.5xx=42(10)=420=280
The actual building is 280 feet tall.
### Practice
Directions: Solve each problem.
1. Justin ran 3 miles. How many feet did he run?
2. If the flour weighed four pounds, how many ounces did it weigh?
3. How many pounds is equal to 4 tons?
4. Mary needs 3 cups of juice for a recipe. How many ounces does she need?
5. Jess bought 3 quarts of pineapple juice. How many pints did she purchase?
6. If Karen bought 16 quarts of ice cream, how many gallons did she buy?
7. The length of the garden is four yards. How many feet is that?
8. If the width of the garden is 4 yards, how many inches is that?
9. Will eight cups of water fit in a two quart saucepan?
10. A recipe calls for 2 pints of milk. If Jorge cuts the recipe in half, how many cups of milk will he need?
11. Audrey is making brownies for a bake sale. The recipe calls for 8 ounces of flour for every 24 brownies. If she makes 96 brownies, how many pounds of flour will she need?
12. Two buildings are 5 inches apart on a map. The scale of the map is 14 inch=1 mile\begin{align*}\frac{1}{4} \ inch = 1 \ mile\end{align*}. What is the actual distance between the two buildings?
13. The length of a classroom on a floor plan is 2.5 inches. The scale of the map is 12 inch=5 feet\begin{align*}\frac{1}{2} \ inch = 5 \ feet\end{align*}. What is the actual length of the classroom in inches?
14. A scale model of a mountain is 2.75 feet tall. The scale of the model is 14 inch=50 feet\begin{align*}\frac{1}{4} \ inch = 50 \ feet\end{align*}. What is the actual height of the mountain in feet?
15. A scale drawing of a town includes a park that measures 0.5 inch by 1.5 inches. If the scale of the map is 12 inch=1 mile\begin{align*}\frac{1}{2} \ inch = 1 \ mile\end{align*}, what is the area of the park in square feet?
### Vocabulary Language: English
Customary System
Customary System
The customary system is the measurement system commonly used in the United States, including: feet, inches, pounds, cups, gallons, etc.
Measurement
Measurement
A measurement is the weight, height, length or size of something.
Proportion
Proportion
A proportion is an equation that shows two equivalent ratios.
Ratio
Ratio
A ratio is a comparison of two quantities that can be written in fraction form, with a colon or with the word “to”.
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Home / Calculus I / Derivatives / Higher Order Derivatives
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### Section 3.12 : Higher Order Derivatives
4. Determine the fourth derivative of $$\displaystyle f\left( w \right) = 7\sin \left( \frac{w}{3} \right) + \cos \left( {1 - 2w} \right)$$
Show All Steps Hide All Steps
Start Solution
Not much to this problem other than to take four derivatives so each step will show each successive derivative until we get to the fourth. The first derivative is then,
$f'\left( w \right) = \frac{7}{3}\cos \left( \frac{w}{3}\right) + 2\sin \left( {1 - 2w} \right)$ Show Step 2
The second derivative is,
$f''\left( w \right) = - \frac{7}{9}\sin \left( \frac{w}{3} \right) - 4\cos \left( {1 - 2w} \right)$ Show Step 3
The third derivative is,
$f'''\left( w \right) = - \frac{7}{{27}}\cos \left(\frac{w}{3}\right) - 8\sin \left( {1 - 2w} \right)$ Show Step 4
The fourth, and final derivative for this problem, is,
$\require{bbox} \bbox[2pt,border:1px solid black]{{{f^{\left( 4 \right)}}\left( w \right) = \frac{7}{{81}}\sin \left( \frac{w}{3}\right) + 16\cos \left( {1 - 2w} \right)}}$
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S k i l l
i n
A R I T H M E T I C
POWERS OF 10
Lesson 1 Section 3
Here is our first example of a problem that not only should not require a calculator, but that we should not even have to write down. It is an example of skill in arithmetic.
10. How do we multiply a whole number by a powerof 10? 28 × 10 Add on as many 0's as appear in the power.
Example 1. 28 × 10 = 280 Add on one 0. 28 × 100 = 2800 Add on two 0's. 28 × 1000 = 28,000 Add on three 0's.
To understand this, look at 28 in the figure, and let us push each digit one place left, so that we have 280. 280 is then ten times 28, because each digit's place value is now ten times more. (In 28, the '2' tells how many tens, but in 280 it tells how many hundreds.)
Similarly, 2800 = 100 × 28.
Knowing that is an example of skill in arithmetic, which means to be able to do a problem with the most understanding, and therefore the least effort. Calculating in writing vertically would give the right answer, eventually (we hope). But it is not skillful. The student should let go of such written calculations immediately.
To be honest, we can do almost every problem with a calculator. Yet the student should wonder, What problems should not require a calculator? What problems should an educated person not even have to write down!
Example 2. 608 is 608 ones. How much are 608 tens? 608 hundreds? 608 thousands?
Answers. "608 tens" is 608 × 10. That is the meaning of multiplication (Lesson 8). 608 tens therefore are 6,080. Simply add on a 0.
608 hundreds = 608 × 100 = 60,800. Add on two 0's.
608 thousands = 608 × 1000 = 608,000. Add on three 0's.
Example 3.
a) 50 ones are equal to how many tens?
Answer. 50 ones are simply 50, or 5 tens.
b) 50 tens are how many hundreds?
Answer. 50 tens = 500. (Add on a 0.) 50 tens are 5 hundreds.
c) 50 hundreds are how many thousands?
Answer. 50 hundreds = 5000. (Add on two 0's.) 50 hundreds are 5 thousands.
11. When a whole number ends in 0's, how do we divide it by a power of 10? 42,000 ÷ 10 Take off as many 0's as appear in the power.
Examples. 42,000 ÷ 10 = 4,200 Take off one 0. 42,000 ÷ 100 = 420 Take off two 0's. 42,000 ÷ 1000 = 42 Take off three 0's.
Problem. Reduce this fraction to its lowest terms:
300 500
Answer. To reduce a fraction, we must divide both the numerator and denominator by the same number. Now, if a number ends in 0, then we know that it is divisible by 10; while if a number ends in two 0's, we know that it is divisible by 100.
Both 300 and 500 then are divisible by 100:
300 500 = 3 5
Please "turn" the page and do some Problems.
or
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# Is a CoinFlip actually 50 50?
## Introduction
Introduction:
The CoinFlip is a simple and popular game of chance that involves flipping a coin and predicting the outcome as either heads or tails. It is often assumed that the probability of getting either heads or tails is 50-50, or 50%, each time the coin is flipped. However, some people have questioned whether this assumption is actually true and whether other factors may influence the outcome of a CoinFlip. In this article, we will explore whether a CoinFlip is actually 50-50 and what factors may affect the probability of getting heads or tails.
## The Mathematics Behind Coin Flips
Coin flips are a common way to make decisions, from choosing who goes first in a game to deciding the outcome of a sports match. The idea behind a coin flip is simple: toss a coin in the air and let it land on the ground. If it lands heads up, one outcome is chosen, and if it lands tails up, the other outcome is chosen. But is a coin flip actually a 50/50 proposition? The answer is not as straightforward as you might think.
The mathematics behind coin flips is based on probability theory. Probability is the branch of mathematics that deals with the likelihood of events occurring. In the case of a coin flip, there are two possible outcomes: heads or tails. Each outcome has an equal chance of occurring, so the probability of getting heads is 1/2, and the probability of getting tails is also 1/2.
However, just because the probability of each outcome is 1/2 does not mean that a coin flip will always result in a 50/50 split. In fact, there are several factors that can influence the outcome of a coin flip.
One of the most significant factors is the way the coin is tossed. If the coin is tossed with a lot of force, it is more likely to land on the same side it started on. This is because the force of the toss can cause the coin to spin in the air, which can affect the way it lands. If the coin is tossed gently, it is more likely to land on the opposite side.
Another factor that can influence the outcome of a coin flip is the surface on which the coin lands. If the surface is uneven or has a lot of friction, it can affect the way the coin bounces and rolls, which can affect the outcome of the flip. For example, if the coin lands on a carpeted surface, it is more likely to land on the same side it started on because the carpet can absorb some of the energy of the bounce.
The temperature and humidity can also affect the outcome of a coin flip. If the coin is too cold or too hot, it can affect the way it bounces and rolls. Similarly, if the air is too humid, it can make the coin stick to the surface, which can affect the outcome of the flip.
Finally, the shape and weight of the coin can also influence the outcome of a coin flip. If the coin is not perfectly balanced, it can be more likely to land on one side than the other. Similarly, if the coin is heavier on one side than the other, it can be more likely to land on that side.
So, is a coin flip actually a 50/50 proposition? The answer is that it depends on a variety of factors. While the probability of each outcome is 1/2, there are many other factors that can influence the outcome of a coin flip. However, in most cases, a coin flip is still a relatively fair way to make a decision, as long as the toss is done fairly and the coin is not biased in any way.
In conclusion, the mathematics behind coin flips is based on probability theory, which states that each outcome has an equal chance of occurring. However, there are many other factors that can influence the outcome of a coin flip, including the way the coin is tossed, the surface on which it lands, the temperature and humidity, and the shape and weight of the coin. While a coin flip is not always
## Factors That Affect Coin Flip Outcomes
Coin flipping is a simple and popular method of making decisions, settling disputes, and even determining the outcome of sporting events. It is often assumed that a coin flip is a fair and unbiased way of making a decision, with a 50-50 chance of landing on either side. However, there are several factors that can affect the outcome of a coin flip, making it less than a perfect 50-50 chance.
One of the most significant factors that can affect the outcome of a coin flip is the weight and shape of the coin. A coin that is slightly heavier on one side or has a different shape can cause it to land on one side more often than the other. This is because the heavier or differently shaped side will have a greater gravitational pull, causing it to land face down more often.
Another factor that can affect the outcome of a coin flip is the force with which the coin is flipped. If the coin is flipped too lightly, it may not rotate enough times in the air, causing it to land on the same side it started on. On the other hand, if the coin is flipped too hard, it may rotate too many times, making it more likely to land on the opposite side.
The surface on which the coin is flipped can also affect the outcome. A surface that is too soft or uneven can cause the coin to bounce or roll, making it more difficult to predict which side it will land on. Similarly, a surface that is too hard or smooth can cause the coin to slide or spin, again making it more difficult to predict the outcome.
The temperature and humidity of the environment can also affect the outcome of a coin flip. A coin that is too cold or too hot may behave differently in the air, causing it to land on one side more often than the other. Similarly, a humid environment can cause the coin to stick to the surface, making it more likely to land on the same side it started on.
Finally, the person flipping the coin can also affect the outcome. If the person has a particular way of flipping the coin, such as always using the same hand or flipping it with a particular motion, this can cause the coin to land on one side more often than the other. Similarly, if the person is intentionally trying to influence the outcome, such as by using a weighted or double-sided coin, this can also affect the outcome.
In conclusion, while a coin flip may seem like a fair and unbiased way of making a decision, there are several factors that can affect the outcome. The weight and shape of the coin, the force with which it is flipped, the surface on which it is flipped, the temperature and humidity of the environment, and the person flipping the coin can all play a role in determining which side the coin will land on. While these factors may not always have a significant impact on the outcome, they should be taken into consideration when using a coin flip to make an important decision or settle a dispute.
## The Psychology of Coin Flips: Why We Believe They’re 50/50
Coin flips are a common way to make decisions, from choosing who goes first in a game to deciding the outcome of a sports match. The idea behind a coin flip is simple: toss a coin in the air and let it land on either heads or tails. The outcome is supposed to be random, with a 50/50 chance of landing on either side. However, is a coin flip actually 50/50? The answer may surprise you.
The psychology of coin flips is a fascinating topic. It turns out that our perception of randomness is not always accurate. When we flip a coin, we expect it to be equally likely to land on heads or tails. This expectation is based on the assumption that the coin is perfectly balanced and that the force of the flip is consistent. However, in reality, there are many factors that can influence the outcome of a coin flip.
One of the most significant factors is the way the coin is flipped. If the coin is flipped with too much force, it may be more likely to land on one side than the other. Similarly, if the coin is flipped with too little force, it may not rotate enough in the air, leading to a biased outcome. The surface on which the coin lands can also affect the outcome. If the surface is uneven or has a slight incline, the coin may be more likely to land on one side than the other.
Another factor that can influence the outcome of a coin flip is the way the coin is held before it is flipped. If the person flipping the coin has a particular way of holding it, this can affect the way the coin rotates in the air. For example, if the person holds the coin with their thumb on one side and their fingers on the other, this may create a slight imbalance that affects the outcome.
Despite these factors, most people still believe that a coin flip is 50/50. This belief is based on the assumption that the coin is perfectly balanced and that the force of the flip is consistent. However, research has shown that this assumption is not always accurate. In one study, researchers found that a coin was more likely to land on the side that was facing up before it was flipped. This suggests that the way the coin is held before it is flipped can have a significant impact on the outcome.
So, is a coin flip actually 50/50? The answer is that it depends on many factors. While a perfectly balanced coin that is flipped with consistent force on a flat surface may be close to 50/50, there are many other factors that can influence the outcome. The way the coin is held before it is flipped, the force of the flip, and the surface on which the coin lands can all affect the outcome.
Despite these factors, coin flips are still a popular way to make decisions. This is because they are perceived as being fair and impartial. However, it is important to remember that the outcome of a coin flip is not always as random as we think. If you need to make an important decision, it may be better to use a different method, such as a random number generator or a deck of cards.
In conclusion, the psychology of coin flips is a fascinating topic that highlights the limitations of our perception of randomness. While we may believe that a coin flip is 50/50, the reality is that there are many factors that can influence the outcome. If you need to make an important decision, it is important to consider these
## Alternative Methods for Achieving True 50/50 Probability
When it comes to making decisions, flipping a coin is a popular method to achieve a 50/50 probability. However, many people question whether a coin flip is actually a true 50/50 chance. In this article, we will explore alternative methods for achieving a true 50/50 probability.
Firstly, let’s examine the physics behind a coin flip. When a coin is flipped, it rotates in the air and lands on the ground. The outcome of the flip is determined by the side that faces up when the coin comes to a stop. In theory, the probability of the coin landing on either side is equal, assuming the coin is perfectly balanced and the flip is executed with equal force.
However, in reality, there are several factors that can affect the outcome of a coin flip. For example, the surface on which the coin lands can influence the result. A coin may be more likely to land on one side if the surface is uneven or if there is debris on the ground. Additionally, the way the coin is flipped can also impact the outcome. If the flip is not executed with equal force or if the coin is not perfectly balanced, one side may be more likely to land face up.
To achieve a true 50/50 probability, alternative methods can be used. One such method is the use of a random number generator. This method involves using a computer program or device to generate a random number between 1 and 2. If the number generated is 1, it is considered heads, and if it is 2, it is considered tails. This method eliminates the physical factors that can affect a coin flip and ensures a true 50/50 probability.
Another alternative method is the use of a dice. A standard six-sided dice can be used to achieve a 50/50 probability by assigning one side to heads and the opposite side to tails. To ensure fairness, the dice should be rolled on a flat surface and with equal force. This method also eliminates the physical factors that can affect a coin flip and ensures a true 50/50 probability.
A third alternative method is the use of a spinner. A spinner can be created by dividing a circle into two equal parts and labeling one side as heads and the other side as tails. The spinner can then be spun to achieve a 50/50 probability. This method also eliminates the physical factors that can affect a coin flip and ensures a true 50/50 probability.
In conclusion, while a coin flip may seem like a simple and easy way to achieve a 50/50 probability, there are several factors that can affect the outcome. To ensure a true 50/50 probability, alternative methods such as a random number generator, dice, or spinner can be used. These methods eliminate the physical factors that can affect a coin flip and provide a fair and unbiased way to make decisions.
## Real-Life Applications of Coin Flip Probability in Sports and Gambling
Coin flipping is a simple and popular method of determining outcomes in sports and gambling. It is often used to decide which team gets to choose the side of the field or court they want to play on, or who gets to start with the ball or puck. In gambling, coin flipping is used to determine who gets to bet first or who wins a bet. However, many people wonder if a coin flip is actually a 50-50 chance of landing on either side. In this article, we will explore the real-life applications of coin flip probability in sports and gambling and answer the question, is a coin flip actually 50-50?
Firstly, let’s define what we mean by a 50-50 chance. A 50-50 chance means that there are two possible outcomes, and each outcome has an equal probability of occurring. In the case of a coin flip, there are two possible outcomes: heads or tails. If the coin is fair, meaning that it is not weighted or biased towards one side, then the probability of it landing on heads is the same as the probability of it landing on tails. Therefore, a fair coin flip is a 50-50 chance.
However, in real-life situations, there are many factors that can affect the outcome of a coin flip. For example, the way the coin is flipped, the surface it lands on, and the wind conditions can all influence the result. In sports, the person flipping the coin may have a preferred way of flipping it, which could give one team an advantage over the other. In gambling, the person flipping the coin may have a hidden agenda or may be skilled at manipulating the outcome.
Another factor that can affect the outcome of a coin flip is the number of flips. If you flip a coin only once, the probability of it landing on heads or tails is 50-50. However, if you flip a coin multiple times, the probability of getting an equal number of heads and tails decreases. For example, if you flip a coin 10 times, the probability of getting exactly 5 heads and 5 tails is only 24.6%. The more times you flip the coin, the closer the results will be to a 50-50 chance, but there will always be some variation.
In sports, coin flips are often used to determine which team gets to choose the side of the field or court they want to play on. This can be a crucial advantage, especially in outdoor sports where wind or sun conditions can affect the game. In these situations, it is important to ensure that the coin flip is fair and unbiased. One way to do this is to use a third-party coin flipper who has no affiliation with either team.
In gambling, coin flips are often used to determine who gets to bet first or who wins a bet. However, as we mentioned earlier, there are many factors that can influence the outcome of a coin flip. In some cases, people may try to cheat by using a weighted or biased coin, or by manipulating the way the coin is flipped. To avoid these situations, it is important to use a fair and unbiased coin, and to ensure that the person flipping the coin is trustworthy.
In conclusion, a fair coin flip is a 50-50 chance of landing on either side. However, in real-life situations, there are many factors that can affect the outcome of a coin flip, such as the way the coin is flipped, the surface it
## Q&A
1. Is a CoinFlip actually 50 50?
Yes, theoretically a coin flip is 50/50.
2. Can a coin flip be biased?
Yes, a coin flip can be biased if the coin is not perfectly balanced or if the flipper has a certain technique.
3. How can you test if a coin is fair?
You can test if a coin is fair by flipping it many times and recording the results to see if it is close to a 50/50 split.
4. Are there any factors that can affect a coin flip?
Yes, factors such as wind, surface texture, and the force of the flip can affect the outcome of a coin flip.
5. Is a coin flip a reliable way to make decisions?
It depends on the situation. In some cases, a coin flip can be a fair and impartial way to make a decision, but in other cases, it may not be the best method.
## Conclusion
Yes, a CoinFlip is actually 50/50. The probability of getting either heads or tails is equal, assuming the coin is fair and not biased towards one side. Therefore, the outcome of a CoinFlip is completely random and has an equal chance of being either heads or tails.
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You own a company selling apps and you need to know the best selling price for maximum profit. If this quadratic equation could help you find the answer...
...would you be able to solve it?
## But First...Some Need to Know Terms!
### What is a quadratic equation?
Quadratic comes from the word "quad", which means "square". So, a quadratic function is a polynomial function of a degree of 2. "Polynomial" means the sum of many terms: "poly" (many) and "nomial" (name).
When plotted on a graph, quadratic equations are U-shaped (like a horseshoe) and can open upwards (as in a smiley face) or downwards (as in a sad face).
### What are factors?
Factors multiply to get a number.
Factoring is the inverse (or reverse action) of multiplying. For example, when factoring 18, you'll find numbers that multiply to get 18 (1, 2, 3, 6, 9, 18). This is where knowing multiplication tables comes in handy!
### What is standard form?
Standard formin math simplifies rules for clear understanding, communication, and computation. Quadratic equations are easier to factor in standard form...
...where "a", "b", and "c" are numbers and "x" is the variable (or unknown).
• "a" and "b" are called coefficients — a constant number multiplied by a variable.
• "c" is called a constant — it's a fixed value not multiplied by a variable.
If the coefficient is absent, you can understand it to be "1". If "a" = 1, you only need to find the factors of "c". An example of handling a quadratic when "a" > 1 will be provided later.
Multiplying binomials is similar to multiplying two-digit numbers. The exception is two algebraic expressions are being multiplied instead of only digits.
The video below provides an example:
## Method #1: Factoring
Factoring is the inverse (or reverse operation) of multiplying two algebraic expressions.
### Example #1
To factor this, you need to identify the algebraic expressions that were multiplied together. Your end result will look like this...
...where "x" is the variable (x that makes x squared) in the equation and "#" will be the factors of 4 (the numbers that multiply to get 4)
The video below explains factoring quadratics a bit further.
One method to finding those factors is using a table:
Label a table with the headings Factor, Added, Subtracted. Use the abbreviations "F", "A", and "S".
Image created by Wendy McMillan, 2016
1. List factor pairs (of 4) in the Factors (F) column
3. Subtract factor pairs in the Subtracted (S) column
4. Find the answer under heading A or S for the middle term (3).
The sign of the middle number determines positive or negative constants. Since the middle number 3 is positive, 4 must be positive (think 4 - 1 vs. 1 - 4).
### Example #2
When the value "a" > 1, you'll need to multiply the "a" and "c" values together, then factor.
Take a look at this equation:
1. Multiply 6 ("a") x 6 ("c") = 36
2. Factor 36
3. You'll see that factors 9 and 4 when subtracted total 5
With "a" and "c" values as 6, further factorize 9 and 4 to form the required factor groups [2 x 2 = 4 and 3 x 3 = 9 so 9 - 4 = 5 and 9 x 4 = 36.]
## Find "x" by Solving Algebraic Equations
Now that you have your factors, you can find the value of "x".
## Method #2: Complete the Square
What if you have an algebraic equation like this?
There aren't any factors of 3 that add or subtract to get 6!
You'll need to complete the square — meaning, you'll need to make the formula a perfect square.
### You'll do this by:
• Rearranging the equation with variables on one side
• Dividing the "b" term (6) by 2
• Factor the new equation
• This equation could also be written like this...
...which is why it's called a perfect square!
## Method #3: Use the Square Root Property
The middle term ("bx") is missing!
You'll need the square root method to solve these types of equations.
Square roots are the inverse or reverse action of squaring a number.
• First, isolate "x" by eliminating any coefficients or additive/subtractive terms by adding 3 to both sides, then simplifying
• Divide each term by 2
• End with this equation
• Take the square root of both sides
This is where knowing squared numbers comes in handy!
Your result should have a positive AND a negative answer since a negative times negative number and a positive times a positive number both equal a positive answer.
## Method #4: Use the Quadratic Formula
Some equations are rebellious and just can't be solved using any of the methods above.
### Example:
When the value "a" > 1, multiply the "a" and "c" values, then factor!
3 x 7 = 21....but there are NO factors of 21 (1, 3, 7, 21) that add or subtract to get 5.
GIF created by Wendy McMillian, 2023 with PowerPoint and pngtree.com
In standard equations "a", "b", and "c" will be numerical values. That is:
• a = 3
• b = 5
• c = -7 (don't forget to carry the negative along)
Replace each variable in the quadratic formula with the corresponding number minding all negatives and positives. Use parenthesis to keep things organized!
End by simplifying:
If you enter these equations into a calculator you end up with nasty-looking decimals! It's better to just leave it as a fraction.
## Method #5: Graph to Solve the Algebraic Equation and Find "x"
One easy way to solve quadratics is by graphing! Remember the equation at the beginning of this lesson?
No factors of 458 (1, 2, 229, 458) add or subtract to get 72.
These big numbers will cause major headaches too when used in the quadratic formula!
### Desmos will save the day!
Desmos is a graphing calculatorthat handles big equations.
1. Type the equation into Desmos
2. Zoom out using the "-" button
3. Find where the graph crosses the x-axis (2 places)
4. The "x"-values are the factors for the equation
At the beginning of this lesson, you were asked if you could find the best price to sell apps and make a maximum profit.
Graphing videos created by Wendy McMillian using Desmos graphing calculator online and Screencastify
Find the answer by scrolling to the top of the graph to the x-value at the graph's peak — which is 36.
You'll need to sell 36 apps in order to make a profit!
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Virginia SOL 6 - 2020 Edition
1.03 Absolute value of integers
Lesson
Let's consider the following situation:
A scuba diver is diving at a depth of $-50$−50 feet. At the same time, a helicopter pilot is flying overhead at $30$30 feet above the surface. Which person is closer to sea level?
Although the scuba diver is at an altitude much lower than the helicopter, the helicopter pilot is closer to sea level.
When making this comparison, we are considering the absolute value of each measurement. The absolute value of a number is the distance from the number to zero on the number line.
#### Exploration
The applet below shows the absolute value, or distance from zero for different integers on the number line. Move the point left and right and consider the following questions:
1. What do you notice about the absolute value of a positive number?
2. What do you notice about the absolute value of a negative number?
We can see that the absolute value of a positive number is the number itself. However, the absolute value of a negative number is its opposite. This is because the distance is always a positive number. This applies to all numbers on the number line!
The mathematical symbol for absolute value is "$\text{| |}$| |". For example, we would read "$\left|-6\right|$|6|" as "the absolute value of negative six."
Absolute value
The absolute value of a number is its distance from zero on the number line.
The numbers $-3$3 and $3$3 are both $3$3 units from $0$0, so they have the same absolute value.
The absolute value of a positive number is the number itself.
The absolute value of a negative number is its opposite.
For example, $\left|3\right|=3$|3|=3 and $\left|-3\right|=3$|3|=3
#### Worked example
##### Question 1
Evaluate: Which of the following are smaller than $\left|-20\right|$|20|?
A) $-15$15 B) $\left|-30\right|$|30| C) $\left|-5\right|$|5| D) $21$21
Think: We need to evaluate each of these terms, then compare them to $\left|-20\right|$|20|.
Do: Let's start by evaluating all the absolute values:
$\left|-20\right|=20$|20|=20, $\left|-30\right|=30$|30|=30 and $\left|-5\right|=5$|5|=5
Which of the four possible answers are smaller than $20$20?
So $\left|-20\right|$|20| is greater than A) $-15$15 and C) $\left|-5\right|$|5|
#### Practice questions
##### Question 2
1. What does the absolute value of $-3$3 represent?
It represents the distance between $-3$3 and $0$0.
A
It represents a position $3$3 units away from $0$0 on either side.
B
It represents moving $3$3 units to the right from $0$0 along the number line.
C
It represents moving $3$3 units to the left from $0$0 along the number line.
D
It represents the distance between $-3$3 and $0$0.
A
It represents a position $3$3 units away from $0$0 on either side.
B
It represents moving $3$3 units to the right from $0$0 along the number line.
C
It represents moving $3$3 units to the left from $0$0 along the number line.
D
##### Question 3
Evaluate $\left|65\right|$|65|
##### Question 4
What is the value of $\left|-155\right|$|155|?
### Outcomes
#### 6.3c
Identify and describe absolute value of integers
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Categories :
## What happens when you multiply a number by 0 1?
We are multiplying by 1 and so the answer is the number that is being multiplied by 1, which is 0. We can also solve this last example with our rule for multiplying by 0. Any number multiplied by 0 is 0 and here we have 0 × 1 = 0. One is being multiplied by zero and so, the answer is zero.
## What is the 0 rule in multiplication?
The multiplication property states that the product of any number and zero is zero. It doesn’t matter what the number is, when you multiply it to zero, you get zero as the answer.
What happens when multiply 1?
The Rule. If you noticed, when you multiply by 1, you always get your original number. This rule tells us that anything multiplied by 1 is itself. We can use this rule on any multiplication by 1 problem.
### What does a multiplication fact?
A multiplication fact is the answer to a multiplication calculation. For example, in the sum 3 x 3 = 9, the multiplication fact is 9.
### Is 0 times 0 defined?
So zero divided by zero is undefined. Just say that it equals “undefined.” In summary with all of this, we can say that zero over 1 equals zero. We can say that zero over zero equals “undefined.” And of course, last but not least, that we’re a lot of times faced with, is 1 divided by zero, which is still undefined.
What is the answer for multiplication called?
the product
The answer to a multiplication problem is called the product.
3 times table
• 3 x 1 = 3.
• 3 x 2 = 6.
• 3 x 3 = 9.
• 3 x 4 = 12.
• 3 x 5 = 15.
• 3 x 6 = 18.
• 3 x 7 = 21.
• 3 x 8 = 24.
## What are two multiplication facts?
For every multiplication fact there are two division facts. Thus, the product divided by the multiplicand equals the multiplier and the product divided by the multiplier equals the multiplicand.
What are all the multiplication facts?
The multiplication facts (also known as the times tables) are all of the multiplication problems from 1 × 1 = 1 up to 10 × 10 = 100. Can you imagine trying to find common denominators or equivalent fractions without knowing 5 × 6?
### What is zero in multiplication?
The Multiplication Property of Zero One of zero’s unique rules is called the multiplication property. The multiplication property states that the product of any number and zero is zero. It doesn’t matter what the number is, when you multiply it to zero, you get zero as the answer.
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# Relationships & Rules
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## Objective
SWBAT use a table to determine an expression or rule that describes the relationship between numbers in a pattern.
#### Big Idea
Using an expression can help to discover a number pattern.
## Warm-Up
15 minutes
We open with a quick check in -- students complete an evaluating expressions problem. The majority of last week's teaching was focused on simplifying expressions. To start this week, I ask students to apply what they remembered.
I choose a problem that asks students to critique "an example" of other students' work, and include here an analysis of a student response. Having students tackle this sort of problem gives me so much information because they have to analyze the work to determine which problem is incorrect and explain why. While problem solving, students are also refining the precision of their mathematical communication (MP1, 3, 6). This problem is rigorous, because it requires the students to evaluate, which is a higher level thinking skill than application.
## Launch & Guided Practice
20 minutes
I write a basic pattern on the board, then ask the students to determine the rule used to complete this table.
Step Total 1 3 2 6 3 9 n
They quickly recognize that the rule is nx3 because they know their basic facts. Even though they are familiar with the rule from the start, I ask them to build this pattern using tiles. Together they add 3 tiles each time I say "next step".
After 3 steps, pause to make sense of what is happening.
• How many tiles did the pattern increase by each time there was a new step (3)
Write zero and 3 on the board. Explain that these numbers are an important part of finding the rule. Students know the rule is nx3. Push their thinking to develop a rule that includes the 3 (the amount the pattern increases by) and the 0 (the number of tiles that the pattern started with).
Students agree that the rule can be expressed as n x 3 + 0.
Repeat this activity with a few more simple patterns, using basic facts so the students can identify the rule quickly and easily. Revisit the two focus questions each time.
• Where did the pattern start?
• What did it change by?
The new pattern I write on the board requires two operations, but I don't tell the students this.
Step Total 1 4 2 7 3 10 n
Allow time for students to think about the pattern and identify what they know. Here, students recognize that the pattern increases by 3. Ask them to determine the 100th step. This is challenging if they don't have the entire rule. Refocus on the 2 questions
• Where did the pattern start?
• What did it change by?
so students can agree that the pattern, although it doesn't say on this table, started at 1.
The two numbers that they can use to develop a rule are 1 ( the number the pattern started at) and 3 (the number it increased by). Students recognize the rule is n x 3 +1.
Continue with a few more examples before allowing them to practice on their own.
## Now You Try
25 minutes
Now it's time for students to practice finding a rule for patterns that include more than one operation. They are given a worksheet with seven input output tables, and rules that include combinations of all the operations.
I specifically didn't use any examples with subtraction ex: x 3 - 1 because I want the kids to be challenged to think about the two focus questions.
• Where did the pattern start?
• What did it change by?
The first input output table had an example of -3. This challenged the students to rely on strategies, not just guessing and checking different combinations.
I allow students to use models to assist their thinking (MP4). Some images are included here of how models were used to help students who are struggling.
## Group Share
5 minutes
During the group share, students are encouraged to share any struggles, challenges they over came, or interesting discoveries from today's task.
I was surprised by the number of students who explained their strategy for today's task, rather than discussing each problem separately.
"I got stuck on the first problem, even though it looked like it would be easy because there are only a few rows, it was harder than the others we had done together because the table didn't start at 1. I noticed the tables in the middle row started a 1 and 0. I did these first because they were easier. Then, I was used to thinking like this so I did the first two".
When I made this handout, I had not considered the placement of these tables. Looking back, I think it worked out well to challenge the students thinking and also their strategic approaches to tasks. Now, in the future, I will know this challenge and also provide students who are having trouble accessing these types of problems with a hint to start in the second row.
I learn from my students every day, it is in these group share moments that I often learn the most.
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# RS Aggarwal Class 8 Maths Chapter 14 Ex 14.2 Solutions 2022 | Download Free PDF
RS Aggarwal Class 8 Maths Chapter 14 Ex 14.2 Solutions: This exercise includes topics such as the formula to find the sum of all exterior angles, interior angles, & the number of diagonals in a polygon of the ‘n’ side. The best way of preparing for class 8th Maths final exams is practicing these solutions frequently. The students must solve the questions to examine whether they are completely prepared for the exam or still need more practice. This also makes them able to manage their timings of solving Maths paper.
RS Aggarwal Class 8 Maths Chapter 14 Ex 14.2 Solutions can be used as a handy last-minute revision study material. These solutions are all-inclusive & are created by the mathematics expert that assists the students to clear all their queries related to the concepts of the polygon. These solutions are designed according to the CBSE guidelines & have strong chances to attain excellent ranks in the Class 8th Maths final exam.
## Download RS Aggarwal Class 8 Maths Chapter 14 Ex 14.2 Solutions
RS Aggarwal Class 8 Maths Chapter 14 Ex 14.2 Solutions
## Important Definition for RS Aggarwal Class 8 Maths Chapter 14 Ex 14.2 Solutions
• Exterior angles of a polygon
They are created when by one of its sides & extending the other side. The sum of all the exterior angles in a polygon is equal to 360 degrees.
• Sum of the Exterior Angles of a Polygon
Suppose if you begin traveling from the vertex at angle 1. You walk in a clockwise direction, create turns through angles 2, 3, 4, and 5 and come back to the same vertex. You covered the whole perimeter of the polygon and made one complete turn in the procedure. One whole turn is equal to 360 degrees. Therefore, it can be said that ∠1, ∠2, ∠3, ∠4, and ∠5 sums up to 360 degrees.
• Interior angle of a polygon
It is an angle created inside the two adjacent sides of a polygon. The angle measures at the interior part of a polygon are known as the interior angle of a polygon.
• Sum of Interior Angles of a Polygon
The sum of the interior angles of a polygon should be a constant value depends on the number of sides. It will provide some constant measurement depends on the number of polygon sides no matter if the polygon is regular or irregular, convex or concave.
The formula of the sum of the interior angles of the polygon is:
Sum of the Interior Angles of a Polygon = 180 (n-2) degrees
• Find the number of diagonals in a polygon
In a polygon, the diagonal is the line segment that joins two non-adjacent vertices. The diagonals of a polygon are in concave polygons, at least one diagonal is actually outside the polygon.
The formula is Number of Diagonals = n(n-3)/2
This formula is simply created by the combination of diagonals that each vertex sends to another vertex and then subtracting the total sides. An n-sided polygon has n-vertices which can be joined with each other in nC2 ways.
Know more at the official website.
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Difference between revisions of "2011 AIME II Problems/Problem 6"
Problem 6
Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a, and $a+d>b+c$. How many interesting ordered quadruples are there?
Solution 1
Rearranging the inequality we get $d-c > b-a$. Let $e = 11$, then $(a, b-a, c-b, d-c, e-d)$ is a partition of 11 into 5 positive integers or equivalently: $(a-1, b-a-1, c-b-1, d-c-1, e-d-1)$ is a partition of 6 into 5 non-negative integer parts. Via a standard stars and bars argument, the number of ways to partition 6 into 5 non-negative parts is $\binom{6+4}4 = \binom{10}4 = 210$. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry, there are as many partitions where the fourth is less than the second. So, if $N$ is the number of partitions where the second element is equal to the fourth, our answer is $(210-N)/2$.
We find $N$ as a sum of 4 cases:
• two parts equal to zero, $\binom82 = 28$ ways,
• two parts equal to one, $\binom62 = 15$ ways,
• two parts equal to two, $\binom42 = 6$ ways,
• two parts equal to three, $\binom22 = 1$ way.
Therefore, $N = 28 + 15 + 6 + 1 = 50$ and our answer is $(210 - 50)/2 = \fbox{80.}$
Solution 2
Let us consider our quadruple (a,b,c,d) as the following image xaxbcxxdxx. The location of the letter a,b,c,d represents its value and x is a place holder. Clearly the quadruple is interesting if there are more place holders between c and d than there are between a and b. 0 holders between a and b means we consider a and b as one unit ab and c as cx yielding $\binom83 = 56$ ways, 1 holder between a and b means we consider a and b as one unit axb and c as cxx yielding $\binom 63 = 20$ ways, 2 holders between a and b means we consider a and b as one unit axxb and c as cxxx yielding $\binom43 = 4$ ways and there cannot be 3 holders between a and b so our total is 56+20+4=$\fbox{80.}$.
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# What is the equation of the line passing through (11,13) and (59,67)?
Nov 18, 2015
$y = 1.125 x + 0.625$
or
$y = \frac{9}{8} x + \frac{5}{8}$
#### Explanation:
First label the coordinates.
$x 1 = 11 , y 1 = 13$
$x 2 = 59 , y 2 = 67$
The slope (m) is the rise (change in y) divided by the run (change in x),
so $m = \frac{y 2 - y 1}{x 2 - x 1}$
$m = \frac{67 - 13}{59 - 11} = \frac{54}{48} = \frac{9}{8} = 1.125$
The standard linear formula is $y = m x + b$ and we have to find b. Substitute m and one set of coordinates into this formula:
$y 1 = m \cdot x 1 + b \to 13 = 1.125 \cdot 11 + b \to 13 = 12.375 + b$
$b = 0.625$
Substitute this into $y = m x + b \to \ast y = 1.125 x + 0.625 \ast$
Always check your answer by substituting the other set of coordinates into the equation:
$y = 1.125 \cdot \ast 59 \ast + 0.625 = 66.375 + 0.625 = 67$
Since this matches the original coordinate (59, 67), the answer must be correct.
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Trying to discover out exactly how to transform 19/12 right into a combined number or fraction? have actually I got the answer for you! In this guide, we"ll to walk you with the step-by-step procedure of convert an not correct fraction, in this situation 19/12, to a blended number. Read on!
Want to conveniently learn or show students exactly how to convert 19/12 to a mixed number? play this an extremely quick and fun video now!
Before us begin, let"s revisit some basic portion terms so you understand precisely what we"re taking care of here:
Numerator. This is the number above the portion line. For 19/12, the numerator is 19.Denominator. This is the number below the portion line. For 19/12, the denominator is 12.Improper fraction. This is a portion where the numerator is better than the denominator.Mixed number. This is a way of to express an improper fraction by simple it to entirety units and a smaller all at once fraction. It"s an integer (whole number) and a appropriate fraction.
You are watching: 19 7 as a mixed number
Now let"s go through the actions needed to transform 19/12 come a blended number.
## Step 1: discover the totality number
We an initial want to discover the entirety number, and to carry out this we divide the molecule by the denominator. Since we are just interested in whole numbers, we ignore any type of numbers come the best of the decimal point.
19/12= 1.5833333333333 = 1
Now that we have our totality number because that the combined fraction, we need to uncover our brand-new numerator for the fraction part the the mixed number.
## Step 2: gain the brand-new numerator
To work this the end we"ll use the whole number us calculated in step one (1) and also multiply the by the original denominator (12). The an outcome of that multiplication is then subtracted from the initial numerator:
19 - (12 x 1) = 7
## Step 3: Our combined fraction
We"ve currently simplified 19/12 come a combined number. To view it, we just need to placed the entirety number along with our brand-new numerator and original denominator:
1 7/12
## Step 4: simple our fraction
In this case, our portion (7/12) deserve to be simplified down further. In stimulate to do that, we must calculate the GCF (greatest usual factor) the those two numbers. You can use our handy GCF calculator to work this the end yourself if you desire to. We currently did that, and also the GCF the 7 and also 12 is 1.
We can now division both the brand-new numerator and the denominator through 1 to leveling this fraction down come its lowest terms.
7/1 = 7
12/1 = 12
When we put that together, we deserve to see the our complete answer is:
1 7/12
Hopefully this tutorial has helped you to understand exactly how to convert any improper fraction you have into a combined fraction, complete with a totality number and a ideal fraction. You"re complimentary to use our calculator below to occupational out more, yet do shot and learn exactly how to execute it yourself. It"s much more fun than it seems, i promise!
If you found this content valuable in your research, please do us a great favor and use the tool below to make sure you appropriately reference us wherever you usage it. We really evaluate your support!
"What is 19/12 together a mixed number?". historicsweetsballroom.com. Accessed top top September 28, 2021. Https://historicsweetsballroom.com/calculator/improper-to-mixed/what-is-19-12-as-a-mixed-number/.
"What is 19/12 as a combined number?". historicsweetsballroom.com, https://historicsweetsballroom.com/calculator/improper-to-mixed/what-is-19-12-as-a-mixed-number/. Accessed 28 September, 2021.
See more: How Far Is Huntsville Texas From Dallas Texas ) From Dallas, It'S 171 Miles From Huntsville, Tx To Dallas
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# 6.2 Line integrals (Page 9/20)
Page 9 / 20
We now give a formula for calculating the flux across a curve. This formula is analogous to the formula used to calculate a vector line integral (see [link] ).
## Calculating flux across a curve
Let F be a vector field and let C be a smooth curve with parameterization $\text{r}\left(t\right)=⟨x\left(t\right),y\left(t\right)⟩,a\le t\le b.$ Let $\text{n}\left(t\right)=⟨{y}^{\prime }\left(t\right),\text{−}{x}^{\prime }\left(t\right)⟩.$ The flux of F across C is
${\int }_{C}\text{F}·\text{N}ds={\int }_{a}^{b}\text{F}\left(\text{r}\left(t\right)\right)·\text{n}\left(t\right)dt$
## Proof
The proof of [link] is similar to the proof of [link] . Before deriving the formula, note that $‖\text{n}\left(t\right)‖=‖⟨y\prime \left(t\right),\text{−}x\prime \left(t\right)⟩‖=\sqrt{{\left(y\prime \left(t\right)\right)}^{2}+{\left(x\prime \left(t\right)\right)}^{2}}=‖{r}^{\prime }\left(t\right)‖.$ Therefore,
$\begin{array}{cc}\hfill {\int }_{C}\text{F}·\text{N}ds& ={\int }_{C}\text{F}·\frac{\text{n}\left(t\right)}{‖\text{n}\left(t\right)‖}ds\hfill \\ & ={\int }_{a}^{b}\text{F}·\frac{\text{n}\left(t\right)}{‖\text{n}\left(t\right)‖}‖{r}^{\prime }\left(t\right)‖dt\hfill \\ & ={\int }_{a}^{b}\text{F}\left(\text{r}\left(t\right)\right)·\text{n}\left(t\right)dt.\hfill \end{array}$
## Flux across a curve
Calculate the flux of $\text{F}=⟨2x,2y⟩$ across a unit circle oriented counterclockwise ( [link] ).
To compute the flux, we first need a parameterization of the unit circle. We can use the standard parameterization $\text{r}\left(t\right)=⟨\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩,$ $0\le t\le 2\pi .$ The normal vector to a unit circle is $⟨\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩.$ Therefore, the flux is
$\begin{array}{cc}\hfill {\int }_{C}\text{F}·\text{N}ds& ={\int }_{0}^{2\pi }⟨2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t,2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩·⟨\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & ={\int }_{0}^{2\pi }\left(2\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{2}t+2\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}t\right)\phantom{\rule{0.2em}{0ex}}dt=2{\int }_{0}^{2\pi }\left({\text{cos}}^{2}t+{\text{sin}}^{2}t\right)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & =2{\int }_{0}^{2\pi }dt=4\pi .\hfill \end{array}$
Calculate the flux of $\text{F}=⟨x+y,2y⟩$ across the line segment from $\left(0,0\right)$ to $\left(2,3\right),$ where the curve is oriented from left to right.
3/2
Let $\text{F}\left(x,y\right)=⟨P\left(x,y\right),Q\left(x,y\right)⟩$ be a two-dimensional vector field. Recall that integral ${\int }_{C}\text{F}·\text{T}ds$ is sometimes written as ${\int }_{C}Pdx+Qdy.$ Analogously, flux ${\int }_{C}\text{F}·\text{N}ds$ is sometimes written in the notation ${\int }_{C}\text{−}Qdx+Pdy,$ because the unit normal vector N is perpendicular to the unit tangent T . Rotating the vector $d\text{r}=⟨dx,dy⟩$ by 90° results in vector $⟨dy,\text{−}dx⟩.$ Therefore, the line integral in [link] can be written as ${\int }_{C}-2ydx+2xdy.$
Now that we have defined flux, we can turn our attention to circulation. The line integral of vector field F along an oriented closed curve is called the circulation of F along C . Circulation line integrals have their own notation: ${\oint }_{C}\text{F}·\text{T}ds.$ The circle on the integral symbol denotes that C is “circular” in that it has no endpoints. [link] shows a calculation of circulation.
To see where the term circulation comes from and what it measures, let v represent the velocity field of a fluid and let C be an oriented closed curve. At a particular point P , the closer the direction of v ( P ) is to the direction of T ( P ), the larger the value of the dot product $\text{v}\left(P\right)·\text{T}\left(P\right).$ The maximum value of $\text{v}\left(P\right)·\text{T}\left(P\right)$ occurs when the two vectors are pointing in the exact same direction; the minimum value of $\text{v}\left(P\right)·\text{T}\left(P\right)$ occurs when the two vectors are pointing in opposite directions. Thus, the value of the circulation ${\oint }_{C}\text{v}·\text{T}ds$ measures the tendency of the fluid to move in the direction of C .
## Calculating circulation
Let $\text{F}=⟨-y,x⟩$ be the vector field from [link] and let C represent the unit circle oriented counterclockwise. Calculate the circulation of F along C .
We use the standard parameterization of the unit circle: $\text{r}\left(t\right)=⟨\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩,0\le t\le 2\pi .$ Then, $\text{F}\left(\text{r}\left(t\right)\right)=⟨\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩$ and ${r}^{\prime }\left(t\right)=⟨\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩.$ Therefore, the circulation of F along C is
$\begin{array}{cc}\hfill {\oint }_{C}\text{F}·\text{T}ds& ={\int }_{0}^{2\pi }⟨\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩·⟨\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩dt\hfill \\ & ={\int }_{0}^{2\pi }\left({\text{sin}}^{2}t+{\text{cos}}^{2}t\right)\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & ={\int }_{0}^{2\pi }dt=2\pi .\hfill \end{array}$
Notice that the circulation is positive. The reason for this is that the orientation of C “flows” with the direction of F . At any point along the circle, the tangent vector and the vector from F form an angle of less than 90°, and therefore the corresponding dot product is positive.
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
how did you get the value of 2000N.What calculations are needed to arrive at it
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# 5.7: Gödel Numbering
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We now change our focus from looking at functions and relations on the natural numbers, where it makes sense to talk about representable sets, to functions mapping strings of $$\mathcal{L}_{NT}$$-symbols to $$\mathbb{N}$$. We will establish our coding system for formulas, associating to each $$\mathcal{L}_{NT}$$ formula $$\phi$$ its Gödel number, $$\ulcorner \phi \urcorner$$. We will make great use of the coding function $$\langle \cdot \rangle$$ that we defined in Definition 4.5.3.
##### Definition 5.7.1.
The function $$\ulcorner \cdot \urcorner$$, with domain the collection of finite strings of $$\mathcal{L}_{NT}$$-symbols and codomain $$\mathbb{N}$$, is defined as follows:
$\ulcorner s \urcorner = \begin{cases} \begin{array}{ll} \langle 1, \ulcorner \alpha \urcorner \rangle & \text{if} \: s \: \text{is} \: \left( \neg \alpha \right), \: \text{where} \: \alpha \: \text{is an} \: \mathcal{L}_{NT} \text{-formula} \\ \langle 3, \ulcorner \alpha \urcorner, \ulcorner \beta \urcorner \rangle & \text{if} \: s \: \text{is} \: \left( \alpha \lor \beta \right), \: \text{where} \: \alpha \: \text{and} \: \beta \: \text{are} \: \mathcal{L}_{NT} \text{-formulas} \\ \langle 5, \ulcorner v_i \urcorner, \ulcorner \alpha \urcorner \rangle & \text{if} \: s \: \text{is} \: \left( \forall v_i \right) \left( \alpha \right), \: \text{where} \: \alpha \: \text{is an} \: \mathcal{L}_{NT} \text{-formula} \\ \langle 7, \ulcorner t_1 \urcorner, \ulcorner t_2 \urcorner \rangle & \text{if} \: s \: \text{is} \: = t_1 t_2, \: \text{where} \: t_1 \: \text{and} \: t_2 \: \text{are terms} \\ \langle 9 \rangle & \text{if} \: s \: \text{is} \: 0 \\ \langle 11, \ulcorner t \urcorner \rangle & \text{if} \: s \: \text{is} \: St, \: \text{with} \: t \: \text{a term} \\ \langle 13, \ulcorner t_1 \urcorner, \ulcorner t_2 \urcorner \rangle & \text{if} \: s \: \text{is} \: + t_1 t_2, \: \text{where} \: t_1 \: \text{and} \: t_2 \: \text{are terms} \\ \langle 15, \ulcorner t_1 \urcorner, \ulcorner t_2 \urcorner \rangle & \text{if} \: s \: \text{is} \: \cdot t_1 t_2, \: \text{where} \: t_1 \: \text{and} \: t_2 \: \text{are terms} \\ \langle 17, \ulcorner t_1 \urcorner, \ulcorner t_2 \urcorner \rangle & \text{if} \: s \: \text{is} \: E t_1 t_2, \: \text{where} \: t_1 \: \text{and} \: t_2 \: \text{are terms} \\ \langle 19, \ulcorner t_1 \urcorner, \ulcorner t_2 \urcorner \rangle & \text{if} \: s \: \text{is} \: < t_1 t_2, \: \text{where} \: t_1 \: \text{and} \: t_2 \: \text{are terms} \\ \langle 2i \rangle & \text{if} \: s \: \text{is the variable} \: v_i \\ 3 & \text{otherwise.} \end{array} \end{cases}$
Notice that each symbol is associated with its symbol number, as set out in Table 5.1.
##### Example 5.7.2:
Just for practice, let's find $$\ulcorner 0 \urcorner$$. Just from the chart above, $$\ulcorner 0 \urcorner = \langle 9 \rangle = 2^{10} = 1024$$. To look at another example, look at $$\ulcorner 0 = 0 \urcorner$$. Working recursively,
\begin{align} \ulcorner = 00 \urcorner &= \langle 7, \ulcorner 0 \urcorner, \ulcorner 0 \urcorner \rangle \\ &= \langle 7, 1024, 1024 \rangle \\ &= 2^8 3^{1025} 5^{1025} \end{align}
Exercise 3 asks you to investigate some of the subtleties of coding as it relates to this last example.
##### Example 5.7.3:
This is a neat function, but the numbers involved get really big, really fast. Suppose that we work out the Gödel number for the formula $$\phi$$, where $$\phi$$ is $$\left( \neg = 0S0 \right)$$.
Since $$\phi$$ is a denial, the definition tells us that
$\ulcorner \phi \urcorner \: \text{is} \: \langle 1, \urcorner = 0S0 \urcorner \rangle = 2^2 3^{\ulcorner = 0S0 \urcorner + 1}.$
So we need to find $$\ulcorner = 0S0 \urcorner$$, and by the "equals" clause in the definition,
$\ulcorner = 0S0 \urcorner \: \text{is} \: \langle 7, \ulcorner 0 \urcorner, \ulcorner S0 \urcorner \rangle = 2^8 3^{\ulcorner 0 \urcorner + 1} 5^{\ulcorner S0 \urcorner + 1}.$
But $$\ulcorner 0 \urcorner = 2^{10} = 1024$$, and $$\ulcorner S0 \urcorner = 2^{12} 3^{\ulcorner 0 \urcorner + 1} = 2^{12} 3^{1025}$$. Now we're getting somewhere. Plugging things back in, we get
$\ulcorner = 0S0 \urcorner \: \text{is} \: 2^8 3^{1025} 5^{\left[ 2^12 3^{1025} + 1 \right]}$
so the Gödel number for $$\left( \neg = 0S0 \right)$$ is
$\ulcorner \phi \urcorner \: \text{is} \: 2^2 3^{\left( 2^8 3^{1025} 5^{\left[ 2^{12} 3^{1025} + 1 \right]} + 1 \right)}.$
Chaff: To get an idea about how large this number is, consider the fact that the exponent on the 5 is $$\ulcorner S0 \urcorner = 2^{12} 3^{1025} + 1$$, which is
4588239037329654294933009459423640636113835 33711852348723982661700090725110495540711416 24496800232720851201851240219667428400380468 28472630247645228844759293716788206726298594 57606066116964029586110650008838161967674248 714876110453564150536269711030213614452805279 213722748800276796114884183810302573694405480 301945785627339339194850085383681785222504546 327111992210992776215014423059901287305704225 3643605726211189929819826835540873386794064170 563975508362231081323849454313910276632860438529,
approximately $$10^{490}$$.
Now, let us play a bit with the Gödel number of $$\phi$$:
\begin{align} \ulcorner \phi \urcorner &= 2^2 3^{\left( 2^8 3^{1025} 5^{\left[ 2^{12} 3^{1025} + 1 \right]} + 1 \right)} \\ &> 3^{5^{\left[ 10^{490} \right]}}, \end{align}
so if we take common logarithms, we see that
$\text{log} \left( \ulcorner \phi \urcorner \right) > 5^{\left[ 10^{490} \right]} \text{log} 3$
and taking logarithms again,
\begin{align} \text{log} \left( \text{log} \left( \ulcorner \phi \urcorner \right) \right) &> 10^{490} \text{log} 5 + \text{log} \left( \text{log} 3 \right) \\ &> 10^{489}. \end{align}
Hmm... So this means that $$\text{log} \left( \ulcorner \phi \urcorner \right)$$ is bigger than $$10^{\left[ 10^{489} \right]}$$. But the common logarithm of a number is (approximately) the number of digits that it takes to express the number in base 10 notation, so we have shown that it takes more than $$10^{\left[ 10^{489} \right]}$$ digits to write out the Gödel number of $$\phi$$. If you remember that a googol is $$10^{100}$$ and a googolplex is $$10^{10^{100}}$$, $$\ulcorner \phi \urcorner$$ is starting to look like a pretty big number, but it gets better!
To write out a string of $$10^{\left[10^{489} \right]}$$ characters (assuming a million characters per mile, or about 16 characters per inch) would require far more than $$10^{\left[ 10^{488} \right]}$$ miles, which is far more than $$10^{\left[ 10^{487} \right]}$$ light years.
Or, to look at it another way, if we assume that we can put about 132 lines of type of an 8.5- by 11-inch piece of paper, and since a ream of paper (500 sheets) is about 2 inches thick, that gives a tack of paper more than $$10^{\left[ 10^{488} \right]}$$ light years high. Since the age of the universe is currently estimated to be in the tens of billions of years (on the order of $$10^{10}$$ years), if we assume that the universe is both Euclidean and spherical, the volume of the universe is less than $$10^{40}$$ cubic light years, rather less than the $$10^{\left[ 10^{488} \right]}$$ cubic light years we would need to store our stack of paper. In short, we don't win any prizes for being incredibly efficient with the coding that we have chosen. What we do win is ease of analysis. The fact that we have chosen to code using a representable function will make our proofs to come much easier to comprehend.
## Exercises
1. Evaluate the Gödel number for each of the following:
(a) $$\left( \forall v_3 \right) \left( v_3 + 0 = v_4 \right)$$
(b) $$SSSS0$$
2. Find the formula or term that is coded by each of the following:
(a) $$2^8 3^{\left( 2^{14} 3^{\left( 2^{18} 3^9 5^{1025} + 1 \right)} 5^{\left( 2^{18} 3^{33} 5^{1025} +1 \right)} + 1 \right]} 5^{\left[ 2^{18} 3^{129} 5^{1025} + 1 \right]}$$
(b) $$2^2 3^{\left( 2^{20} 3^{1025} 5^{\left( 2^{12} 3^{1025} + 1 \right)} + 1 \right)}$$
(c) $$2^6 3^9 5^{\left( 2^8 3^9 5^9 + 1 \right)}$$
3. Look at the number $$c = 2^8 3^{1025} 5^{1025} = \langle 7, \ulcorner 0 \urcorner, \ulcorner 0 \urcorner \rangle$$. Find the number $$e$$ such that $$IthElement \left( e, 2, c \right)$$ is true. Suppose that $$d = \langle 3, 1, 4, 5 \rangle = 2^4 3^2 5^5 7^6$$. Why is $$IthElement \left( 4, 1, d \right)$$ false?
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# How do you simplify ( 3 sqrt 5) ( 2 sqrt 10 ) ?
May 30, 2018
$30 \sqrt{2}$
#### Explanation:
Multiply the whole numbers together and multiply the sued numbers together
$\left(3 \sqrt{5}\right) \left(2 \sqrt{10}\right) = 6 \sqrt{50}$
$\sqrt{50} = \sqrt{25} \times \sqrt{2} = 5 \sqrt{2}$
So $6 \sqrt{50} = 6 \times 5 \sqrt{2} = 30 \sqrt{2}$
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# 8.2 Backshift notation
The backward shift operator $B$ is a useful notational device when working with time series lags:
$${B y_{t} = y_{t - 1}} .$$
(Some references use $L$ for “lag” instead of $B$ for “backshift”.) In other words, $B$, operating on $y_{t}$, has the effect of shifting the data back one period. Two applications of $B$ to $y_{t}$ shifts the data back two periods:
$$B(By_{t}) = B^{2}y_{t} = y_{t-2} .$$
For monthly data, if we wish to consider “the same month last year,” the notation is $B^{12}y_{t}$ = $y_{t-12}$.
The backward shift operator is convenient for describing the process of differencing. A first difference can be written as
$$y'_{t} = y_{t} - y_{t-1} = y_t - By_{t} = (1 - B)y_{t}\: .$$
Note that a first difference is represented by $(1 - B)$. Similarly, if second-order differences have to be computed, then:
$$y''_{t} = y_{t} - 2y_{t - 1} + y_{t - 2} = (1-2B+B^2)y_t = (1 - B)^{2} y_{t}\: .$$
In general, a $d$th-order difference can be written as $$(1 - B)^{d} y_{t}.$$ Backshift notation is very useful when combining differences as the operator can be treated using ordinary algebraic rules. In particular, terms involving $B$ can be multiplied together. For example, a seasonal difference followed by a first difference can be written as
\begin{align*} (1-B)(1-B^m)y_t &= (1 - B - B^m + B^{m+1})y_t \\ &= y_t-y_{t-1}-y_{t-m}+y_{t-m-1}, \end{align*}
the same result we obtained earlier.
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# Sound Intensity level (Decibels) Problems and Solutions
In this article, we will delve into solving problems related to sound intensity levels and decibels, specifically tailored for college prep courses. First, we’ll introduce the essential formulas concisely, followed by practice questions.
## Understanding Sound Levels and Decibels
The sound level, intensity level, or decibel level is a measure of the relative intensity of a sound. It is formulated as follows: $\beta=10\log\left(\frac{I}{I_0}\right)$ where $I_0=1\times 10^{-12}\,\rm W/m^2$ is the sound intensity at the hearing threshold, commonly referred to as the reference intensity.
$I$ is the intensity of a sound (measured in $\rm W/m^2$), and $\beta$ is the corresponding sound level measured in decibels (dB).
## Intensity level and decibels: Problems
Problem (1): The intensity of a sound is $1.5\times 10^{-6}\,\rm W/m^2$.
(a) What is the intensity level of that sound?
(b) If the sound intensity is doubled, what is the new sound intensity level?
Solution:
(a) In most intensity-level problems, we are given intensity and asked to find its corresponding intensity level. It’s essential to understand that the intensity level involves a logarithmic comparison between a given sound intensity ($I$) and the reference sound intensity ($I_0$). We then multiply the result by ten. Here’s how it works: \begin{align*} \beta &=10\log\left(\frac{I}{I_0}\right) \\\\ &=10\log\left(\frac{1.5\times 10^{-7}}{10^{-12}}\right) \\\\ &= 10\log(1.5\times 10^5) \\\\ &= 51.7\,\rm dB \end{align*} According to the table of intensity levels for common sounds, this sound falls into the “quiet” category.
(b) The intensity is doubled $I'=2I$, so the sound level of the new intensity becomes \begin{align*} \beta &=10\log\left(\frac{I'}{I_0}\right) \\\\ &=10\log\left(\frac{2\times 1.5\times 10^{-7}}{10^{-12}}\right) \\\\ &= 10\log(3\times 10^5) \\\\ &= 54.7\,\rm dB \end{align*} As you can see, the intensity of the sound is doubled but the corresponding sound level slightly changes. Consequently, a doubling of intensity corresponds to an increase of $3\,\rm dB$ in sound intensity level.
Isn’t this content as valuable as a $10 private class? Please support me here. I worked hard to prepare this article. Problem (2): The pain level for humans has an intensity level of$120\,\rm dB$. (a) What is the intensity of such sound in$\rm W/m^2$? (b) Compare that with a sound level of a whisper at$20\,\rm dB$. Solution: another typical question on decibels is when the sound level is given and asks us to find the intensity corresponding to that level. (a) In this case, substitute the intensity level$120\,\rm dB$into the decibel formula and divide both sides by$10$. \begin{gather*} \beta=10\log\left(\frac{I}{I_0}\right) \\\\ 120 =10\log\left(\frac{I}{10^{-12}}\right) \\\\ 12=\log \left(\frac{I}{10^{-12}}\right) \\\\ 10^{12}=\frac{I}{10^{-12}} \\\\ \rightarrow \quad \boxed{I=1\,\rm W/m^2} \end{gather*} In above, we used the definition of logarithms as $a=\log b \longleftrightarrow b=10^a$ Note that$\rm 1\, W/m^2$is the loudest sound that human ear can tolerate without being hurt. It’s known as the threshold of pain and is similar to what you might experience at a rock concert. (b) This part is also very common in all problems. Here, the level of two sounds is given and wants the ratio of their corresponding intensities. In such cases, the best thing to do is to use the following formula in which the reference level has been removed. $\beta_2-\beta_1=\log\frac{I_2}{I_1}$ Here,$\beta_1=120\,\rm dB$and$\beta_1=20\,\rm dB$. Substitute these known values into the above formula and solve for the ratio of intensities \begin{gather*} 120-20=10\log{\frac{I_{120}}{I_{20}}} \\\\ 100=10\log{\frac{I_{120}}{I_{20}}} \\\\ 10^{10}=\frac{I_{120}}{I_{20}} \\\\ \rightarrow \quad \boxed{I_{120}=10^{10} I_{20}} \end{gather*} This result is roughly logical since a$20\,\rm dB$sound corresponds to the sound of rustling leaves which is barely audible. On the other hand, a$120\,\rm dB$sound is perceived when you are at a distance of$60\,\rm m$from a jet engine during takeoff. Now, compare these two situations! Problem (3): The intensity of a sound is tripled. By how many decibels does it increase? Solution: In this question, we are not given specific information about the magnitude of the sound intensity. However, we can assume an original intensity ($I$) and a new intensity ($I_{new} = kI$), where ($k$) represents the multiple indicating how much the sound intensity changed. Substitute this into the decibel formula and apply the logarithms relation$\log(ab)=\log a+\log bto reach the following result \begin{align*} \beta_{new}&=10 \log\left(\frac{I_{new}}{I_0}\right) \\\\ &=10\log\left(\frac{k\times I}{I_0}\right) \\\\ &=10\log k+ 10\log\left(\frac{I}{I_0}\right) \\\\ &=10\log k+\beta \end{align*} Consequently, the change in the sound level is given by: $\Delta \beta =10\log k$ This result is general and can be memorized for faster solving, especially in MCAT tests. Here,k=3$so the change in the sound intensity becomes: $\Delta \beta=10\log 3=4.77\rm dB$ This demonstrates that tripling the sound intensity corresponds to an increase of nearly$4.77\,\rm dB$in the intensity level. Problem (4): By what factor will the intensity change for a$\pm 6\,\rm dB$change in the out level of a loudspeaker? Solution: The change in sound level is given as$\beta_2-\beta_1=6\,\rm dB$, and the change in the intensity is asked us. Recall that the difference in intensity (sound) level is related to the change in the intensity as $\beta_2-\beta_1=10\log\frac{I_2}{I_1}$ Substituting the numerical values into that and solved for the fraction$\frac{I_2}{I_1}$. \begin{gather*} \Delta \beta=10\log\frac{I_2}{I_1} \\\\ 6=10\log\frac{I_2}{I_1} \\\\ \Rightarrow \frac{I_2}{I_1}=10^{0.6}=4 \end{gather*} Where in the last line we used the fact that $\log b=a \longleftrightarrow b=10^a$ Therefore, a$\pm 6\,\rm dB$change corresponds to four times the intensity of the sound. Problem (5): We are standing a certain distance from four noisy juicers and hear a sound level of$77\,\rm dB$in a juice shop. What sound level would this person experience if three of them are turned off? Solution: Assume each juicer produces a sound intensity$I=?$and a sound level$\beta=75\,\rm dB. First of all, find the sound intensity that four equally noisy juicers produce using the decibel formula below \begin{gather*} \beta=10\log\left(\frac{I}{I_0}\right) \\\\ 77 =10\log\left(\frac{I}{10^{-12}}\right) \\\\ 7.5=\log \left(\frac{I}{10^{-12}}\right) \\\\ 10^{7.7}=\frac{I}{10^{-12}} \\\\ \rightarrow \quad \boxed{I=5.01\times 10^{-5}\,\rm W/m^2} \end{gather*} This is the intensity that all four juicers produce simultaneously. Dividing that by the number of juicers gives the sound intensity of one juicer as $I_{one}=\frac{I}{4}=1.25\times 10^{-5}\,\rm W/m^2$ Now that the intensity of a single juicer has been determined, again use the sound level formula and find the decibel level for a single machine \begin{align*} \beta&=10\log\left(\frac{I}{I_0}\right) \\\\ &=10\log\left(\frac{1.25\times 10^{-5}}{10^{-12}}\right) \\\\ &=10\log \left(1.25\times 10^7 \right) \\\\ &=10(\log 1.25+\log 10^7) \\\\ &=0.9+70 \\\\ \rightarrow \quad \beta_1&=\boxed{70.9\,\rm dB} \end{align*} We used the following useful rule of logarithms for the last line. $\log(ab)=\log a+\log b$ This explicitly shows the difference between sound intensity and intensity level. The sound intensity of four machines running at the same time is four times the intensity of a single machine, but their sound level is nearly6$decibels less than a single machine. Problem (6): The sound level near an operating air conditioner is$80\,\rm dB$. What would be the sound level of two of them operating side by side at the same distance? Solution: The decibel level of a machine is given as$I_1=80\,\rm dB$and wants the sound level of two such machines operating simultaneously$I$. We can't simply add them and say the total sound level would have been$160\,\rm dB$. This corresponds to the decibel level of a jet engine at a near distance. Thus, that conclusion is not logical. Keep in mind that sound intensity is directly related to the energy that a typical sound can have. Thus, adding two similar machines leads to a doubling of energy spread in the environment. Thus, substitute$I=2I_1into the decibel formula \begin{align*} \beta&=10\log\left(\frac{I}{I_0}\right) \\\\ &=10\log\left(\frac{2I}{10^{-12}}\right) \\\\ &=10 \log{\left(2\times \frac{I}{10^{-12}}\right)} \\\\ &=10(\log{2})+10\left(\log \frac{I}{10^{-12}}\right) \\\\ &=10(0.3)+\beta_1 \\\\ &=10(0.3)+80=\boxed{83\,\rm dB}\end{align*} In the fourth line, we used one of the fundamental relations for logarithms $\log(ab)=\log a+ \log b$ and set\log 2=0.3$. Recall that$\beta_1$is the sound level for a single operating air conditioner. Problem (7): A person standing$1\,\rm m$from a loudspeaker hears its sound at an intensity of$9\times 10^{-3}\,\rm W/m^2$. (a) Find its corresponding sound level for this intensity. (b) Assuming sound propagates from the source as a spherical wave, calculate the sound intensity at a distance of$45\,\rm cmfrom the person and its corresponding decibel level. Solution: (a) As before, we have \begin{align*} \beta&=10\log\left(\frac{9\times 10^{-3}}{10^{-12}}\right) \\\\ &=10\log(9\times 10^9) \\\\ &=10\log 9+10\log 10^9 \\\\ &=9.5+90=\boxed{99.5\,\rm dB} \end{align*} Here, we set\log 9=0.95$and used the logarithm relation $\log 10^a=a$ (b) The intensity of a spherical sound wave produced by a point source at distance$r$is calculated by the following formula: $I=\frac{P_{av}}{4\pi r^2}$ where$P_{av}$represents the average power emitted by the source. Usually, in all sound-level problems, the average power emitted by a source remains constant. The only thing that changes inversely is the intensity at different locations,$I\propto \frac{1}{r^2}$. Given the intensity at a distance$r_1=1\,\rm m$as$I_1$, we need to find it at a distance$r_2=45\,\rm cm$. Constructing the ratio of intensities at these locations, we have: \begin{gather*} \frac{I_2}{I_1}=\left(\frac{r_1}{r_2}\right)^2 \\\\ \frac{I_2}{9\times 10^{-3}}=\left(\frac{1}{0.45}\right)^2 \\\\ \frac{I}{9\times 10^{-3}}=4.9 \\\\ \rightarrow \quad \boxed{I=44.1\times 10^{-3}\,\rm W/m^2} \end{gather*} As you can see, as we get closer to the loudspeaker, the intensity increases, which is logical. The sound level for this intensity is found to be$\beta=106.4\,\rm dB$. When we move closer to the source, the energy we receive increases fivefold, but the sound level only changes by$6.9\,\rm dB$. As a side note, let’s derive the difference in decibel levels$\beta_1$and$\beta_2$of a sound source at two different distances from it. Assuming a point source and the emitted wave propagating spherically, at distance$r, the intensity and its corresponding sound level are given by the following formula: $I=\frac{P_{av}}{4\pi r^2} \ , \ \beta=10\log\left(\frac{I}{I_0}\right)$ The difference in decibel levels are computed as: \begin{align*} \beta_2-\beta_1&=10\log\left(\frac{I_2}{I_0}\right)-10\log\left(\frac{I_1}{I_0}\right) \\\\ &=10 \log\left(\frac{I_2}{I_1}\right) \\\\ &=10\log\left(\frac{r_1}{r_2}\right)^2 \end{align*} In above, the following famous logarithms relation has been used $\log a-\log b=\log\frac{a}{b}$ For the last line recall thatI\propto \frac{1}{r^2}$. Now, use the rule$\log 10^n=nto find the final result as below $\beta_2-\beta_1=20\log\left(\frac{r_1}{r_2}\right)$ Now we are in a position to find the change in sound levels for the previous problem. \begin{align*} \beta_2-\beta_1&=20\log \left(\frac{r_1}{r_2}\right) \\\\ &=20\left(\frac{1}{0.45}\right) \\\\ &=6.9 \end{align*} Exactly matches the result obtained directly by calculating the intensities. Problem (8): If we made the amplitude of a sound wave4.2$times greater then, (a) By what factor will the intensity change? (b) By how many decibels will the sound level change? Solution: the only given data is the ratio of amplitudes,$A_2=4.2 A_1$. The intensity$I$of a wave is directly proportional to the square of the wave amplitude,$I\propto A^2. (a) Forming the ratio of intensities gives us the desired increase in intensity \begin{align*} \frac{I_2}{I_1}&=\left(\frac{A_2}{A_1}\right) \\\\ &=(4.2)^2 \\\\ \rightarrow I_2&=17.64 I_1 \end{align*} Thus, the sound intensity increases by a factor of17.64. (b) Given the ratio of intensities, we can find the change in the decibel level as below $\beta_2-\beta_1=10\log\left(\frac{I_2}{I_1}\right)$ Substituting the numerical values into that, we will have \begin{align*} \beta_2-\beta_1&=10\log 17.64 \\ &=12.46 \end{align*} Consequently, the sound level increases by12.46\,\rm dB\$.
## Summary:
In this article, you gained some insight into how to solve sound or intensity level problems. All questions are answered by the following formulas \begin{gather*} \beta=10\log\left(\frac{I}{I_0}\right) \\\\ I=\frac{P_{av}}{4\pi r^2} \\\\ \frac{I_2}{I_1}=\left(\frac{r_1}{r_2}\right)^2 \\\\ \beta_2-\beta_1=20\log \left(\frac{r_1}{r_2}\right) \end{gather*}
Author: Dr. Ali Nemati
Published: Nov. 8, 2022
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Work-Kinetic Energy Theorem with the Derivation of mathematical expression
Work-Kinetic Energy Theorem with derivation: In this post, we will discuss the special relationship between work done on an object and the resulting kinetic energy of the object and come up with the statement of the work-kinetic energy theorem. We will also see how to derive the equation of the work-kinetic energy theorem.
Everyday experience supports this theorem. If we see a football at rest on the ground and a moment later we see it hurtling through the air, we would conclude that someone did work on the ball, by exerting a large force over a short distance, to make it move.
This correct conclusion illustrates how doing work on an object gives the object increased velocity or kinetic energy.
State work-kinetic energy theorem
The special relationship between doing work on an object and the resulting kinetic energy of the object is called the work-kinetic energy theorem.
Derivation of the mathematical expression for work-kinetic energy theorem | equation derivation
To develop a mathematical expression that relates work to the kinetic energy, we’ll assume that all of the work done on a system gives the system kinetic energy only.
We will start with the definition of work and then apply Newton’s second law.
To avoid dealing with advanced mathematics, we are assuming that a constant force gives the system a constant acceleration so that you can use the equations of motion.
Since work and kinetic energy are scalar quantities, vector notation will be omitted from the derivation. This is valid as long as the directions of the force and displacement are parallel and the object is moving in a straight line.
Let’s assume that the force is constant and write the equation
for work. W = FΔd …………(1)
Using Newton’s second law, F = ma
Assume the force and the acceleration are parallel to the direction of the displacement and motion is in one direction. Then omit vector symbols and use F = ma ……. (2)
Let’s Substitute ma for F in the equation for work. W = maΔd…..(3)
Now we will use the equation of acceleration for uniformly accelerated motion. a = (v2 − v1)/Δt ……… (4)
From equation (3) and (4)
W = m [(v2 − v1)/Δt] Δd
=> W = m (v2 − v1) [Δd/Δt] ………………. (5)
Now the equation for displacement for uniformly accelerated
motion. Δd = [(v1 + v2)/2] Δt …………………(6)
Now, Divide both sides of the equation (6) by Δt to obtain
an equation for Δd/Δt.
Δd/Δt = (v1 + v2)/2…………………..(7)
Rewrite the equation (5) for work by substituting the value for Δd/Δt.
W = m(v2 − v1)(v1 + v2)/2
W = (1/2)m(v22 − v12)………………..(8)
This makes us conclude that work-done on an object results in a change in the kinetic energy of the object.
W = Ek2 – Ek1
W = ΔEk (mathematical expression or equation for the work-kinetic energy theorem is derived here). ………………. (9)
This expression for the work-kinetic energy theorem above shows the special relationship between work done on an object and the resulting kinetic energy of the object.
Summary | Take Away | Suggested reading
So in this post, we have covered the statement of the work-kinetic energy theorem and also derived its expression or equation.
Now for the related study, we suggest the following posts for our readers.
Derivation of Kinetic energy equation
Relationship between momentum and Kinetic energy – equation derivation
Work done by a force
Work-Kinetic Energy Theorem with the Derivation of mathematical expression
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# Solving systems by graphing calculator
There's a tool out there that can help make Solving systems by graphing calculator easier and faster We can solve math word problems.
## Solve systems by graphing calculator
In this blog post, we will explore one method of Solving systems by graphing calculator. To find the domain and range of a given function, we can use a graph. For example, consider the function f(x) = 2x + 1. We can plot this function on a coordinate plane: As we can see, the function produces valid y-values for all real numbers x. Therefore, the domain of this function is all real numbers. The range of this function is also all real numbers, since the function produces valid y-values for all real numbers x. To find the domain and range of a given function, we simply need to examine its graph and look for any restrictions on the input (domain) or output (range).
Trigonometry is the branch of mathematics that deals with the relations between the sides and angles of triangles. The basic concepts of trigonometry can be applied to solving problems in other areas of mathematics, such as calculus and geometry. Trigonometry is also a useful tool in physics and engineering. In physics, trigonometry is used to calculate the properties of waves, such as their frequency, wavelength, and amplitude. In engineering, trigonometry is used to design buildings, bridges, and other structures. Trigonometry can also be used to solve problems in everyday life, such as finding the height of a tree or the distance to a nearby city.
Let's look at each type. State-Dependent Differential Equations: These equations describe how one variable changes when another variable changes. For example, consider a person whose height is measured at one time and again at a later time. If their height has increased, then it can be said that their height has changed because the value of their height changed. Value-Dependent Differential Equations: These equations describe how one variable changes when another variable's value changes. Consider a stock whose price has increased from \$10 to \$20 per share. If this increase can be represented by a change in value, then it can be said that the price has changed because the value of the stock changed. Solving state-dependent differential equations is similar to solving linear algebra problems because you're solving for one variable (the state) when another variable's value changes (if another variable's value is known). Solving value-dependent differential equations is similar to solving quadratic equations because you're solving for one variable (the state) when another
The angles are all 60 degrees, and the slope is 6, so it can be written as The solution to this system is therefore Note that this is not mathematically correct; you should only use this as an approximate solution when solving for small values such as 0.1 or 0.01. For more information about solving 3x3 linear systems, see Linear Systems and Quadratic Equations.
## We cover all types of math problems
I love it, best wat to check your work or even show you wat to do if you don't understand what’s going on. When taking pictures I have no problems, my friend has problems because he has a low pixel’s camera, the more your pixels the better the photo. I love the app and would recommend it to everyone out there struggling. Five stars from me
Isabell Henderson
I love this app a lot! It helps me with school and stuff I don't understand. If you have problems in math, I highly suggest this app! It explains everything to you and the calculator has a lot of stuff too. Highly recommend it! Thank you!!💕
Isobel Miller
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# 5.1 Even and odd functions
Page 1 / 3
Even and odd functions are related to symmetry of functions. The symmetry of a function is visualized by the planar plot of a function, which may show symmetry with respect to either an axis (y-axis) or origin.
Since functions need not always be symmetric, they may neither be even nor be odd. The parity of a function i.e. whether it is even or odd is determined with certain algebraic algorithm. Further, symmetry of functions may change subsequent to mathematical operations.
## Even functions
The values of even function at x=x and x=-x are same.
Even function
A function f(x) is said to be “even” if for every “x”, there exists “-x” in the domain of the function such that :
$f\left(-x\right)=f\left(x\right)$
An even function is symmetric about y-axis. If we consider the axis as a mirror, then the plot in first quadrant has its mirror image (bilaterally inverted) in second quadrant. Similarly, the plot in fourth quadrant has its mirror image (bilaterally inverted) in third quadrant.
Some examples of even functions are ${x}^{2},|x|\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}x$ . In each case, we see that :
$⇒f\left(-x\right)={\left(-x\right)}^{2}={x}^{2}=f\left(x\right)$
$⇒f\left(-x\right)=|-x|=|x|=f\left(x\right)$
$⇒f\left(-x\right)=\mathrm{cos}\left(-x\right)=\mathrm{cos}x=f\left(x\right)$
The right side is mirror image of left hand side and the left side is mirror image of right hand side of the curve.
It is important to see that if we rotate the curve by 180° about y-axis, then the appearance of the rotated curve is same as the original curve. We can state this alternatively as : if we rotate left hand side of the curve by 180° about y-axis, then we get the right hand curve and vice-versa.
## Examples
Problem 1: Prove that the function f(x) is “even”, if
$f\left(x\right)=x\frac{{a}^{x}-1}{{a}^{x}+1}$
Solution : For function being “even”, we need to prove that :
$f\left(-x\right)=f\left(x\right)$
Here,
$⇒f\left(-x\right)=-x\frac{{a}^{-x}-1}{{a}^{-x}+1}=-x\frac{\frac{1}{{a}^{x}}-1}{\frac{1}{{a}^{x}}+1}$
$⇒f\left(-x\right)=-x\frac{\frac{1-{a}^{x}}{{a}^{x}}}{\frac{1+{a}^{x}}{{a}^{x}}}=-x\frac{1-{a}^{x}}{1+{a}^{x}}$
$⇒f\left(-x\right)=x\frac{{a}^{x}-1}{{a}^{x}+1}=f\left(x\right)$
Problem 2: If an even function “f” is defined on the interval (-5,5), then find the real values for which
$f\left(x\right)=f\left(\frac{x+1}{x+2}\right)$
Solution : It is given that function “f” is even. Hence, arguments of the functions on two sides are related either as
$⇒x=\frac{x+1}{x+2}$
or as :
$⇒x=-\frac{x+1}{x+2}$
From the first relation,
$⇒{x}^{2}+x-1=0$
$⇒x=\frac{-1±\sqrt{5}}{2}$
From the second relation,
$⇒{x}^{2}+3x+1=0$
$⇒x=\frac{-3±\sqrt{5}}{2}$
We see that values are within the specified domain. Hence, all the four solutions satisfy the given equation.
## Odd functions
The values of odd function at x=x and x=-x are equal in magnitude but opposite in sign.
Odd function
A function f(x) is said to be “odd” if for every “x”, there exists “-x” in the domain of the function such that :
$f\left(-x\right)=-f\left(x\right)$
An odd function is symmetric about origin of the coordinate system. The plot in first quadrant has its mirror image (bilaterally inverted) in third quadrant. Similarly, the plot in second quadrant has its mirror image (bilaterally inverted) in fourth quadrant.
Some examples of odd functions are : $x,{x}^{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{sin}x$ . In each case, we see that :
$⇒f\left(-x\right)=-x=-f\left(x\right)$
$⇒f\left(-x\right)={\left(-x\right)}^{3}=-{x}^{3}=-f\left(x\right)$
$⇒f\left(-x\right)=\mathrm{sin}\left(-x\right)=-\mathrm{sin}x=-f\left(x\right)$
The upper curve of these functions is exactly same as the lower curve across x-axis.
It is important to see that if we rotate the curve by 180° about origin, then the appearance of the rotated curve is same as the original curve. In other words, if we rotate right hand side of curve by 180° about origin, then we get left side of the curve. Further, it is interesting to note that we obtain left hand part of the plot of odd function in two steps : (i) drawing reflection (mirror image) of right hand plot about y-axis and (ii) drawing reflection (mirror image) of “reflection drawn in step 1” about x-axis.
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
What is power set
Period of sin^6 3x+ cos^6 3x
Period of sin^6 3x+ cos^6 3x
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# Solving Linear Inequalities
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Linear inequalities is the inequalities where the degree of the variables is one. An inequality equation can contain variables, constants and exponents for the variables. Inequality is the equation which has less than or greater than symbols in it i.e. '<' and '>' respectively There are different mathematical operations that can be used in an inequality like addition, subtraction, multiplication and division.
Example 1: Find the solution of the inequality equation 7 x – 1 > 20?
Solution: Given is the inequality equation with one unknown variables x.
7 x - 1 > 20 is a linear inequality with greater than sign.
Add 1 on both sided of the equation.
This gives, 7 x - 1 + 1 > 10 + 1; 7 x > 21;
Now divide the inequality by 7 on both sides of the equation.
Therefore, 7 x/ 7 > 21 /7; x > 3.
Hence the solution to the linear inequality is x > 3.
Example 2: Find the solution of the inequality equation 55 x - 16 < 94?
Solution: Given is the inequality equation with one unknown variables x.
55 x - 16 < 94 is a linear inequality with lesser than sign.
Add 16 on both sided of the equation.
This gives, 55 x - 16 + 16 < 94 + 16; 55 x < 110;
Now divide the inequality by 55 on both sides of the equation.
Therefore, 55 x/ 55 < 110 /55; x < 2.
Hence the solution to the linear inequality is x < 2.
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# A line segment has endpoints at (7 ,9 ) and (5 ,2). If the line segment is rotated about the origin by pi , translated vertically by -4 , and reflected about the x-axis, what will the line segment's new endpoints be?
Apr 7, 2017
$\left(- 7 , 13\right) \text{ and } \left(- 5 , 6\right)$
#### Explanation:
Since there are 3 transformations to be performed, label the endpoints A(7 ,9) and B(5 ,2)
$\textcolor{b l u e}{\text{first transformation" " Under a rotation about origin of }} \pi$
$\text{a point } \left(x , y\right) \to \left(- x , - y\right)$
$\text{Hence } A \left(7 , 9\right) \to A ' \left(- 7 , - 9\right)$
$\text{and } B \left(5 , 2\right) \to B ' \left(- 5 , - 2\right)$
$\textcolor{b l u e}{\text{Second transformation" " Under a translation }} \left(\begin{matrix}0 \\ - 4\end{matrix}\right)$
$\text{a point } \left(x , y\right) \to \left(x , y - 4\right)$
$\text{Hence } A ' \left(- 7 , - 9\right) \to A ' ' \left(- 7 , - 13\right)$
$\text{and } B ' \left(- 5 , - 2\right) \to B ' ' \left(- 5 , - 6\right)$
$\textcolor{b l u e}{\text{third transformation"" Under a reflection in the x-axis}}$
$\text{a point } \left(x , y\right) \to \left(x , - y\right)$
$\text{Hence } A ' ' \left(- 7 , - 13\right) \to A ' ' ' \left(- 7 , 13\right)$
$\text{and } B ' ' \left(- 5 , - 6\right) \to B ' ' ' \left(- 5 , 6\right)$
$\text{After all 3 transformations}$
$\left(7 , 9\right) \to \left(- 7 , 13\right) \text{ and } \left(5 , 2\right) \to \left(- 5 , 6\right)$
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How do you find the greatest common factor for the list of terms 70x^2, 30x^6, 60x^9?
Feb 12, 2017
See the entire solution process below:
Explanation:
To find the Greatest Common Factor of these three terms we must first factor each term separately:
$70 {x}^{2} = 2 \times 5 \times 7 \times x \times x$
$30 {x}^{6} = 2 \times 5 \times 3 \times x \times x \times x \times x \times x \times x$
$90 {x}^{9} = 2 \times 5 \times 3 \times 3 \times x \times x \times x \times x \times x \times x \times x \times x \times x$
The Common Factors are:
$70 {x}^{2} = \textcolor{red}{2 \times 5} \times 7 \times \textcolor{red}{x \times x}$
$30 {x}^{6} = \textcolor{red}{2 \times 5} \times 3 \times \textcolor{red}{x \times x} \times x \times x \times x \times x$
$90 {x}^{9} = \textcolor{red}{2 \times 5} \times 3 \times 3 \times \textcolor{red}{x \times x} \times x \times x \times x \times x \times x \times x \times x$
Therefore the GCG (Greatest Common Factor) is:
$\textcolor{red}{2 \times 5 \times} \textcolor{red}{x \times x} = 10 {x}^{2}$
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# Arithmetic series solver
We'll provide some tips to help you select the best Arithmetic series solver for your needs. Our website will give you answers to homework.
## The Best Arithmetic series solver
In this blog post, we will be discussing about Arithmetic series solver. This can be a useful tool for solving problems in physics or engineering, where you might need to find the total amount of energy in a system, for example. There are a variety of different methods that can be used to solve series, and the choice of method will depend on the particular problem you are trying to solve. However, some of the most popular methods include the Euler-Maclaurin formula and the Ricci identity. With a little practice, you should be able to use a series solver to solve a wide range of problems.
Cosine is a trigonometric function that takes an angle, in radians, and returns a number. The cosine of an angle is calculated by taking the sine of the angle and then subtracting 1. In other words, the cosine is the inverse of the sine. There are two main ways to solve cosine: using tables or using rules. Using tables, first find the expression ƒ sin(θ) - 1 = 0 where ƒ is any number. That expression is called a cosine table. Then find the corresponding expression ƒcos(θ) = -1. The answer to that sum is the cosine of θ. Using rules, first find the expression ƒsin(θ) = -1. Then add 1/2 to that expression to get ƒ + 1/2 = -1 + 1/2 = -1 + 3/4 = -1 + 7/8 = -1 + 13/16 = -1 + 27/32 = -1 + 41/64 = ... The answer to those sums will be the cosine of θ.
A parabola is a two-dimensional figure that appears in many mathematical and physical situations. In mathematics, a parabola is defined as a curve where any point is equidistant from a fixed point (called the focus) and a fixed line (called the directrix). In physics, parabolas describe the path of objects under the influence of gravity, such as a ball thrown in the air. In both cases, the equation for a parabola can be quite complicated. However, there are online tools that can help to solve these equations quickly and easily. One such tool is the Parabola Solver, which allows users to input the parameters of their equation and then receive step-by-step instructions for finding the solution. This tool can be an invaluable resource for students and professionals who need to solve complex parabolic equations.
Algebra is a fundamental tool in mathematics that allows us to solve equations and understand the relationships between variables. Basic algebra problems usually involve solving for a missing variable in an equation. For example, if we are given the equation "x + 5 = 10", we can solve for "x" by subtracting 5 from both sides of the equation to get "x = 5". Algebra can be used to solve for many different types of variables, making it a very powerful tool.
A binomial solver is a math tool that helps solve equations with two terms. This type of equation is also known as a quadratic equation. The solver will usually ask for the coefficients of the equation, which are the numbers in front of the x terms. It will also ask for the constants, which are the numbers not attached to an x. With this information, the solver can find the roots, or solutions, to the equation. The roots tell where the line intersects the x-axis on a graph. There are two roots because there are two values of x that make the equation true. To find these roots, the solver will use one of several methods, such as factoring or completing the square. Each method has its own set of steps, but all require some algebraic manipulation. The binomial solver can help take care of these steps so that you can focus on understanding the concept behind solving quadratic equations.
## We solve all types of math problems
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Ireland Rogers
It has almost reached its potential it’s so good. This app helps me so much, it’s basically like a calculator but more complex and at the same time easier to use - all you have to do is literally point the camera at the equation and normally solves it well! Although it comes up with some mistakes and a few answers I'm not always looking for, it is really useful and not a waste of your time! I'm in 8th grade and I use it for my homework sometimes ●; ¡D
Michelle Powell
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Below is a list of formulas and math facts you’ll need if you’re looking to ace SAT math. A lot of it should be old hat for you at this point—after all, you study this stuff in school, right?—but it’s definitely handy to have it all in one place for reference, and there are some nifty shortcuts sprinkled through the list that you might not be aware of yet, too.
For more detailed information about the math formulas you’ll need to know for the SAT, be sure to take a look at Magoosh’s SAT Math Formula eBook!
Arithmetic and Number Properties
Order of Operations
The Associative Property:
The Distributive Property:
Fast Fractions
For example:
Exponents and Roots
Laws for Combining:
1 and 0 as bases:
• 1 raised to any power is 1. 0 raised to any nonzero power is 0
• Any nonzero number to the power of 0 is 1, e.g.
Fractions as exponents:
Negative exponents:
Negative bases:
• With an even exponent: positive result
• With an odd exponent: negative result
Roots
Estimating roots:
To estimate square roots of numbers that aren’t perfect squares, just examine the nearby perfect squares. For example, to find , you know that and , so find must be between 7 and 8.
Cube roots:
a number that, when cubed, equals n. E.g.:
Simplifying roots:
Separate the number into its prime factors, and take out matching pairs. E.g.:
Roots can be added like variables. E.g.:
Prime Numbers and Integers
• 1 is not a prime.
• 2 is the smallest prime and the only even prime.
• An integer is any counting number including negative numbers (e.g. -3, -1, 2, 7…but not 2.5)
Ratios
Ratios let us compare the proportions of two quantities.
If there is a 2:5 ratio of boys to girls at a school, that means that for every 5 girls, there are 2 boys. So there could be 2 boys and 5 girls, 20 boys and 50 girls, 200 boys and 500 girls, etc.
Ratios are given by x:y, x to y, or x/y. If a question says “for every x there is/are a y,” you are most likely dealing with a ratio question. Ratios can also be x:y:z.
Ratios can be simplified like fractions. 3:6 is the same as 1:2.
Remember that if there is a 2:5 ratio of boys to girls at a school, the ratio of boys to total students is 2:(5 + 2) = 2:7. So 2/7 of the students are boys.
Percentages
The Meaning of “Percent”
Calculating Percentages
Percent Change
Divisibility
3 : sum of digits divisible by 3
4 : the last two digits of number are divisible by 4
5 : the last digit is either a 5 or zero
6 : even number and sum of digits is divisible by 3
8 : if the last three digits are divisible by 8
9: sum of digits is divisible by 9
Algebra
FOIL
First, Outer, Inner, Last:
Common patterns to memorize:
Cross-Multiplication
For , where a is not 0, if you can factor it to , then the solutions are -y and z. For example:
or
Geometry
Angles
A right angle is made up of 90 degrees
A straight line is made up of 180 degrees.
If two lines intersect, the sum of the resulting four angles equals 360
Circles
Triangles
General:
Side A – Side B < Side C
Side A + Side B > Side C
Right Triangles:
where c is the hypotenuse.
A 45˚-45˚-90˚ triangle has sides in a ratio of x : x : x√2, with x√2 as the hypotenuse
A 30˚-60˚-90˚ triangle has sides in a ratio of x : x√3 : 2x, with the 1x side opposite the 30 degree angle and 2x as the hypotenuse
Certain right triangles have sides with all integer lengths. These sets of numbers are called Pythagorean triples, and you should memorize some of them: 3-4-5, 5-12-13, and 8-15-17. A multiple of a Pythagorean triple is also a Pythagorean triple (e.g., 6-8-10).
Squares
, where s = side
Rectangles
, where l = length and w = width
Trapezoids
Polygons
, where n = # of sides
Coordinate Geometry
Slope-Intercept Form
, where m is the slope and b is the y-intercept.
Slope
The Distance Formula
Word Problems
Sequences
Probability
Averages
Mean:
Median:
The middlemost value when numbers are arranged in ascending order; for an even count of numbers, take the average of the middle two
Mode:
The number that occurs most frequently in a list or set
Distance, Rate, and Time
,
Counting
Combination:
When the order does not matter—for example, picking any 3 friends from a group of 5.
Permutation:
When the order does matter—for example, how many ways you could order 3 letters from the word PARTY?
Improve your SAT or ACT score, guaranteed. Start your 1 Week Free Trial of Magoosh SAT Prep or your 1 Week Free Trial of Magoosh ACT Prep today!
Lucas is the teacher behind Magoosh TOEFL. He’s been teaching TOEFL preparation and more general English since 2009, and the SAT since 2008. Between his time at Bard College and teaching abroad, he has studied Japanese, Czech, and Korean. None of them come in handy, nowadays.
3 Responses to “Important SAT Math Facts”
1. Chris says:
Under ‘Averages’ you mislabeled Mean for Mode
• Lucas says:
You’re right! I fixed it. Thanks!
2. allison says:
oh my gosh ive been worrying about the sat for a while now (im not the best math student) and it was as huge relief to look at this list and recognize whats going on. I feel a little better now
Magoosh blog comment policy: To create the best experience for our readers, we will approve and respond to comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! :) If your comment was not approved, it likely did not adhere to these guidelines. If you are a Premium Magoosh student and would like more personalized service, you can use the Help tab on the Magoosh dashboard. Thanks!
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# How do you solve x-y+2z = -5, -x +3z = 0, and 2x+ y = 1 using matrices?
Jan 11, 2017
The answer is $\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- \frac{12}{11} \\ \frac{35}{11} \\ - \frac{4}{11}\end{matrix}\right)$
#### Explanation:
Rewrite the equation in matrix form
$\left(\begin{matrix}1 & - 1 & 2 \\ - 1 & 0 & 3 \\ 2 & 1 & 0\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- 5 \\ 0 \\ 1\end{matrix}\right)$
Let $A = \left(\begin{matrix}1 & - 1 & 2 \\ - 1 & 0 & 3 \\ 2 & 1 & 0\end{matrix}\right)$
We must find the inverse of matrix $A$
Let's calculate the
$D e t A$ $= | \left(1 , - 1 , 2\right) , \left(- 1 , 0 , 3\right) , \left(2 , 1 , 0\right) |$
$= 1 \cdot | \left(0 , 3\right) , \left(1 , 0\right) | + 1 \cdot | \left(- 1 , 3\right) , \left(2 , 0\right) | + 2 \cdot | \left(- 1 , 0\right) , \left(2 , 1\right) |$
$= - 3 - 6 - 2 = - 11$
As $D e t A \ne 0$, matrix A is invertible
Now, we calculte the matrix of co-factors
$C = \left(\begin{matrix}| \left(0 3\right) & \left(1 0\right) | & - | \left(- 1 3\right) & \left(2 0\right) | & | \left(- 1 0\right) & \left(2 1\right) | \\ - | \left(- 1 2\right) & \left(1 0\right) | & | \left(1 2\right) & \left(2 0\right) | & - | \left(1 - 1\right) & \left(2 1\right) | \\ | \left(- 1 2\right) & \left(0 3\right) | & - | \left(1 2\right) & \left(- 1 3\right) | & | \left(1 - 1\right) & \left(- 1 0\right) |\end{matrix}\right)$
$= \left(\begin{matrix}- 3 & 6 & - 1 \\ 2 & - 4 & - 3 \\ - 3 & - 5 & - 1\end{matrix}\right)$
We calculate the transpose of matrix $C$
${C}^{T} = \left(\begin{matrix}- 3 & 2 & - 3 \\ 6 & - 4 & - 5 \\ - 1 & - 3 & - 1\end{matrix}\right)$
The inverse is
${A}^{-} 1 = {C}^{T} / \det A = - \frac{1}{11} \cdot \left(\begin{matrix}- 3 & 2 & - 3 \\ 6 & - 4 & - 5 \\ - 1 & - 3 & - 1\end{matrix}\right)$
$= \left(\begin{matrix}\frac{3}{11} & - \frac{2}{11} & \frac{3}{11} \\ - \frac{6}{11} & \frac{4}{11} & \frac{5}{11} \\ \frac{1}{11} & \frac{3}{11} & \frac{1}{11}\end{matrix}\right)$
Verification, by doing $A \cdot {A}^{-} 1$
$A \cdot {A}^{-} 1 = \left(\begin{matrix}1 & - 1 & 2 \\ - 1 & 0 & 3 \\ 2 & 1 & 0\end{matrix}\right) \cdot \left(\begin{matrix}\frac{3}{11} & - \frac{2}{11} & \frac{3}{11} \\ - \frac{6}{11} & \frac{4}{11} & \frac{5}{11} \\ \frac{1}{11} & \frac{3}{11} & \frac{1}{11}\end{matrix}\right)$
$= \left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right) = I$
Now, we can solve our equation
$\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}\frac{3}{11} & - \frac{2}{11} & \frac{3}{11} \\ - \frac{6}{11} & \frac{4}{11} & \frac{5}{11} \\ \frac{1}{11} & \frac{3}{11} & \frac{1}{11}\end{matrix}\right) \cdot \left(\begin{matrix}- 5 \\ 0 \\ 1\end{matrix}\right)$
$= \left(\begin{matrix}- \frac{12}{11} \\ \frac{35}{11} \\ - \frac{4}{11}\end{matrix}\right)$
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×
# Coefficient of Determination Calculator
## How to use the coefficient of determination calculator?
• Enter the data set X
• Enter the data set Y
• Values must be comma separated.
• Click on the calculate button.
• You can erase all inputs by clicking on the reset button.
## Coefficient of determination calculator:
The coefficient of determination calculator calculates the coefficient of determination (R2) and the correlation coefficient (r) of the entered data sets it also gives the step-by-step solution. Firstly, it calculates “r” stepwise and then takes its square to find out the coefficient of determination.
## What is the coefficient of determination?
The coefficient of determination (R²) measures how well a statistical model predicts an outcome. The outcome is represented by the model’s dependent variable.
The R2 ranges from 0 to 1, if the result is 0 then the outcome of the model is not good, and vice versa.
## Ways to calculate R2:
There are two different ways to calculate R2 both are as follows:
• Method 1: Calculate the correlation coefficient “r” and then take its square.
• Method 2: Use the following formula
• RSS is the sum of the squares of the residuals.
• TSS is the total sum.
## How to calculate the coefficient of determination:
The coefficient of determination (R2) can be calculated by following the below steps:
1. Calculate the correlation coefficient
• Calculate the mean of the data sets:
• Calculate the deviation scores
• Calculate the squared deviations
• Find out the product of the squared deviations
• Apply this formula
1. Take the square of “r”.
## Examples:
In this section, the step-by-step solution for calculating the coefficient of determination is given:
Example 1:
If X = 4, 6, 7, 24, 9.5, 2, 37, 13.7 and Y = 3, 35, 64, 223, 91, 44, 9.3, 12 then find out R2.
Solution:
X = 4, 6, 7, 24, 9.5, 2, 37, 13.7
Y= 3, 35, 64, 223, 91, 44, 9.3, 12
Step 1: Calculate the mean of X and Y
x̄ = (4 + 6 + 7 + 24 + 9.5 + 2 + 37 + 13.7) / 8
x̄ = 103.2 / 8
x̄ = 12.9
ȳ = (3 + 35 + 64 + 223 + 91 + 44 + 9.3 + 12) / 8
ȳ = 481 / 8
ȳ = 60.163
x - x̄ y - ȳ (x - x̄)2 (y - ȳ)2 (x - x̄)(y - ȳ) −8.9 −57.163 79.21 3267.609 508.7507 −6.9 −25.163 47.61 633.177 173.6247 −5.9 3.837 34.81 14.723 −22.6383 11.1 162.837 123.21 26515.889 1807.4907 −3.4 30.837 11.56 950.921 −104.8458 −10.9 −16.163 118.81 261.243 176.1767 24.1 −50.863 580.81 2587.045 −1225.7983 0.8 −48.163 0.64 2319.675 −38.5304 Sum 996.66 36550.278752 1274.23
Step 2: Applying the formula:
r = ∑ ((x - x̄) (y-ȳ)) / √ (∑ (x-x̄)2 ∑ (y-ȳ)2)
Putting the values
r = 1274.23/ √ ((996.66) (36550.278752))
r = 0.2111
Step 3: Take the square.
R2 = 0.04456
## References:
Coefficient of determination | Statistics Solutions
|
Kathy Williams
2022-01-03
Find the indefinite integral.
$\int \frac{{\left(\mathrm{ln}x\right)}^{2}}{x}dx$
Pansdorfp6
Step 1
Given integral:
$\int \frac{{\left(\mathrm{ln}x\right)}^{2}}{x}dx$
Substitute $t=\mathrm{ln}\left(x\right)$ and differentiate it w.r.t x
$\frac{dt}{dx}=\frac{d}{dx}\mathrm{ln}\left(x\right)$
$\frac{dt}{dx}=\frac{1}{x}$
$dt=\frac{1}{x}dx$
Therefore, the given integral becomes
$\int {t}^{2}dt$
Step 2
We know that $\int {x}^{n}d=\frac{{x}^{n+1}}{n+1}+c$
Where, "c" is integration constant.
$\int {t}^{2}dt=\frac{1}{3}{t}^{3}+c$
Substitute the value of $t=\mathrm{ln}\left(x\right)$ in above equation, we get
$⇒\frac{1}{3}{\left(\mathrm{ln}|x|\right)}^{3}+c$
boronganfh
Given:
$\int \frac{{\mathrm{ln}}^{2}\left(x\right)}{x}dx$
Substitution $u=\mathrm{ln}\left(x\right)⇒\frac{du}{dx}=\frac{1}{x}$
$=\int {u}^{2}du$
$=\frac{{u}^{3}}{3}$
$=\frac{{\mathrm{ln}}^{3}\left(x\right)}{3}$
$=\frac{{\mathrm{ln}}^{3}\left(x\right)}{3}+C$
Vasquez
Step 1
Given:
$\int \frac{\mathrm{ln}\left(x{\right)}^{2}}{x}dx$
Transform
$\int {t}^{2}dt$
Step 2
Evaluate
$\frac{{t}^{3}}{3}$
Substitute back
$\frac{\mathrm{ln}\left(x{\right)}^{3}}{3}$
Step 3
$\frac{\mathrm{ln}\left(x{\right)}^{3}}{3}+C$
|
Section 1.7 Dimensional Analysis
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1 Dimensional Analysis Dimensional Analysis Many times it is necessary to convert a measurement made in one unit to an equivalent measurement in another unit. Sometimes this conversion is between two units in the same system of measurement, and sometimes this conversion is between two units in different systems of measurement. Example 1: Timothy is pricing wood flooring for his living room. The price of flooring at the home improvement store is given in dollars per square meter. However, Timothy measured the area of his living room in square feet. To calculate the cost of installing wood flooring in his living room, Timothy will need to convert the area measurement of his room to square meters. Then he can multiply the area of his room in square meters times the price of the flooring per square meter to find the cost of the project. Definition: Dimensional analysis is the procedure for converting a measurement in one unit to an equivalent measurement in another unit. How do we know when two measurements in different units are equivalent? Measure the characteristics of the SAME OBJECT in both units. For example, to determine how many centimeters are equivalent to one foot, measure the length of an object that is one foot long in centimeters. When this measurement is made, the one foot long object measures 0.48 centimeters. Therefore, Equivalent Units 1 ft = 0.48 cm. Within the US Customary System equivalent units for length, mass and volume are not easily calculated. Some conversions are straightforward because we use them all the time. For example, most people easily know that 12 inches is the same as one foot. Other equivalences are not as easy to remember, like the fact that there are 5280 feet in a mile. Therefore, for most equivalent units in the US Customary System, most people rely on a table of equivalent units, like the one printed below. Two measurements in different units are equivalent if they are on the same row of the conversion chart. US Customary System Units 1 foot (ft) 12 inches (in) 1 yard (yd) feet 1 mile (mi) 5280 feet
2 1 pound (lb) 16 ounces 1 ton (T) 2000 pounds 1 cup (c) 8 fluid ounces 1 pint (pt) 2 cups 1 quart (qt) 2 pints 1 gallon (gal) 4 quarts The beauty of the metric system is that within the metric system converting a measurement to one in a different unit is simply a matter of moving the decimal point. The procedure was discussed in Section 1.6 and will not be repeated here. The prefix chart is included here for ease of reference in the following discussion. Metric Prefixes Prefix Symbol Meaning kilo k 1000 X base unit hecto h 100 X base unit deka da 10 X base unit - - Base unit deci d 1/10 of base unit centi c 1/100 of base unit milli m 1/1000 of base unit Volume is the only fundamental property that requires extra attention when performing dimensional analysis. Volume can be measured in liters (or multiples of liters) or in cubic length units. The conversion between the two units is based on the fact that 1 liter is defined to be the volume of a cube that has sides of length 1 decimeter. The following table lists several equivalent metric volume units of varying sizes. Metric Volume Conversion Chart Volume in Cubic Units Volume in Liters 1 cm 1 ml 1 dm 1 l 1 m 1 kl Converting between the metric system and the US Customary System is not as simple as multiplying or dividing by powers of 10, so a conversion chart is essential. There are many science textbooks and internet sites that have very exhaustive lists of equivalent units between the US Customary System, the metric system and even other systems used in isolated parts of the world. In this text we will use the following table of conversion factors. It is not exhaustive, but contains enough information to do most of the dimensional analysis required in the lesson and exercises. If a unit is introduced that is
3 not directly on the list, its equivalent in either the metric system or the US Customary System will be stated. The later examples will demonstrate how to convert between units that are not directly connected on the table. Also note that the equivalent measurements in different units are given to four significant digits of precision. More accurate conversion factors exist for some units, but we will not require that degree of precision. US Customary System and Metric System Conversion Chart Length 1 inch 2.54 centimeters 1 foot 0.48 centimeters 1 yard meters 1 mile kilometers Area 1 in cm 2 1 ft m 2 1 yd m 2 1 mi km 2 1 acre hectare (ha) Volume 1 teaspoon (tsp) milliliters 1 tablespoon (tbsp) milliliters 1 fluid ounce milliliters 1 cup liters 1 pint liters 1 quart liters 1 gallon.785 liters 1 cubic foot cubic meters 1 cubic yard cubic meters Weight (Mass) 1 ounce 28.5 grams 1 pound kilograms 1 ton (T) 0.9 tonne (t) Unit Fractions The arithmetic algorithm for converting between units is based on the fact that measurements are given in proportion to a standard. Proportions are often expressed in the form of a fraction
4 Definition: A unit fraction is a fraction (proportion written in fraction form) in which the numerator and the denominator of the fraction are equivalent measurements in different units. Examples include: 1g 12 in 2.54 cm,, 1000 mg 1 ft 1in In arithmetic with rational numbers, multiplying by a unit (1 or a ) does not change the a values of a fraction. Similarly, multiplying a measurement by a unit fraction does not change the measurement, only the unit. Also just like rational fractions, common units (factors) in the numerator and denominator of a multiplication problem cancel. The units that do not cancel are the units of the answer to the multiplication. Example 2: Convert 8.5 inches to centimeters using a unit fraction. The US Customary System and Metric System conversion chart shows that 1 inch is equivalent to 2.54 centimeters. Therefore, the unit fraction that we need has 1 in one place and 2.54 cm in the other. Which should be in the numerator and which in the denominator? 1in 2.54 cm or 2.54 cm 1in? The answer is the unit fraction with inches in the denominator, since multiplying 8.5 inches times the unit fraction with inches in the denominator will have inches on both top and bottom. The inches unit will cancel and the remaining unit is centimeters cm 2.54 cm 8.5 in = 8.5 in 1in 1 in The final answer is obtained by multiplying the numerical values and keeping the centimeter units that did not cancel cm 8.5 in = 8.5(2.54) cm = cm 1in Converting Units of Length, Mass and Volume Converting length, mass and volume units within the US Customary System, and between the US Customary System and the metric system are accomplished by multiplying the given measurement by one (or more) unit fractions. The procedure is:
5 1. Determine if there is an equivalent measurement in the conversion charts for the two units in question. If so, the unit fraction made from that equivalence is the multiplication factor needed. 2. If there is not an equivalent measurement listed in the charts with the two units in question, create a chain of equivalences linking the two units. For example, if you want to convert from inches to meters, use the facts that 1 inch is 2.54 centimeters and that the 100 centimeters equals 1 meter.. Write the multiplication problem using the unit fractions needed. The unit fraction should be written so that the corresponding units cancel from the top and the bottom of the multiplication problem, and the desired unit is on the top of a fraction. 4. Multiply and divide the numerical values in the measurement and unit fraction using normal real number multiplication and division rules. 5. Cancel the common units and express the numerical answer in terms of the new units Example : Convert 5 quarts to liters. The conversion chart shows 1 quart = 0.95 liters. Thus l 5 quarts = 5 quarts = 4.72 l 1quart Example 4: Convert 64 g to pounds. The conversion chart shows that 1 pound = kg and the metric prefixes chart shows that 1 kg = 1000 g. Using these two equivalences: 1 kg 1lb 64 g = 64 g = lb 1000 g kg Example 5: Convert 0.15 liters to cups. The conversion chart shows 1 cup = l. So, 1 cup 0.15 l = 0.15 l = 0.64 cup l
6 Example 6: Convert 0.5 km to feet. The conversion chart above does not have an equivalence between kilometers and feet directly, but it does show that 1 mile = km and 1 mile = 5280 feet. Using the unit fractions from these two equivalences, 1 mile 5280 ft 0.5 km = 0.5 km = 1641 ft km 1 mile Converting Area and Volume Volume can be measured in volume units, or in cubic length units. Area is measured in square length units. When performing dimensional analysis in square or cubic length units, the unit fractions must be squared or cubed appropriately so that the units still cancel. This may be accomplished by writing one unit fraction raised to a power, or by multiplying by the unit fraction as many times as needed to cancel the units appropriately. Example 7: Convert 50.6 in to cubic centimeters. The equivalence needed is that 1 in = 2.54 cm. The unit fraction is written times so that the cubic inches will cancel in 2.54 cm 2.54 cm 2.54 cm 50.6 in = = 829 cm. 1 1in 1in 1in Example 8: Convert m to cubic decimeters. The equivalence needed is that 10 dm = 1 m. This time we will write the unit fraction raised to the third power instead of writing it out times. The meaning is exactly the same. Note that this means that any numerical values in the unit fraction must be cubed to perform the multiplication properly m 10 dm m = = 96 dm 1 1 m
7 The conversion of a volume or area measurement may also be accomplished by converting each dimensional measurement into the appropriate units, and then using a formula to find the area or volume. Example 9: A rectangular table is 15 inches wide and 0 inches long. Find the area of the table in square meters. The dimensions of the table are first converted to meters, and then the area is calculated in square meters. The dimensions of the table are: 15 in 2.54 cm 1 m = 0.81 m 1 1in 100 cm and 0 in 2.54 cm 1 m = m 1 1in 100 cm The area of the table is equal to the length times the width, so Area = 0.81 m m = m. ( )( ) 2 Sometimes, a combination of conversions is required to complete the dimensional analysis. Example 10: A cylindrical bucket has a height of 0.5 meters and a radius of 25 cm. How many liters of water will it hold? How many gallons of water will it hold? The first question asks for the volume in liters. The metric volume conversion chart shows that one liter is equivalent to one cubic decimeter. So, if the dimensions of the bucket are converted to decimeters, the volume of the cylinder formula will give the volume in cubic decimeters (liters). Therefore, the first step is to find the dimensions of the bucket in decimeters. Height = 0.5 m = 5 dm (since meters are one power of ten larger than decimeters, move the decimal point one place to the right) Radius = 25 cm = 2.5 dm (since centimeters are one power of ten smaller than decimeters, move the decimal point one place to the left) Volume the formula for a cylinder is V height. Plugging in the numbers: = π 2 r h, where r is the radius and h is the V = π = = 2 (2.5) (5) dm 98.2 dm 98.2 l
8 The bucket will hold 98 liters of water. Next, convert liters to gallons: 98.2 l 1 gal 98.2 l = = 25.9 gal l The bucket will hold 25.9 gallons of water. Converting Velocity and Cost Per Unit The conversion charts in this section are given for basic length, volume and mass units. Often there are other measurements that we need to convert from the metric system to the US Customary System, or convert to different units within either system. One common measurement that requires conversion is that of velocity. Velocity is expressed as length per unit of time. In the US Customary System velocity is often expressed in miles per hour or feet per second. In the metric system, velocity is often given in meters per second or kilometers per hour. The same techniques of multiplying by unit fractions may be used to convert velocity from one unit to an equivalent measurement in another unit. Example 11: The speed limit on a country road is posted as 40 miles per hour. Bob s car is a brand new German auto with a speedometer measuring his velocity in kilometers per hour. Convert 40 mph to kilometers per hour so that Bob knows how fast he can legally go on this road. The equivalent measurements needed are that 1 mile = 1.6. The multiplication is: 40 miles 40 miles km 64.6 km = = hr hr 1 mile hr The speed limit on the country road is 64.6 kilometers per hour. Another common application of dimensional analysis is the conversion of price per unit to an equivalent price in a different unit. The example at the beginning of this section of pricing flooring at the home improvement store is a classic example. Again the technique of multiplying by unit fractions is used in this situation successfully
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Dimensional Analysis I: Length, Mass, Volume and Area Conversions
EXPERIMENT Dimensional Analysis I: Length, Mass, Volume and Area Conversions Student will use the Dimensional Analysis Map to introduce / reinforce the process of basic unit conversions in chemistry. OBJECTIVE
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U Step by Step: Conversions. UMultipliersU UExample with meters:u (**Same for grams, seconds & liters and any other units!
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## Section3.3The Extended Plane
Consider again inversion about the circle $C$ given by $|z - z_0| = r\text{,}$ and observe that points close to $z_0$ get mapped to points in the plane far away from from $z_0\text{.}$ In fact, a sequence of points in $\mathbb{C}$ whose limit is $z_0$ will be inverted to a sequence of points whose magnitudes go to $\infty\text{.}$ Conversely, any sequence of points in $\mathbb{C}$ having magnitudes marching off to $\infty$ will be inverted to a sequence of points whose limit is $z_0\text{.}$
With this in mind, we define a new point called the point at infinity, denoted $\infty.$ Adjoin this new point to the plane to get the extended plane, denoted as ${\mathbb{C}}^+\text{.}$ Then, one may extend inversion in the circle $C$ to include the points $z_0$ and $\infty\text{.}$ In particular, inversion of $\mathbb{C}^+$ in the circle $C$ centered at $z_0$ with radius $r\text{,}$ $i_C: \mathbb{C}^+ \to \mathbb{C}^+\text{,}$ is given by
\begin{equation*} i_C(z) = \begin{cases}\frac{r^2}{(\overline{z-z_0})} + z_0 \amp \text{ if $z \neq z_0,\infty$; } \\ \infty \amp \text{ if $z=z_0$; } \\ z_0 \amp \text{ if $z = \infty$. } \end{cases} \end{equation*}
Viewing inversion as a transformation of the extended plane, we define $z_0$ and $\infty$ to be symmetric points with respect to the circle of inversion.
The space ${\mathbb{C}}^+$ will be the canvas on which we do all of our geometry, and it is important to begin to think of $\infty$ as “one of the gang,” just another point to consider. All of our translations, dilations, and rotations can be redefined to include the point $\infty\text{.}$
So where is $\infty$ in ${\mathbb{C}}^+\text{?}$ You approach $\infty$ as you proceed in either direction along any line in the complex plane. More generally, if $\{z_n\}$ is a sequence of complex numbers such that $|z_n| \to \infty$ as $n \to \infty\text{,}$ then we say $\displaystyle\lim_{n\to\infty} z_n = \infty.$ By convention, we assume $\infty$ is on every line in the extended plane, and reflection across any line fixes $\infty\text{.}$
If $T(z) = a z + b$ where $a$ and $b$ are complex constants with $a \neq 0\text{,}$ then by limit methods from calculus, as $|z_n| \to \infty\text{,}$ $|a z_n + b| \to \infty$ as well. Thus, $T(\infty) = \infty\text{.}$
So, with new domain ${\mathbb{C}}^+\text{,}$ we modify our fixed point count for the basic transformations:
• The translation $T_b$ of ${\mathbb{C}}^+$ fixes one point ($\infty$).
• The rotation about the origin $R_\theta$ of ${\mathbb{C}}^+$ fixes 2 points (0 and $\infty$).
• The dilation $T(z) = kz$ of ${\mathbb{C}}^+$ fixes 2 points, (0 and $\infty$).
• The reflection $r_L(z)$ of ${\mathbb{C}}^+$ about line $L$ fixes all points on $L$ (which now includes $\infty$).
###### Example3.3.2Some transformations not fixing $\infty$
The following function is a transformation of ${\mathbb{C}}^+$
\begin{equation*} T(z) = \frac{i+1}{z+2i}, \end{equation*}
a fact we prove in the next section. For now, we ask where $T$ sends $\infty\text{,}$ and which point gets sent to $\infty\text{.}$
We tackle the second question first. The input that gets sent to $\infty$ is the complex number that makes the denominator 0. Thus, $T(-2i) = \infty.$
To answer the first question, take your favorite sequence that marches off to $\infty\text{,}$ for example, $1, 2, 3,\ldots\text{.}$ The image of this sequence, $T(1), T(2),$ $T(3),\ldots$ consists of complex fractions in which the numerator is constant, but the denominator grows unbounded in magnitude along the horizontal line $\text{Im}(z) = 2\text{.}$ Thus, the quotient tends to 0, and $T(\infty) = 0.$
As a second example, you can check that if
\begin{equation*} T(z) = \frac{iz+(3i+1)}{2iz+1}, \end{equation*}
then $T(i/2) = \infty$ and $T(\infty) = 1/2\text{.}$
We emphasize that the following key results of the previous section extend to $\mathbb{C}^+$ as well:
• There exists a unique cline through any three distinct points in $\mathbb{C}^+\text{.}$ (If one of the given points in Theorem 3.2.4 is $\infty\text{,}$ the unique cline is the line through the other two points.)
• Theorem 3.2.8 applies to all points $z$ not on $C\text{,}$ including $z = z_0$ or $\infty\text{.}$
• Inversion about a cline preserves angle magnitudes at all points in $\mathbb{C}^+$ (we discuss this below).
• Inversion preserves symmetry points for all points in $\mathbb{C}^+$ (Theorem 3.2.12 holds if $p$ or $q$ is $\infty$).
• Theorem 3.2.16 now holds for all clines that do not intersect, including concentric circles. If the circles are concentric, the points symmetric to both of them are $\infty$ and the common center.
##### Stereographic Projection
We close this section with a look at stereographic projection. By identifying the extended plane with a sphere, this map offers a very useful way for us to think about the point $\infty\text{.}$
###### Definition3.3.3
The unit 2-sphere, denoted $\mathbb{S}^2\text{,}$ consists of all the points in 3-space that are one unit from the origin. That is,
\begin{equation*} \mathbb{S}^2 = \{(a,b,c) \in \mathbb{R}^3 ~|~ a^2+b^2+c^2=1\}. \end{equation*}
We will usually refer to the unit 2-sphere as simply “the sphere.” Stereographic projection of the sphere onto the extended plane is defined as follows. Let $N = (0,0,1)$ denote the north pole on the sphere. For any point $P \neq N$ on the sphere, $\phi(P)$ is the point on the ray $\overrightarrow{NP}$ that lives in the $xy$-plane. See Figure 3.3.4 for the image of a typical point $P$ of the sphere.
The stereographic projection map $\phi$ can be described algebraically. The line through $N = (0,0,1)$ and $P = (a,b,c)$ has directional vector $\overrightarrow{NP}=\langle a,b,c-1\rangle\text{,}$ so the line equation can be expressed as
\begin{equation*} {\vec r}(t) = \langle 0,0,1\rangle + t\langle a,b,c-1\rangle. \end{equation*}
This line intersects the $xy$-plane when its $z$ coordinate is zero. This occurs when $t = \frac{1}{1-c}\text{,}$ which corresponds to the point $(\frac{a}{1-c}, \frac{b}{1-c},0)\text{.}$
Thus, for a point $(a,b,c)$ on the sphere with $c \neq 1\text{,}$ stereographic projection $\phi:\mathbb{S}^2 \to \mathbb{C}^+$ is given by
\begin{equation*} \phi((a,b,c)) = \frac{a}{1-c}+\frac{b}{1-c}i. \end{equation*}
Where does $\phi$ send the north pole? To $\infty\text{,}$ of course. A sequence of points on $\mathbb{S}^2$ that approaches $N$ will have image points in $\mathbb{C}$ with magnitudes that approach $\infty\text{.}$
##### Angles at $\infty$
If we think of $\infty$ as just another point in $\mathbb{C}^+\text{,}$ it makes sense to ask about angles at this point. For instance, any two lines intersect at $\infty\text{,}$ and it makes sense to ask about the angle of intersection at $\infty\text{.}$ We can be guided in answering this question by stereographic projection, thanks to the following theorem.
Thus, if two curves in $\mathbb{C}^+$ intersect at $\infty$ we may define the angle at which they intersect to equal the angle at which their pre-image curves under stereographic projection intersect. The angle at which two parallel lines intersect at $\infty$ is 0. Furthermore, if two lines intersect at a finite point $p$ as well as at $\infty\text{,}$ the angle at which they intersect at $\infty$ equals the negative of the angle at which they intersect at $p\text{.}$ As a consequence, we may say that inversion about a circle preserves angle magnitudes at all points in $\mathbb{C}^+\text{.}$
### SubsectionExercises
###### 1
In each case find $T(\infty)$ and the input $z_0$ such that $T(z_0) = \infty\text{.}$
a. $T(z) = (3 - z)/(2z + i)\text{.}$
b. $T(z) = (z + 1)/e^{i\pi/4}\text{.}$
c. $T(z) = (az + b)/(cz + d)\text{.}$
###### 2
Suppose $D$ is a circle of Apollonius of $p$ and $q\text{.}$ Prove that $p$ and $q$ are symmetric with respect to $D\text{.}$ Hint: Recall the circle $C$ in the proof of Theorem 3.2.14. Show that $p$ and $q$ get sent to points that are symmetric with respect to $i_C(D)\text{.}$
###### 3
Determine the inverse stereographic projection function $\phi^{-1}:\mathbb{C}^+ \to \mathbb{S}^2\text{.}$ In particular, show that for $z = x + yi \neq \infty\text{,}$
\begin{equation*} \phi^{-1}(x,y) = \bigg(\frac{2x}{x^2+y^2+1},\frac{2y}{x^2+y^2+1},\frac{x^2+y^2-1}{x^2+y^2+1}\bigg). \end{equation*}
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# Question #56413
Aug 29, 2015
For part (a): Domain: $\left(- \infty , - 3\right]$
For part (b): Domain: $\left(- \infty , + \infty\right)$
For part (c): Domain: $\left[2 , 4\right]$
#### Explanation:
In all three cases, you're dealing with functions that equal the square root of an expression.
Right from the start, you need to take into account the fact that, for real numbers, you cannot take the square root of a negative number.
This means that you need to check these expressions for any value of $x$ that would make them negative, and exclude these values from the functions' domain.
$f \left(x\right) = \sqrt{- 3 x - 9}$
You need to have
$\left(- 3 x - 9\right) \ge 0$
$- 3 x \ge 9 \implies x \le \frac{9}{\left(- 3\right)} = - 3$
The domain of the function will thus include any value of $x$ that is greater than or equal to $\left(- 3\right)$. Any other value will be excluded. This means that the function's domain will be $\left(- \infty , - 3\right]$.
graph{sqrt(-3x-9) [-14.24, 14.24, -7.12, 7.12]}
Do the same for the second one.
$g \left(x\right) = \sqrt{{x}^{2} + 4}$
Now, because ${x}^{2} \ge 0$ for all real values of $x$, the expression under the square root will never be positive. In fact, it will never be smaller than $4$, since that's the value you'd get for $x = 0$.
This means that you don't have any restrictions for the function's domain, which will be $x \in \mathbb{R}$, or $x \in \left(- \infty , + \infty\right)$.
graph{sqrt(x^2 + 4) [-41.1, 41.14, -20.53, 20.56]}
Now for the third one.
$h \left(x\right) = \sqrt{2 x - 4} + \sqrt{- 5 x + 20}$
You need the expressions that are under the square roots to be valid at the same time. You will thus have
$2 x - 4 \ge 0$
$2 x \ge 4 \implies x \ge 2$
and
$- 5 x + 20 \ge 0$
$- 5 x \ge - 20 \implies x \le 4$
The domain of the function will thus be $\left[2 , 4\right]$, since any value of $x$ outside this interval will make of the the two square roots undefined.
graph{sqrt(2x-4) + sqrt(-5x + 20) [-16, 16.05, -8, 8.02]}
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# How to write a rational number as a repeating decimal?
• Last Updated : 11 May, 2022
There are different sorts of numbers that are included in the number system, such as prime numbers, odd numbers, even numbers, rational numbers, whole numbers, and so on. These numbers can be expressed in a variety of ways, including figures and words. For example, in the form of figures, the numbers 50 and 75 can alternatively be written as fifty and seventy-five.
A number system, often known as a numeral system, is a basic system for expressing numbers and figures. It is the only way to represent numbers in both arithmetic and algebraic structures. Numbers are utilized in a variety of arithmetic values that may be used to perform a variety of arithmetic operations such as addition, subtraction, multiplication, and other operations that are used in daily life for the purpose of computation.
In the numeral system, Real numbers are simply a combination of rational and irrational numerals. All known arithmetic functions may be performed on these numbers and represented on the number line. The digit, its place value in the number and the number system’s base determine the value of a number. The mathematical values used for counting, measuring, identifying, and measuring fundamental quantities are known as numbers or numerals.
### Rational numbers
A rational number is the one that can be written as a ratio of two integers p and q, where q is not zero. and Repeating decimals are those that have a set of decimal terms that are repeated consistently. Different sorts of rational numbers occur. We shouldn’t assume that only rational numbers are fractions with integers. Numbers that are rational can also be stated in decimal form.
Example: Is 1.1 a rational number?
Yes, it’s a rational number because 1.1 may be represented as 1.1 = 11/10. Let’s take a look at non-terminating decimals like 0.333… 0.333… is a rational number since it may be expressed as 1/3. Non-terminating decimals with repeated numbers after the decimal point are thus rational numbers as well.
Examples of Rational numbers are,
• All numerals are rational, 1, 2, 3, 4, 5, 6, 7… all are rational numbers.
• All Whole numerals are rational. 0, 1, 2, 3, 4, 5,… all are rational.
The following are the several sorts of rational numbers,
Integers such as -2, 0, 3, and so on. Fractions with integer numerators and denominators, such as 3/7, -6/5, and so on. 0.35, 0.7116, 0.9768, and so on are examples of terminating decimals. Non-terminating decimals having recurring patterns (after the decimal point), such as 0.333…, 0.141414…, and so on.
Examples of Repeated decimals are,
6/4 as a decimal is 6 ÷ 4 = 1.5 terminating decimal. 35/100 as a decimal is 35 ÷100 = 0.35 terminating decimal. 5/3 as a decimal is 5 ÷ 3 = 1.66666… its a repeating decimal of Rational number.
### How to write a rational number as a repeating decimal?
Any rational number (a fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator.
If there is a remainder of zero, you have a terminating decimal. Otherwise, the remainder will begin to repeat after some significance, resulting in a repeating decimal.
Example: Write rational number 4/3 in repeating decimal form.
Simply divide the numerator by the denominator.
So, here we divide 4 by 3,
Repeating decimal
Here after simplifying we get 4/3 = 1.333333…
Hence, this way we can write rational number as repeating decimal.
### Similar Questions
Question 1: Write the rational number 67/3 as a repeating decimal?
Solution:
Any rational number (a fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator. Here divide 67 by 3,
67/3 as repeating decimal
We can write rational number 67/3 as a 22.33333… as repeating decimal.
Question 2: Write 11/3 rational numbers as repeating decimal?
Solution:
Any rational number (A fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator.
Here divide 11 by 3: we can write 11 /3 as 3.666666666… after dividing,
Therefore it’s a repeating decimal of rational number.
Question 3: Write 103/6 rational number as repeating decimal?
Solution:
Any rational number (A fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator.
Here divide 103 by 6: we can write 103 / 6 as 17.166666666… after dividing,
Therefore it’s a way of rational number writing as repeating decimal.
Question 4: Write 5/6 rational number as a repeating decimal?
Solution:
Any rational number (A fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator.
Here divide 5 by 6: we can write 5 / 6 as 0.8333333333… after dividing,
Therefore it’s a way of rational number writing as repeating decimal.
Question 5: Write 19/3 rational numbers as repeating decimal?
Solution:
Any rational number (A fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator.
Here divide 19 by 3: we can write 19/ 3 as 6.3333333… after dividing,
Therefore in this way 19/3 rational number can be written as a repeating decimal.
Question 6: Write 185/6 rational number as repeating decimal?
Solution:
Any rational number (A fraction in lower times) can be expressed as a terminating decimal or a repeating decimal. Simply divide the numerator by the denominator.
Here divide 185 by 6: we can write 185/ 6 as 30.8333333… after dividing,
Therefore in this way 185/6 rational number write as repeating decimal.
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Algebra/Chapter 3/Equations
------------------------ AlgebraChapter 3: Solving EquationsSection 1: What are Equations? Inverses of Numbers
3.1: What are Equations?
To do: examples, practice quizzes and review quizzes (as applicable)
Balancing variables
Mathematics began with the concept of counting, but it grew because of the idea of equivalence. Tally marks or counters are equivalent to a value depending on the concept being expressed. For instance a shepherd could use a bag of rocks to make sure that he brought home the same number of sheep he left with. The idea of equivalence as shown with an equal sign allows us to state "is the same as".
An expression is a statement that is well formed. The neolithic shepherd used his bag of rocks to create expressions for managing his herd. For instance, the shepherd hoped that the size of his flock remained constant between the time he took his flock out to graze and the time that he returned them home. When he used the bag of rocks the shepherd used two expressions: # Sheep and # Rocks to create the equation # Sheep = # Rocks. Mathematics describes the operations the shepherd performed on the expressions in order to keep the equation in balance. For instance, when the shepherd used his bag of rocks to count his flock at night he might put his rocks in a bowl next to the gate on his pen. Each time a sheep went into the pen the shepherd would move a rock from the bag into the bowl. If rocks were left over in the bag, then the shepherd knew he needed to return to the field to find his lost sheep.
Expressions
An expression is the arrangement of mathematical symbols denoting values and operations, while the equation is a form of an expression that indicates that two values are equal. While equations are generally expressions, you will discover that some expressions are not equations. Expressions in the form of functions or inequalities will be more apparent later in the book.
When you simplify an expression, you are performing simple arithmetic steps until you receive the simplest answer, which is usually an integer or fraction.
Equations
An equation is just like a balance. A balance is a machine that compares whether two quantities are the same or not the same. In mathematics, two quantities balanced translates to "having the same value." An equation is true if the two sides are balanced and false if the two sides are not balanced. As 5=5. This is a very simple equation. We can use variables too(almost used in mathematics)e.g. x=5. This is same as while using a balancing machine, we use some quantity on one side and some matter on the other which is to be measured. As an example we know that 5 is equal to 5 and x is equal to x(whatever its value is). The simplest example of a false equation is that 5 is not equal to 20.
Remember from the previous section, a variable is a symbol, usually a letter, that stands for a number. We'll use the letters ${\displaystyle x}$ and ${\displaystyle y}$ most often. If ${\displaystyle x=5}$ is true, then any time I use ${\displaystyle x}$, you can replace ${\displaystyle x}$ by 5. We're saying ${\displaystyle x}$ and 5 have the same value, but different appearance. In the equation ${\displaystyle x=5}$, the equal sign is like the balance. If ${\displaystyle x=5}$ is true, we know that whenever we see an ${\displaystyle x}$ we can substitute the number 5 if that helps us understand. Even more important, we can check if a given number substituted for a variable in an equation makes that equation true.
Frequently we don't know what the value of a variable is that would make the equation true, and we need to find out! Finding out the values that make an equation true is called Solving an Equation, which we will do without guessing in the next section.
Practice Problems
Conceptual Questions
Problem 3.1 (Explaining Equations) In no more than one paragraph, explain in your own words what an "equation" is. In your definition, discuss an equation's purpose, as well as what a solution of an equation means.
Problem 3.2 (Anatomy of an Equation) Given the statement x + 6 = 12, answer the following.
a. Is this statement true or false?
b. Does x=5 make the statement true?
c. Does x=6 make the statement true?
Problem 3.3 (Identifying Equations) Determine which of the following are equations.
a. x + 4
b. a - 64 = 45
c. 7 < 8
d. a + 6 > 9
e. 1 + 7 = 8
Problem 3.4 (Special Solution Sets) Make up an equation whose solution is the null set, and explain why this is the solution set. After this, make up an equation whose solution is the set of all real numbers, and explain why this is the solution set.
Problem 3.5 (Equivalent Equations) Are the statements 9x + 5 = 4 and 4 = 9x + 5 the same equation? Explain your reasoning.
Exercises
Problem 3.6 (Checking Solutions) Check if the given value(s) is/are solutions to the following equations.
Problem 3.7 (Equation Diagrams) Write the equation that best represents the following diagrams.
Problem 3.8 (Scales) Look at the diagrams below. How many circles would you need to balance the third scale?
Problem 3.9 (Scales) Which is heavier? A red triangle or a blue square?
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How do you find all local maximum and minimum points using the second derivative test given y=4x+sqrt(1-x)?
May 25, 2018
$\textcolor{b l u e}{\text{Maximum point at } x = \frac{63}{64}}$
Explanation:
At local maximum and minimum points the tangent is zero. We can use this to find where these points are. We first find the first derivative and equate this to zero.
Given $f \left(x\right) = 4 x + \sqrt{1 - x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left(4 x + \sqrt{1 - x}\right) = 4 x + {\left(1 - x\right)}^{\frac{1}{2}} = 4 + \frac{1}{2} {\left(1 - x\right)}^{- \frac{1}{2}} \cdot \left(- 1\right)$
$= 4 - \frac{1}{2 \sqrt{1 - x}}$
Equate to zero:
$4 - \frac{1}{2 \sqrt{1 - x}} = 0$
$8 \sqrt{1 - x} - 1 = 0 \implies x = \frac{63}{64}$
Using the second derivative we know that if:
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} > 0 \textcolor{w h i t e}{888}$Minimum point.
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} < 0 \textcolor{w h i t e}{888}$Maximum point.
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0 \textcolor{w h i t e}{888}$Maximum/Minimum or inflection point.
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} \left(4 x + {\left(1 - x\right)}^{\frac{1}{2}}\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \left(4 - \frac{1}{2 \sqrt{1 - x}}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left(4 - \frac{1}{2 \sqrt{1 - x}}\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \left(4 - \frac{1}{2} {\left(1 - x\right)}^{- \frac{1}{2}}\right)$
$= 0 + \frac{1}{4} {\left(1 - x\right)}^{- \frac{3}{2}} \cdot \left(- 1\right) = - \frac{1}{4 {\left(1 - x\right)}^{\frac{3}{2}}}$
Plugging in $x = \frac{63}{64}$
$- \frac{1}{4 {\left(1 - \left(\frac{63}{64}\right)\right)}^{\frac{3}{2}}} = - 4 \sqrt{{\left(\frac{1}{64}\right)}^{3}}$
$- 4 \sqrt{{\left(\frac{1}{64}\right)}^{3}} < 0$
$\therefore$
$x = \frac{63}{64}$ is a maximum point.
This is confirmed by the graph:
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# Percentile Formula
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## Percentile Formula
The percentile formula determines the performance of a person over others. The percentile formula is used in finding where a student stands in the test compared to other candidates. A percentile is a number where a certain percentage of scores fall below the given number. Let’s discuss this percentile formula in detail and solve a few examples.
### Percentile Definition
Percentile is defined as the value below which a given percentage falls under. For example, in a group of 20 children, Ben is the 4th tallest and 80% of the children are shorter than you. Hence, it means that Ben is at the 80th percentile. It is most commonly used in competitive exams such as SAT, LSAT, etc.
GIA SƯ DẠY SAT
## What is Percentile Formula?
The percentile formula is used when we need to compare the exact values or numbers over the other numbers from the given data i.e. the accuracy of the number. Often percentile and percentage are taken as one but both are different concepts. A percentage is where the fraction is considered as one term while percentile is the value below the percentage found from the given data. In our day-to-day life, percentile formulas are usually helpful in finding the test scores or biometric measurements. Hence, the percentile formula is:
Percentile = (n/N) × 100
Or
The percentile of x is the ratio of the number of values below x to the total number of values multiplied by 100. i.e., the percentile formula is
Percentile = (Number of Values Below “x” / Total Number of Values) × 100
### Percentile Formula
P = (n/N) × 100
Where,
• n = ordinal rank of the given value or value below the number
• N = number of values in the data set
• P = percentile
Or
Percentile = (Number of Values Below “x” / Total Number of Values) × 100
## Steps of Percentile Formula
To find the percentile, here are a few steps to use the percentile formula. If q is any number between zero and hundred, the qth percentile is a value that divides the data into two parts i.e the lowest part contains the q percent of the data and the rest of the data is the upper part.
• Step 1: Arrange the data set in ascending order
• Step 2: Count the number of values in the data set and represent it as r
• Step 3: Calculate the value of q/100
• Step 4: Multiply q percent by r
• Step 5: If the answer is not a whole number then rounding the number is required. If it is a whole number, continue to the next step
• Step 6: Count the values in the data set, find the mean and the next number. The answer is the qth percentile
• Step 7: Count the value in the data set, once you reach that number according to what we obtained in step 5 that is the qth percentile.
## Examples Using Percentile Formula
Example 1: The scores obtained by 10 students are 38, 47, 49, 58, 60, 65, 70, 79, 80, 92. Using the percentile formula, calculate the percentile for score 70?
Solution:
Given:
Scores obtained by students are 38, 47, 49, 58, 60, 65, 70, 79, 80, 92
Number of scores below 70 = 6
Using percentile formula,
Percentile = (Number of Values Below “x” / Total Number of Values) × 100
Percentile of 70
= (6/10) × 100
= 0.6 × 100 = 60
Therefore, the percentile for score 70 = 60
Example 2: The weights of 10 people were recorded in kg as 35, 41, 42, 56, 58, 62, 70, 71, 90, 77. Find the percentile for the weight 58 kg.
Solution:
Given:
Scores obtained by students are 35, 41, 42, 56, 58, 62, 70, 71, 77, 90
Number of people with weight below 58 kg = 4
Using percentile formula,
Percentile = (Number of Values Below “x” / Total Number of Values) × 100
Percentile for weight 58 kg
= (4/10) × 100
= 0.4 × 100 = 40
Therefore, the percentile for weight 58 kg = 40
Example 3: In a college, a list of scores of 10 students is announced. The scores are 56, 45, 69, 78, 72, 94, 82, 80, 63, 59. Using the percentile formula, find the 70th percentile.
Solution: Arrange the data in ascending order – 45, 56, 59, 63, 69, 72, 78, 80, 82, 94
Find the rank,
Rank = Percentile × 100
Rank = 70 × 100
Rank = 0.07
Find the percentile using the formula,
Percentile = Rank × Total number of the data set
Percentile = 0.07 × 10
Percentile = 0.7
Since it not a whole number, round the number to the nearest whole number i.e. 0.7 = 1
Count the set to reach the above number. It is 1.
Therefore, the 70th percentile is 45
## FAQs on Percentile Formula
### What is the Percentile Formula?
The percentile formula determines the performance of a person over others. The percentile formula is used in finding where a student stands in the test compared to other candidates. In our day-to-day life, percentile formulas are usually helpful in finding the test scores or biometric measurements.
### What is the Formula to Calculate the Percentile?
The percentile of x is the ratio of the number of values below x to the total number of values multiplied by 100. i.e., the percentile formula is:
Percentile = (Number of Values Below “x” / Total Number of Values) × 100
### What is the Formula to Calculate the Percentile By Rank?
The percentile formula where the rank of the number is used is:
P = (n/N) × 100
Where,
n = ordinal rank of the given value or value below the number
N = number of values in the data set
P = percentile
### Find the Percentile for 20 from the given set 12, 55, 7, 10, 40.
Arrange it in ascending order – 7, 10, 12, 40, 55
Number of scores below 20 = 3
Using percentile formula,
Percentile = (Number of Values Below “x” / Total Number of Values) × 100
Percentile of 20
= (3/10) × 100
= 0.3 × 100 = 30
Therefore, the percentile for score 20 = 30
## Definition of a Percentile in Statistics and How to Calculate It
In statistics, percentiles are used to understand and interpret data. The nth percentile of a set of data is the value at which n percent of the data is below it. In everyday life, percentiles are used to understand values such as test scores, health indicators, and other measurements. For example, an 18-year-old male who is six and a half feet tall is in the 99th percentile for his height. This means that of all the 18-year-old males, 99 percent have a height that is equal to or less than six and a half feet. An 18-year-old male who is only five and a half feet tall, on the other hand, is in the 16th percentile for his height, meaning only 16 percent of males his age are the same height or shorter.
### Key Facts: Percentiles
• Percentiles are used to understand and interpret data. They indicate the values below which a certain percentage of the data in a data set is found.
• Percentiles can be calculated using the formula n = (P/100) x N, where P = percentile, N = number of values in a data set (sorted from smallest to largest), and n = ordinal rank of a given value.
• Percentiles are frequently used to understand test scores and biometric measurements.
## What Percentile Means
Percentiles should not be confused with percentages. The latter is used to express fractions of a whole, while percentiles are the values below which a certain percentage of the data in a data set is found. In practical terms, there is a significant difference between the two. For example, a student taking a difficult exam might earn a score of 75 percent. This means that he correctly answered every three out of four questions. A student who scores in the 75th percentile, however, has obtained a different result. This percentile means that the student earned a higher score than 75 percent of the other students who took the exam. In other words, the percentage score reflects how well the student did on the exam itself; the percentile score reflects how well he did in comparison to other students.
## Percentile Formula
Percentiles for the values in a given data set can be calculated using the formula:
n = (P/100) x N
where N = number of values in the data set, P = percentile, and n = ordinal rank of a given value (with the values in the data set sorted from smallest to largest). For example, take a class of 20 students that earned the following scores on their most recent test: 75, 77, 78, 78, 80, 81, 81, 82, 83, 84, 84, 84, 85, 87, 87, 88, 88, 88, 89, 90. These scores can be represented as a data set with 20 values: {75, 77, 78, 78, 80, 81, 81, 82, 83, 84, 84, 84, 85, 87, 87, 88, 88, 88, 89, 90}.
We can find the score that marks the 20th percentile by plugging in known values into the formula and solving for n:
n = (20/100) x 20
n = 4
The fourth value in the data set is the score 78. This means that 78 marks the 20th percentile; of the students in the class, 20 percent earned a score of 78 or lower.
## Deciles and Common Percentiles
Given a data set that has been ordered in increasing magnitude, the median, first quartile, and third quartile can be used split the data into four pieces. The first quartile is the point at which one-fourth of the data lies below it. The median is located exactly in the middle of the data set, with half of all the data below it. The third quartile is the place where three-fourths of the data lies below it.
The median, first quartile, and third quartile can all be stated in terms of percentiles. Since half of the data is less than the median, and one-half is equal to 50 percent, the median marks the 50th percentile. One-fourth is equal to 25 percent, so the first quartile marks the 25th percentile. The third quartile marks the 75th percentile.
Besides quartiles, a fairly common way to arrange a set of data is by deciles. Each decile includes 10 percent of the data set. This means that the first decile is the 10th percentile, the second decile is the 20th percentile, etc. Deciles provide a way to split a data set into more pieces than quartiles without splitting the set into 100 pieces as with percentiles.
## Applications of Percentiles
Percentile scores have a variety of uses. Anytime that a set of data needs to be broken into digestible chunks, percentiles are helpful. They are often used to interpret test scores—such as SAT scores—so that test-takers can compare their performance to that of other students. For example, a student might earn a score of 90 percent on an exam. That sounds pretty impressive; however, it becomes less so when a score of 90 percent corresponds to the 20th percentile, meaning only 20 percent of the class earned a score of 90 percent or lower.
Another example of percentiles is in children’s growth charts. In addition to giving a physical height or weight measurement, pediatricians typically state this information in terms of a percentile score. A percentile is used in order to compare the height or weight of a child to other children of the same age. This allows for an effective means of comparison so that parents can know if their child’s growth is typical or unusual.
## How to Calculate Percentile Step by Step (and Examples)
When scoring tests or other sets of important values, the raw number is not necessarily enough to relate the scores to one another. You might score a 75 on a college exam, but that only tells you your score and not how it compares to the scores of your classmates. This is when percentiles are useful. To get a full understanding of your score, you need to be able to relate it to the others to determine if your score is higher than, lower than, or equal to the average score for that test among your classmates. This article will discuss how percentiles are defined and how to calculate them.
## What is a percentile?
A percentile is a term used in statistics to express how a score compares to other scores in the same set. While there is technically no standard definition of percentile, it’s typically communicated as the percentage of values that fall below a particular value in a set of data scores.
Percentiles are commonly used to report values from norm-referenced tests (in which the average is determined by comparing a set of results in the same group) as the percentages of scores that fall below those of the average of the set. For example, a male child age 12 with a weight of 130 pounds is at the 90th percentile of weight for males of that age, which indicates that he weighs more than 90 percent of other 12-year-old boys.
In statistical terms, there are three separate definitions of percentile. They are:
1. Greater than: The kth percentile is the lowest score in a data set that is greater than a percentage (k) of the scores. For example, if k = .25, you’d be trying to identify the lowest score that is greater than 25% of scores in the data set.
2. Greater than or equal to: The k**th percentile is the lowest score in the data set that is greater than or equal to a percentage (k) of the scores. For example, if k = .25, you’d be looking for a value that is greater or equal to 25% of the scores.
3. Weighted average: In this method, the kth percentile is the weighted average of the percentiles calculated in the two definitions above. This method allows for numbers to be more neatly rounded and defines the median of the set as the 50th percentile.
Percentiles are useful as they can tell you how one value compares to other values in the data set. Typically, if value n is at the kth percentile, then n is greater than k**% of the values in the set. One of the most common examples to express this is by looking at test scores. Calculating where your score lies in the data set (your score and the scores of the other test-takers), can help you figure out how your score compares to the rest of the scores.
Another example is tracking the weight of children compared to other children of the same age. This can help you determine whether your child’s weight falls into an acceptable range for its age, or if you need to take steps to help them gain or lose weight to get closer to an average weight.
### Percentile terms
Percentiles can also be used to split your dataset into portions to measure dispersion and identify the average of the values (known as the central tendency). Certain meaningful percentiles are referred to by their own terms. Here are some:
### Median
The 50th percentile, in which half the values of a data set are above the 50th percentile, and half are below.
### Quartile
Values that split the data set into quarters based on percentiles.
• The first quartile is referred to as Q1 or the lower quartile. This value is the 25th percentile, in which the lower quarter of the values fall below the 25th percentile while three quarters are above it.
• The second quartile, or Q2, is the value at the 50th percentile. This is the median of the data set.
• Q3, the third quartile, is referred to as the ‘upper quartile’ and is the value of the 75th percentile, meaning only 25% of values in the set are above this value.
### Interquartile range
Used to measure the dispersion of values, this is meant to show the middle half of the data. One quarter of the data values sit above this number, and one quarter falls below. The IQR is calculated by finding the difference between the first quartile and the third quartile (Q3 – Q1). The larger the IQR, the more spread out the values.
## How to calculate percentile
Follow these steps to calculate the kth percentile:
1. Rank the values in the data set in order from smallest to largest.
2. Multiply k (percent) by n (total number of values in the data set). This is the index. You’ll refer to this in the next steps as the position of a value in your data set (first, second, third…).
3. If the index is not a round number, round it up (or down, if it’s closer to the lower number) to the nearest whole number.
4. Use your ranked data set to find your percentile. Refer to the value that correlates with the index number, as determined in step 3. Since the value for the kth percentile must be greater than the values that precede the index, the next ranked value would be the kth percentile. Similarly, when using the ‘greater than or equal to’ method, steps 1-3 remain the same, but this time we would include the index value. The kth percentile is then calculated by taking the average of the index value in your data set and the next ranked value.
## Calculating percentile example
Using the data set below, here’s an example of calculating the 60th percentile:
1. Rank the values in the data set in order from smallest to largest, as shown below.
2. Calculate the index. To find the 60th percentile using the data set below, multiply k (.6) by n (8) to reach an index of 4.8.
3. Round the index to the nearest whole number (5).
4. To calculate percentile according to the ‘greater than’ method, count the values in your data set from smallest to largest until you reach the number ranked 5th, as determined in step 3. Since the value for the 60th percentile must be greater than the first five values, the 6th ranked value would be the kth (60th) percentile. In this data set, that value is 67.
5. Using the ‘greater than or equal to’ method, steps 1-3 above remain the same, but this time we would include the fifth-ranked value, or 56, in this data set. The kth (60th) percentile is then calculated by taking the average of that value in your data set (56) and the next ranked value (67). (56 + 67) / 2 = 61.5
There are several methods that can be used to calculate a percentile and these steps demonstrate one of the simplest ways to do so.
### Percentile range
A percentile range is expressed as the difference between any two specified percentiles. In this example, the 10-90 percentile range will be used. To find the 10-90 percentile range of the sample data set above, follow these steps:
#### 1. Follow the steps above to calculate the 10th percentile.
• (.1 x 8)=.8 (round to 1)
• K=33 (greater than) and k=30 (greater than or equal to)
• Average. (33 + 30) / 2 = 31.5
#### 2. Find the 90th percentile following the steps above.
• (.9 x 8)=7.2 (round to 7)
• K=72 (greater than), k=68 (greater than or equal to)
• Average (72 + 68) / 2 = 70
#### 3. Subtract the 10th percentile from the 90th percentile.
• 70 – 31.5 = 38
## Formula for Percentile
The Percentile Formula is given as,
Another formula to find the percentile is given by:
Here,
n = Ordinal rank of the given value or value below the number
N = Number of values in the data set
P = Percentile
Rank = Percentile/100
Ordinal rank for Percentile value = Rank × Total number of values in the list
### Solved Example
Question 1:
The scores for student are 40, 45, 49, 53, 61, 65, 71, 79, 85, 91. What is the percentile for score 71?
Solution:
Given,
No. of. scores below 71 = 6
Total no. of. scores = 10
The formula for percentile is given as, Percentile = (Number of Values Below “x” / Total Number of Values) × 100
Percentile of 71
= (6/10) × 100
= 0.6 × 100 = 60
Question 2:
Consider the list {50, 45, 60, 25, 30}. Find the 5th, 30th, 40th, 50th and 100th percentiles of the list given.
Solution:
Given list – 50, 45, 60, 25, 30
Ordered list – 25, 30, 45, 50, 60
N = 5
### What is Percentile?
Mostly we can define percentile as a number where a certain percentage of scores fall below that given number. Percentile and percentage are often confused, but both are different concepts. The percentage is used to express fractions of a whole, while percentiles are the values below which a certain percentage of the data in a data set is found. If you want to know where you stand compared to the rest of the crowd, you need a statistic that reports relative standing, and that statistic is called a percentile.
For example, you are the fourth tallest person in a group of 20.80% of people who are shorter than you. That means you are at the 80th percentile.
If your height is 5.4inch then “5.4 inch” is the 80th percentile height in that group.
Percentile Formula is given as –
Where, n =ordinal rank of a given value
N = number of values in the data set,
P = Percentile
### Formula for Percentile
In mathematics, we use the term percentile to understand and interpret data. It is also used informally to indicate that a certain percentage falls below that percentile value. Percentiles play a vital role in everyday life in understanding values such as test scores, health indicators, and other measurements. For example, if we score in the 25th percentile, then 25% of test-takers are below this score. Here 25 is called the percentile rank. Percentile divides a data set into 100 equal parts. A percentile is a measurement that tells us what percent of the total frequency of a data set was at or below that measure. As an example, let us consider a student’s percentile in some exams.
If on this test, a given student scored in the 60th percentile on the quantitative section, she scored at or better than 60% of the other students. Further, if a total of 500 students took the test, then this student scored at or better than 500 × 0.60 = 300 students out of 500 students. In other words, 200 students scored better than that particular student.
Therefore, percentiles are used to understand and interpret the given data. They also indicate the values below which a certain percentage of the data in a data set is found. Percentiles are frequently useful to understand the test scores and biometric measurements in our day-to-day routine.
Percentile is calculated by the ratio of the number of values below ‘x’ to the total number of values.
The Percentile Formula is given as,
### Procedure to Calculate Kth Percentile
The kth percentile is a value in a data set that divides the data into two parts. The lower part contains k percent of the data, and the upper part contains the rest of the data.
The following are the steps to calculate the kth percentile (where k is any number between zero and one hundred).
Step 1: Arrange all data values in the data set in ascending order.
Step 2: Count the number of values in the data set where it is represented as ‘n’.
Step 3: calculate the value of k/100, where k = any number between zero and one hundred.
Step 4: Multiply ‘k’ percent by ‘n’.The resultant number is called an index.
Step 5: If the resultant index is not a whole number then round to the nearest whole number, then go to Step 7. If the index is a whole number, then go to Step 6.
Step 6: Count the values in your data set from left to right until you reach the number. Then find the mean for that corresponding number and the next number. The resultant value is the kth percentile of your data set.
Step 7: Count the values in your data set from left to right until you reach the number. The obtained value will be the kth percentile of your data set.
### Solved Examples
Percentile formula example.
Example 1: Let us consider the percentile example problem: In a college, a list of grades of 15 students has been declared. Their grades are given as: 85, 34, 42, 51, 84, 86, 78, 85, 87, 69, 74, 65. Find the 80th percentile?
Solution:
Step 1:
Arrange the data in ascending order.
Let us arrange in Ascending Order = 34, 42, 51, 65, 69, 74, 78, 84, 85, 85, 86, 87.
Step 2:
Find Rank,
k = 0.80
Step 3:
Find 80th percentile,
80th percentile = 0.80 ×
12
= 9.6
Step 4:
Since it is not a whole number, round to the nearest whole number.
Therefore, 9.6 is rounded to 10.
Now, Count the values from left to right in the given data set until you reach the number 10.
From the given data set, the 10th number is 85.
Therefore, the 80th percentile of the given data set is 85
Example 2: The scores for students are 40, 45, 49, 53, 61, 65, 71, 79, 85, 91. What is the percentile for score 71?
Solution: We have,
No. of. scores below 71 = 6 Total no. of. scores = 10
The formula for percentile is given as,
= 0.6 × 100 = 60
Example 3: If in the test, a student scored 60th percentile on the quantitative section and if 500 students took the test, then this student scores at or better than will be?
Solution: 500 ×0.60=300
Thus Percentile also indicates the values below which a certain percentage of the data in a data set is found.
1. What is the Percentile Rank Formula?
Percentile Rank is a usual term mainly used in statistics which is arrived from Percentile. Percentile (also referred to as Centile) is the percentage of scores that range between 0 and 100 which is less than or equal to the given set of distribution. Percentiles divide any distribution into 100 equal parts. This is predominantly used for the interpretation of scores with different ranges across various tests. The percentile Rank formula (PR) is arrived at based on the total number of ranks and the number of ranks below and above percentile.
A Percentile Rank formula is given by:
Where,
• M = Number of Ranks below x
• R = Number of Ranks equals x
• Y = Total Number of Ranks
2. What is the Percentile Range?
A percentile range is a difference between two specified percentiles. These could theoretically be any two percentiles, but the 10-90 percentile range is the most common. To find the 10-90 percentile range:
1. Calculate the 10th percentile using the percentile formula.
2. Calculate the 90th percentile using the percentile formula.
3. Subtract the 10th percentile from the 90th percentile.
3. What is the Formula for Percentile?
In most common terms it is defined as the simplest and straightforward way of calculating percentile. With the help of the formula one can easily calculate the score of rank which can be equal to, or less than the given score. It can easily give you an idea of a perfect score.
4. What is the Percentile Rank Formula?
In general terms, percentile ranks are frequently expressed as a number between 1 and 99, with 50 being the average. Thus to calculate the Percentile Rank the formula which is discovered is R = P / 100 (N + 1) where; R represents the rank order of the score, P represents the percentile rank and N represents the number of scores that is in the distribution.
5. What is percentile used for?
Surprising the Percentile will tell you how to compare values with other values. It will give you an understanding of relative standing value. Besides, it will also give you an idea of calculating different mathematical problems by providing you with the concept of using different methods to solve the equations.
### Solved Examples for Percentile Formula
Q.1: The scores for some candidates in a test are 40, 45, 49, 53, 61, 65, 71, 79, 85, 91. What will be the percentile for the score 71? Use the percentile formula.
Solution: Given parameters:
No. of. Scores below the value “71” i.e. n = 6
Total no. of. Scores i.e. N = 10
The formula for percentile is given as below:
Thus, Percentile of 71 will be 60.
Q.2: The scores for some candidates in a test are 40, 45, 49, 53, 61, 65, 71, 79, 85, 91. What will be the score with a percentile value of 90?
Solution: It is obvious that we need to find the score “x” with percentile 90 for the given data set.
Thus, No. of. Scores below the value “x” i.e. n =?
Total no. of. Scores i.e. N = 10
Given percentile value, P = 90
The formula for percentile is given as below:
Thus 9th value in the data set i.e. 85 will be the score with 90 percentile.
Example 1:
If the scores of a set of students in a math test are 20 , 30 , 15 and 75 what is the percentile rank of the score 30?
Arrange the numbers in ascending order and give the rank ranging from 1 to the lowest to 4 to the highest.
Note that, if the percentile rank R is an integer, the P th percentile would be the score with rank R
when the data points are arranged in ascending order.
If R is not an integer, then the P
th percentile is calculated as shown.
Let I be the integer part and be the decimal part of D of R . Calculate the scores with the ranks I and I+1 . Multiply the difference of the scores by the decimal part of R . The P th percentile is the sum of the product and the score with the rank I
.Example 2:
Determine the 35 th percentile of the scores 7,3,12,15,14,4 and 20
.Arrange the numbers in ascending order and give the rank ranging from 1 to the lowest to 7 to the highest.
The integer part of R is 2 , calculate the score corresponding to the ranks 2 and 3 . They are 4 and 7 . The product of the difference and the decimal part is 0.45(7−4)=1.35
.
Therefore, the 35 th percentile is 2+1.35=3.35 .
Math Formulas ⭐️⭐️⭐️⭐️⭐
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### Iff
Prove that if n is a triangular number then 8n+1 is a square number. Prove, conversely, that if 8n+1 is a square number then n is a triangular number.
### Relative Powers
Weekly Problem 10 - 2007
The square of a number is 12 more than the number itself. The cube of the number is 9 times the number. What is the number?
### Summing Squares
Discover a way to sum square numbers by building cuboids from small cubes. Can you picture how the sequence will grow?
##### Stage: 4 Challenge Level:
Adithya from Lexden Primary School noted that there is a pattern to the gap between square numbers, with the gap increasing by two each time. Adithya also noticed a pattern in column A; this was also spotted by James from Wilson's School, Niteesh from Vidyashilp Academy, Dan and Linden from Wilson's School, and James from Eltham College. Here is a picture of Linden's spreadsheet showing the pattern:
Column A contains squares of odd numbers. Column D contains squares of even numbers which are not divisible by four. Column H contains squares of multiples of four.
An anonymous solver sent us these thoughts:
Firstly I noticed that all of the squares of odd numbers were in the left-hand column, column A. All of the squares of even multiples of 4 were in the right-hand column, column H, while all of the squares of other even numbers are in column D.
I can explain these by saying that the squares of multiples of $4$ are always multiples of $8$. The squares of any other even numbers are always multiples of $4$, while the squares of odd numbers do not all have a common factor and therfore are only divisble by one.
I noticed that the final digit of the odd squares was always $1$, $5$ or $9$ and for the even squares always $4$, $6$ or $0$. I also noticed that the first square number was in column A, then the second in column D, the third back in column A and the fourth in column H. I also noticed that in column A, which has the squares of odd numbers, the rows between the squares increases by $1$ every time, for example there are $0$ rows between $1$ and $9$, $1$ row between $9$ and $25$, $2$ rows between $25$ and $49$ etcetera. This therefore tells me that the difference between odd numbers is also always a multiple of $8$.
I can also see that when each odd square is represented in the form $2n+1$ $n$ is always a multiple of $4$ ($2\times4+1=9, 2\times40+1=81$ etc.) I can use examples to demonstrate this, for example $13\times13=169$ which not only ends in $9$ but is also $2\times84+1$, and $84$ is a multiple of $4$. Not only this, the difference between $169$ and the previous odd square is $48$, which is a multiple of $8$, just like the pattern suggested. The patterns also work for the next square numbers, i.e. $14\times14=196$ and $15\times15=225$.
Lots of fantastic ideas - can it be proved that these patterns will continue?
Herschel from the European School of Varese, and Simon who didn't give his school both sent us algebraic explanations for some of the patterns. Here is Herschel's solution:
The square numbers show a definite pattern:
$1^2=1$ is in column A.
$2^2=4$ is in column D.
$3^2=9$ is in column A.
$4^2=16$ is in column H.
This pattern (A,D,A,H) repeats itself throughout the series, albeit with larger gaps (blank rows) as the numbers increase.
This can be proven algebraically:
Suppose we call $n$ any multiple of $4$ We can write any whole number as either $n$, $n+1$, $n+2$ or $n+3$. Since $n$ is a multiple of $4$, we can re-write $n$ as $4a$, so that all whole numbers are either $4a$, $4a+1$, $4a+2$ or $4a+3$.
$(4a)^2 = 16a^2$, which is $8\times(2a^2)$, so it is a multiple of $8$ and hence appears in column H.
$(4a+1)^2 = 16a^2+8a+1$ which is $8\times(2a^2+a) + 1$, so it is $1$ more than a multiple of $8$ and hence appears in column A.
$(4a+2)^2 = 16a^2+16a+4$ which is $8\times(2a^2+2a) + 4$, so it is $4$ more than a multiple of $8$ and hence appears in column D.
$(4a+3)^2 = 16a^2+24a+9$ which is $8\times(2a^2+3a+1) + 1$, so it is $1$ more than a multiple of $8$ and hence appears in column A.
After $4a+3$, the next number is $4a+4$ which is $4(a+1)$, so we increase $a$ by $1$ and continue the trend. This means the pattern of highlighted squares is H,A,D,A (or A,D,A,H since we ignore $0^2=0$) and this continues forever.
Well done to all those who submitted solutions!
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Project Euler
Permuted multiples
Problem 52
It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.
Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.
Unbelievably, I was able to do this by hand in less than 30 minutes.
First, the number has to be at least 6 digits to be able to satisfy the criteria because each of the multiples are a permutation of the original number. Since we are looking for the smallest, I start with 6 digits and thinking that I will have to worry about 7 digits later. Turn out the answer is a 6 digits integer.
Next, the first digit must be 1 or it will become 7 digits at 5x if it is greater than 1.
` x: 1 _ _ _ _ _`
Now, if we have only 6 digits, and the first digit is 1, the last digit must be 7 because 2,4,5,6,8 will give us a zero at some point for the unit digit which is impossible for a 6 digits integer to start with zero as one of the permutation, and the multiple of 3 or 9 will never produce 1 as the last digit in any of the multiples.
` x: 1 _ _ _ _ 7`
In fact, since we know the last digit is 7, we also know that 1,2,4,5,7, and 8 are the digits we must work with because those are the unit digit produced by 7’s multiples. The order of those last digits from 1x to 6x are 7, 4, 1, 8, 5, and 2. We also know that the order of the first digits must be 1, 2, 4, 5, 7, 8 because the first digit must grow in sequence.
``` x: 1 _ _ _ _ 7
2x: 2 _ _ _ _ 4
3x: 4 _ _ _ _ 1
4x: 5 _ _ _ _ 8
5x: 7 _ _ _ _ 5
6x: 8 _ _ _ _ 2```
Now, working from 1x to 2x, we know that the 2nd digit must not carry because 1st digit must be 2 at 2x. Therefore, 2 or 4 must be the 2nd digit in the original number. We also know that 4 can not be in front of 5, 7, and 8 because 4×2+1 = 9, which is not one of the digit we are working with. In another word, 4 must be in front of 2. Therefore
``` x: 1 4 2 _ _ 7
2x: 2 8 5 _ _ 4```
Both combination of 5, 8 work from 1x to 2x. However, if the original 5th digit place is 8 and last digit is 7, then it will produce 3×8+2 = 6 at 3x, which is not one of the digit we are working with again. Therefore,
` x: 1 4 2 8 5 7`
Continue to multiply x up to 6x will find that this number works and it is the smallest positive integer that works.
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# 7.10: Map Quest 5
Difficulty Level: At Grade Created by: CK-12
Look at the map and list of facts below. Can you figure out the distances in miles between the cities using a ruler? Can you name the cities using the facts? In this concept we will learn how to read and work with maps.
• Fact 1: Hilton to Time is 20 miles.
• Fact 2: Time to North Shore by way of Hilton is 80 miles.
• Fact 3: Westway to North Shore is 20 miles less than Westway to Raleigh.
• Fact 4: Time to Pepper is 20 miles longer than Pepper to Edmonton.
• Fact 5: Edmonton is 80 miles from Raleigh.
### Guidance
In order to solve the problem above, use the problem solving steps.
• Start by describing what information is given.
• Then, identify what your job is. In these problems, your job will be to figure out where each city is on the map.
• Next, make a plan for how you will solve. In these problems, measure distances with a ruler and use the facts to figure out which city is which.
• Then, solve the problem.
• Finally, check your solution. Verify that the measured distances between the cities work with the facts.
#### Example A
Use your ruler and the key to figure out distances between cities. Use the facts to name the cities.
• Fact 1: Portland to Ellsworth by way of Surrey is 500 miles.
• Fact 2: Surrey is 300 miles from Ellsworth.
• Fact 3: Felton to Dayton is 50 miles less than Felton to Granada.
• Fact 4: Simpson to Dayton round trip is 200 miles.
Solution:
We can use problem solving steps to help.
$& \mathbf{Describe:} && \text{The map shows 7 cities. One is Surrey. The key shows that one inch stands for}\\&&& \text{100 miles. The facts give information about location and distances between cities.}\\& \mathbf{My \ Job:} && \text{Use the facts. Name the cities.}\\& \mathbf{Plan:} && \text{Measure distances in inches. Use the key to figure out actual distances. Use the}\\&&& \text{facts to name the cities.}\\& \mathbf{Solve:} && \text{Fact} \ 2: \ \ \text{B is Ellsworth}\\&&& \text{Fact} \ 1: \ \ \text{A is Portland}\\&&& \text{Fact} \ 3: \ \ \text{D is Felton, E is Dayton, C is Granada}\\&&& \text{Fact} \ 4: \ \ \text{F is Simpson}\\& \mathbf{Check:} && \text{Fact} \ 1: \ \ \text{Portland to Ellsworth by way of Surrey is} \ 5 \ \text{inches or} \ 500 \ \text{miles.}\\&&& \text{Fact} \ 2: \ \ \text{Surrey to Ellsworth is } \ 3 \ \text{inches, or} \ 3 \times 100, \ \text{or} \ 300 \ \text{miles.}\\&&& \text{Fact} \ 3: \ \ \text{Felton to Dayton is} \ 1.5 \ \text{inches, or} \ 1.5 \times 100, \ \text{or} \ 150 \ \text{miles.}\\&&& \qquad \qquad \ \text{Felton to Granada is} \ 2 \ \text{inches or,} \ 2 \ \times 100, \ \text{or} \ 200 \ \text{miles.}\\&&& \qquad \qquad \ 200 - 150 = 50 \ \text{miles}\\&&& \text{Fact} \ 4: \ \ \text{Simpson to Dayton roundtrip is} \ 2 \ \text{inches, or} \ 200 \ \text{miles.}$
#### Example B
Use your ruler and the key to figure out distances between cities. Use the facts to name the cities.
• Fact 1: Regis to Hoosier round trip is 70 miles less than the round trip from Regis to Bandoff.
• Fact 2: Sandford to Tulsa is the same distance as Eckard to Tulsa.
• Fact 3: Eckard to Adams is 35 miles more than Adams to Hoosier.
Solution:
We can use problem solving steps to help.
$& \mathbf{Describe:} && \text{The map shows 7 cities. One is Regis. The key shows that one inch stands for}\\&&& \text{70 miles. The facts give information about location and distances between cities.}\\& \mathbf{My \ Job:} && \text{Use the facts. Name the cities.}\\& \mathbf{Plan:} && \text{Measure distances in inches. Use the key to figure out actual distances. Use the}\\&&& \text{facts to name the cities.}\\& \mathbf{Solve:} && \text{Fact} \ 1: \ \ \text{F is Bandoff and E is Hoosier}\\&&& \text{Fact} \ 2: \ \ \text{B is Tulsa}\\&&& \text{Fact} \ 3: \ \ \text{A is Sanford, C is Eckard, D is Adams}\\& \mathbf{Check:} && \text{Fact} \ 1: \ \ \text{Regis to Hoosier round trip is } \ 5 \ \text{inches or} \ 350 \ \text{miles.}\\&&& \qquad \ \text{Regis to Bandoff round trip is } \ 4 \ \text{inches or} \ 280 \ \text{miles.}\\&&& \text{Fact} \ 2: \ \ \text{Sandford to Tulsa and Eckard to Tulsa are both} \ 1 \ \text{inch, or} \ 1 \times 70, \ \text{or} \ 70 \ \text{miles.}\\&&& \text{Fact} \ 3: \ \ \text{Eckard to Adams is} \ 2.5 \ \text{inches, or} \ 2.5 \times 70, \ \text{or} \ 175 \ \text{miles.}\\&&& \qquad \qquad \ \text{Adams to Hoosier is} \ 2 \ \text{inches or,} \ 2 \ \times 70, \ \text{or} \ 140 \ \text{miles.}\\&&& \qquad \qquad \ 175 - 140 = 35$
#### Concept Problem Revisited
• Fact 1: Hilton to Time is 20 miles.
• Fact 2: Time to North Shore by way of Hilton is 80 miles.
• Fact 3: Westway to North Shore is 20 miles less than Westway to Raleigh.
• Fact 4: Time to Pepper is 20 miles longer than Pepper to Edmonton.
• Fact 5: Edmonton is 80 miles from Raleigh.
We can use problem solving steps to help.
$& \mathbf{Describe:} && \text{The map shows 7 cities. One is Hilton. The key shows that one inch stands for}\\&&& \text{40 miles. The facts give information about location and distances between cities.}\\& \mathbf{My \ Job:} && \text{Use the facts. Name the cities.}\\& \mathbf{Plan:} && \text{Measure distances in inches. Use the key to figure out actual distances. Use the}\\&&& \text{facts to name the cities.}\\& \mathbf{Solve:} && \text{Fact} \ 1: \ \ \text{F is Time}\\&&& \text{Fact} \ 2: \ \ \text{E is North Shore}\\&&& \text{Fact} \ 3: \ \ \text{D is Westway}\\&&& \text{Fact} \ 4: \ \ \text{A is Pepper and B is Edmonton}\\&&& \text{Fact} \ 5: \ \ \text{C is Raleigh}\\& \mathbf{Check:} && \text{Fact} \ 1: \ \ \text{Hilton to Time is} \ \frac{1}{2} \ \text{inch or} \ 20 \ \text{miles.}\\&&& \text{Fact} \ 2: \ \ \text{Time to North Shore by way of Hilton is} \ 2 \ \text{inches, or} \ 2 \times 40, \ \text{or} \ 80 \ \text{miles.}\\&&& \text{Fact} \ 3: \ \ \text{Westway to North Shore is} \ 2 \ \text{inches, or} \ 2 \times 40, \ \text{or} \ 80 \ \text{miles.}\\&&& \qquad \qquad \ \text{Westway to Raleigh is} \ 2 \ \frac{1}{2} \ \text{inches or,} \ 2 \ \frac{1}{2} \times 40, \ \text{or} \ 100 \ \text{miles.}\\&&& \qquad \qquad \ 100 - 80 = 20 \ \text{miles}\\&&& \text{Fact} \ 4: \ \ \text{Time to Pepper is} \ 1 \ \text{inch, or} \ 40 \ \text{miles.}\\&&& \qquad \qquad \ \text{Pepper to Edmonton is} \ \frac{1}{2} \ \text{inch, or} \ 20 \ \text{miles.}\\&&& \qquad \qquad \ 40 - 20 = 20 \ \text{miles}\\&&& \text{Fact} \ \ 5: \ \text{Edmonton to Raleigh is} \ 2 \ \text{inches, or} \ 2 \times 40, \ \text{or} \ 80 \ \text{miles.}$
### Vocabulary
A map is a picture that represents an area of land. A ruler is a device that measures and finds the distance between two points.
### Guided Practice
1. Use your ruler and the key to figure out distances between cities. Use the facts to name the cities.
• Fact 1: Sample to Telegraph to Antonville is 180 miles.
• Fact 2: Sample to Readville round trip is 100 miles.
• Fact 3: Readville is 60 miles from Antonville.
• Fact 4: Diablo to Carlton is 15 miles less than Diablo to Bedford.
• Fact 5: Bedford to Sample to Readville is 60 miles.
2. Use your ruler and the key to figure out distances between cities. Use the facts to name the cities.
• Fact 1: Brickyard to Yerba is 300 miles
• Fact 2: Yerba to Brio is the same distance as Yerba to Tempe.
• Fact 3: Superstition is 30 miles closer to Tempe than to Yerba.
• Fact 4: Brickyard to Danyo is 90 miles.
• Fact 5: Littleton is 30 miles from Danyo.
3. Use your ruler and the key to figure out distances between cities. Use the facts to name the cities.
• Fact 1: Trenton to Storrs is 20 miles more than Storrs to Wentworth.
• Fact 2: Wentworth to Storrs is 20 miles.
• Fact 3: Storrs to Leopold round trip is 120 miles
• Fact 4: New Bertson to Leopold to Carlton is 70 miles.
• Fact 5: Halland to Leopold is 10 miles more than Leopold to Carlton.
1. A is Bedford, B is Sample, C is Readville, D is Antonville, E is Carlton, F is Diablo
2. A is Tempe, B is Superstition, C is Yerba, D is Brio, E is Danyo, F is Littleton
3. A is Wentworth, B is Carlton, C is Leopold, D is Halland, E is New Bertson, F is Trenton
### Practice
1. Use your ruler and the key to figure out distances between cities. Use the facts to name the cities.
• Fact 1: Grinder to Riverwalk to Sienna is 225 miles.
• Fact 2: Sienna to Riverwalk round trip is 270 miles.
• Fact 3: Hartman to Grinder is 180 miles.
• Fact 4: Snake City to Sienna is the same distance as Sienna to Briarwood.
• Fact 5: Ringville to Hartman is 45 miles.
2. Use your ruler and the key to figure out distances between cities. Use the facts to name the cities.
• Fact 1: Calder to Whitton to Ballard is 360 miles.
• Fact 2: Ballard roundtrip to Nicksville is 400 miles.
• Fact 3: Calder to Exeter is 40 miles more than Exeter to Reston.
• Fact 4: Reston to Franklin is 240 miles.
• Fact 5: Nicksville to Franklin is 120 miles.
3. Use your ruler and the key to figure out distances between cities. Use the facts to name the cities.
• Fact 1: Girard to Browning by way of Dallas is 125 miles.
• Fact 2: Girard to Dallas is the same distance as Crowville to Browning.
• Fact 3: Houston to Coventry is 75 miles less than Browning to Girard.
• Fact 4: Coventry to Winston is 100 miles.
4. Use your ruler and the key to figure out distances between cities. Use the facts to name the cities.
• Fact 1: Redwood to Mateo is 75 miles.
• Fact 2: Mateo to Evergreen is the same distance as Evergreen to Longwood.
• Fact 3: Redwood to Berkview is 100 miles.
5. Use your ruler and the key to figure out distances between cities. Use the facts to name the cities.
• Fact 1: Albany to Queens is 180 miles.
• Fact 2: Venice to Albany is 90 miles.
• Fact 3: Albany to Sacrey is 240 miles.
• Fact 4: Sacrey to Denton is the same distance as Albany to Venice.
Jan 18, 2013
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## College Algebra (11th Edition)
$1.0791$
$\bf{\text{Solution Outline:}}$ Use the Laws of Logarithms to find an equivalent expression for the given expression, $\log_{10} 12 ,$ which uses the given logarithmic values, $\log_{10} 2= 0.3010$ and/or $\log_{10} 3= 0.4771 .$ $\bf{\text{Solution Details:}}$ The given expression can be expressed as \begin{array}{l}\require{cancel} \log_{10} (4\cdot3) \\\\= \log_{10} (2^2\cdot3) .\end{array} Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the given expression is equivalent to \begin{array}{l}\require{cancel} \log_{10} 2^2+\log_{10} 3 .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2\log_{10} 2+\log_{10} 3 .\end{array} Substituting the known values of the logarithmic expressions results to \begin{array}{l}\require{cancel} 2(0.3010) 2+0.4771 \\\\= 0.6020 2+0.4771 \\\\= 1.0791 .\end{array}
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# 5.4: Writing Proportions
Difficulty Level: At Grade Created by: CK-12
## Introduction
The Longest Books
Candice loves to read really long books. The first one that she selected was over 800 pages. The next one was 825 pages. She is not a very fast reader, but she enjoys all of the details that are put in really long books. Often, she walks through the bookstore looking at books that have tons of pages, then when she finds one that seems to be the right size, she takes it off the shelf to see if she is interested in the topic.
After 12 weeks, Candice had finished reading two books. She was very proud of herself given that she had read over 1600 pages in all. She had finished the first and second books in about the same length of time.
Given this, how long did it take Candice to read 1 book?
Because Candice read the two books in the same amount of time, you can use a proportion to solve this problem. Proportions are formed by equal ratios. Use this lesson to learn about proportions then you will be able figure out the time that it took Candice to read 1 book.
What You Will Learn
By the end of this lesson, you will be able to demonstrate the following:
• Recognize Proportions as a statement of equivalent ratios.
• Solve proportions with the variable in the numerator using algebra.
• Recognize equivalent proportions representing the same situation with the variable in the numerator.
• Model and solve real-world problems using proportions.
Teaching Time
I. Recognize Proportions as a Statement of Equivalent Ratios
A ratio represents a comparison between two quantities. We can write ratios in fraction form, using a colon or using the word “to”.
We also learned that equivalent ratios are two ratios that are equal. The numbers in the ratios may not be the same, but the comparison of quantities is the same.
Equivalent ratios are directly related to proportions.
What is a proportion?
A proportion states that two ratios are equivalent. Here is an example of a proportion.
12=24\begin{align*}\frac{1}{2} = \frac{2}{4}\end{align*}
This proportion shows that the ratios 12\begin{align*}\frac{1}{2}\end{align*} and 24\begin{align*}\frac{2}{4}\end{align*} are equivalent. In other words, a proportion is made up of two equivalent ratios.
In the example above, we knew all of the parts of the two ratios that made up the proportion. Sometimes, we will know three of the numbers, but not all four of them. In this instance a variable is in the place of the missing number.
Look at this proportion.
12=n12\begin{align*}\frac{1}{2} = \frac{n}{12}\end{align*}
Notice that the first term of the second ratio––its numerator––is a variable. Suppose we wanted to find the value of this variable. We could do that by using proportional reasoning.
Proportional reasoning is figuring out a missing value in a proportion by thinking about the relationship between the numbers in the two ratios.
Take a look at this example.
Example
Use proportional reasoning to solve for n: 12=n12\begin{align*}n: \ \frac{1}{2} = \frac{n}{12}\end{align*}.
To figure this out, we need to figure out a relationship between either numerators or denominators. The proportion does not show the relationship between the first terms in the ratios––the numerators of the fractions. However, we can determine the relationship between the second terms in the ratios––the denominators of the fractions.
We can ask ourselves: “what number, when multiplied by 2, results in 12?”
Since 2×6=12\begin{align*}2 \times 6 = 12\end{align*}, we can multiply both the numerator and the denominator of 12\begin{align*}\frac{1}{2}\end{align*} by 6 to find the value of n\begin{align*}n\end{align*}.
12=1×62×6=612=n12\begin{align*}\frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12} = \frac{n}{12}\end{align*}
This shows that in this ratio the second term (the denominator) of the ratio is 12, the first term (the numerator) is 6.
The value of n\begin{align*}n\end{align*} is 6.
Good for you! Mental math is very helpful when using proportional reasoning. When you can figure out the relationship between numbers, then you can solve for the missing value of the variable.
Let’s look at another example.
Example
Use proportional reasoning to solve for x: 1535=x7\begin{align*}x: \ \frac{15}{35} = \frac{x}{7}\end{align*}.
Which relationship can we use to figure out the variable? This proportion does not show the relationship between the first terms in the ratios––the numerators of the fractions. We need to find the relationship between the second terms in the ratios––the denominators of the fractions.
We can ask ourselves, “what number can we divide 35 by to get 7?”
Since 35÷5=7\begin{align*}35 \div 5 = 7\end{align*}, we can divide both the numerator and the denominator of 1535\begin{align*}\frac{15}{35}\end{align*} by 5 to find the value of x\begin{align*}x\end{align*}.
1535=15÷535÷5=37=x7\begin{align*}\frac{15}{35} = \frac{15 \div 5}{35 \div 5} = \frac{3}{7} = \frac{x}{7}\end{align*}
This shows that in this ratio the second term (the denominator) of the ratio is 7, the first term (the numerator) is 3.
The value of x\begin{align*}x\end{align*} is 3.
Example
Use proportional reasoning to solve for z: z9=3236\begin{align*}z: \ \frac{z}{9} = \frac{32}{36}\end{align*}.
The proportion does not show the relationship between the first terms in the ratios––the numerators of the fractions. We need to find the relationship between the second terms in the ratios––the denominators of the fractions.
We can ask ourselves “what number, when multiplied by 9, results in 36?” The answer is 4.
9×4z9=36=z×49×4=3236.\begin{align*}9 \times 4 &= 36\\ \frac{z}{9} &= \frac{z \times 4}{9 \times 4} = \frac{32}{36}.\end{align*} From this, we can see that z×4=32\begin{align*}z \times 4 = 32\end{align*}.
We must ask ourselves, “what number, when multiplied by 4, results in 32?”
8×4=32\begin{align*}8 \times 4 = 32\end{align*}, so z=8\begin{align*}z = 8\end{align*}.
5H. Lesson Exercises
Use proportional reasoning to find each unknown.
1. 23=x6\begin{align*}\frac{2}{3} = \frac{x}{6}\end{align*}
2. 1224=24z\begin{align*}\frac{12}{24} = \frac{24}{z}\end{align*}
3. y5=1420\begin{align*}\frac{y}{5} = \frac{14}{20}\end{align*}
II. Solve Proportions with the Variable in the Numerator Using Algebra
Sometimes, it is difficult to use proportional reasoning alone to figure out the missing variable in a proportion. When this happens, we can use algebra to figure out the missing variable.
Using algebra involves thinking of multiplication or division in relationship to the values in the proportion. Sometimes, you will be multiplying as we did in the last section and sometimes, you will be dividing. Either way, we will be using our algebraic thinking to figure out the variable.
Example
Find the value of x: x10=2430\begin{align*}x: \ \frac{x}{10} = \frac{24}{30}\end{align*}
Here we can begin by looking at the relationship between the denominators. You can see that 10×3=30\begin{align*}10 \times 3 = 30\end{align*}.
Now we aren’t looking for the multiple of the first numerator, but we have the second numerator. We have to do the inverse operation of multiplication to figure out the value of the variable.
We have been given the value 24 as a numerator. The inverse operation for multiplication is division.
We can divide 24 by 3 to find the other numerator.
24÷3=8\begin{align*}24 \div 3 = 8\end{align*}
The value of x\begin{align*}x\end{align*} is 8.
5I. Lesson Exercises
Use inverse operations to find the value of each unknown variable.
1. 25=6x\begin{align*}\frac{2}{5} = \frac{6}{x}\end{align*}
2. a9=2036\begin{align*}\frac{a}{9} = \frac{20}{36}\end{align*}
3. 4b=2436\begin{align*}\frac{4}{b} = \frac{24}{36}\end{align*}
III. Recognize Equivalent Proportions Representing the Same Situation with the Variable in the Numerator
Sometimes, we will have equivalent proportions. This means that two proportions are equal to each other. When this happens, you will have a total of four ratios, but each of them will be equivalent. Let’s look at an example.
Example
1236=24=612\begin{align*}\frac{1}{2} &= \frac{2}{4}\\ \frac{3}{6} &= \frac{6}{12}\end{align*}
Each of these ratios is the same quantity. Therefore, they are all equivalent.
They have been set up as two proportions, but they are all equivalent proportions.
Why would we have equivalent proportions?
Well, you can think of real-life situations that would represent equivalent proportions. Real-life situations that might have equivalent proportions are found in situations like cooking.
Example
Tania and her mother are making cookies for a party. They aren’t sure if how many people are coming to the party. They aren’t sure if they should double, triple or quadruple the ingredients. The ratio of sugar to flour is 13\begin{align*}\frac{1}{3}\end{align*}. How will this change if they double, triple or quadruple the recipe?
Now think about this. We can figure out that the ratios will all be equivalent because we are starting with the same measurement. We are beginning with 13\begin{align*}\frac{1}{3}\end{align*}. To double it we multiply each value by 2. To triple we multiply by 3, to quadruple by four.
1339=26=412\begin{align*}\frac{1}{3} &= \frac{2}{6}\\ \frac{3}{9} &= \frac{4}{12}\end{align*}
Notice that the relationships between the ratios in the second proportion are not multiples of each other or divisible by each other, but if viewed as a fraction reduce to the same number in simplest form.
IV. Model and Solve Real-World Problems Using Proportions
Sometimes, the best way to solve a real-world problem is to set up and solve a proportion. Remember that because a proportion uses ratios, they are helpful when comparing quantities.
Example
At the vet's office, the ratio of cats to dogs in the waiting room is 2 to 3. If there are 6 dogs in the waiting room, what is the number of cats in the waiting room?
One way to set up a proportion for this problem would be to write two equivalent ratios, comparing cats to dogs. We will need to use a variable to represent the number of cats in the second ratio.
The ratio of cats to dogs is 2 to 3. So, we can express this ratio as a fraction.
catsdogs=23\begin{align*}\frac{cats}{dogs} = \frac{2}{3}\end{align*}
Let's write a second ratio comparing cats to dogs in the waiting room. We know that there are a total of 6 dogs in the waiting room. We don't know the total number of cats. So, we can use the variable c\begin{align*}c\end{align*}, to represent the unknown number of cats and set up a second equivalent ratio.
catsdogs=c6\begin{align*}\frac{cats}{dogs} = \frac{c}{6}\end{align*}
Since these two ratios are equivalent, we can put them together to form a proportion.
23=c6.\begin{align*}\frac{2}{3} = \frac{c}{6}.\end{align*}
To find the total number of cats in the vet's office, solve this proportion for c\begin{align*}c\end{align*}. The proportion does not show the relationship between the first terms in the ratios––the numerators of the fractions.
We need to find the relationship between the second terms in the ratios––the denominators of the fractions.
We can ask ourselves: what number, when multiplied by 3, results in 6?
Since 3×2=6\begin{align*}3 \times 2 = 6\end{align*}, we can multiply by 2 to find the value of c\begin{align*}c\end{align*}.
23=2×23×2=46=c6\begin{align*}\frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6} = \frac{c}{6}\end{align*}
So, the value of c\begin{align*}c\end{align*} is 4. That means that there are 4 cats in the waiting room.
## Real-Life Example Completed
The Longest Books
Here is the original problem once again. Reread it, underline the important information and figure out the proportion.
Candice loves to read really long books. The first one that she selected was over 800 pages. The next one was 825 pages. She is not a very fast reader, but she enjoys all of the details that are put in really long books. Often, she walks through the bookstore looking at books that have tons of pages, then when she finds one that seems to be the right size, she takes it off the shelf to see if she is interested in the topic.
After 12 weeks, Candice had finished reading two books. She was very proud of herself given that she had read over 1600 pages in all. She had finished the first and second books in about the same length of time.
Given this, how long did it take Candice to read 1 book?
First, let’s write a proportion to show that it took Candice 12 weeks to read two books.
12 weeks2 books\begin{align*}\frac{12 \ weeks}{2 \ books}\end{align*}
Next, she read the two books in the same amount of time, so we can set up a proportion to show that these ratios are equal.
12 weeks2 books=x1 book\begin{align*}\frac{12 \ weeks}{2 \ books} = \frac{x} {1 \ book}\end{align*}
Now we can look at the relationship between the denominators and determine the missing numerator.
2÷2=1\begin{align*}2 \div 2 = 1\end{align*}
We can do that to the numerator.
12÷2=6\begin{align*}12 \div 2 = 6\end{align*}
It took Candice 6 weeks to read 1 book.
## Vocabulary
Here are the vocabulary words that are found in this lesson.
Ratio
a comparison between two quantities. Ratios can be written in fraction form, using a colon or with the word “to”.
Equivalent Ratios
two ratios that are equal
Proportion
When two ratios are equal, they form a proportion.
Proportional Reasoning
deducing the relationship between the numerators or the denominators of a proportion. Anytime you have a proportion, there is some kind of relationship between the values.
## Time to Practice
Directions: Look at each pair of ratios. Tell whether or not these ratios form a proportion.
1. 12\begin{align*}\frac{1}{2}\end{align*} and 48\begin{align*}\frac{4}{8}\end{align*}
2. \begin{align*}\frac{3}{7}\end{align*} and \begin{align*}\frac{6}{14}\end{align*}
3. \begin{align*}\frac{5}{2}\end{align*} and \begin{align*}\frac{10}{6}\end{align*}
4. \begin{align*}\frac{3}{1}\end{align*} and \begin{align*}\frac{9}{3}\end{align*}
5. \begin{align*}\frac{2}{9}\end{align*} and \begin{align*}\frac{1.5}{4.5}\end{align*}
6. \begin{align*}\frac{4}{9}\end{align*} and \begin{align*}\frac{8}{10}\end{align*}
7. \begin{align*}\frac{1}{4}\end{align*} and \begin{align*}\frac{5}{20}\end{align*}
8. \begin{align*}\frac{3}{4}\end{align*} and \begin{align*}\frac{9}{10}\end{align*}
Directions: Use proportional reasoning to find the value of the variable in each proportion.
9. \begin{align*}\frac{1}{4} = \frac{a}{20}\end{align*}
10. \begin{align*}\frac{15}{30} = \frac{x}{2}\end{align*}
11. \begin{align*}\frac{2}{9} = \frac{n}{63}\end{align*}
12. \begin{align*}\frac{z}{7} = \frac{12}{21}\end{align*}
13. \begin{align*}\frac{3}{5} = \frac{t}{60}\end{align*}
14. \begin{align*}\frac{k}{72} = \frac{5}{12}\end{align*}
Directions: Use proportional reasoning to solve each series of problems.
15. There are 12 quarts of water in a sink. Cliff wants to know how many gallons of water are in the sink. Using this unit conversion 1 gallon = 4 quarts, Cliff writes two ratios that compare gallons to quarts. Then he writes this proportion to represent \begin{align*}g\end{align*}, the number of gallons of water in the sink.
\begin{align*}\frac{g}{12} = \frac{1}{4}\end{align*}
His sister Nikki says he could also have used this proportion to represent g, the number of gallons of water in the sink.
\begin{align*}\frac{g}{1} = \frac{4}{12}\end{align*}
Do the two proportions both have equal ratios? If not, which proportion could be used to find the actual number of gallons of water in the sink?
16. A machine that produces binder clips has been operating for 5 minutes. Mr. Wilson knows that the machine can produce 15 binder clips in 1 minute. So, Mr. Wilson writes two proportions that can be used to represent \begin{align*}n\end{align*}, the number of binder clips that the machine produced in 5 minutes.
a) \begin{align*}\frac{n}{5} = \frac{15}{1}\end{align*} and b) \begin{align*}\frac{n}{15} = \frac{5}{1}\end{align*}
Do the two proportions have equal ratios? If not, which proportion––a or b––could be used to find \begin{align*}n\end{align*}, the number of binder clips produced in 5 minutes?
Directions: The ratio of green tiles to orange tiles in a bag is 3 to 8. There are 48 orange tiles in the bag.
17. Set up a proportion that could be used to find g, the number of green tiles in the bag.
18. How many green tiles are in the bag?
Directions: Marty bought 2 pounds of ham for $11 at a deli. 19. Set up a proportion that could be used to find p, the number of pounds of ham Marty could have bought if he had spent$33.
20. How many pounds of ham could he have bought if he had spent \$33?
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# A helicopter flies 65km [N32°E] then 42km[E 21°] determine the total displacement of the helicopter.
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## To determine the total displacement of the helicopter, we can break down the two displacements into their horizontal and vertical components.
1. First, let's analyze the displacement of 65km [N32°E]:
- The direction N32°E means the displacement is 32° east of north (clockwise from north direction).
- We can split this displacement into its horizontal (east-west) and vertical (north-south) components using trigonometry:
- Horizontal component = 65km * cos(32°)
- Vertical component = 65km * sin(32°)
2. Next, let's analyze the displacement of 42km [E21°]:
- The direction E21° means the displacement is 21° east of east (counterclockwise from east direction).
- Since this displacement is purely horizontal, it only affects the east-west component.
3. Now, let's calculate the horizontal and vertical components for each displacement:
- Displacement 1:
- Horizontal component = 65km * cos(32°)
- Vertical component = 65km * sin(32°)
- Displacement 2:
- Horizontal component = 42km * cos(21°)
- Vertical component = 0 (since it is a purely horizontal displacement)
4. Finally, to determine the total displacement, we can add the horizontal and vertical components separately:
- Total horizontal displacement = Horizontal component of Displacement 1 + Horizontal component of Displacement 2
- Total vertical displacement = Vertical component of Displacement 1 + Vertical component of Displacement 2
By calculating the total horizontal and vertical displacements, you can find the magnitude and direction of the total displacement vector using the Pythagorean theorem and trigonometry.
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## All angles are measured CCW from +x-axis.
Disp. = 65km[58o] + 42km[21o].
(65*Cos58+65*sin58) + (42*Cos21+42*sin21 =
(34.44+55.12i) + (39.21+15.05i) =
73.65 + 70.17i = 101.7km[43.6o].
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# How do you integrate int 1/(x^2sqrt(x^2-7)) by trigonometric substitution?
Nov 25, 2016
$\frac{\sqrt{{x}^{2} - 7}}{7 x} + C$
#### Explanation:
$I = \int \frac{\mathrm{dx}}{{x}^{2} \sqrt{{x}^{2} - 7}}$
Apply the substitution $x = \sqrt{7} \sec \theta , \mathrm{dx} = \sqrt{7} \sec \theta \tan \theta d \theta$.
$\textcolor{w h i t e}{I} = \int \frac{\sqrt{7} \sec \theta \tan \theta d \theta}{7 {\sec}^{2} \theta \sqrt{7 {\sec}^{2} \theta - 7}}$
$\textcolor{w h i t e}{I} = \frac{1}{7} \int \frac{\tan \theta d \theta}{\sec \theta \sqrt{{\sec}^{2} \theta - 1}}$
$\textcolor{w h i t e}{I} = \frac{1}{7} \int \frac{d \theta}{\sec} \theta$
$\textcolor{w h i t e}{I} = \frac{1}{7} \int \cos \theta d \theta$
$\textcolor{w h i t e}{I} = \frac{1}{7} \sin \theta$
$\textcolor{w h i t e}{I} = \frac{1}{7} \tan \frac{\theta}{\sec} \theta$
$\textcolor{w h i t e}{I} = \frac{1}{7} \frac{\sqrt{{\sec}^{2} \theta - 1}}{\sec} \theta$
Using $x = \sqrt{7} \sec \theta \implies \sec \theta = \frac{x}{\sqrt{7}}$:
$\textcolor{w h i t e}{I} = \frac{1}{7} \frac{\sqrt{{x}^{2} / 7 - 1}}{\frac{x}{\sqrt{7}}}$
$\textcolor{w h i t e}{I} = \frac{1}{7} \frac{\frac{1}{\sqrt{7}} \sqrt{{x}^{2} - 7}}{\frac{x}{\sqrt{7}}}$
$\textcolor{w h i t e}{I} = \frac{\sqrt{{x}^{2} - 7}}{7 x} + C$
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# 2.4 Practice A Algebra 2 Answers: Solving For Variables
Algebra 2 can be difficult for some students. One of the most challenging topics is solving for variables. This lesson covers how to solve for variables and includes a practice exercise with answers. It is important to understand how to solve for variables in order to move forward in Algebra 2.
## Understanding Variables and Expressions
In Algebra 2, variables are used to represent unknown numbers. An expression is a combination of numbers, variables, and operations. Expressions can be used to represent relationships between numbers and to describe real-world situations. For example, the expression “2x + 3” is a combination of a number (2), a variable (x), and an operation (+). This expression could be used to represent the relationship between two numbers, such as the cost of two items that cost \$2 each, plus an additional \$3.
## Solving for Variables
Solving for variables means finding the value of the variable in an expression. This is done by isolating the variable on one side of the equation. To do this, you must use the rules of Algebra 2, such as the order of operations and combining like terms. For example, to solve for x in the expression “2x + 3”, you must use the order of operations to isolate the variable. The first step is to subtract 3 from both sides of the equation. This leaves “2x = -3”. The next step is to divide both sides by 2, which results in “x = -1.5”. This means that the value of x in the expression “2x + 3” is -1.5.
## 2.4 Practice A Algebra 2 Answers
The following are the answers to the 2.4 Practice A Algebra 2 exercise. The exercise asks students to solve for the variables in each expression. The answers are as follows:
• a) x = -3
• b) y = -2
• c) z = -7
• d) n = 2
• e) x = 4
• f) m = -5
• g) y = 5
• h) z = 8
• i) n = -3
• j) x = -2
##
Solving for variables is an important concept in Algebra 2. It is important to understand how to solve for variables in order to progress in the course. The 2.4 Practice A Algebra 2 exercise helps students practice this skill. The answers to the exercise are provided above for reference.
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# Math Teaching Strategies for Teaching Division in the 3rd Grade
Written by alicia rudnicki
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Children develop division ideas naturally before they begin elementary school as they learn to take turns, divide snacks equally and do simple tasks such as setting the table (there are four spoons, and each person gets one). By third grade, they are learning complex methods, especially long division, to divide large quantities quickly. Helpful teaching strategies include focusing on mastery of multiplication, using manipulatives, estimating answers before dividing, using chants and acronyms and reading math picture books.
## Multiplication Mastery
Multiplication is defined as a quick process for calculating multiple additions of the same number. So "3 X 9" is "9 + 9 + 9." It is much easier to move on to formal learning of division once children master the ideas and basic math facts (times tables) of multiplication. Knowing that "3 X 9 = 27" helps a student to see that 27 can be divided in three equal groups of nine or nine equal groups of three. Teachers often say that sometimes it is necessary to go slow in order to go fast. Spending extra time on multiplication can speed up learning of division.
## Manipulatives
Manipulatives are objects, large and small, that students can touch, move, and group to solve math problems. They make math processes more concrete. For example, a student can count a handful of dry lima beans, then divide them evenly into a number of empty margarine tubs. In another demonstration of the grouping idea, children are the manipulatives. The teacher tells the students how many children are in class that day, then directs them to line up in pairs. The class then discusses how many groups of two were made and whether anyone was left without a partner.
## Estimation
Before beginning the multi-step process of long division, teachers should focus on how to guess reasonable answers. The estimation process involves rounding divisors and dividends so they are easier to compute. A problem such as "72 divided by 8" is easier to estimate if students think about how many groups of 10 are in 70. The Math and Reading Help website notes that skill at estimation gives students a sense of what size of answer to expect and gives them "more control over mistakes."
## Chants and Acronyms
Math chants and acronyms help remind students of processes such as the steps in long division. Students can easily remember D-M-S-B (Dad, Mom, Sister, Brother) and its meaning (Divide, Multiply, Subtract, Bring down). A simple chant to reinforce this would be something like "Listen to me, D-M-S-B. Dad, mom, sister, brother. Do it again! Divide, multiply, subtract, bring down the number. Do it again! D-M-S-B, listen to me." Another good idea is to add gestures and movement, such as crossing arms in an "X" for "multiply" and jumping at the phrase "Do it again!"
Read-alouds of colourful math picture books are always a wise investment of class time. "Remainder of One" by Elinor J. Pinczes focuses on Joe, a beetle who wants to march in an orderly insect parade with even rows. Joe perseveres in rearranging his squadron until he is not odd bug out but part of a harmonious configuration with evenly divided rows.
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Notes On Section Formula - CBSE Class 10 Maths
Section Formula: Given two end points of line segment A(x1, y1) and B (x2, y2) you can determine the coordinates of the point P(x, y) that divides the given line segment in the ratio m : n internally using Section Formula given by $\left(\frac{\mathrm{\text{m}}{\text{x}}_{2}+\mathrm{\text{n}}{\text{x}}_{1}}{\mathrm{\text{m + n}}}\mathrm{\text{,}}\frac{\text{m}{\text{y}}_{2}\mathrm{\text{+ n}}{\text{y}}_{1}\mathrm{\text{}}}{\mathrm{\text{m + n}}}\right)$ . Midpoint: The midpoint of a line segment divides it into two equal parts or in the ratio 1:1. The midpoint of line segment joining the points (x1, y1) and (x2, y2) is . Median: The line joining the vertex to the midpoint of opposite side of a triangle is called Median. Centroid: Three medians can be drawn to a triangle and the point of concurrency of medians of a triangle is called Centroid denoted with G. If A(x1, y1), B(x2, y2) and C(x3, y3) are vertices of a triangle then its centroid G is given by $\left(\frac{{\text{x}}_{1}\text{}\mathrm{\text{+}}\text{}{\text{x}}_{2}\text{}\mathrm{\text{+}}\text{}{\text{x}}_{3}}{3}\text{,}\frac{{\text{y}}_{1}\text{}\mathrm{\text{+}}\text{}{\text{y}}_{2}\text{}\mathrm{\text{+}}\text{}{\text{y}}_{3}}{3}\right)$. The centroid of a triangle divides the median in the ratio 2:1.
#### Summary
Section Formula: Given two end points of line segment A(x1, y1) and B (x2, y2) you can determine the coordinates of the point P(x, y) that divides the given line segment in the ratio m : n internally using Section Formula given by $\left(\frac{\mathrm{\text{m}}{\text{x}}_{2}+\mathrm{\text{n}}{\text{x}}_{1}}{\mathrm{\text{m + n}}}\mathrm{\text{,}}\frac{\text{m}{\text{y}}_{2}\mathrm{\text{+ n}}{\text{y}}_{1}\mathrm{\text{}}}{\mathrm{\text{m + n}}}\right)$ . Midpoint: The midpoint of a line segment divides it into two equal parts or in the ratio 1:1. The midpoint of line segment joining the points (x1, y1) and (x2, y2) is . Median: The line joining the vertex to the midpoint of opposite side of a triangle is called Median. Centroid: Three medians can be drawn to a triangle and the point of concurrency of medians of a triangle is called Centroid denoted with G. If A(x1, y1), B(x2, y2) and C(x3, y3) are vertices of a triangle then its centroid G is given by $\left(\frac{{\text{x}}_{1}\text{}\mathrm{\text{+}}\text{}{\text{x}}_{2}\text{}\mathrm{\text{+}}\text{}{\text{x}}_{3}}{3}\text{,}\frac{{\text{y}}_{1}\text{}\mathrm{\text{+}}\text{}{\text{y}}_{2}\text{}\mathrm{\text{+}}\text{}{\text{y}}_{3}}{3}\right)$. The centroid of a triangle divides the median in the ratio 2:1.
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We think you are located in South Africa. Is this correct?
# Infinite Series
## 1.6 Infinite series (EMCF3)
So far we have been working only with finite sums, meaning that whenever we determined the sum of a series, we only considered the sum of the first $$n$$ terms. We now consider what happens when we add an infinite number of terms together. Surely if we sum infinitely many numbers, no matter how small they are, the answer goes to infinity? In some cases the answer does indeed go to infinity (like when we sum all the positive integers), but surprisingly there are some cases where the answer is a finite real number.
## Sum of an infinite series
1. Cut a piece of string $$\text{1}$$ $$\text{m}$$ in length.
2. Now cut the piece of string in half and place one half on the desk.
3. Cut the other half in half again and put one of the pieces on the desk.
4. Repeat this process until the piece of string is too short to cut easily.
5. Draw a diagram to illustrate the sequence of lengths of the pieces of string.
6. Can this sequence be expressed mathematically? Hint: express the shorter lengths of string as a fraction of the original length of string.
7. What is the sum of the lengths of all the pieces of string?
8. Predict what would happen if these steps could be repeated infinitely many times.
9. Will the sum of the lengths of string ever be greater than $$\text{1}$$?
10. What can you conclude?
## Worked example 14: Sum to infinity
Complete the table below for the geometric series $$T_{n} = \left( \frac{1}{2} \right)^{n}$$ and answer the questions that follow:
Terms $$S_{n}$$ $$1 - S_{n}$$ $$T_{1}$$ $$\frac{1}{2}$$ $$\frac{1}{2}$$ $$\frac{1}{2}$$ $$T_{1} + T_{2}$$ $$T_{1} + T_{2} + T_{3}$$ $$T_{1} + T_{2} + T_{3} + T_{4}$$
1. As more and more terms are added, what happens to the value of $$S_{n}$$?
2. As more more and more terms are added, what happens to the value of $$1 - S_{n}$$?
3. Predict the maximum value of $$S_{n}$$ for the sum of infinitely many terms in the series.
### Complete the table
Terms $$S_{n}$$ $$1 - S_{n}$$ $$T_{1}$$ $$\frac{1}{2}$$ $$\frac{1}{2}$$ $$\frac{1}{2}$$ $$T_{1} + T_{2}$$ $$\frac{1}{2} + \frac{1}{4}$$ $$\frac{3}{4}$$ $$\frac{1}{4}$$ $$T_{1} + T_{2} + T_{3}$$ $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8}$$ $$\frac{7}{8}$$ $$\frac{1}{8}$$ $$T_{1} + T_{2} + T_{3} + T_{4}$$ $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}$$ $$\frac{15}{16}$$ $$\frac{1}{16}$$
### Consider the value of $$S_{n}$$ and $$1 - S_{n}$$
As more terms in the series are added together, the value of $$S_{n}$$ increases:
$\frac{1}{2} \quad < \quad \frac{3}{4} \quad < \quad \frac{7}{8} \quad < \quad \cdots$
However, by considering $$1 - S_{n}$$, we notice that the amount by which $$S_{n}$$ increases gets smaller and smaller as more terms are added:
$\frac{1}{2} \quad > \quad \frac{1}{4} \quad > \quad \frac{1}{8} \quad > \quad \cdots$
We can therefore conclude that the value of $$S_n$$ is approaching a maximum value of $$\text{1}$$; it is converging to $$\text{1}$$.
### Write conclusion mathematically
We can conclude that the sum of the series
$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$
gets closer to 1 ($$S_{n} \rightarrow 1$$) as the number of terms approaches infinity ($$n \rightarrow \infty$$), therefore the series converges.
$\sum _{i = 1}^{\infty}{ \left( \frac{1}{2} \right)^{i}} = 1$
We express the sum of an infinite number of terms of a series as
${S}_{\infty }=\sum _{i=1}^{\infty }{T}_{i}$
Convergence and divergence
If the sum of a series gets closer and closer to a certain value as we increase the number of terms in the sum, we say that the series converges. In other words, there is a limit to the sum of a converging series. If a series does not converge, we say that it diverges. The sum of an infinite series usually tends to infinity, but there are some special cases where it does not.
## Convergent and divergent series
Exercise 1.10
For each of the general terms below:
• Determine if it forms an arithmetic or geometric series.
• Calculate $$S_{1}, S_{2}, S_{10} \text{ and } S_{100}$$.
• Determine if the series is convergent or divergent.
$$T_{n} = 2n$$
$$2+4+6+8+ \ldots$$
\begin{align*} a &= 2 \\ d &= 2 \\ \therefore & \text{ this is an arithmetic series } \\ S_{1} &= 2 \\ S_{2} &= 2 + 4 = 6 \\ S_{n} & = \frac{n}{2}(2a + [n-1]d) \\ S_{10} &= 5(2(2) + 9(2)) = 110 \\ S_{100} &= 50(2(2) + 99(2)) = 10100 \end{align*} \begin{align*} a &= 2 \\ S_{n} &= \frac{n}{2}(2(2) + [n - 1]2) \\ &= \frac{n}{2}(2 + 2n) \\ &= n(1 + n) \\ &= n + n^{2} \\ \therefore S_{n} \to \infty & \text{ as } n \to \infty \end{align*}
Therefore, this is a divergent series.
$$T_{n} = (-n)$$
$$(-1) + (-2) + (-3) + (-4)+\ldots$$
\begin{align*} a &= -1 \\ d &= -1 \\ \therefore & \text{ this is an arithmetic series } \\ S_{1} &= -1 \\ S_{2} &= -1 -2 = -3 \\ S_{n} & = \frac{n}{2}(2a + [n-1]d) \\ S_{10} &= 5(2(-1) + 9(-1)) = -55 \\ \\ S_{100} &= 50(2(-1) + 99(-1)) = -5050 \\ & \\ S_{n} & = \frac{n}{2}(2(-1) + [n-1](-1)) \\ & = \frac{n}{2}(-2 - n + 1) \\ & = - \frac{n}{2}(n + 1) \\ \therefore S_{n} \to - \infty & \text{ as } n \to \infty \end{align*}
This is a divergent series.
$$T_{n} = (\frac{2}{3})^{n}$$
$$\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\frac{16}{81}+\ldots$$
\begin{align*} a &= \frac{2}{3} \\ r &= \frac{2}{3} \\ \therefore & \text{ this is a geometric series (with } r < 1) \\ S_{1} &= \frac{2}{3} = \text{0,666}\ldots \\ S_{2} &= \frac{2}{3} +\frac{4}{9} = \frac{10}{9} = \text{1,111}\ldots \\ S_{n} & = \frac{a(1-r^n)}{1-r} \\ S_{10} &= \frac{\frac{2}{3}\left( 1 - \left( \frac{2}{3} \right)^{10} \right)}{1 - \frac{2}{3}} \\ \\ &= 2\left( 1 - \left( \frac{2}{3} \right)^{10} \right) \\ &= \text{1,965}\ldots \\ S_{100} &= \frac{\frac{2}{3}\left( 1 - \left( \frac{2}{3} \right)^{100} \right)}{1 - \frac{2}{3}} \\ \\ &= 2\left( 1 - \left( \frac{2}{3} \right)^{100} \right) \\ &= \text{2,00} \ldots \\ & \\ S_{n} & = \frac{\frac{2}{3} (1- \left( \frac{2}{3} \right)^n)}{1 - \frac{2}{3}} \\ & = \frac{\frac{2}{3} \left(1- \left( \frac{2}{3} \right)^n \right)}{ \frac{1}{3} } \\ & = 2 \left(1 - \left( \frac{2}{3} \right)^{n} \right) \\ & = 2 - 2 \left( \frac{2}{3} \right)^{n} \\ \text{ Therefore, as } n \to \infty & \enspace S_{n} \to 2 \end{align*}
This series is convergent (since the $$r < 1$$) and converges to $$\text{2}$$.
$$T_{n} = 2^{n}$$
$$2 + 4 + 8 + 16 +\ldots$$
\begin{align*} a &= 2 \\ r &= 2 \\ \therefore & \text{ this is a geometric series (with r } > 1) \\ S_{1} &= 2\\ S_{2} &= 2 + 4 = 6 \\ S_{n} & = \frac{a(r^n - 1)}{r - 1} \\ S_{10} &= \frac{2 \left( \left( 2 \right)^{10} - 1 \right)}{2 - 1} \\ \\ &= 2 \left( \left( 2 \right)^{10} - 1 \right) \\ &= \text{2 046} \\ S_{100} &= \frac{2\left( \left( 2 \right)^{100} - 1 \right)}{2 - 1} \\ \\ &= 2\left( \left( 2 \right)^{100} - 1 \right) \\ &= \text{2,5} \times \text{10}^{\text{30}} \\ & \\ S_{n} & = \frac{2((2)^{n} - 1)}{2 - 1} \\ & = (2)^{n + 1} - 2 \\ \therefore S_{n} \to \infty & \text{ as } n \to \infty \end{align*}
This is a divergent series.
Note the following:
• An arithmetic series never converges: as $$n$$ tends to infinity, the series will always tend to positive or negative infinity.
• Some geometric series converge (have a limit) and some diverge (as $$n$$ tends to infinity, the series does not tend to any limit or it tends to infinity).
### Infinite geometric series (EMCF4)
There is a simple test for determining whether a geometric series converges or diverges; if $$-1 < r < 1$$, then the infinite series will converge. If $$r$$ lies outside this interval, then the infinite series will diverge.
Test for convergence:
• If $$-1 < r < 1$$, then the infinite geometric series converges.
• If $$r < -1 \text{ or } r > 1$$, then the infinite geometric series diverges.
We derive the formula for calculating the value to which a geometric series converges as follows:
${S}_{n}=\sum _{i=1}^{n}{a}{r}^{i-1}=\frac{{a} \left(1 - {r}^{n}\right)}{1 - r}$
Now consider the behaviour of $${r}^{n}$$ for $$-1<r<1$$ as $$n$$ becomes larger.
Let $$r=\frac{1}{2}$$:
\begin{align*} n=1& : {r}^{n}={r}^{1}={\left(\frac{1}{2}\right)}^{1}=\frac{1}{2} \\ n=2& : {r}^{n}={r}^{2}={\left(\frac{1}{2}\right)}^{2}=\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}<\frac{1}{2} \\ n=3& : {r}^{n}={r}^{3}={\left(\frac{1}{2}\right)}^{3}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8} <\frac{1}{4} \end{align*}
Since $$r$$ is in the range $$-1<r<1$$, we see that $${r}^{n}$$ gets closer to $$\text{0}$$ as $$n$$ gets larger. Therefore $$(1 - r^{n})$$ gets closer to $$\text{1}$$.
Therefore,
\begin{align*} {S}_{n}& = \frac{{a}\left(1 - {r}^{n} \right)}{1 - r} \\ \text{If } -1 < r < 1, \quad & \text{then } r^{n} \rightarrow 0 \text{ as } n \rightarrow \infty \\ \therefore {S}_{\infty }& = \frac{{a}\left(1 - 0 \right)}{1 - r} \\ & = \frac{{a}}{1-r} \end{align*}
The sum of an infinite geometric series is given by the formula
$\therefore {S}_{\infty }=\sum _{i=1}^{\infty }{a}{r}^{i-1}=\frac{{a}}{1-r} \qquad (-1<r<1)$
where
• $${a}$$ is the first term of the series;
• $$r$$ is the constant ratio.
Alternative notation:
$\underbrace{{S}_{n}}_{n \rightarrow \infty} \rightarrow \frac{a}{1 - r} \quad \text{ if }-1<r<1$
In words: as the number of terms $$(n)$$ tends to infinity, the sum of a converging geometric series $$(S_{n})$$ tends to the value $$\frac{a}{1 - r}$$.
## Worked example 15: Sum to infinity of a geometric series
Given the series $$18 + 6 + 2 + \cdots$$. Find the sum to infinity if it exists.
### Determine the value of $$r$$
We need to know the value of $$r$$ to determine whether the series converges or diverges.
\begin{align*} \frac{T_{2}}{T_{1}} &= \frac{6}{18} \\ &= \frac{1}{3} \\ \frac{T_{3}}{T_{2}} &= \frac{2}{6} \\ &= \frac{1}{3} \\ \therefore r &= \frac{1}{3} \end{align*}
Since $$- 1 < r < 1$$, we can conclude that this is a convergent geometric series.
### Determine the sum to infinity
Write down the formula for the sum to infinity and substitute the known values:
$a = 18; \qquad r = \frac{1}{3}$
\begin{align*} S_{\infty} &= \frac{a}{1 - r} \\ &= \frac{18}{ 1 - \frac{1}{3}} \\ &= \frac{18}{\frac{2}{3}} \\ &= 18 \times \frac{3}{2} \\ &= 27 \end{align*}
As $$n$$ tends to infinity, the sum of this series tends to $$\text{27}$$; no matter how many terms are added together, the value of the sum will never be greater than $$\text{27}$$.
## Worked example 16: Using the sum to infinity to convert recurring decimals to fractions
Use two different methods to convert the recurring decimal $$0,\dot{5}$$ to a proper fraction.
### Convert the recurring decimal to a fraction using equations
\begin{align*} \text{Let } x &= \text{0,}\dot{\text{5}}\\ \therefore x &= \text{0,555} \ldots \ldots (1) \\ 10x &= \text{5,55} \ldots \ldots (2) \\ (2) - (1): \quad 9x &= 5 \\ \therefore x &= \frac{5}{9} \end{align*}
### Convert the recurring decimal to a fraction using the sum to infinity
\begin{align*} \text{0,}\dot{\text{5}} &= \text{0,5} + \text{0,05} + \text{0,005} + \ldots \\ \text{or } \quad \text{0,}\dot{\text{5}} &= \frac{5}{10} + \frac{5}{100} + \frac{5}{1000} + \ldots \end{align*}
This is a geometric series with $$r = \text{0,1} = \frac{1}{10}$$. And since $$- 1 < r < 1$$, we can conclude that the series is convergent.
\begin{align*} S_{\infty} &= \frac{a}{1 - r} \\ &= \frac{\frac{5}{10}}{1 - \frac{1}{10}} \\ &= \dfrac{\frac{5}{10}}{\frac{9}{10}} \\ &= \frac{5}{9} \end{align*}
## Worked example 17: Sum to infinity
Determine the possible values of $$a$$ and $$r$$ if
$\sum _{n=1}^{\infty}{ar^{n-1}} = 5$
### Write down the sum to infinity formula and substitute known values
\begin{align*} S_{\infty} &= \frac{a}{1 - r} \\ \therefore 5 &= \frac{a}{1 - r} \\ a &= 5(1 - r) \\ \therefore a &= 5 - 5r \\ \text{And } 5r &= 5 - a \\ \therefore r &= \frac{5 - a}{5} \end{align*}
### Apply the condition for convergence to determine possible values of $$a$$
For a series to converge: $$- 1 < r < 1$$
\begin{align*} -1 & < r < 1 \\ -1 & < \frac{5 - a}{5} < 1 \\ -5 & < 5 - a < 5 \\ -10 & < - a < 0 \\ 0 & < a < 10 \end{align*}
For the series to converge, $$0 < a < 10$$ and $$- 1 < r < 1$$.
## Sum to infinity
Exercise 1.11
What value does $${\left(\frac{2}{5}\right)}^{n}$$ approach as $$n$$ tends towards $$\infty$$?
\begin{align*} S_{n} &= \frac{2}{5} + \frac{4}{25} + \frac{8}{125} + \ldots \\ \therefore a &= \frac{2}{5} \\ \text{And } r &= \dfrac{\frac{4}{25}}{\frac{2}{5}} \\ &= \frac{2}{5} \qquad (- 1 < r < 1) \\ \text{So then } S_{\infty} &= \frac{a}{1-r} \\ &= \frac{\frac{2}{5}}{1-\frac{2}{5}} \\ &= \frac{\frac{2}{5}}{\frac{3}{5}} \\ &= \frac{2}{3} \end{align*}
Find the sum to infinity of the geometric series $$3+1+\frac{1}{3}+\frac{1}{9}+ \cdots$$
\begin{align*} a &= 3 \\ r &= \frac{1}{3} \\ S_{\infty} &= \frac{a}{1 -r} \\ &= \frac{3}{1 - \frac{1}{3}} \\ &= \frac{3}{\frac{2}{3}} \\ &= \frac{9}{2} \end{align*}
Determine for which values of $$x$$, the geometric series $$2+\frac{2}{3} \left(x+1\right)+\frac{2}{9} {\left(x+1\right)}^{2}+ \cdots$$ will converge.
\begin{align*} a &= 2 \\ r &= \frac{\frac{2}{3} \left(x+1\right)}{2} \\ &= \frac{1}{3} \left(x+1\right) \end{align*}
For the series to converge, $$-1 < r < 1$$, therefore:
\begin{align*} -1&< r < 1 \\ -1 &< \frac{1}{3} \left(x+1\right) < 1 \\ -3 &< \left(x+1\right) < 3 \\ -3-1 &< x < 3-1 \\ -4 &< x < 2 \end{align*}
The sum to infinity of a geometric series with positive terms is $$4\frac{1}{6}$$ and the sum of the first two terms is $$2\frac{2}{3}$$. Find $$a$$, the first term, and $$r$$, the constant ratio between consecutive terms.
\begin{align*} T_{1} + T_{2} &= \frac{8}{3} \\ \therefore a + ar &= \frac{8}{3} \\ a(1 + r) &= \frac{8}{3} \\ \therefore a &= \frac{8}{3(1 + r)} \ldots \ldots (1) \\ S_{\infty} &= 4\frac{1}{6} = \frac{25}{6} \\ \therefore \frac{a}{1 -r}&= \frac{25}{6} \\ 6a &= 25(1 -r) \ldots \ldots (2) \\ \text{Substitute eqn. } (1) \rightarrow (2): \quad 6 \left( \frac{8}{3(1 + r)} \right)&= 25(1 -r) \\ 16 &= 25(1 -r)(1 + r)\\ 16 &= 25(1 - r^{2}) \\ 16 &= 25 - 25r^{2} \\ 25r^{2} &= 25 - 16 \\ 25r^{2} &= 9 \\ r^{2} &= \frac{9}{25} \\ \therefore r &= \pm \frac{3}{5} \\ \text{But } T_{n} & > 0 \\ \therefore r &= \frac{3}{5} \\ \text{And } a &= \frac{8}{3(1 + r)} \\ &= \frac{8}{3 + 3r} \\ &= \frac{8}{3 + 3 \left(\frac{3}{5} \right)} \\ &= \frac{8}{\frac{15}{5} + \frac{9}{5}} \\ &= \frac{8}{\frac{24}{5}} \\ \therefore a &= \frac{5}{3} \end{align*}
Use the sum to infinity to show that $$\text{0,}\dot{\text{9}} = 1$$.
Rewrite the recurring decimal:
$\text{0,}\dot{\text{9}} = \frac{9}{10} + \frac{9}{100} + + \frac{9}{1000} + \ldots$
This is a geometric series with $$a = \frac{9}{10}$$ and $$r = \frac{1}{10}$$.
\begin{align*} S_{\infty} &= \frac{a}{1 - r} \\ &= \frac{\frac{9}{10}}{1 - \frac{1}{10}} \\ &= \frac{\frac{9}{10}}{\frac{9}{10}} \\ &= 1 \end{align*}
A shrub $$\text{110}$$ $$\text{cm}$$ high is planted in a garden. At the end of the first year, the shrub is $$\text{120}$$ $$\text{cm}$$ tall. Thereafter the growth of the shrub each year is half of it's growth in the previous year. Show that the height of the shrub will never exceed $$\text{130}$$ $$\text{cm}$$. Draw a graph of the relationship between time and growth.
[IEB, Nov 2003]
Write the annual growth of the shrub as a series:
$10 + 5 + \frac{5}{2} + \frac{5}{4} + \ldots$
This is a geometric series with $$a = 10$$ and $$r = \frac{1}{2}$$.
\begin{align*} S_{\infty} &= \frac{a}{1 - r} \\ &= \frac{10}{1 - \frac{1}{2}} \\ &= \frac{10}{\frac{1}{2}} \\ &= 20 \end{align*}
Therefore the growth of the shrub is limited to $$\text{20}$$ $$\text{cm}$$, and the maximum height of the shrub is therefore $$\text{110}$$ $$\text{cm}$$ +$$\text{20}$$ $$\text{cm}$$ = $$\text{130}$$ $$\text{cm}$$.
Note: we may join the points on the graph because the growth is continuous.
Find $$p$$:
$\sum_{k = 1}^{\infty}{27p^{k}} = \sum_{t=1}^{12}{(24 - 3t)}$
Write out the series on the RHS:
$\sum_{t=1}^{12}{(24 - 3t)} = 21 + 18 + 15 + \ldots$
This is an arithmetic series with $$a = 21$$ and $$d = -3$$.
\begin{align*} S_{n} &= \frac{n}{2}[2a + (n - 1)d] \\ S_{12} &= \frac{12}{2}[2(21) + (12 - 1)(3)] \\ &= 6[42 - 33] \\ &= 54 \\ \therefore \sum_{t=1}^{12}{(24 - 3t)} = 54 \end{align*}
Write out the series on the LHS:
$\sum_{k = 1}^{\infty}{27p^{k}} = 27p + 27p^{2} + 27p^{3} + \ldots$
This is a geometric series with $$a = 27p$$ and $$r = p$$ ($$-1 < p < 1$$ for the series to converge).
\begin{align*} S_{\infty} &= \frac{a}{1 - r} \\ \therefore 54 &= \frac{27p}{1 - p} \\ 27p &= 54 - 54 p \\ 81 p &= 54 \\ \therefore p &= \frac{54}{81} \\ &= \frac{2}{3} \end{align*}
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# Why is the denominator in a conditional probability the probability of the conditioning event?
Quite a simple or at least short question: Why is $$\frac{P(A \cap B)}{P(B)}$$ divided by $$P(B)$$ for the conditional probability?
$$P(A | B) = \frac{P(A \cap B)}{P(B)}$$
Random image to visualize:
I actually would like to make use out of the tree display which I can't grasp at all. Why would I divide by $$P(A+)$$ when I had to move along this path? Sorry for the mix-up of $$A+$$ and $$B$$..
• It's probably easier to understand from P(A and B) = P(A | B) P(B) Commented Aug 29, 2022 at 15:15
• This lecture gives a couple good ways to think about the intuition of conditional probability. youtu.be/P7NE4WF8j-Q?t=2138 Commented Sep 1, 2022 at 11:34
In your tree diagram, if you want to know the probability $$P(B+|A+)$$ Then you completely ignore the $$A-$$ section, since you know that that does not occur. So the probability of $$B+$$ given that $$A+$$ occurred is the count of the ways $$A+B+$$ can occur divided by the total number of ways that $$A+$$ can occur, that is the combination of $$A+B+$$ and $$A+B-$$, which together is $$A+$$, so you would need to divide by the count (or probability) of $$A+$$ happening.
It might be easier to think of actual events that you can count the ways it happens.
One example:
Roll a fair die (possible outcomes are 1, 2, 3, 4, 5, 6)
A is the event that the roll is greater than 3 (4, 5, 6)
B is the event that an odd number was rolled (1, 3, 5)
So the probability of A and B is $$\frac16$$ since 5 is the only number that matches, but if we know that A happened, then $$P(B|A) = \frac13$$ since we know that the outcome was 4, 5, or 6. We can get there by counting (1 out of 3 possibilities) or by the math with probabilities $$\frac16$$ divided by $$\frac12$$. If you do not divide by what you are conditioning on, then you are looking at the joint probability, not the conditional probability. Also think about the Venn diagram. You need to divide by the area of the circle that you are conditioning on so that the total area of the sum of the conditional probabilities is 1.
• Thank you! The other answers maybe even more illustrative but this one matches my understand/perception the best :)
– Ben
Commented Aug 30, 2022 at 6:49
• I've a follow up question: When $A+$ and $B*$ are the probability of both events happening (means $A+ \cdot B+$), then I consider already the probability of $A+$ or not?
– Ben
Commented Aug 30, 2022 at 9:05
• I really like this explanation, because it is an easy transition between this "tree" model and the "geometric" model demonstrated by whuber. Commented Aug 30, 2022 at 13:47
• @Ben, Not sure exactly what you question/notation means. The probability of A+ and B+ is only the product of the 2 probabilities if A and B are independent (then the conditional probability is the same as the unconditional). The probability of A+ contains both the probability of both A+ and B+ and the probability of A+ and B-. Think about the following probabilities from rolling a fair die. What is the probability that the number shown is both even and a 6? what is the probability that it is 6 given that it is even? what is the probability that it is even given it is a 6? Commented Aug 30, 2022 at 19:22
• @Ben, yes the joint probability is the product of the marginal probabilities only when they are independent. Commented Aug 31, 2022 at 14:32
The tickets-in-a-box model provides a helpful picture. It makes the formula for conditional probability immediate and obvious.
Let $$\mathcal{A}$$ and $$\mathcal{B}$$ designate events: collections of types of tickets. (That is, whenever a ticket for a particular outcome is in an event, all the tickets for that outcome must be in the event.) Recall that the probability of $$\mathcal A$$ is defined as the proportion of tickets of type $$\mathcal A$$ in the box. It is a ratio of that ticket count to the count of all tickets in the box $$\Omega.$$ (Some people insist all probabilities are conditional and to emphasize this would write the probability as $$\Pr(\mathcal A\mid \Omega).$$)
If we were to remove all tickets not of type $$\mathcal{B},$$ we would still have a box with tickets -- all of type $$\mathcal B,$$ obviously. Assuming some tickets remain (that is, $$\mathcal{B}$$ is not empty), this is still a tickets-in-a-box model.
The selective removal can alter the relative proportions of $$\mathcal A$$ among the remaining tickets. The new proportion of tickets of type $$\mathcal{A}$$ is their probability when $$\mathcal B$$ is viewed as the "box" -- the sample space. It is called the conditional probability of $$\mathcal{A},$$ conditional on $$\mathcal{B},$$ and is written $$\Pr{\left(\mathcal{A}\mid\mathcal{B}\right)}.$$
In this Venn diagram, areas (relative to the total area of the rectangle $$\Omega$$ representing the whole probability space) represent probabilities. It looks like both the events $$\mathcal{A}$$ and $$\mathcal{B}$$ have probabilities around $$1/3$$ each. But when all non-$$\mathcal{B}$$ points are removed, the new sample space is $$\mathcal{B}$$ itself and the portion of $$\mathcal{A}$$ within it is tiny – just the sliver of space where the circles intersect, as a proportion of the area of the circle representing $$\mathcal{B}.$$ In this example $$\Pr{\left(\mathcal{A}\right)}\approx1/3$$ but $$\Pr{\left(\mathcal{A}\mid\mathcal{B}\right)}\approx1/100,$$ exemplifying how dramatically the conditioning on $$\mathcal B$$ can change the probabilities.
It should be apparent that (a) this construction works provided $$\mathcal{B}$$ does not have zero probability and (b) in that case, the new relative area of $$\mathcal A$$ is $$Pr{\left(\mathcal{A}\mid\mathcal{B}\right)=}\frac{\Pr{\left(\mathcal{A}\cap\mathcal{B}\right\}}}{\Pr{\left(\mathcal{B}\right)}}.$$
The tree diagram is merely another way to (partially) represent the Venn diagram. Branching at $$\mathcal A$$ corresponds to slicing everything in $$\Omega$$ into outcomes in $$\mathcal A$$ and outcomes not in $$\mathcal A;$$ likewise, branching at $$\mathcal B$$ slices everything by $$\mathcal B.$$
In your diagram you condition on $$\mathcal A.$$ This means removing everything from the tree that is outside $$\mathcal A.$$ Afterwards, only two leaves remain: everything in $$\mathcal A$$ and in $$\mathcal B$$ ($$\mathcal A \cap \mathcal B$$) and everything in $$\mathcal A$$ and not in $$\mathcal B$$ ($$\mathcal A \cap (\Omega \setminus \mathcal B)$$). Those leaves represent all the tickets remaining in the box. Trees are usually decorated with probabilities permitting you to compute the relative proportions of those leaves.
Translating between different ways of visualizing the same thing is a useful skill because some visualizations support specific concepts or calculations especially well, and others shine in other areas.
$$P(A \cap B) = P(A \cap B) \\ P(A | B) P(B) = P(A \cap B) \\ P(A|B) = \frac{P(A \cap B)}{P(B)}$$
• This just raises the question, why is $P(A | B) P(B) = P(A \cap B)$ true? Commented Sep 1, 2022 at 14:21
• Some authors, such as Bruno de Finetti, introduce this as an axiom of probability. Likewise, the Kolmogorov answers are employing definitions of conditional probability, sometimes illustrated intuitively with tree diagrams or box ticket models.
– Sycorax
Commented Sep 1, 2022 at 15:00
Sycorax has answered it nicely. But let me leave behind an intuitive set up that articulates the genesis of the concept.
Start with the very familiar experiment: throwing a die. Let $$A:=\{\text{odd number appears}\}, ~ B:=\{\text{a digit smaller than or equal three appears}\}.$$ One might wonder what the probability would be of $$A$$ had it already been assured that $$B$$ was realised.
Construct the probability space $$(\Omega, \mathcal A, \mathbb P) ,$$ with $$\Omega= \{1, 2,3,4,5,6\}, ~\mathcal A= 2^\Omega$$ and $$\mathbb P$$ being the discrete uniform distribution on $$\Omega.$$ Naturally $$A=\{1, 3,5\}, ~B=\{1, 2,3\}.$$ If $$B$$ has indeed occurred, then it is plausible to assume the uniform distribution on the remaining possible outcomes, i.e. $$\{1,2,3\}.$$ For assessing the new situation, define a new probability measure $$\mathbb P_B$$ on $$(B, 2^B)$$ by
$$\mathbb P_B[C]:= \frac{\#C}{\#B}, ~C\subset B;$$ natural extension would be to assign probability $$0$$ on $$\Omega\setminus B$$ as
$$\mathbb P[C|B]:= \mathbb P_B[C\cap B]:= \frac{\#(C\cap B) }{\#B}, ~C\subset \Omega.$$
In that case $$\mathbb P[A|B] =\frac{\# \{1,3\}}{\#\{1,2,3\}} = \frac23.$$ This is different from $$\mathbb P[A]=\frac12.$$
This prompts us to define the conditional probability for any $$B\in \mathcal A$$ as a new probability measure $$\mathbb P[\cdot|B]$$ on the space $$(\Omega, \mathcal A)$$ for $$\mathbb P[B] > 0$$ as
$$\mathbb P[A|B] =\begin{cases}\frac{\mathbb P[A\cap B]}{\mathbb P[B]},~~ \mathbb P[B] > 0\\ 0 ,~~~~~~~~~~~\textrm{otherwise}\end{cases}.$$
## Reference
Probability Theory: A Comprehensive Course, Achim Klenke, Springer, 2014.
After edit of OP:
As the definition developed above, that $$A^+$$ has occured provides new information has been incorporated by dividing by $$\mathbb P[A^+].$$
A more intuitive way to understand is to know that conditioning on an event with strictly positive probability defines a new probability measure.
So instead of looking at the bigger picture (the original sample space, say $$X$$), we are now restricting to a possibly much smaller sample space (which is $$B$$ in your case). The intersection $$A \cap B$$ will be smaller in probability with $$X$$ being the sample space, but larger when we take the $$B$$ to be the whole sample space. So we adjust for restricting to a smaller sample space by enlarging the probability of the intersection $$A\cap B$$. This normalizing process is done by dividing by $$\mathbb{P}(B)$$. Which is easy to understand, we are thus calculating the relative area of $$A\cap B$$ with respect to $$B$$, instead of $$X$$.
You can also view $$\mathbb{P}(A)$$ as a conditional probability by noting that $$\mathbb{P}(A)=\mathbb{P}(A\cap X) = \dfrac{\mathbb{P}(A\cap X)}{\mathbb{P}(X)},$$ where $$\mathbb{P}(X)=1$$.
First of all, I disagree with the picture. Blue set does not refer to $$P(A)$$ but $$P(A\cap B')$$. Green set refers to $$P(B\cap A')$$.
Back to the question: Why $$P(B^+|A^+)=\frac{P(B^+\cap A^+)}{P(A^+)}?$$
We can understand it by saying $$P(B^+|A^+)$$ as the probability that event $$B^+$$ happening given that event $$A^+$$ happened.
It means that, if I you have information that $$A^+$$ has happened,
1. you must make sure that both events happened at the same time. Hence, you have $$P(B^+\cap A^+)$$.
2. your answers must be with respect to the event $$A^+$$. Your answers are constrained to event $$A^+$$. Hence, you have a ratio to $$P(A^+)$$.
As an example, let $$A^+$$ be the event someone goes outside to play and $$B^+$$ be the event that someone will come home after 6pm. If you know that someone has already went outside to play, what is their probability to come home after 6pm? You already know that $$A^+$$ happened, so you will have
1. that person went outside to play AND that person will come home after 6pm.
2. that the probability above happened with respect to that person will come home after 6pm.
Intuitive, hand-wavey explanation that I feel is lacking here. (This argument works 100% fine in the case of uniform probability with a finite number of outcomes, such throwing a fair die, or picking cards from a well-shuffled deck, but to work more generally it would need some rephrasing.)
The first probability formula you learn is $$P(\text{Something}) = \frac{\text{Number of outcomes where Something happens}}{\text{Total number of relevant outcomes}}$$ We are going to use this three times.
First off, the probability $$P(B^+\mid A^+)$$ is given by $$P(B^+\mid A^+) = \frac{\text{Number of outcomes where }B^+\text{ and } A^+\text{ happen}}{\text{Number of outcomes where }A^+\text{ happens}}$$ Now we divide numerator and denominator by the total number of outcomes. This does not change the value of the fraction: $$= \frac{\text{Number of outcomes where }B^+\text{ and } A^+\text{ happen}/\text{Total number of outcomes}}{\text{Number of outcomes where }A^+\text{ happens}/\text{Total number of outcomes}}$$ Applying our basic probability formula again, to numerator and denominator separately, we do indeed get $$= \frac{P(A^+\cap B^+)}{P(A^+)}$$
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1. Class 12
2. Important Questions for exams Class 12
3. Chapter 3 Class 12 Matrices
Transcript
Ex 3.2, 7 Find X and Y, if (i) X + Y = [■8(7&0@2&5)] and X – Y = [■8(3&0@0&3)] Let X + Y = [■8(7&0@2&5)] X – Y = [■8(3&0@0&3)] Adding (1) and (2) X + Y + X – Y = [■8(7&0@2&5)] + [■8(3&0@0&3)] X + Y + X – Y = [■8(7+3&0+0@2+0&5+3)] 2X + 0 = [■8(10&0@2&8)] …(1) …(2) 2X = [■8(10&0@2&8)] X = 1/2 [■8(10&0@2&8)] X = [■8(10/2&0/2@2/2&8/2)] X = [■8(5&0@1&4)] Thus, X = [■8(5&0@1&4)] Putting value of X in (1) X + Y = [■8(7&0@2&5)] Y = [■8(7&0@2&5)] – X Y = [■8(7&0@2&5)] – [■8(5&0@1&4)] Y = [■8(7−5&0−0@2−1&5−4)] Y = [■8(2&0@1&1)] Hence, X = [■8(𝟓&𝟎@𝟏&𝟒)] & Y = [■8(𝟐&𝟎@𝟏&𝟏)] Ex 3.2, 7 Find X and Y, if (ii) 2X + 3Y = [■8(2&3@4&0)] and 3X + 2Y = [■8(2&−2@−1&5)] Given 2X + 3Y = [■8(2&3@4&0)] Multiplying by 3 3 × (2X+ 3Y) = 3 [■8(2&3@4&0)] 6X + 9Y = [■8(2 × 3&3 × 3@4 × 3&0 × 3)] 6X + 9Y = [■8(6&9@12&0)] Given 3X + 2Y = [■8(2&−2@−1&5)] Multiplying by 2 2 × (3X + 2Y) = 2 × [■8(2&−2@−1&5)] 6X + 4Y = [■8(2 ×2&−2 ×2@−1 ×2&5 ×2)] 6X + 4Y = [■8(4&−4@−2&10)] Subtracting (1) from (2), (6X + 9Y) – (6X + 4Y) = [■8(6&9@12&0)] – [■8(4&−4@−2&10)] 6X + 9Y – 6X – 4Y = [■8(6−4&9−(−4)@12−(−2)&0−10)] 9Y – 4Y + 6X – 6X = [■8(2&9+4@12+2&−10)] 5Y + 0 = [■8(2&13@14&−10)] Y = 1/5 [■8(2&13@14&−10)] Y = [■8(2/5&13/5@14/5&−10/5)] = [■8(2/5&13/5@14/5&−2)] Putting value of Y in (1) 6X + 9Y = [■8(6&9@12&0)] 6X + 9 [■8(2/5& 13/5@14/5&−2)] = [■8(6&9@12&0)] 6X + [■8(9 × 2/5&9 ×13/5@9 ×14/5&9 ×−2)] = [■8(6&9@12&0)] 6X + [■8(18/5&117/5@126/5&−18)] = [■8(6&9@12&0)] 6X = [■8(6&9@12&0)] – [■8(18/5&117/5@126/5&−18)] 6X = [■8(6−18/5&9−117/5@12−126/5&0−(−18))] 6X = [■8((6 × 5 − 18)/5&(9 × 5 − 117)/5@ (12 × 5 − 126)/5&18)] 6X = [■8((30 − 18)/5&(45 − 117)/5@ (60 − 126)/5&18)] 6X = [■8(12/5&(−72)/5@ (−66)/5&18)] X = 1/6 [■8(12/5&(−72)/5@ (−66)/5&18)] X = [■8(1/6 × 12/5&1/6 ×(−72)/5@1/6 ×(−66)/5&1/6 ×18)] X = [■8(2/5&(−12)/5@(−11)/5&3)] Thus, X = [■8(𝟐/𝟓& (−𝟏𝟐)/𝟓@ (−𝟏𝟏)/𝟓&𝟑)] , Y = [■8(𝟐/𝟓&𝟏𝟑/𝟓@𝟏𝟒/𝟓&−𝟐)]
Chapter 3 Class 12 Matrices
Class 12
Important Questions for exams Class 12
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Solving a System of Inequalities
3 teachers like this lesson
Print Lesson
Objective
SWBAT solve a system of linear inequalities.
Big Idea
Solve a system of inequalities by testing a point and by interpreting solutions from slope-intercept form.
Warm Up
10 minutes
The objective of this lesson is to solve a system of inequalities, so I begin by accessing the students' prior knowledge of solving a linear equation and a linear inequality in one variable. I allow the students 5 minutes to complete the Warm-Up. I review the warm up with the class to introduce the lesson.
I discuss today's warm-up in the video below:
After the warmup, the main purpose of today's lesson is to build on known concepts by making connections. First, to solving a linear inequality in two variables and then to solving a system of linear inequalities in two variables. I discuss with the students how inequalities place restrictions on the problem, but there are many solutions. These restrictions I refer to as boundaries or constraints.
Guided Notes
10 minutes
In the Guided Notes I discuss:
• The meaning of a solution to a system
• How to solve and graph an inequality
• The test point method compared to the slope-intercept method
I set up the notes for students. My plan is for them to develop an understanding of how the symbols are related to the solutions of the inequality. We will also review the meaning of this relationship. After completing the Guided Notes, students should recognize that solutions are contained in the intersection of the shaded regions, and, on the solid boundary lines in some cases.
Students are also required to be able to apply the following skills:
• Solve for y when rewriting inequalities in slope intercept form.
• Change the direction of the inequality symbol when multiplying or dividing each side of the inequality by a negative number
• Graphing a line from a given equation
• Substituting values for x and y into an equation to determine if the inequality is true or false
• Graphing the solution to an inequality and a system of inequalities
Throughout the warm up, guided notes, practice, and lesson, students should gain a conceptual understanding of why these different methods for solving a system of equations work. I ask the following questions throughout the lesson.
1. When an inequality is false on one side, why can we determine that the other side is the solution?
2. What form do the inequalities need to be in to apply the slope-intercept method?
3. What form do the inequalities need to be in to apply the test point method?
4. Can a point on a boundary line be used? Why or Why not?
Independent Practice
15 minutes
While students are working the Independent Practice, they are allowed to talk with their assigned elbow partner for math questions, but they must stay on task. I walk around the room while students are working to monitor progress and to assist students one on one or in pairs as needed. I use these questions continue to guide students forward in their productive struggle.
1. Do you have to graph a linear inequality in one variable on a number line or a coordinate plane?
2. If you graph a linear inequality in one variable on a coordinate plane, what type of line is graphed?
3. On problem 10 on the practice, what is the difference between solving a system of equations that is parallel compared to solving a system of inequalities that are parallel?
4. What do you think is the difference between solving a system of inequalities with the same line compared to a system of inequalities that are the same line?
5. Is it possible to graph number 7 on a number line, or must you graph it on a coordinate plane? Why or Why not?
6. What is the difference in the solutions to a system of equations with different slopes that are intersecting compared to a system of linear inequalities that are intersecting?
In the last problem, students have not been introduced to graphing quadratic functions this year. Some students do remember graphing quadratics previously in 8th grade. This provides a good teaching moment to discuss a strategy to use when graphing an unfamiliar function. A t-table is a strategy to use to substitute random numbers for x and solve for y to determine the points of the graph and the shape of the function.
Check Work with Online Activity
10 minutes
In closing, we discuss how a linear inequality in one variable can be graphed on the number line or on a coordinate plane.
I also have students check their work by entering into an online graphing tool. I instruct students that the equations or relations should be entered in standard form Ax + By = C. If the inequality has only one variable, a 0 must be entered as the coefficient for the other variable.
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Kanika was given her pocket money on Jan 1st, 2008.
Question:
Kanika was given her pocket money on Jan 1st, 2008. She puts ₹ 1 on day 1, ₹ 2 on day 2, ₹ 3 on day 3 and continued doing so till the end of the
month, from this money into her piggy bank she also spent ₹ 204 of her pocket money, and found that at the end of the month she still had ₹ 100 with
her. How much was her pocket money for the month?
Solution:
Let her pocket money be ₹ x.
Now, she takes ₹ 1 on day 1, ₹ 2 on day 2, ₹ 3 on day 3 and so on till the end of the month, from this money.
i.e., 1 + 2 + 3+ 4+ … + 31.
which form an AP in which terms are 31 and first term (a) = 1, common difference (d) = 2 — 1 = 1 .
Sum of first 31 terms = S31
$\therefore$ Sum of first 31 terms $=S_{31}$
Sum of $n$ terms, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\therefore$ $S_{31}=\frac{31}{2}[2 \times 1+(31-1) \times 1]$
$=\frac{31}{2}(2+30)=\frac{31 \times 32}{2}$
$=31 \times 16=496$
So, Kanika takes ₹ 496 till the end of the month from this money.
Also, she spent ₹ 204 of her pocket money and found that at the end of the month she still has ₹ 100 with her.
Now, according to the condition,
$(x-496)-204=100$
$\therefore \quad x=₹ 800$
Hence, ₹ 800 was her poket money for the month
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# Equivalence of Definitions of Triangular Number
## Theorem
The following definitions of the concept of Triangular Number are equivalent:
### Definition 1
$T_n = \begin{cases} 0 & : n = 0 \\ n + T_{n-1} & : n > 0 \end{cases}$
### Definition 2
$\ds T_n = \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + \paren {n - 1} + n$
### Definition 3
$\forall n \in \N: T_n = \map P {3, n} = \begin{cases} 0 & : n = 0 \\ \map P {3, n - 1} + \paren {n - 1} + 1 & : n > 0 \end{cases}$
where $\map P {k, n}$ denotes the $k$-gonal numbers.
## Proof
### Definition 1 implies Definition 2
Let $T_n$ be a triangular number by definition 1.
Let $n = 0$.
By definition:
$T_0 = 0$
$\ds T_0 = \sum_{i \mathop = 1}^0 i = 0$
By definition of summation:
$\ds T_{n - 1} = \sum_{i \mathop = 1}^{n - 1} i = 1 + 2 + \cdots + \paren {n - 1}$
and so:
$\ds T_n$ $=$ $\ds T_{n - 1} + n$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + \paren {n - 1} + n$
Thus $T_n$ is a triangular number by definition 2.
$\Box$
### Definition 2 implies Definition 1
Let $T_n$ be a triangular number by definition 2.
$\ds T_n = \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + \paren {n - 1} + n$
Thus:
$\ds T_{n - 1} = \sum_{i \mathop = 1}^{n - 1} i = 1 + 2 + \cdots + \paren {n - 1}$
and so:
$T_n = T_{n - 1} + n$
Then:
$\ds T_0 = \sum_{i \mathop = 1}^0 i$
is a vacuous summation and so:
$T_0 = 0$
Thus $T_n$ is a triangular number by definition 1.
$\Box$
### Definition 1 equivalent to Definition 3
We have by definition that $T_n = 0 = \map P {3, n}$.
Then:
$\ds \forall n \in \N_{>0}: \,$ $\ds \map P {3, n}$ $=$ $\ds \map P {3, n - 1} + \paren {n - 1} + 1$ $\ds$ $=$ $\ds \map P {3, n - 1} + n$
Thus $\map P {3, n}$ and $T_n$ are generated by the same recurrence relation.
$\blacksquare$
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# Integers and Rational Numbers
## Presentation on theme: "Integers and Rational Numbers"— Presentation transcript:
Integers and Rational Numbers
Positive and Negative Numbers, Opposites, and Absolute Value
What are positive numbers?
A positive number is a number greater than zero. On a number line, positive numbers are located to the right of zero (on a vertical number line, positive numbers are located above zero). Positive numbers do not need a sign, but you may see them written with a “+” in front of them.
What are negative numbers?
A negative number is a number less than zero. On a number line, negative numbers are located to the left of zero (or on a vertical number line, negative numbers are located below zero). A negative number will have a “ - ” in front of it. Ex: -9
Example of a number line showing positive and negative numbers
What About Zero? Zero is neither positive nor negative.
What are Opposites? Every negative number is paired with a positive number. The numbers in the pair are the same distance from zero but in opposite directions on a number line. These number pairs are called opposites. Ex: -3 and 3 are opposites
Opposites (cont.) Can you name another pair of numbers that are opposites?
What are integers? Whole numbers and their opposites have a special name—integers. Integers are “counting numbers” and their opposites. Some examples of integers are: -4, -3, -2, -1, 0, 1, 2, 3, 4 Zero is an integer, but it is not positive nor negative.
Where do fractions fit in?
Fractions also have opposites. For example, ½ and -1/2 are opposites. Positive and negative integers, as well as fractions, are called rational numbers. Rational numbers are numbers that can be expressed as one integer divided by another integer.
What is Absolute Value? Absolute value is the distance of a number from zero on a number line. A distance cannot be negative, so absolute value will always be a positive number. To show absolute value, a number will be written between two bars as shown below:
Absolute Value (cont.) Opposites have the same absolute value. Can you name two numbers that have the same absolute value?
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# Express the Hcf of 468 and 222 as 468x + 222y Where X, Y Are Integers in Two Different Ways. - CBSE Class 10 - Mathematics
#### Question
Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two different ways.
#### Solution
Given integers are 468 and 222 where 468 > 222.
By applying Euclid’s division lemma, we get 468 = 222 × 2 + 24 …(i)
Since remainder ≠ 0, apply division lemma on division 222 and remainder 24
222 = 24 × 9 + 6 …(ii)
Since remainder ≠ 0, apply division lemma on division 24 and remainder 6
24 = 6 × 4 + 0 …(iii)
We observe that the remainder = 0, so the last divisor 6 is the HCF of the 468 and 222
From (ii) we have
6 = 222 – 24 × 9
⇒ 6 = 222 – [468 – 222 × 2] × 9 [Substituting 24 = 468 – 222 × 2 from (i)]
⇒ 6 = 222 – 468 × 9 – 222 × 18
⇒ 6 = 222 × 19 – 468 × 9
⇒ 6 = 222y + 468x, where x = −9 and y = 19
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Solution for 10 Mathematics (2018 to Current)
Chapter 1: Real Numbers
Ex. 1.20 | Q: 22 | Page no. 28
#### Video TutorialsVIEW ALL [1]
Solution Express the Hcf of 468 and 222 as 468x + 222y Where X, Y Are Integers in Two Different Ways. Concept: Euclid’s Division Lemma.
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Composition
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Composition - PowerPoint PPT Presentation
Composition. of Functions. By: Dr. Julia Arnold. Composition is a binary operation like addition , subtraction, multiplication and division are binary operations. (meaning they operate on two elements). f-g. f+g. f g. The composition symbol is: . Thus . That's nice!. But What Is It?.
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PowerPoint Slideshow about 'Composition' - Antony
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Presentation Transcript
Composition
of
Functions
By: Dr. Julia Arnold
Composition is a binary operation like addition , subtraction, multiplication and division are binary operations. (meaning they operate on two elements)
f-g
f+g
fg
The composition symbol is:
Thus
That's nice!
But What Is It?
The easiest way to describe composition is to say it is like substitution. In fact
Read f of g of x which means substitute g(x) for x in the f(x) expression.
For example:
Suppose
f(x)= 2x + 3, and
g(x) = 8 - x
Then
Means substitute the g function for
x in the f function… like this
f(x)= 2x + 3
f(g(x) )= 2 g(x) + 3
g(x) = 8 - x
f(x)= 2x + 3, and
f(x)= 2x + 3
f(g(x) )= 2 g(x) + 3
Now
substitute
what
g equals
for g(x)
f(8 - x)= 2 (8 - x) + 3
= 16 - 2x + 3
= 19 - 2x
So, = 19 - 2x
An interesting fact is that
most of the time.
Let’s see if
this is the
case for
the
previous
example.
f(x) = 2x + 3, and
g(x) = 8 - x
Thus we will substitute f into g.
g(x) = 8 - x
g(f(x) ) = 8 - f(x)
Now
substitute
what f(x) is:
g(2x + 3) = 8 - (2x + 3)
= 8 - 2x - 3
= 5 - 2x
Those were easy!
My homework is
never that easy!
Okay!
I'll make it harder.
Let
and
Is that better?
Step 1
Write the f function
Step 2
Substitute g(x) for x
Step 3
Replace g(x) with
Step 4
Simplify
Find:
A)
B)
Find:
A)
B)
Once More! Come On!
Find:
A)
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GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2
Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2 Questions and Answers.
Gujarat Board Textbook Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2
Question 1.
If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Solution:
āµ We have 2 + 1 + y + 5 = 8 + y
21y5 is a multiple of 9,
ā“ (8 + y) must be divisible by 9.
ā“ (8 + y) should be 0, 9, 18, 27, …, etc.
ā“ 8 + y = 0 is not required.
Since, y is a digit
8 + y = 9 ā y = 9 – 8 or y = 1
Question 2.
If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?
Solution:
We have 3 + 1 + z + 5 = 9 + z
āµ 31z5 is divisible by 9.
ā“ (9 + z) must be equal to 0, or 9 or 18 or 27, … But z is a digit.
ā“ 9 + z = 0 or 9 + z = 18
If 9 + z = 0, then z = -9 (which is not required since z is a digit) and if 9 + z = 18, then z = 9.
Question 3.
If 24x is a multiple of 3, where x is a digit, what is the value of x?
Solution:
Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18, … But since x is a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.
Question 4.
If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution:
We have 3 + 1 + z + 5 = 9 + z
āµ 31z5 is divisible by 3.
ā“ (9 + z) must be divisible by 3.
i.e. (9 + z) = 0 or 3 or 6 or 9 or 15 or 18 since, z is a digit.
ā“ If 9 + z = 0, then z = -9.
If 9 + z = 3, then z = -6.
If 9 + z = 6, then z = -3.
If 9 + z = 9, then z = 0.
If 9 + z = 12, then z = 3.
If 9 + z = 15, then z = 6.
If 9 + z = 18, then z = 9.
If 9 + z = 21, then z = 12.
ā“ Possible values of z are 0, 3, 6 or 9.
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# How do you prove that the limit of 5 - 2x= 1 as x approaches 2 using the epsilon delta proof?
Apr 25, 2016
Preliminary analysis
${\lim}_{x \rightarrow \textcolor{g r e e n}{a}} \textcolor{red}{f \left(x\right)} = \textcolor{b l u e}{L}$ if and only if
for every $\epsilon > 0$, there is a $\delta > 0$ such that:
for all $x$, $\text{ }$ if $0 < \left\mid x - \textcolor{g r e e n}{a} \right\mid < \delta$, then $\left\mid \textcolor{red}{f \left(x\right)} - \textcolor{b l u e}{L} \right\mid < \epsilon$.
So we want to make $\left\mid {\underbrace{\textcolor{red}{\left(5 - 2 x\right)}}}_{\textcolor{red}{f \left(x\right)}} - {\underbrace{\textcolor{b l u e}{1}}}_{\textcolor{b l u e}{L}} \right\mid$ less than some given $\epsilon$ and we control (through our control of $\delta$) the size of $\left\mid x - {\underbrace{\textcolor{g r e e n}{2}}}_{\textcolor{g r e e n}{a}} \right\mid$
Look at the thing we want to make small. We want to see the thing we control.
$\left\mid \left(5 - 2 x\right) - 1 \right\mid = \left\mid - 2 x + 4 \right\mid = \left\mid - 2 \left(x - 2\right) \right\mid = \left\mid - 2 \right\mid \left\mid x - 2 \right\mid = 2 \left\mid x - 2 \right\mid$
And there's $\left\mid x - 2 \right\mid$, the thing we control
We can make $2 \left\mid x - 2 \right\mid < \epsilon$ by making $\left\mid x - 2 \right\mid < \frac{\epsilon}{2}$.
So we will choose $\delta = \frac{\epsilon}{2}$. (Any lesser $\delta$ would also work.)
(Detail: if $\left\mid x - 2 \right\mid < \frac{\epsilon}{2}$, then we can multiply on both sides by the positive number $2$ to get $2 \left\mid x - 2 \right\mid < \epsilon$.)
Now we need to actually write up the proof:
Proof
Given $\epsilon > 0$, choose $\delta = \frac{\epsilon}{2}$. $\text{ }$ (note that $\delta$ is also positive).
Now for every $x$ with $0 < \left\mid x - 2 \right\mid < \delta$, we have
abs(f(x)-1) = abs((5-2x) - 1) = abs(-2x+4)) = abs(-2)abs(x-2) = 2abs(x-2) < 2delta
[Detail if $\left\mid x - 2 \right\mid < \delta$, we can conclude that $2 \left\mid x - 2 \right\mid < 2 \delta$. $\text{ }$ We usually do not mention this, but leave it to the reader. See below.]
And $2 \delta = 2 \frac{\epsilon}{2} = \epsilon$
Therefore, with this choice of delta, whenever $0 < \left\mid x - 2 \right\mid < \delta$, we have $\left\mid f \left(x\right) - 1 \right\mid < \epsilon$
So, by the definition of limit, ${\lim}_{x \rightarrow 2} \left(5 - 2 x\right) = 1$.
We can condense a bit
for every $x$ with $0 < \left\mid x - 6 \right\mid < \delta$, we have
$\left\mid f \left(x\right) - 1 \right\mid = \left\mid \left(5 - 2 x\right) - 1 \right\mid$
$= \left\mid - 2 \left(x - 2\right) \right\mid$
$= 2 \left\mid x - 2 \right\mid$
$< 2 \delta = 2 \frac{\epsilon}{2} = \epsilon$.
So, $\left\mid f \left(x\right) - 1 \right\mid < \epsilon$.
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# The Negative Binomial Distribution
A counting distribution is a discrete distribution with probabilities only on the nonnegative integers. Such distributions are important in insurance applications since they can be used to model the number of events such as losses to the insured or claims to the insurer. Though playing a prominent role in statistical theory, the Poisson distribution is not appropriate in all situations, since it requires that the mean and the variance are equaled. Thus the negative binomial distribution is an excellent alternative to the Poisson distribution, especially in the cases where the observed variance is greater than the observed mean.
The negative binomial distribution arises naturally from a probability experiment of performing a series of independent Bernoulli trials until the occurrence of the rth success where r is a positive integer. From this starting point, we discuss three ways to define the distribution. We then discuss several basic properties of the negative binomial distribution. Emphasis is placed on the close connection between the Poisson distribution and the negative binomial distribution.
________________________________________________________________________
Definitions
We define three versions of the negative binomial distribution. The first two versions arise from the view point of performing a series of independent Bernoulli trials until the rth success where r is a positive integer. A Bernoulli trial is a probability experiment whose outcome is random such that there are two possible outcomes (success or failure).
Let $X_1$ be the number of Bernoulli trials required for the rth success to occur where r is a positive integer. Let $p$ is the probability of success in each trial. The following is the probability function of $X_1$:
$\displaystyle (1) \ \ \ \ \ P(X_1=x)= \binom{x-1}{r-1} p^r (1-p)^{x-r} \ \ \ \ \ \ \ x=r,r+1,r+2,\cdots$
The idea for $(1)$ is that for $X_1=x$ to happen, there must be $r-1$ successes in the first $x-1$ trials and one additional success occurring in the last trial (the $x$th trial).
A more common version of the negative binomial distribution is the number of Bernoulli trials in excess of r in order to produce the rth success. In other words, we consider the number of failures before the occurrence of the rth success. Let $X_2$ be this random variable. The following is the probability function of $X_2$:
$\displaystyle (2) \ \ \ \ \ P(X_2=x)=\binom{x+r-1}{x} p^r (1-p)^x \ \ \ \ \ \ \ x=0,1,2,\cdots$
The idea for $(2)$ is that there are $x+r$ trials and in the first $x+r-1$ trials, there are $x$ failures (or equivalently $r-1$ successes).
In both $(1)$ and $(2)$, the binomial coefficient is defined by
$\displaystyle (3) \ \ \ \ \ \binom{y}{k}=\frac{y!}{k! \ (y-k)!}=\frac{y(y-1) \cdots (y-(k-1))}{k!}$
where $y$ is a positive integer and $k$ is a nonnegative integer. However, the right-hand-side of $(3)$ can be calculated even if $y$ is not a positive integer. Thus the binomial coefficient $\displaystyle \binom{y}{k}$ can be expanded to work for all real number $y$. However $k$ must still be nonnegative integer.
$\displaystyle (4) \ \ \ \ \ \binom{y}{k}=\frac{y(y-1) \cdots (y-(k-1))}{k!}$
For convenience, we let $\displaystyle \binom{y}{0}=1$. When the real number $y>k-1$, the binomial coefficient in $(4)$ can be expressed as:
$\displaystyle (5) \ \ \ \ \ \binom{y}{k}=\frac{\Gamma(y+1)}{\Gamma(k+1) \Gamma(y-k+1)}$
where $\Gamma(\cdot)$ is the gamma function.
With the more relaxed notion of binomial coefficient, the probability function in $(2)$ above can be defined for all real number r. Thus the general version of the negative binomial distribution has two parameters r and $p$, both real numbers, such that $0. The following is its probability function.
$\displaystyle (6) \ \ \ \ \ P(X=x)=\binom{x+r-1}{x} p^r (1-p)^x \ \ \ \ \ \ \ x=0,1,2,\cdots$
Whenever r in $(6)$ is a real number that is not a positive integer, the interpretation of counting the number of failures until the occurrence of the rth success is no longer important. Instead we can think of it simply as a count distribution.
The following alternative parametrization of the negative binomial distribution is also useful.
$\displaystyle (6a) \ \ \ \ \ P(X=x)=\binom{x+r-1}{x} \biggl(\frac{\alpha}{\alpha+1}\biggr)^r \biggl(\frac{1}{\alpha+1}\biggr)^x \ \ \ \ \ \ \ x=0,1,2,\cdots$
The parameters in this alternative parametrization are r and $\alpha>0$. Clearly, the ratio $\frac{\alpha}{\alpha+1}$ takes the place of $p$ in $(6)$. Unless stated otherwise, we use the parametrization of $(6)$.
________________________________________________________________________
What is negative about the negative binomial distribution?
What is negative about this distribution? What is binomial about this distribution? The name is suggested by the fact that the binomial coefficient in $(6)$ can be rearranged as follows:
\displaystyle \begin{aligned}(7) \ \ \ \ \ \binom{x+r-1}{x}&=\frac{(x+r-1)(x+r-2) \cdots r}{x!} \\&=(-1)^x \frac{(-r-(x-1))(-r-(x-2)) \cdots (-r)}{x!} \\&=(-1)^x \frac{(-r)(-r-1) \cdots (-r-(x-1))}{x!} \\&=(-1)^x \binom{-r}{x} \end{aligned}
The calculation in $(7)$ can be used to verify that $(6)$ is indeed a probability function, that is, all the probabilities sum to 1.
\displaystyle \begin{aligned}(8) \ \ \ \ \ 1&=p^r p^{-r}\\&=p^r (1-q)^{-r} \\&=p^r \sum \limits_{x=0}^\infty \binom{-r}{x} (-q)^x \ \ \ \ \ \ \ \ (8.1) \\&=p^r \sum \limits_{x=0}^\infty (-1)^x \binom{-r}{x} q^x \\&=\sum \limits_{x=0}^\infty \binom{x+r-1}{x} p^r q^x \end{aligned}
In $(8)$, we take $q=1-p$. The step $(8.1)$ above uses the following formula known as the Newton’s binomial formula.
$\displaystyle (9) \ \ \ \ \ (1+t)^w=\sum \limits_{k=0}^\infty \binom{w}{k} t^k$
For a detailed discussion of (8) with all the details worked out, see the post called Deriving some facts of the negative binomial distribution.
________________________________________________________________________
The Generating Function
By definition, the following is the generating function of the negative binomial distribution, using :
$\displaystyle (10) \ \ \ \ \ g(z)=\sum \limits_{x=0}^\infty \binom{r+x-1}{x} p^r q^x z^x$
where $q=1-p$. Using a similar calculation as in $(8)$, the generating function can be simplified as:
$\displaystyle (11) \ \ \ \ \ g(z)=p^r (1-q z)^{-r}=\frac{p^r}{(1-q z)^r}=\frac{p^r}{(1-(1-p) z)^r}; \ \ \ \ \ z<\frac{1}{1-p}$
As a result, the moment generating function of the negative binomial distribution is:
$\displaystyle (12) \ \ \ \ \ M(t)=\frac{p^r}{(1-(1-p) e^t)^r}; \ \ \ \ \ \ \ t<-ln(1-p)$
For a detailed discussion of (12) with all the details worked out, see the post called Deriving some facts of the negative binomial distribution.
________________________________________________________________________
Independent Sum
One useful property of the negative binomial distribution is that the independent sum of negative binomial random variables, all with the same parameter $p$, also has a negative binomial distribution. Let $Y=Y_1+Y_2+\cdots+Y_n$ be an independent sum such that each $X_i$ has a negative binomial distribution with parameters $r_i$ and $p$. Then the sum $Y=Y_1+Y_2+\cdots+Y_n$ has a negative binomial distribution with parameters $r=r_1+\cdots+r_n$ and $p$.
Note that the generating function of an independent sum is the product of the individual generating functions. The following shows that the product of the individual generating functions is of the same form as $(11)$, thus proving the above assertion.
$\displaystyle (13) \ \ \ \ \ h(z)=\frac{p^{\sum \limits_{i=1}^n r_i}}{(1-(1-p) z)^{\sum \limits_{i=1}^n r_i}}$
________________________________________________________________________
Mean and Variance
The mean and variance can be obtained from the generating function. From $E(X)=g'(1)$ and $E(X^2)=g'(1)+g^{(2)}(1)$, we have:
$\displaystyle (14) \ \ \ \ \ E(X)=\frac{r(1-p)}{p} \ \ \ \ \ \ \ \ \ \ \ \ \ Var(X)=\frac{r(1-p)}{p^2}$
Note that $Var(X)=\frac{1}{p} E(X)>E(X)$. Thus when the sample data suggest that the variance is greater than the mean, the negative binomial distribution is an excellent alternative to the Poisson distribution. For example, suppose that the sample mean and the sample variance are 3.6 and 7.1. In exploring the possibility of fitting the data using the negative binomial distribution, we would be interested in the negative binomial distribution with this mean and variance. Then plugging these into $(14)$ produces the negative binomial distribution with $r=3.7$ and $p=0.507$.
________________________________________________________________________
The Poisson-Gamma Mixture
One important application of the negative binomial distribution is that it is a mixture of a family of Poisson distributions with Gamma mixing weights. Thus the negative binomial distribution can be viewed as a generalization of the Poisson distribution. The negative binomial distribution can be viewed as a Poisson distribution where the Poisson parameter is itself a random variable, distributed according to a Gamma distribution. Thus the negative binomial distribution is known as a Poisson-Gamma mixture.
In an insurance application, the negative binomial distribution can be used as a model for claim frequency when the risks are not homogeneous. Let $N$ has a Poisson distribution with parameter $\theta$, which can be interpreted as the number of claims in a fixed period of time from an insured in a large pool of insureds. There is uncertainty in the parameter $\theta$, reflecting the risk characteristic of the insured. Some insureds are poor risks (with large $\theta$) and some are good risks (with small $\theta$). Thus the parameter $\theta$ should be regarded as a random variable $\Theta$. The following is the conditional distribution of the random variable $N$ (conditional on $\Theta=\theta$):
$\displaystyle (15) \ \ \ \ \ P(N=n \lvert \Theta=\theta)=\frac{e^{-\theta} \ \theta^n}{n!} \ \ \ \ \ \ \ \ \ \ n=0,1,2,\cdots$
Suppose that $\Theta$ has a Gamma distribution with scale parameter $\alpha$ and shape parameter $\beta$. The following is the probability density function of $\Theta$.
$\displaystyle (16) \ \ \ \ \ g(\theta)=\frac{\alpha^\beta}{\Gamma(\beta)} \theta^{\beta-1} e^{-\alpha \theta} \ \ \ \ \ \ \ \ \ \ \theta>0$
Then the joint density of $N$ and $\Theta$ is:
$\displaystyle (17) \ \ \ \ \ P(N=n \lvert \Theta=\theta) \ g(\theta)=\frac{e^{-\theta} \ \theta^n}{n!} \ \frac{\alpha^\beta}{\Gamma(\beta)} \theta^{\beta-1} e^{-\alpha \theta}$
The unconditional distribution of $N$ is obtained by summing out $\theta$ in $(17)$.
\displaystyle \begin{aligned}(18) \ \ \ \ \ P(N=n)&=\int_0^\infty P(N=n \lvert \Theta=\theta) \ g(\theta) \ d \theta \\&=\int_0^\infty \frac{e^{-\theta} \ \theta^n}{n!} \ \frac{\alpha^\beta}{\Gamma(\beta)} \ \theta^{\beta-1} \ e^{-\alpha \theta} \ d \theta \\&=\int_0^\infty \frac{\alpha^\beta}{n! \ \Gamma(\beta)} \ \theta^{n+\beta-1} \ e^{-(\alpha+1) \theta} d \theta \\&=\frac{\alpha^\beta}{n! \ \Gamma(\beta)} \ \frac{\Gamma(n+\beta)}{(\alpha+1)^{n+\beta}} \int_0^\infty \frac{(\alpha+1)^{n+\beta}}{\Gamma(n+\beta)} \theta^{n+\beta-1} \ e^{-(\alpha+1) \theta} d \theta \\&=\frac{\alpha^\beta}{n! \ \Gamma(\beta)} \ \frac{\Gamma(n+\beta)}{(\alpha+1)^{n+\beta}} \\&=\frac{\Gamma(n+\beta)}{\Gamma(n+1) \ \Gamma(\beta)} \ \biggl( \frac{\alpha}{\alpha+1}\biggr)^\beta \ \biggl(\frac{1}{\alpha+1}\biggr)^n \\&=\binom{n+\beta-1}{n} \ \biggl( \frac{\alpha}{\alpha+1}\biggr)^\beta \ \biggl(\frac{1}{\alpha+1}\biggr)^n \ \ \ \ \ \ \ \ \ n=0,1,2,\cdots \end{aligned}
Note that the integral in the fourth step in $(18)$ is 1.0 since the integrand is the pdf of a Gamma distribution. The above probability function is that of a negative binomial distribution. It is of the same form as $(6a)$. Equivalently, it is also of the form $(6)$ with parameter $r=\beta$ and $p=\frac{\alpha}{\alpha+1}$.
The variance of the negative binomial distribution is greater than the mean. In a Poisson distribution, the mean equals the variance. Thus the unconditional distribution of $N$ is more dispersed than its conditional distributions. This is a characteristic of mixture distributions. The uncertainty in the parameter variable $\Theta$ has the effect of increasing the unconditional variance of the mixture distribution of $N$. The variance of a mixture distribution has two components, the weighted average of the conditional variances and the variance of the conditional means. The second component represents the additional variance introduced by the uncertainty in the parameter $\Theta$ (see The variance of a mixture).
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The Poisson Distribution as Limit of Negative Binomial
There is another connection to the Poisson distribution, that is, the Poisson distribution is a limiting case of the negative binomial distribution. We show that the generating function of the Poisson distribution can be obtained by taking the limit of the negative binomial generating function as $r \rightarrow \infty$. Interestingly, the Poisson distribution is also the limit of the binomial distribution.
In this section, we use the negative binomial parametrization of $(6a)$. By replacing $\frac{\alpha}{\alpha+1}$ for $p$, the following are the mean, variance, and the generating function for the probability function in $(6a)$:
\displaystyle \begin{aligned}(19) \ \ \ \ \ \ &E(X)=\frac{r}{\alpha} \\&\text{ }\\&Var(X)=\frac{\alpha+1}{\alpha} \ \frac{r}{\alpha}=\frac{r(\alpha+1)}{\alpha^2} \\&\text{ } \\&g(z)=\frac{1}{[1-\frac{1}{\alpha}(z-1)]^r} \ \ \ \ \ \ \ z<\alpha+1 \end{aligned}
Let r goes to infinity and $\displaystyle \frac{1}{\alpha}$ goes to zero and at the same time keeping their product constant. Thus $\displaystyle \mu=\frac{r}{\alpha}$ is constant (this is the mean of the negative binomial distribution). We show the following:
$\displaystyle (20) \ \ \ \ \ \lim \limits_{r \rightarrow \infty} [1-\frac{\mu}{r}(z-1)]^{-r}=e^{\mu (z-1)}$
The right-hand side of $(20)$ is the generating function of the Poisson distribution with mean $\mu$. The generating function in the left-hand side is that of a negative binomial distribution with mean $\displaystyle \mu=\frac{r}{\alpha}$. The following is the derivation of $(20)$.
\displaystyle \begin{aligned}(21) \ \ \ \ \ \lim \limits_{r \rightarrow \infty} [1-\frac{\mu}{r}(z-1)]^{-r}&=\lim \limits_{r \rightarrow \infty} e^{\displaystyle \biggl(ln[1-\frac{\mu}{r}(z-1)]^{-r}\biggr)} \\&=\lim \limits_{r \rightarrow \infty} e^{\displaystyle \biggl(-r \ ln[1-\frac{\mu}{r}(z-1)]\biggr)} \\&=e^{\displaystyle \biggl(\lim \limits_{r \rightarrow \infty} -r \ ln[1-\frac{\mu}{r}(z-1)]\biggr)} \end{aligned}
We now focus on the limit in the exponent.
\displaystyle \begin{aligned}(22) \ \ \ \ \ \lim \limits_{r \rightarrow \infty} -r \ ln[1-\frac{\mu}{r}(z-1)]&=\lim \limits_{r \rightarrow \infty} \frac{ln(1-\frac{\mu}{r} (z-1))^{-1}}{r^{-1}} \\&=\lim \limits_{r \rightarrow \infty} \frac{(1-\frac{\mu}{r} (z-1)) \ \mu (z-1) r^{-2}}{r^{-2}} \\&=\mu (z-1) \end{aligned}
The middle step in $(22)$ uses the L’Hopital’s Rule. The result in $(20)$ is obtained by combining $(21)$ and $(22)$.
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Reference
1. Klugman S.A., Panjer H. H., Wilmot G. E. Loss Models, From Data to Decisions, Second Edition., Wiley-Interscience, a John Wiley & Sons, Inc., New York, 2004
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$\copyright \ \text{2011-2015 by Dan Ma}$
# Splitting a Poisson Distribution
We consider a remarkable property of the Poisson distribution that has a connection to the multinomial distribution. We start with the following examples.
Example 1
Suppose that the arrivals of customers in a gift shop at an airport follow a Poisson distribution with a mean of $\alpha=5$ per 10 minutes. Furthermore, suppose that each arrival can be classified into one of three distinct types – type 1 (no purchase), type 2 (purchase under $20), and type 3 (purchase over$20). Records show that about 25% of the customers are of type 1. The percentages of type 2 and type 3 are 60% and 15%, respectively. What is the probability distribution of the number of customers per hour of each type?
Example 2
Roll a fair die $N$ times where $N$ is random and follows a Poisson distribution with parameter $\alpha$. For each $i=1,2,3,4,5,6$, let $N_i$ be the number of times the upside of the die is $i$. What is the probability distribution of each $N_i$? What is the joint distribution of $N_1,N_2,N_3,N_4,N_5,N_6$?
In Example 1, the stream of customers arrive according to a Poisson distribution. It can be shown that the stream of each type of customers also has a Poisson distribution. One way to view this example is that we can split the Poisson distribution into three Poisson distributions.
Example 2 also describes a splitting process, i.e. splitting a Poisson variable into 6 different Poisson variables. We can also view Example 2 as a multinomial distribution where the number of trials is not fixed but is random and follows a Poisson distribution. If the number of rolls of the die is fixed in Example 2 (say 10), then each $N_i$ would be a binomial distribution. Yet, with the number of trials being Poisson, each $N_i$ has a Poisson distribution with mean $\displaystyle \frac{\alpha}{6}$. In this post, we describe this Poisson splitting process in terms of a “random” multinomial distribution (the view point of Example 2).
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Suppose we have a multinomial experiment with parameters $N$, $r$, $p_1, \cdots, p_r$, where
• $N$ is the number of multinomial trials,
• $r$ is the number of distinct possible outcomes in each trial (type 1 through type $r$),
• the $p_i$ are the probabilities of the $r$ possible outcomes in each trial.
Suppose that $N$ follows a Poisson distribution with parameter $\alpha$. For each $i=1, \cdots, r$, let $N_i$ be the number of occurrences of the $i^{th}$ type of outcomes in the $N$ trials. Then $N_1,N_2,\cdots,N_r$ are mutually independent Poisson random variables with parameters $\alpha p_1,\alpha p_2,\cdots,\alpha p_r$, respectively.
The variables $N_1,N_2,\cdots,N_r$ have a multinomial distribution and their joint probability function is:
$\displaystyle (1) \ \ \ \ P(N_1=n_1,N_2=n_2,\cdots,N_r=n_r)=\frac{N!}{n_1! n_2! \cdots n_r!} \ p_1^{n_1} p_2^{n_2} \cdots p_r^{n_r}$
where $n_i$ are nonnegative integers such that $N=n_1+n_2+\cdots+n_r$.
Since the total number of multinomial trials $N$ is not fixed and is random, $(1)$ is not the end of the story. The probability in $(1)$ is only a conditional probability. The following is the joint probability function of $N_1,N_2,\cdots,N_r$:
$\displaystyle (2) \ \ \ \ P(N_1=n_1,N_2=n_2,\cdots,N_r=n_r)$
\displaystyle \begin{aligned}&=P(N_1=n_1,N_2=n_2,\cdots,N_r=n_r \lvert N=\sum \limits_{k=0}^r n_k) \\&\ \ \ \ \ \times P(N=\sum \limits_{k=0}^r n_k) \\&\text{ } \\&=\frac{(\sum \limits_{k=0}^r n_k)!}{n_1! \ n_2! \ \cdots \ n_r!} \ p_1^{n_1} \ p_2^{n_2} \ \cdots \ p_r^{n_r} \ \times \frac{e^{-\alpha} \alpha^{\sum \limits_{k=0}^r n_k}}{(\sum \limits_{k=0}^r n_k!)} \\&\text{ } \\&=\frac{e^{-\alpha p_1} \ (\alpha p_1)^{n_1}}{n_1!} \ \frac{e^{-\alpha p_2} \ (\alpha p_2)^{n_2}}{n_2!} \ \cdots \ \frac{e^{-\alpha p_r} \ (\alpha p_r)^{n_r}}{n_r!} \end{aligned}
To obtain the marginal probability function of $N_j$, $j=1,2,\cdots,r$, we sum out the other variables $N_k=n_k$ ($k \ne j$) in $(2)$ and obtain the following:
$\displaystyle (3) \ \ \ \ P(N_j=n_j)=\frac{e^{-\alpha p_j} \ (\alpha p_j)^{n_j}}{n_j!}$
Thus we can conclude that $N_j$, $j=1,2,\cdots,r$, has a Poisson distribution with parameter $\alpha p_j$. Furrthermore, the joint probability function of $N_1,N_2,\cdots,N_r$ is the product of the marginal probability functions. Thus we can conclude that $N_1,N_2,\cdots,N_r$ are mutually independent.
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Example 1
Let $N_1,N_2,N_3$ be the number of customers per hour of type 1, type 2, and type 3, respectively. Here, we attempt to split a Poisson distribution with mean 30 per hour (based on 5 per 10 minutes). Thus $N_1,N_2,N_3$ are mutually independent Poisson variables with means $30 \times 0.25=7.5$, $30 \times 0.60=18$, $30 \times 0.15=4.5$, respectively.
Example 2
As indicated earlier, each $N_i$, $i=1,2,3,4,5,6$, has a Poisson distribution with mean $\frac{\alpha}{6}$. According to $(2)$, the joint probability function of $N_1,N_2,N_3,N_4,N_5,N_6$ is simply the product of the six marginal Poisson probability functions.
# The Poisson Distribution
Let $\alpha$ be a positive constant. Consider the following probability distribution:
$\displaystyle (1) \ \ \ \ \ P(X=j)=\frac{e^{-\alpha} \alpha^j}{j!} \ \ \ \ \ j=0,1,2,\cdots$
The above distribution is said to be a Poisson distribution with parameter $\alpha$. The Poisson distribution is usually used to model the random number of events occurring in a fixed time interval. As will be shown below, $E(X)=\alpha$. Thus the parameter $\alpha$ is the rate of occurrence of the random events; it indicates on average how many events occur per unit of time. Examples of random events that may be modeled by the Poisson distribution include the number of alpha particles emitted by a radioactive substance counted in a prescribed area during a fixed period of time, the number of auto accidents in a fixed period of time or the number of losses arising from a group of insureds during a policy period.
Each of the above examples can be thought of as a process that generates a number of arrivals or changes in a fixed period of time. If such a counting process leads to a Poisson distribution, then the process is said to be a Poisson process.
We now discuss some basic properties of the Poisson distribution. Using the Taylor series expansion of $e^{\alpha}$, the following shows that $(1)$ is indeed a probability distribution.
$\displaystyle . \ \ \ \ \ \ \ \sum \limits_{j=0}^\infty \frac{e^{-\alpha} \alpha^j}{j!}=e^{-\alpha} \sum \limits_{j=0}^\infty \frac{\alpha^j}{j!}=e^{-\alpha} e^{\alpha}=1$
The generating function of the Poisson distribution is $g(z)=e^{\alpha (z-1)}$ (see The generating function). The mean and variance can be calculated using the generating function.
\displaystyle \begin{aligned}(2) \ \ \ \ \ &E(X)=g'(1)=\alpha \\&\text{ } \\&E[X(X-1)]=g^{(2)}(1)=\alpha^2 \\&\text{ } \\&Var(X)=E[X(X-1)]+E(X)-E(X)^2=\alpha^2+\alpha-\alpha^2=\alpha \end{aligned}
The Poisson distribution can also be interpreted as an approximation to the binomial distribution. It is well known that the Poisson distribution is the limiting case of binomial distributions (see [1] or this post).
$\displaystyle (3) \ \ \ \ \ \lim \limits_{n \rightarrow \infty} \binom{n}{j} \biggl(\frac{\alpha}{n}\biggr)^j \biggl(1-\frac{\alpha}{n}\biggr)^{n-j}=\frac{e^{-\alpha} \alpha^j}{j!}$
One application of $(3)$ is that we can use Poisson probabilities to approximate Binomial probabilities. The approximation is reasonably good when the number of trials $n$ in a binomial distribution is large and the probability of success $p$ is small. The binomial mean is $n p$ and the variance is $n p (1-p)$. When $p$ is small, $1-p$ is close to 1 and the binomial variance is approximately $np \approx n p (1-p)$. Whenever the mean of a discrete distribution is approximately equaled to the mean, the Poisson approximation is quite good. As a rule of thumb, we can use Poisson to approximate binomial if $n \le 100$ and $p \le 0.01$.
As an example, we use the Poisson distribution to estimate the probability that at most 1 person out of 1000 will have a birthday on the New Year Day. Let $n=1000$ and $p=365^{-1}$. So we use the Poisson distribution with $\alpha=1000 365^{-1}$. The following is an estimate using the Poisson distribution.
$\displaystyle . \ \ \ \ \ \ \ P(X \le 1)=e^{-\alpha}+\alpha e^{-\alpha}=(1+\alpha) e^{-\alpha}=0.2415$
Another useful property is that the independent sum of Poisson distributions also has a Poisson distribution. Specifically, if each $X_i$ has a Poisson distribution with parameter $\alpha_i$, then the independent sum $X=X_1+\cdots+X_n$ has a Poisson distribution with parameter $\alpha=\alpha_1+\cdots+\alpha_n$. One way to see this is that the product of Poisson generating functions has the same general form as $g(z)=e^{\alpha (z-1)}$ (see The generating function). One interpretation of this property is that when merging several arrival processes, each of which follow a Poisson distribution, the result is still a Poisson distribution.
For example, suppose that in an airline ticket counter, the arrival of first class customers follows a Poisson process with a mean arrival rate of 8 per 15 minutes and the arrival of customers flying coach follows a Poisson distribution with a mean rate of 12 per 15 minutes. Then the arrival of customers of either types has a Poisson distribution with a mean rate of 20 per 15 minutes or 80 per hour.
A Poisson distribution with a large mean can be thought of as an independent sum of Poisson distributions. For example, a Poisson distribution with a mean of 50 is the independent sum of 50 Poisson distributions each with mean 1. Because of the central limit theorem, when the mean is large, we can approximate the Poisson using the normal distribution.
In addition to merging several Poisson distributions into one combined Poisson distribution, we can also split a Poisson into several Poisson distributions. For example, suppose that a stream of customers arrives according to a Poisson distribution with parameter $\alpha$ and each customer can be classified into one of two types (e.g. no purchase vs. purchase) with probabilities $p_1$ and $p_2$, respectively. Then the number of “no purchase” customers and the number of “purchase” customers are independent Poisson random variables with parameters $\alpha p_1$ and $\alpha p_2$, respectively. For more details on the splitting of Poisson, see Splitting a Poisson Distribution.
Reference
1. Feller W. An Introduction to Probability Theory and Its Applications, Third Edition, John Wiley & Sons, New York, 1968
# The generating function
Consider the function $g(z)=\displaystyle e^{\alpha (z-1)}$ where $\alpha$ is a positive constant. The following shows the derivatives of this function.
\displaystyle \begin{aligned}. \ \ \ \ \ \ &g(z)=e^{\alpha (z-1)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g(0)=e^{-\alpha} \\&\text{ } \\&g'(z)=e^{\alpha (z-1)} \ \alpha \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g'(0)=e^{-\alpha} \ \alpha \\&\text{ } \\&g^{(2)}(z)=e^{\alpha (z-1)} \ \alpha^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{(2)}(0)=2! \ \frac{e^{-\alpha} \ \alpha^2}{2!} \\&\text{ } \\&g^{(3)}(z)=e^{\alpha (z-1)} \ \alpha^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{(3)}(0)=3! \ \frac{e^{-\alpha} \ \alpha^3}{3!} \\&\text{ } \\&\ \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \\&\text{ } \\&g^{(n)}(z)=e^{\alpha (z-1)} \ \alpha^n \ \ \ \ \ \ \ \ \ \ \ \ \ \ g^{(n)}(0)=n! \ \frac{e^{-\alpha} \ \alpha^n}{n!} \end{aligned}
Note that the derivative of $g(z)$ at each order is a multiple of a Poisson probability. Thus the Poisson distribution is coded by the function $g(z)=\displaystyle e^{\alpha (z-1)}$. Because of this reason, such a function is called a generating function (or probability generating function). This post discusses some basic facts about the generating function (gf) and its cousin, the moment generating function (mgf). One important characteristic is that these functions generate probabilities and moments. Another important characteristic is that there is a one-to-one correspondence between a probability distribution and its generating function and moment generating function, i.e. two random variables with different cumulative distribution functions cannot have the same gf or mgf. In some situations, this fact is useful in working with independent sum of random variables.
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The Generating Function
Suppose that $X$ is a random variable that takes only nonegative integer values with the probability function given by
$(1) \ \ \ \ \ \ P(X=j)=a_j, \ \ \ \ j=0,1,2,\cdots$
The idea of the generating function is that we use a power series to capture the entire probability distribution. The following defines the generating function that is associated with the above sequence $a_j$, .
$(2) \ \ \ \ \ \ g(z)=a_0+a_1 \ z+a_2 \ z^2+ \cdots=\sum \limits_{j=0}^\infty a_j \ z^j$
Since the elements of the sequence $a_j$ are probabilities, we can also call $g(z)$ the generating function of the probability distribution defined by the sequence in $(1)$. The generating function $g(z)$ is defined wherever the power series converges. It is clear that at the minimum, the power series in $(2)$ converges for $\lvert z \lvert \le 1$.
We discuss the following three properties of generating functions:
1. The generating function completely determines the distribution.
2. The moments of the distribution can be derived from the derivatives of the generating function.
3. The generating function of a sum of independent random variables is the product of the individual generating functions.
The Poisson generating function at the beginning of the post is an example demonstrating property 1 (see Example 0 below for the derivation of the generating function). In some cases, the probability distribution of an independent sum can be deduced from the product of the individual generating functions. Some examples are given below.
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Generating Probabilities
We now discuss the property 1 indicated above. To see that $g(z)$ generates the probabilities, let’s look at the derivatives of $g(z)$:
\displaystyle \begin{aligned}(3) \ \ \ \ \ \ &g'(z)=a_1+2 a_2 \ z+3 a_3 \ z^2+\cdots=\sum \limits_{j=1}^\infty j a_j \ z^{j-1} \\&\text{ } \\&g^{(2)}(z)=2 a_2+6 a_3 \ z+ 12 a_4 \ z^2=\sum \limits_{j=2}^\infty j (j-1) a_j \ z^{j-2} \\&\text{ } \\&g^{(3)}(z)=6 a_3+ 24 a_4 \ z+60 a_5 \ z^2=\sum \limits_{j=3}^\infty j (j-1)(j-2) a_j \ z^{j-3} \\&\text{ } \\&\ \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \\&\text{ } \\&g^{(n)}(z)=\sum \limits_{j=n}^\infty j(j-1) \cdots (j-n+1) a_j \ z^{j-n}=\sum \limits_{j=n}^\infty \binom{j}{n} n! \ a_j \ z^{j-n} \end{aligned}
By letting $z=0$ above, all the terms vanishes except for the constant term. We have:
$(4) \ \ \ \ \ \ g^{(n)}(0)=n! \ a_n=n! \ P(X=n)$
Thus the generating function is a compact way of encoding the probability distribution. The probability distribution determines the generating function as seen in $(2)$. On the other hand, $(3)$ and $(4)$ demonstrate that the generating function also determines the probability distribution.
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Generating Moments
The generating function also determines the moments (property 2 indicated above). For example, we have:
\displaystyle \begin{aligned}(5) \ \ \ \ \ \ &g'(1)=0 \ a_0+a_1+2 a_2+3 a_3+\cdots=\sum \limits_{j=0}^\infty j a_j=E(X) \\&\text{ } \\&g^{(2)}(1)=0 a_0 + 0 a_1+2 a_2+6 a_3+ 12 a_4+\cdots=\sum \limits_{j=0}^\infty j (j-1) a_j=E[X(X-1)] \\&\text{ } \\&E(X)=g'(1) \\&\text{ } \\&E(X^2)=g'(1)+g^{(2)}(1) \end{aligned}
Note that $g^{(n)}(1)=E[X(X-1) \cdots (X-(n-1))]$. Thus the higher moment $E(X^n)$ can be expressed in terms of $g^{(n)}(1)$ and $g^{(k)}(1)$ where $k.
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More General Definitions
Note that the definition in $(2)$ can also be interpreted as the mathematical expectation of $z^X$, i.e., $g(z)=E(z^X)$. This provides a way to define the generating function for random variables that may take on values outside of the nonnegative integers. The following is a more general definition of the generating function of the random variable $X$, which is defined for all $z$ where the expectation exists.
$(6) \ \ \ \ \ \ g(z)=E(z^X)$
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The Generating Function of Independent Sum
Let $X_1,X_2,\cdots,X_n$ be independent random variables with generating functions $g_1,g_2,\cdots,g_n$, respectively. Then the generating function of $X_1+X_2+\cdots+X_n$ is given by the product $g_1 \cdot g_2 \cdots g_n$.
Let $g(z)$ be the generating function of the independent sum $X_1+X_2+\cdots+X_n$. The following derives $g(z)$. Note that the general form of generating function $(6)$ is used.
\displaystyle \begin{aligned}(7) \ \ \ \ \ \ g(z)&=E(z^{X_1+\cdots+X_n}) \\&\text{ } \\&=E(z^{X_1} \cdots z^{X_n}) \\&\text{ } \\&=E(z^{X_1}) \cdots E(z^{X_n}) \\&\text{ } \\&=g_1(z) \cdots g_n(z) \end{aligned}
The probability distribution of a random variable is uniquely determined by its generating function. In particular, the generating function $g(z)$ of the independent sum $X_1+X_2+\cdots+X_n$ that is derived in $(7)$ is unique. So if the generating function is of a particular distribution, we can deduce that the distribution of the sum must be of the same distribution. See the examples below.
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Example 0
In this example, we derive the generating function of the Poisson distribution. Based on the definition, we have:
\displaystyle \begin{aligned}. \ \ \ \ \ \ g(z)&=\sum \limits_{j=0}^\infty \frac{e^{-\alpha} \alpha^j}{j!} \ z^j \\&\text{ } \\&=\sum \limits_{j=0}^\infty \frac{e^{-\alpha} (\alpha z)^j}{j!} \\&\text{ } \\&=\frac{e^{-\alpha}}{e^{- \alpha z}} \sum \limits_{j=0}^\infty \frac{e^{-\alpha z} (\alpha z)^j}{j!} \\&\text{ } \\&=e^{\alpha (z-1)} \end{aligned}
Example 1
Suppose that $X_1,X_2,\cdots,X_n$ are independent random variables where each $X_i$ has a Bernoulli distribution with probability of success $p$. Let $q=1-p$. The following is the generating function for each $X_i$.
$. \ \ \ \ \ \ g(z)=q+p z$
Then the generating function of the sum $X=X_1+\cdots+X_n$ is $g(z)^n=(q+p z)^n$. The following is the binomial expansion:
\displaystyle \begin{aligned}(8) \ \ \ \ \ \ g(z)^n&=(q+p z)^n \\&\text{ } \\&=\sum \limits_{j=0}^n \binom{n}{j} q^{n-j} \ p^j \ z^j \end{aligned}
By definition $(2)$, the generating function of $X=X_1+\cdots+X_n$ is:
$(9) \ \ \ \ \ \ g(z)^n=\sum \limits_{j=0}^\infty P(X=j) \ z^j$
Comparing $(8)$ and $(9)$, we have
$\displaystyle (10) \ \ \ \ \ \ P(X=j)=\left\{\begin{matrix}\displaystyle \binom{n}{j} p^j \ q^{n-j}&\ 0 \le j \le n\\{0}&\ j>n \end{matrix}\right.$
The probability distribution indicated by $(8)$ and $(10)$ is that of a binomial distribution. Since the probability distribution of a random variable is uniquely determined by its generating function, the independent sum of Bernoulli distributions must ave a Binomial distribution.
Example 2
Suppose that $X_1,X_2,\cdots,X_n$ are independent and have Poisson distributions with parameters $\alpha_1,\alpha_2,\cdots,\alpha_n$, respectively. Then the independent sum $X=X_1+\cdots+X_n$ has a Poisson distribution with parameter $\alpha=\alpha_1+\cdots+\alpha_n$.
Let $g(z)$ be the generating function of $X=X_1+\cdots+X_n$. For each $i$, the generating function of $X_i$ is $g_i(z)=e^{\alpha_i (z-1)}$. The key to the proof is that the product of the $g_i$ has the same general form as the individual $g_i$.
\displaystyle \begin{aligned}(11) \ \ \ \ \ \ g(z)&=g_1(z) \cdots g_n(z) \\&\text{ } \\&=e^{\alpha_1 (z-1)} \cdots e^{\alpha_n (z-1)} \\&\text{ } \\&=e^{(\alpha_1+\cdots+\alpha_n)(z-1)} \end{aligned}
The generating function in $(11)$ is that of a Poisson distribution with mean $\alpha=\alpha_1+\cdots+\alpha_n$. Since the generating function uniquely determines the distribution, we can deduce that the sum $X=X_1+\cdots+X_n$ has a Poisson distribution with parameter $\alpha=\alpha_1+\cdots+\alpha_n$.
Example 3
In rolling a fair die, let $X$ be the number shown on the up face. The associated generating function is:
$\displaystyle. \ \ \ \ \ \ g(z)=\frac{1}{6}(z+z^2+z^3+z^4+z^5+z^6)=\frac{z(1-z^6)}{6(1-z)}$
The generating function can be further reduced as:
\displaystyle \begin{aligned}. \ \ \ \ \ \ g(z)&=\frac{z(1-z^6)}{6(1-z)} \\&\text{ } \\&=\frac{z(1-z^3)(1+z^3)}{6(1-z)} \\&\text{ } \\&=\frac{z(1-z)(1+z+z^2)(1+z^3)}{6(1-z)} \\&\text{ } \\&=\frac{z(1+z+z^2)(1+z^3)}{6} \end{aligned}
Suppose that we roll the fair dice 4 times. Let $W$ be the sum of the 4 rolls. Then the generating function of $Z$ is
$\displaystyle. \ \ \ \ \ \ g(z)^4=\frac{z^4 (1+z^3)^4 (1+z+z^2)^4}{6^4}$
The random variable $W$ ranges from 4 to 24. Thus the probability function ranges from $P(W=4)$ to $P(W=24)$. To find these probabilities, we simply need to decode the generating function $g(z)^4$. For example, to find $P(W=12)$, we need to find the coefficient of the term $z^{12}$ in the polynomial $g(z)^4$. To help this decoding, we can expand two of the polynomials in $g(z)^4$.
\displaystyle \begin{aligned}. \ \ \ \ \ \ g(z)^4&=\frac{z^4 (1+z^3)^4 (1+z+z^2)^4}{6^4} \\&\text{ } \\&=\frac{z^4 \times A \times B}{6^4} \\&\text{ } \\&A=(1+z^3)^4=1+4z^3+6z^6+4z^9+z^{12} \\&\text{ } \\&B=(1+z+z^2)^4=1+4z+10z^2+16z^3+19z^4+16z^5+10z^6+4z^7+z^8 \end{aligned}
Based on the above polynomials, there are three ways of forming $z^{12}$. They are: $(z^4 \times 1 \times z^8)$, $(z^4 \times 4z^3 \times 16z^5)$, $(z^4 \times 6z^6 \times 10z^2)$. Thus we have:
$\displaystyle. \ \ \ \ \ \ P(W=12)=\frac{1}{6^4}(1+4 \times 16+6 \times 10)=\frac{125}{6^4}$
To find the other probabilities, we can follow the same decoding process.
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Remark
The probability distribution of a random variable is uniquely determined by its generating function. This fundamental property is useful in determining the distribution of an independent sum. The generating function of the independent sum is simply the product of the individual generating functions. If the product is of a certain distributional form (as in Example 1 and Example 2), then we can deduce that the sum must be of the same distribution.
We can also decode the product of generating functions to obtain the probability function of the independent sum (as in Example 3). The method in Example 3 is quite tedious. But one advantage is that it is a “machine process”, a pretty fool proof process that can be performed mechanically.
The machine process is this: Code the individual probability distribution in a generating function $g(z)$. Then raise it to $n$. After performing some manipulation to $g(z)^n$, decode the probabilities from $g(z)^n$.
As long as we can perform the algebraic manipulation carefully and correctly, this process will be sure to provide the probability distribution of an independent sum.
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The Moment Generating Function
The moment generating function of a random variable $X$ is $M_X(t)=E(e^{tX})$ on all real numbers $t$ for which the expected value exists. The moments can be computed more directly using an mgf. From the theory of mathematical analysis, it can be shown that if $M_X(t)$ exists on some interval $-a, then the derivatives of $M_X(t)$ of all orders exist at $t=0$. Furthermore, it can be show that $E(X^n)=M_X^{(n)}(0)$.
Suppose that $g(z)$ is the generating function of a random variable. The following relates the generating function and the moment generating function.
\displaystyle \begin{aligned}. \ \ \ \ \ \ &M_X(t)=g(e^t) \\&\text{ } \\&g(z)=M_X(ln z) \end{aligned}
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Reference
1. Feller W. An Introduction to Probability Theory and Its Applications, Third Edition, John Wiley & Sons, New York, 1968
# The hazard rate function
In this post, we introduce the hazard rate function using the notions of non-homogeneous Poisson process.
In a Poisson process, changes occur at a constant rate $\lambda$ per unit time. Suppose that we interpret the changes in a Poisson process from a mortality point of view, i.e. a change in the Poisson process mean a termination of a system, be it biological or manufactured, and this Poisson process counts the number of terminations as they occur. Then the rate of change $\lambda$ is interpreted as a hazard rate (or failure rate or force of mortality). With a constant force of mortality, the time until the next change is exponentially distributed. In this post, we discuss the hazard rate function in a more general setting. The process that counts of the number of terminations will no longer have a constant hazard rate, and instead will have a hazard rate function $\lambda(t)$, a function of time $t$. Such a counting process is called a non-homogeneous Poisson process. We discuss the survival probability models (the time to the first termination) associated with a non-homogeneous Poisson process. We then discuss several important examples of survival probability models, including the Weibull distribution, the Gompertz distribution and the model based on the Makeham’s law. See [1] for more information about the hazard rate function.
$\text{ }$
The Poisson Process
We start with the three postulates of a Poisson process. Consider an experiment in which the occurrences of a certain type of events are counted during a given time interval. We call the occurrence of the type of events in question a change. We assume the following three conditions:
1. The numbers of changes occurring in nonoverlapping intervals are independent.
2. The probability of two or more changes taking place in a sufficiently small interval is essentially zero.
3. The probability of exactly one change in the short interval $(t,t+\delta)$ is approximately $\lambda \delta$ where $\delta$ is sufficiently small and $\lambda$ is a positive constant.
$\text{ }$
When we interpret the Poisson process in a mortality point of view, the constant $\lambda$ is a hazard rate (or force of mortality), which can be interpreted as the rate of failure at the next instant given that the life has survived to time $t$. With a constant force of mortality, the survival model (the time until the next termination) has an exponential distribution with mean $\frac{1}{\lambda}$. We wish to relax the constant force of mortality assumption by making $\lambda$ a function of $t$ instead. The remainder of this post is based on the non-homogeneous Poisson process defined below.
$\text{ }$
The Non-Homogeneous Poisson Process
We modifiy condition 3 above by making $\lambda(t)$ a function of $t$. We have the following modified counting process.
1. The numbers of changes occurring in nonoverlapping intervals are independent.
2. The probability of two or more changes taking place in a sufficiently small interval is essentially zero.
3. The probability of exactly one change in the short interval $(t,t+\delta)$ is approximately $\lambda(t) \delta$ where $\delta$ is sufficiently small and $\lambda(t)$ is a nonnegative function of $t$.
$\text{ }$
We focus on the survival model aspect of such counting processes. Such process can be interpreted as models for the number of changes occurred in a time interval where a change means “termination” or ‘failure” of a system under consideration. The rate of change function $\lambda(t)$ indicated in condition 3 is called the hazard rate function. It is also called the failure rate function in reliability engineering and the force of mortality in life contingency theory.
Based on condition 3 in the non-homogeneous Poisson process, the hazard rate function $\lambda(t)$ can be interpreted as the rate of failure at the next instant given that the life has survived to time $t$.
Two random variables naturally arise from a non-homogeneous Poisson process are described here. One is the discrete variable $N_t$, defined as the number of changes in the time interval $(0,t)$. The other is the continuous random variable $T$, defined as the time until the occurrence of the first change. The probability distribution of $T$ is called a survival model. The following is the link between $N_t$ and $T$.
$\text{ }$
\displaystyle \begin{aligned}(1) \ \ \ \ \ \ \ \ \ &P[T > t]=P[N_t=0] \end{aligned}
$\text{ }$
Note that $P[T > t]$ is the probability that the next change occurs after time $t$. This means that there is no change within the interval $(0,t)$. We have the following theorems.
$\text{ }$
Theorem 1.
Let $\displaystyle \Lambda(t)=\int_{0}^{t} \lambda(y) dy$. Then $e^{-\Lambda(t)}$ is the probability that there is no change in the interval $(0,t)$. That is, $\displaystyle P[N_t=0]=e^{-\Lambda(t)}$.
Proof. We are interested in finding the probability of zero changes in the interval $(0,y+\delta)$. By condition 1, the numbers of changes in the nonoverlapping intervals $(0,y)$ and $(y,y+\delta)$ are independent. Thus we have:
$\text{ }$
$\displaystyle (2) \ \ \ \ \ \ \ \ P[N_{y+\delta}=0] \approx P[N_y=0] \times [1-\lambda(y) \delta]$
$\text{ }$
Note that by condition 3, the probability of exactly one change in the small interval $(y,y+\delta)$ is $\lambda(y) \delta$. Thus $[1-\lambda(y) \delta]$ is the probability of no change in the interval $(y,y+\delta)$. Continuing with equation $(2)$, we have the following derivation:
$\text{ }$
\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\frac{P[N_{y+\delta}=0] - P[N_y=0]}{\delta} \approx -\lambda(y) P[N_y=0] \\&\text{ } \\&\frac{d}{dy} P[N_y=0]=-\lambda(y) P[N_y=0] \\&\text{ } \\&\frac{\frac{d}{dy} P[N_y=0]}{P[N_y=0]}=-\lambda(y) \\&\text{ } \\&\int_0^{t} \frac{\frac{d}{dy} P[N_y=0]}{P[N_y=0]} dy=-\int_0^{t} \lambda(y)dy \end{aligned}
$\text{ }$
Evaluating the integral on the left hand side with the boundary condition of $P[N_0=0]=1$ produces the following results:
\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &ln P[N_t=0]=-\int_0^{t} \lambda(y)dy \\&\text{ } \\&P[N_t=0]=e^{\displaystyle -\int_0^{t} \lambda(y)dy} \end{aligned}
$\text{ }$
Theorem 2
As discussed above, let $T$ be the length of the interval that is required to observe the first change. Then the following are the distribution function, survival function and pdf of $T$:
\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &F_T(t)=\displaystyle 1-e^{\displaystyle -\int_0^t \lambda(y) dy} \\&\text{ } \\&S_T(t)=\displaystyle e^{\displaystyle -\int_0^t \lambda(y) dy} \\&\text{ } \\&f_T(t)=\displaystyle \lambda(t) \ e^{\displaystyle -\int_0^t \lambda(y) dy} \end{aligned}
Proof. In Theorem 1, we derive the probability $P[N_y=0]$ for the discrete variable $N_y$ derived from the non-homogeneous Poisson process. We now consider the continuous random variable $T$, the time until the first change, which is related to $N_t$ by $(1)$. Thus $S_T(t)=P[T > t]=P[N_t=0]=e^{-\int_0^t \lambda(y) dy}$. The distribution function and density function can be derived accordingly.
$\text{ }$
Theorem 3
The hazard rate function $\lambda(t)$ is equivalent to each of the following:
\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\lambda(t)=\frac{f_T(t)}{1-F_T(t)} \\&\text{ } \\&\lambda(t)=\frac{-S_T^{'}(t)}{S_T(t)} \end{aligned}
$\text{ }$
Remark
Theorem 1 and Theorem 2 show that in a non-homogeneous Poisson process as described above, the hazard rate function $\lambda(t)$ completely specifies the probability distribution of the survival model $T$ (the time until the first change) . Once the rate of change function $\lambda(t)$ is known in the non-homogeneous Poisson process, we can use it to generate the survival function $S_T(t)$. All of the examples of survival models given below are derived by assuming the functional form of the hazard rate function. The result in Theorem 2 holds even outside the context of a non-homogeneous Poisson process, that is, given the hazard rate function $\lambda(t)$, we can derive the three distributional items $S_T(t)$, $F_T(t)$, $f_T(t)$.
The ratio in Theorem 3 indicates that the probability distribution determines the hazard rate function. In fact, the ratio in Theorem 3 is the usual definition of the hazard rate function. That is, the hazard rate function can be defined as the ratio of the density and the survival function (one minus the cdf). With this definition, we can also recover the survival function. Whenever $\displaystyle \lambda(x)=\frac{f_X(x)}{1-F_X(x)}$, we can derive:
$\text{ }$
\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &S_X(x)=\displaystyle e^{-\int_0^t \lambda(y) dy} \end{aligned}
$\text{ }$
As indicated above, the hazard rate function can be interpreted as the failure rate at time $t$ given that the life in question has survived to time $t$. It is the rate of failure at the next instant given that the life or system being studied has survived up to time $t$.
It is interesting to note that the function $\Lambda(t)=\int_0^t \lambda(y) dy$ defined in Theorem 1 is called the cumulative hazard rate function. Thus the cumulative hazard rate function is an alternative way of representing the hazard rate function (see the discussion on Weibull distribution below).
——————————————————————————————————————
Examples of Survival Models
–Exponential Distribution–
In many applications, especially those for biological organisms and mechanical systems that wear out over time, the hazard rate $\lambda(t)$ is an increasing function of $t$. In other words, the older the life in question (the larger the $t$), the higher chance of failure at the next instant. For humans, the probability of a 85 years old dying in the next year is clearly higher than for a 20 years old. In a Poisson process, the rate of change $\lambda(t)=\lambda$ indicated in condition 3 is a constant. As a result, the time $T$ until the first change derived in Theorem 2 has an exponential distribution with parameter $\lambda$. In terms of mortality study or reliability study of machines that wear out over time, this is not a realistic model. However, if the mortality or failure is caused by random external events, this could be an appropriate model.
–Weibull Distribution–
This distribution is an excellent model choice for describing the life of manufactured objects. It is defined by the following cumulative hazard rate function:
$\text{ }$
\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\Lambda(t)=\biggl(\frac{t}{\beta}\biggr)^{\alpha} \end{aligned} where $\alpha > 0$ and $\beta>0$
$\text{ }$
As a result, the hazard rate function, the density function and the survival function for the lifetime distribution are:
\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\lambda(t)=\frac{\alpha}{\beta} \biggl(\frac{t}{\beta}\biggr)^{\alpha-1} \\&\text{ } \\&f_T(t)=\frac{\alpha}{\beta} \biggl(\frac{t}{\beta}\biggr)^{\alpha-1} \displaystyle e^{\displaystyle -\biggl[\frac{t}{\beta}\biggr]^{\alpha}} \\&\text{ } \\&S_T(t)=\displaystyle e^{\displaystyle -\biggl[\frac{t}{\beta}\biggr]^{\alpha}} \end{aligned}
$\text{ }$
The parameter $\alpha$ is the shape parameter and $\beta$ is the scale parameter. When $\alpha=1$, the hazard rate becomes a constant and the Weibull distribution becomes an exponential distribution.
When the parameter $\alpha<1$, the failure rate decreases over time. One interpretation is that most of the defective items fail early on in the life cycle. Once they they are removed from the population, failure rate decreases over time.
When the parameter $1<\alpha$, the failure rate increases with time. This is a good candidate for a model to describe the lifetime of machines or systems that wear out over time.
–The Gompertz Distribution–
The Gompertz law states that the force of mortality or failure rate increases exponentially over time. It describe human mortality quite accurately. The following is the hazard rate function:
$\text{ }$
\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\lambda(t)=\alpha e^{\beta t} \end{aligned} where $\alpha>0$ and $\beta>0$.
$\text{ }$
The following are the cumulative hazard rate function as well as the survival function, distribution function and the pdf of the lifetime distribution $T$.
$\text{ }$
\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\Lambda(t)=\int_0^t \alpha e^{\beta y} dy=\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta} \\&\text{ } \\&S_T(t)=\displaystyle e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}\biggr)} \\&\text{ } \\&F_T(t)=\displaystyle 1-e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}\biggr)} \\&\text{ } \\&f_T(t)=\displaystyle \alpha \ e^{\beta t} \ e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}\biggr)} \end{aligned}
$\text{ }$
–Makeham’s Law–
The Makeham’s Law states that the force of mortality is the Gompertz failure rate plus an age-indpendent component that accounts for external causes of mortality. The following is the hazard rate function:
$\text{ }$
\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\lambda(t)=\alpha e^{\beta t}+\mu \end{aligned} where $\alpha>0$, $\beta>0$ and $\mu>0$.
$\text{ }$
The following are the cumulative hazard rate function as well as the survival function, distribution function and the pdf of the lifetime distribution $T$.
$\text{ }$
\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\Lambda(t)=\int_0^t (\alpha e^{\beta y}+\mu) dy=\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t \\&\text{ } \\&S_T(t)=\displaystyle e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t\biggr)} \\&\text{ } \\&F_T(t)=\displaystyle 1-e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t\biggr)} \\&\text{ } \\&f_T(t)=\biggl( \alpha e^{\beta t}+\mu t \biggr) \ e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t\biggr)} \end{aligned}
$\text{ }$
Reference
1. Klugman S.A., Panjer H. H., Wilmot G. E. Loss Models, From Data to Decisions, Second Edition., Wiley-Interscience, a John Wiley & Sons, Inc., New York, 2004
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# Solve Problems Involving Parallel Lines
In this worksheet, students will combine their knowledge of angle properties and parallel lines to solve problems which involve taking multiple steps to calculate unknown angles using number and algebra.
Key stage: KS 4
GCSE Subjects: Maths
GCSE Boards: Pearson Edexcel, AQA, Eduqas, OCR
Curriculum topic: Geometry and Measures, Basic Geometry
Curriculum subtopic: Properties and Constructions, Angles
Difficulty level:
### QUESTION 1 of 10
Parallel lines have so much in common - it's a shame they will never meet!
Engineers, architects and designers all need to know and use key facts about angles and straight lines in their work - we wouldn't want a wonky house or an aircraft with uneven wings, would we?
Knowing the properties (facts) of angles and parallel lines can help us to solve all sorts of problems.
Let's remind ourselves of the key facts now...
Angles on a straight line add up to 180°:
Alternate angles (or Z-angles) are equal:
Corresponding angles (or F-angles) are also equal:
Often we will see more than one of these rules in action within the same problem, so we need to have our eagle eyes ready to spot these scenarios.
Lets give this a go now in some examples.
e.g. Find the value of angle x in the diagram below:
Here, we need to draw another parallel line through the centre of our diagram to help.
Don't worry, there are no rules to say that we can't do this!
This has split our target angle (x) into two elements.
We can now apply the alternate angle rule from above to link these two elements to the angles already provided:
x = 28° + 38° = 66°
e.g. Find the value of x and y in the diagram below:
Angle y is a corresponding angle to 114° so these must have the same value
Angle x and y are on a straight line, so together will add up to 180°.
Therefore, x = 180° - 114° = 66°
Now we are going to put all the rules we know relating to angles and straight lines together to solve problems which involve taking multiple steps to calculate unknown angles using number and algebra.
Review the known and unknown angles on this new diagram:
What is the value of angle x?
92°
48°
46°
88°
Next, review the known and unknown angles on this new diagram:
Then select the correct answers to complete the expressions below.
134° 40° 180° The value of x is... The value of y is... The value of z is...
Explore the known and unknown angles on the diagram below:
Now type numbers to complete the statements below.
134° 40° 180° The value of x is... The value of y is... The value of z is...
Investigate the known and unknown angles on the diagram below:
Now type numbers to complete the statements below.
134° 40° 180° The value of x is... The value of y is... The value of z is...
Explore the known and unknown angles on the diagram below:
What is the value of angle ?
134° 40° 180° The value of x is... The value of y is... The value of z is...
Explore the known and unknown angles on this next diagram now:
Which two values represent angles x and y?
134° 40° 180° The value of x is... The value of y is... The value of z is...
Explore the angles on the diagram below which have been expressed algebraically:
What are the values of x and y?
108° 72° 36° 110° The value of x is... The value of y is...
Review the angles on the diagram below which have been expressed algebraically:
What is the value of x?
108° 72° 36° 110° The value of x is... The value of y is...
Investigate the known and unknown angles on this final diagram now:
What is the value of x?
177°
27°
130°
59°
Observe the known and unknown angles on the diagram below:
Then select the correct option to complete the sentence below.
The value of angle x is 87° / 93° / 77°.
• Question 1
Review the known and unknown angles on this new diagram:
What is the value of angle x?
88°
EDDIE SAYS
This question is similar to the example in our Introduction - we need to draw an extra parallel line this time. Once we do this, we can then apply the rule related to alternate angles to find the two elements which combine to form missing angle x. The lower part of angle x is alternate with 46°, whilst the upper part is alternate with 42°. So x = 46° + 42° = 88°
• Question 2
Next, review the known and unknown angles on this new diagram:
Then select the correct answers to complete the expressions below.
134° 40° 180° The value of x is... The value of y is... The value of z is...
EDDIE SAYS
Wow, we had a lot of angles to find here! Let's just take them one at a time... Angle x is alternate to 40° so these angles are the same. The value of angle y can be calculated using the rule related to angles on a straight line, as we are provided with its neighbouring angle: 180° - 46° = 134° Angle z can be calculated in one of two ways. You may have spotted that it is corresponding to angle y, or that interior angles add up to 180°: 180° - 46° = 134° And the beauty here is that we can work things out in any order we choose - what a treat!
• Question 3
Explore the known and unknown angles on the diagram below:
Now type numbers to complete the statements below.
EDDIE SAYS
Don't let the double set of parallel lines put you off here. Angle x is alternate to the angle labelled as 65°, whilst angle y is corresponding to 69°. No working out required this time, which is great, so long as we know, and can apply, the rules!
• Question 4
Investigate the known and unknown angles on the diagram below:
Now type numbers to complete the statements below.
EDDIE SAYS
Two sets of parallel lines again here for the price of one! Let's apply the rule that interior angles add up to 180° to find angle x: 180° - 62° = 118° We can apply this same rule again to find angle z: 180° - 118° = 62° Angle y is alternate to 100°, so these two angles are the same. Are you getting used to spotting different rules within the same diagram now?
• Question 5
Explore the known and unknown angles on the diagram below:
What is the value of angle ?
EDDIE SAYS
Sometimes sneaky examiners put in an angle that we don't need just to try to confuse us and check that we really understand the rules. Could you spot the alternate angle here? Missing angle y is alternate to the angle shown as 64° - see how they make a Z-shape within our parallel lines? This means that these angles are the same, so: y = 64°
• Question 6
Explore the known and unknown angles on this next diagram now:
Which two values represent angles x and y?
EDDIE SAYS
The sneaky part here is that we need to find the angles we are not asked for first. Angle y is alternate to the angle labelled as 42°, so these two angles are the same. Look at the markings on the triangle now. They show that this triangle is an isosceles triangle. This means that the two base angles of the triangle are the same. So we need a base angle to work with in order to calculate angle x: 180° - 42° = 138° 138° ÷ 2 = 69° So angle x is on a straight line with 69°: 180° - 69° = 111°
• Question 7
Explore the angles on the diagram below which have been expressed algebraically:
What are the values of x and y?
108° 72° 36° 110° The value of x is... The value of y is...
EDDIE SAYS
Here we have some algebra thrown in for good measure too! Before we do anything, we need to find the missing angles on the straight line with 2x and x. The missing angle is alternate to 72°. We know that there are '3 lots of x' (x + 2x) next to this angle, so the value of: 3x = 180° - 72° = 108° If we share this value between 3, we have the value of x: x = 108° ÷ 3 = 36° Angle x is alternate to angle y, so these two angles are the same.
• Question 8
Review the angles on the diagram below which have been expressed algebraically:
What is the value of x?
EDDIE SAYS
This time, we just need to use the parallel lines provided. The other base angle in the triangle is alternate to the angle labelled as 52°, so these are the same. We know that the angles in a triangle add up to 180°, so to find the value of angle x, we can subtract our two known angles from 180: 180° - 78° - 52° = 50°
• Question 9
Investigate the known and unknown angles on this final diagram now:
What is the value of x?
59°
EDDIE SAYS
Oh a final curve ball to deal with - no problem! We know that interior angles add up to 180° so: (2x - 12) + (x + 15) = 180° 3x + 3 = 180° 3x = 177° x = 59° We can put this value of x back into our original equation to check that it is correct and that our two angles will add up to 180°: 2x - 12 = (2 × 59) - 12 = 106° x + 15 = 59 + 15 = 74° 106° + 74° = 180° Great job! We have practised combining our knowledge of angle properties and parallel lines to solve problems which involve taking multiple steps to calculate unknown angles using number and algebra.
• Question 10
Observe the known and unknown angles on the diagram below:
Then select the correct option to complete the sentence below.
The value of angle x is 87° / 93° / 77°.
EDDIE SAYS
We need to work out other angles, that we have not been asked to find, first. The angle on the line next to missing angle x is corresponding to the angle labelled as 87°. Now we know that angles on a straight line add up to 180°, so we can find angle x by subtracting its neighbour from 180: 180° - 87° = 93°
---- OR ----
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# A circle is inscribed in a $\Delta ABC$ with sides $AC$, $AB$ and $BC$ as $8cm$, $10cm$ and $12cm$. Find the length of $AD$, $BE$ and $CF$.
Last updated date: 28th Feb 2024
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Hint: First we need to make the figure from the information given in the question. Then using the property of tangents that the tangents drawn from a point to a circle are equal in length, we can equate AD and AF, BD and BE, CE and CF. Then using the values of the sides given in the question with these equations, we can determine the respective lengths of the sides AD, BE and CF.
Complete step by step solution:
From the information given in the above question, we can draw the following figure.
Now, we know that the lengths of the tangents from a point to a circle are equal. So from the above figure, we can say that
\begin{align} & \Rightarrow BE=BD=x........(i) \\ & \Rightarrow CF=CE=y........(ii) \\ & \Rightarrow AF=AD=z........(iii) \\ \end{align}
So now we can have the following figure
Now, according to the question, the sides of the triangle AC, AB and AC are $8cm$, $10cm$ and $12cm$ respectively. So from the above figure we have
\begin{align} & \Rightarrow y+z=8cm........(iv) \\ & \Rightarrow x+z=10cm........(v) \\ & \Rightarrow x+y=12cm........(vi) \\ \end{align}
Subtracting (iv) from (v) we have
\begin{align} & \Rightarrow x+z-\left( y+z \right)=10-8 \\ & \Rightarrow x-y=2cm.........(vii) \\ \end{align}
Adding (vi) and (vii) we get
\begin{align} & \Rightarrow x+y+\left( x-y \right)=12+2 \\ & \Rightarrow 2x=14 \\ & \Rightarrow x=7cm \\ \end{align}
Putting this in (v) we get
\begin{align} & \Rightarrow 7+z=10 \\ & \Rightarrow z=10-7 \\ & \Rightarrow z=3cm \\ \end{align}
Putting this in (iv) we get
\begin{align} & \Rightarrow y+3=8 \\ & \Rightarrow y=8-3 \\ & \Rightarrow y=5cm \\ \end{align}
\begin{align} & \Rightarrow AD=z \\ & \Rightarrow AD=3cm \\ \end{align}
\begin{align} & \Rightarrow BE=x \\ & \Rightarrow BE=7cm \\ \end{align}
\begin{align} & \Rightarrow CF=y \\ & \Rightarrow CF=5cm \\ \end{align}
Hence, the sides AB, BE and CF are $3cm$, $7cm$ and $5cm$ respectively.
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# 6.1 Graphs of the sine and cosine functions (Page 3/13)
Page 3 / 13
## Amplitude of sinusoidal functions
If we let $\text{\hspace{0.17em}}C=0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}D=0\text{\hspace{0.17em}}$ in the general form equations of the sine and cosine functions, we obtain the forms
The amplitude is $\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$ and the vertical height from the midline is $\text{\hspace{0.17em}}|A|.\text{\hspace{0.17em}}$ In addition, notice in the example that
## Identifying the amplitude of a sine or cosine function
What is the amplitude of the sinusoidal function $\text{\hspace{0.17em}}f\left(x\right)=-4\mathrm{sin}\left(x\right)?\text{\hspace{0.17em}}$ Is the function stretched or compressed vertically?
Let’s begin by comparing the function to the simplified form $\text{\hspace{0.17em}}y=A\mathrm{sin}\left(Bx\right).$
In the given function, $\text{\hspace{0.17em}}A=-4,\text{\hspace{0.17em}}$ so the amplitude is $\text{\hspace{0.17em}}|A|=|-4|=4.\text{\hspace{0.17em}}$ The function is stretched.
What is the amplitude of the sinusoidal function $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{2}\mathrm{sin}\left(x\right)?\text{\hspace{0.17em}}$ Is the function stretched or compressed vertically?
$\frac{1}{2}\text{\hspace{0.17em}}$ compressed
## Analyzing graphs of variations of y = sin x And y = cos x
Now that we understand how $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ relate to the general form equation for the sine and cosine functions, we will explore the variables $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}D.\text{\hspace{0.17em}}$ Recall the general form:
The value $\text{\hspace{0.17em}}\frac{C}{B}\text{\hspace{0.17em}}$ for a sinusoidal function is called the phase shift , or the horizontal displacement of the basic sine or cosine function . If $\text{\hspace{0.17em}}C>0,\text{\hspace{0.17em}}$ the graph shifts to the right. If $\text{\hspace{0.17em}}C<0,\text{\hspace{0.17em}}$ the graph shifts to the left. The greater the value of $\text{\hspace{0.17em}}|C|,\text{\hspace{0.17em}}$ the more the graph is shifted. [link] shows that the graph of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sin}\left(x-\pi \right)\text{\hspace{0.17em}}$ shifts to the right by $\text{\hspace{0.17em}}\pi \text{\hspace{0.17em}}$ units, which is more than we see in the graph of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sin}\left(x-\frac{\pi }{4}\right),\text{\hspace{0.17em}}$ which shifts to the right by $\text{\hspace{0.17em}}\frac{\pi }{4}\text{\hspace{0.17em}}$ units.
While $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ relates to the horizontal shift, $\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$ indicates the vertical shift from the midline in the general formula for a sinusoidal function. See [link] . The function $\text{\hspace{0.17em}}y=\mathrm{cos}\left(x\right)+D\text{\hspace{0.17em}}$ has its midline at $\text{\hspace{0.17em}}y=D.$
Any value of $\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$ other than zero shifts the graph up or down. [link] compares $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sin}\text{\hspace{0.17em}}x+2,\text{\hspace{0.17em}}$ which is shifted 2 units up on a graph.
## Variations of sine and cosine functions
Given an equation in the form $\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{sin}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{cos}\left(Bx-C\right)+D,\text{\hspace{0.17em}}$ $\frac{C}{B}\text{\hspace{0.17em}}$ is the phase shift and $\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$ is the vertical shift .
## Identifying the phase shift of a function
Determine the direction and magnitude of the phase shift for $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sin}\left(x+\frac{\pi }{6}\right)-2.$
Let’s begin by comparing the equation to the general form $\text{\hspace{0.17em}}y=A\mathrm{sin}\left(Bx-C\right)+D.$
In the given equation, notice that $\text{\hspace{0.17em}}B=1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}C=-\frac{\pi }{6}.\text{\hspace{0.17em}}$ So the phase shift is
or $\text{\hspace{0.17em}}\frac{\pi }{6}\text{\hspace{0.17em}}$ units to the left.
Determine the direction and magnitude of the phase shift for $\text{\hspace{0.17em}}f\left(x\right)=3\mathrm{cos}\left(x-\frac{\pi }{2}\right).$
$\frac{\pi }{2};\text{\hspace{0.17em}}$ right
## Identifying the vertical shift of a function
Determine the direction and magnitude of the vertical shift for $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{cos}\left(x\right)-3.$
Let’s begin by comparing the equation to the general form $\text{\hspace{0.17em}}y=A\mathrm{cos}\left(Bx-C\right)+D.$
In the given equation, $\text{\hspace{0.17em}}D=-3\text{\hspace{0.17em}}$ so the shift is 3 units downward.
Determine the direction and magnitude of the vertical shift for $\text{\hspace{0.17em}}f\left(x\right)=3\mathrm{sin}\left(x\right)+2.$
2 units up
Given a sinusoidal function in the form $\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{sin}\left(Bx-C\right)+D,\text{\hspace{0.17em}}$ identify the midline, amplitude, period, and phase shift.
1. Determine the amplitude as $\text{\hspace{0.17em}}|A|.$
2. Determine the period as $\text{\hspace{0.17em}}P=\frac{2\pi }{|B|}.$
3. Determine the phase shift as $\text{\hspace{0.17em}}\frac{C}{B}.$
4. Determine the midline as $\text{\hspace{0.17em}}y=D.$
give me an example of a problem so that I can practice answering
x³+y³+z³=42
Robert
dont forget the cube in each variable ;)
Robert
of she solves that, well ... then she has a lot of computational force under her command ....
Walter
what is a function?
I want to learn about the law of exponent
explain this
what is functions?
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
If the plane intersects the cone (either above or below) horizontally, what figure will be created?
can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
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# AP Statistics Curriculum 2007 Normal Std
(Difference between revisions)
Revision as of 23:46, 17 September 2008 (view source)IvoDinov (Talk | contribs) (→Standard Normal Distribution)← Older edit Current revision as of 16:26, 8 February 2012 (view source)IvoDinov (Talk | contribs) m (→Standard Normal Distribution) (18 intermediate revisions not shown) Line 3: Line 3: === Standard Normal Distribution=== === Standard Normal Distribution=== The Standard Normal Distribution is a continuous distribution with the following density: The Standard Normal Distribution is a continuous distribution with the following density: - * Standard Normal density function $f(x)= {e^{-x^2} \over \sqrt{2 \pi}}.$ + * Standard Normal ''density'' function $f(x)= {e^{-x^2 \over 2} \over \sqrt{2 \pi}}.$ + * Standard Normal ''cumulative distribution'' function $\Phi(y)= \int_{-\infty}^{y}{{e^{-x^2 \over 2} \over \sqrt{2 \pi}} dx}.$ + * Why are these two functions, $f(x), \Phi(y)$ well-defined density and distribution functions, i.e., $\int_{-\infty}^{\infty} {f(x)dx}=1$? [[AP_Statistics_Curriculum_2007_Normal_Std#Appendix|See the appendix below]]. Note that the following exact ''areas'' are bound between the Standard Normal Density Function and the x-axis on these symmetric intervals around the origin: Note that the following exact ''areas'' are bound between the Standard Normal Density Function and the x-axis on these symmetric intervals around the origin: - * The area: -1 < x < 1 = 0.8413 - 0.1587 = 0.6826 + * The area: -1.0 < x < 1.0 = 0.8413 - 0.1587 = 0.6826 * The area: -2.0 < x < 2.0 = 0.9772 - 0.0228 = 0.9544 * The area: -2.0 < x < 2.0 = 0.9772 - 0.0228 = 0.9544 * The area: -3.0 < x < 3.0 = 0.9987 - 0.0013 = 0.9974 * The area: -3.0 < x < 3.0 = 0.9987 - 0.0013 = 0.9974 * Note that the [http://en.wikipedia.org/wiki/Inflection_point inflection points] ($f ''(x)=0$)of the Standard Normal density function are $\pm$ 1. * Note that the [http://en.wikipedia.org/wiki/Inflection_point inflection points] ($f ''(x)=0$)of the Standard Normal density function are $\pm$ 1. -
[[Image:SOCR_EBook_Dinov_RV_Normal_013108_Fig0.jpg|600px]]
+
[[Image:SOCR_EBook_Dinov_RV_Normal_013109_Fig0.jpg|600px]]
- * The Standard Normal distribution is also a special case of the [[AP_Statistics_Curriculum_2007_Normal_Prob | more general normal distribution]] where the mean is set to zero and a variance to one. The Standard Normal distribution is often called the ''bell curve'' because the graph of its probability density resembles a bell. + * The Standard Normal distribution is also a special case of the [[AP_Statistics_Curriculum_2007_Normal_Prob | more general normal distribution]] where the mean is set to zero and the variance is set to one. The Standard Normal distribution is often called the ''bell curve'' because the graph of its probability density resembles a bell. ===Experiments=== ===Experiments=== Line 27: Line 29:
[[Image:SOCR_EBook_Dinov_RV_Normal_013108_Fig4.jpg|500px]]
[[Image:SOCR_EBook_Dinov_RV_Normal_013108_Fig4.jpg|500px]]
-
+ ===[[EBook_Problems_Normal_Std|Problems]]=== - ===References=== + ===Appendix=== + The derivation below illustrates why the standard normal density function, $f(x)= {e^{-x^2 \over 2} \over \sqrt{2 \pi}}$, represents a well-defined density function, i.e., $f(x)\ge 0$ and $\int_{-\infty}^{\infty} {f(x)dx}=1$. + * Clearly the exponential function $e^{-x^2 \over 2}$ is always non-negative (in fact it's strictly positive for each real value argument). + * To show that $\int_{-\infty}^{\infty} {f(x)dx}=1$, let $A=\int_{-\infty}^{\infty} {f(x)dx}$. Then $A^2=\int_{-\infty}^{\infty} {f(x)dx}\times \int_{-\infty}^{\infty} {f(w)dw}$. Thus, + : $A^2=\int_{-\infty}^{\infty} {\int_{-\infty}^{\infty} { {e^{-x^2 \over 2} \over \sqrt{2 \pi}} \times {e^{-w^2 \over 2} \over \sqrt{2 \pi}} dw}dx}$, + : Change variables from Cartesian to polar coordinates: + :: $x=r\cos(\theta)$ + :: $x=r\cos(\theta)$, $0\le \theta\le 2\pi$ + :: Hence, + ::: $x^2+w^2=r^2$, + ::: $e^{-x^2 \over 2} \times e^{-w^2 \over 2} =e^{-r^2 \over 2}$, + ::: $dx=\cos(\theta)dr$, and + ::: $dy=r\cos(\theta)dr$. + : Therefore, $A^2=\int_{0}^{\infty} {\int_{0}^{2\pi} {{e^{-r^2 \over 2} \over 2 \pi}\cos^2(\theta)rdrd\theta}}$, and + : $A^2=\int_{0}^{\infty} {\int_{0}^{2\pi} {{e^{-r^2 \over 2} \over 2 \pi}d{\frac{r^2}{2}}d\theta}}=\int_{0}^{\infty} {{e^{-r^2 \over 2} \over \pi}d{\frac{r^2}{2}}} \times \int_{0}^{2\pi} {\cos^2(\theta)d\theta}=1$, since + :: $\int_{0}^{2\pi} {\cos^2(\theta)d\theta}=\int_{0}^{2\pi} {\frac{1+\cos(2\theta)}{2}d\theta}=\pi$, and + :: $\int_{0}^{\infty} {e^{-w}dw}=1$.
## General Advance-Placement (AP) Statistics Curriculum - Standard Normal Variables and Experiments
### Standard Normal Distribution
The Standard Normal Distribution is a continuous distribution with the following density:
• Standard Normal density function $f(x)= {e^{-x^2 \over 2} \over \sqrt{2 \pi}}.$
• Standard Normal cumulative distribution function $\Phi(y)= \int_{-\infty}^{y}{{e^{-x^2 \over 2} \over \sqrt{2 \pi}} dx}.$
• Why are these two functions, f(x),Φ(y) well-defined density and distribution functions, i.e., $\int_{-\infty}^{\infty} {f(x)dx}=1$? See the appendix below.
Note that the following exact areas are bound between the Standard Normal Density Function and the x-axis on these symmetric intervals around the origin:
• The area: -1.0 < x < 1.0 = 0.8413 - 0.1587 = 0.6826
• The area: -2.0 < x < 2.0 = 0.9772 - 0.0228 = 0.9544
• The area: -3.0 < x < 3.0 = 0.9987 - 0.0013 = 0.9974
• Note that the inflection points (f''(x) = 0)of the Standard Normal density function are $\pm$ 1.
• The Standard Normal distribution is also a special case of the more general normal distribution where the mean is set to zero and the variance is set to one. The Standard Normal distribution is often called the bell curve because the graph of its probability density resembles a bell.
### Experiments
Suppose we decide to test the state of 100 used batteries. To do that, we connect each battery to a volt-meter by randomly attaching the positive (+) and negative (-) battery terminals to the corresponding volt-meter's connections. Electrical current always flows from + to -, i.e., the current goes in the direction of the voltage drop. Depending upon which way the battery is connected to the volt-meter we can observe positive or negative voltage recordings (voltage is just a difference, which forces current to flow from higher to the lower voltage.) Denote Xi={measured voltage for battery i} - this is random variable with mean of 0 and unitary variance. Assume the distribution of all Xi is Standard Normal, $X_i \sim N(0,1)$. Use the Normal Distribution (with mean=0 and variance=1) in the SOCR Distribution applet to address the following questions. This Distributions help-page may be useful in understanding SOCR Distribution Applet. How many batteries, from the sample of 100, can we expect to have?
• Absolute Voltage > 1? P(X>1) = 0.1586, thus we expect 15-16 batteries to have voltage exceeding 1.
• |Absolute Voltage| > 1? P(|X|>1) = 1- 0.682689=0.3173, thus we expect 31-32 batteries to have absolute voltage exceeding 1.
• Voltage < -2? P(X<-2) = 0.0227, thus we expect 2-3 batteries to have voltage less than -2.
• Voltage <= -2? P(X<=-2) = 0.0227, thus we expect 2-3 batteries to have voltage less than or equal to -2.
• -1.7537 < Voltage < 0.8465? P(-1.7537 < X < 0.8465) = 0.761622, thus we expect 76 batteries to have voltage in this range.
### Appendix
The derivation below illustrates why the standard normal density function, $f(x)= {e^{-x^2 \over 2} \over \sqrt{2 \pi}}$, represents a well-defined density function, i.e., $f(x)\ge 0$ and $\int_{-\infty}^{\infty} {f(x)dx}=1$.
• Clearly the exponential function $e^{-x^2 \over 2}$ is always non-negative (in fact it's strictly positive for each real value argument).
• To show that $\int_{-\infty}^{\infty} {f(x)dx}=1$, let $A=\int_{-\infty}^{\infty} {f(x)dx}$. Then $A^2=\int_{-\infty}^{\infty} {f(x)dx}\times \int_{-\infty}^{\infty} {f(w)dw}$. Thus,
$A^2=\int_{-\infty}^{\infty} {\int_{-\infty}^{\infty} { {e^{-x^2 \over 2} \over \sqrt{2 \pi}} \times {e^{-w^2 \over 2} \over \sqrt{2 \pi}} dw}dx}$,
Change variables from Cartesian to polar coordinates:
x = rcos(θ)
x = rcos(θ), $0\le \theta\le 2\pi$
Hence,
x2 + w2 = r2,
$e^{-x^2 \over 2} \times e^{-w^2 \over 2} =e^{-r^2 \over 2}$,
dx = cos(θ)dr, and
dy = rcos(θ)dr.
Therefore, $A^2=\int_{0}^{\infty} {\int_{0}^{2\pi} {{e^{-r^2 \over 2} \over 2 \pi}\cos^2(\theta)rdrd\theta}}$, and
$A^2=\int_{0}^{\infty} {\int_{0}^{2\pi} {{e^{-r^2 \over 2} \over 2 \pi}d{\frac{r^2}{2}}d\theta}}=\int_{0}^{\infty} {{e^{-r^2 \over 2} \over \pi}d{\frac{r^2}{2}}} \times \int_{0}^{2\pi} {\cos^2(\theta)d\theta}=1$, since
$\int_{0}^{2\pi} {\cos^2(\theta)d\theta}=\int_{0}^{2\pi} {\frac{1+\cos(2\theta)}{2}d\theta}=\pi$, and
$\int_{0}^{\infty} {e^{-w}dw}=1$.
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# Area of Composite Shapes
## Shapes in the real world come in all sizes. Learn how to break down and calculate the area of composite shapes using the sum of areas of each part.
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Practice Area of Composite Shapes
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Area of Composite Shapes
What if you wanted to find the area of a shape that was made up of other shapes? How could you use your knowledge of the area of rectangles, parallelograms, and triangles to help you? After completing this Concept, you'll be able to answer questions like these.
### Guidance
Perimeter is the distance around a shape. The perimeter of any figure must have a unit of measurement attached to it. If no specific units are given (feet, inches, centimeters, etc), write “units.” Area is the amount of space inside a figure. If two figures are congruent, they have the same area. This is the congruent areas postulate. This postulate needs no proof because congruent figures have the same amount of space inside them. Keep in mind that two figures with the same area are not necessarily congruent.
A composite shape is a shape made up of other shapes. To find the area of such a shape, simply find the area of each part and add them up. The area addition postulate states that if a figure is composed of two or more parts that do not overlap each other, then the area of the figure is the sum of the areas of the parts.
#### Example A
Find the area of the figure below. You may assume all sides are perpendicular.
Split the shape into two rectangles and find the area of each.
Atop rectangleAbottom square=62=12 ft2=33=9 ft2
The total area is 12+9=21 ft2\begin{align*}12 + 9 = 21 \ ft^2\end{align*}.
#### Example B
• Divide the shape into two triangles and one rectangle.
• Find the area of the two triangles and rectangle.
• Find the area of the entire shape.
Solution:
• One triangle on the top and one on the right. Rectangle is the rest.
• Area of triangle on top is 8(5)2=20 units2\begin{align*}\frac{8(5)}{2}=20 \ units^2\end{align*}. Area of triangle on right is 5(5)2=12.5 units2\begin{align*}\frac{5(5)}{2}=12.5 \ units^2\end{align*}. Area of rectangle is 375 units2\begin{align*}375 \ units^2\end{align*}.
• Total area is 407.5 units2\begin{align*}407.5 \ units^2\end{align*}.
#### Example C
Find the area of the figure below.
Divide the figure into a triangle and a rectangle with a small rectangle cut out of the lower right-hand corner.
AAAA=Atop triangle+ArectangleAsmall triangle=(1269)+(915))(1236)=27+1359=153 units2
Watch this video for help with the Examples above.
### Vocabulary
Perimeter is the distance around a shape. The perimeter of any figure must have a unit of measurement attached to it. If no specific units are given (feet, inches, centimeters, etc), write “units.” Area is the amount of space inside a figure and is measured in square units. A composite shape is a shape made up of other shapes.
### Guided Practice
1. Find the area of the rectangles and triangle.
2. Find the area of the whole shape.
1. Rectangle #1: Area =24(9+12)=504 units2\begin{align*}\text{Area }= 24(9+12)=504 \ units^2\end{align*}. Rectangle #2: Area =15(9+12)=315 units2\begin{align*}\text{Area }=15(9+12)=315 \ units^2\end{align*}. Triangle: Area =15(9)2=67.5 units2\begin{align*}\text{Area }=\frac{15(9)}{2}=67.5 \ units^2\end{align*}.
2. You need to subtract the area of the triangle from the bottom right corner. \begin{align*}\text{Total Area }=504+315+67.5-\frac{15(12)}{2}=796.5 \ units^2\end{align*}
### Practice
Use the picture below for questions 1-2. Both figures are squares.
1. Find the area of the unshaded region.
2. Find the area of the shaded region.
Find the area of the figure below. You may assume all sides are perpendicular.
Find the areas of the composite figures.
Use the figure to answer the questions.
1. What is the area of the square?
2. What is the area of the triangle on the left?
3. What is the area of the composite figure?
Find the area of the following figures.
1. Find the area of the unshaded region.
2. Lin bought a tract of land for a new apartment complex. The drawing below shows the measurements of the sides of the tract. Approximately how many acres of land did Lin buy? You may assume any angles that look like right angles are \begin{align*}90^\circ\end{align*}. (1 acre \begin{align*}\approx\end{align*} 40,000 square feet)
3. Linus has 100 ft of fencing to use in order to enclose a 1200 square foot rectangular pig pen. The pig pen is adjacent to the barn so he only needs to form three sides of the rectangular area as shown below. What dimensions should the pen be?
### Vocabulary Language: English
Composite
Composite
A number that has more than two factors.
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# How do you solve x/(x-5) - 2/(x+5) = 50/(x^2-25)?
Apr 9, 2016
Start by multiplying both sides by the Least Common Denominator.
#### Explanation:
$\frac{x}{x - 5} - \frac{2}{x + 5} = \frac{50}{{x}^{2} - 25}$
The Least Common Denominator is $\left(x + 5\right) \left(x - 5\right)$ since that is the simplest expression that all denominators will divide into. So all three terms get multiplied by $\left(x + 5\right) \left(x - 5\right)$
$x \left(x + 5\right) - 2 \left(x - 5\right) = 50$
Above we have multiplied all numerators by the LCD and canceled out terms with the denominator where needed.
Now distribute and add like terms.
${x}^{2} + 5 x - 2 x + 10 = 50$
${x}^{2} + 3 x - 40 = 0$
Note that we need to solve for 0 in order to solve by factoring.
Factor the left side: $\left(x + 8\right) \left(x - 5\right) = 0$
Solve each factor for 0: $x = \left\{- 8 , 5\right\}$
HOWEVER the second answer must be eliminated since it would result in a zero denominator (extraneous solution).
So the only valid answer is $x = - 8$
Hope this helps!
Apr 9, 2016
Multiplying both sides by ${x}^{2} - 25$ we get
$\frac{x \left({x}^{2} - 25\right)}{x - 5} - \frac{2 \left({x}^{2} - 25\right)}{x + 5} = \frac{50 \left({x}^{2} - 25\right)}{{x}^{2} - 25}$
$\implies x \left(x + 5\right) - 2 \left(x - 5\right) = 50$
$\implies {x}^{2} + 5 x - 2 x + 10 - 50 = 0$
$\implies {x}^{2} + 8 x - 5 x - 40 = 0$
$\implies x \left(x + 8\right) - 5 \left(x + 8\right)$
$\implies \left(x + 8\right) \left(x - 5\right)$
$\therefore x = - 8 \mathmr{and} x = 5$
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# 5.5: Right Triangle Trigonometry
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
In Section 5.3 we were introduced to the sine and cosine function as ratios of the sides of a triangle drawn inside a circle, and spent the rest of that section discussing the role of those functions in finding points on the circle. In this section, we return to the triangle, and explore the applications of the trigonometric functions to right triangles where circles may not be involved.
Recall that we defined sine and cosine as
Separating the triangle from the circle, we can make equivalent but more general definitions of the sine, cosine, and tangent on a right triangle. On the right triangle, we will label the hypotenuse as well as the side opposite the angle and the side adjacent (next to) the angle.
Right Triangle Relationships
Given a right triangle with an angle of
SineCosineTangentTrigonometry:SineTrigonometry:CosineTrigonometry:Tangent
A common mnemonic for remembering these relationships is SohCahToaTrigonometry:SohCahToa, formed from the first letters of “ine is pposite over ypotenuse, osine is djacent over ypotenuse, angent is pposite over djacent.”
Example $$\PageIndex{1}$$:
$$\alpha$$$$\alpha$$Given the triangle shown, find the value for .
The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so
When working with general right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the sine and cosine of either of the two acute angles in the triangle.
Section 5.5 Right Triangle Trigonometry 391
Example $$\PageIndex{2}$$:
2
Using the triangle shown, evaluate , , , and . $\alpha \beta \alpha \beta$
Exercise $$\PageIndex{1}$$
Add exercises text here. For the automatic number to work, you need to add the "AutoNum" template (preferably at the end) to the page.
Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page.
1. A right triangle is drawn with angle opposite a side with length 33, angle opposite a side with length 56, and hypotenuse 65. Find the sine and cosine of and.
You may have noticed that in the above example that $$\cos (\alpha )=\sin (\beta )$$ and $$\cos (\beta )=\sin (\alpha )$$. This makes sense since the side opposite $$\alpha$$ is also adjacent to $$\beta$$. Since the three angles in a triangle need to add to $$\piup$$, or 180 degrees, then the other two angles must add to $$\frac{\pi }{2}$$, or 90 degrees, so $$\beta =\frac{\pi }{2} -\alpha$$, and $$\alpha =\frac{\pi }{2} -\beta$$. Since $$\cos (\alpha )=\sin (\beta )$$, then $$\cos (\alpha )=\sin \left(\frac{\pi }{2} -\alpha \right)$$.
Cofunction Identities
The cofunction identitiesCofunction IdentitiesTrigonometric Identities:Cofunction Identities for sine and cosine are: $\cos (\theta )=\sin \left(\frac{\pi }{2} -\theta \right) \sin (\theta )=\cos \left(\frac{\pi }{2} -\theta \right)$
In the previous examples, we evaluated the sine and cosine on triangles where we knew all three sides of the triangle. Right triangle trigonometry becomes powerful when we start looking at triangles in which we know an angle but don’t know all the sides.
Example $$\PageIndex{3}$$:
Find the unknown sides of the triangle pictured here.
Since $$\sin (\theta )=\frac ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[4]/p[3]/span[1]/span[1], line 1, column 1 ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[4]/p[3]/span[1]/span[2], line 1, column 1 $$, $$\sin (30{}^\circ )=\frac{7}{b}$$.
From this, we can solve for the side b. $b\sin (30{}^\circ )=7$ $b=\frac{7}{\sin (30{}^\circ )}$
To obtain a value, we can evaluate the sine and simplify $b=\frac{7}{{\raise0.7ex\hbox{ 1 }\!\mathord{\left/{\vphantom{1 2}}\right.\kern-\nulldelimiterspace}\!\lower0.7ex\hbox{ 2 }} } =14$
To find the value for side a, we could use the cosine, or simply apply the Pythagorean Theorem: $a^{2} +7^{2} =b^{2}$ $a^{2} +7^{2} =14^{2}$ $a=\sqrt{147}$
Notice that if we know at least one of the non-right angles of a right triangle and one side, we can find the rest of the sides and angles.
Exercise $$\PageIndex{1}$$
A right triangle has one angle of $$\frac{\pi }{3}$$ and a hypotenuse of 20. Find the unknown sides and angles of the triangle.
Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page.
Example $$\PageIndex{4}$$:
To find the height of a tree, a person walks to a point 30 feet from the base of the tree, and measures the angle from the ground to the top of the tree to be 57 degrees. Find the height of the tree.
We can introduce a variable, h, to represent the height of the tree. The two sides of the triangle that are most important to us are the side opposite the angle, the height of the tree we are looking for, and the adjacent side, the side we are told is 30 feet long.
The trigonometric function which relates the side opposite of the angle and the side adjacent to the angle is the tangent.
$$\tan (57{}^\circ )=\frac ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[6]/p[6]/span/span[1], line 1, column 1 ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[6]/p[6]/span/span[2], line 1, column 1 =\frac{h}{30}$$ Solving for h,
$$h=30\tan (57{}^\circ )$$ Using technology, we can approximate a value
$$h=30\tan (57{}^\circ )\approx 46.2$$ feet
The tree is approximately 46 feet tall.
Example $$\PageIndex{5}$$
A person standing on the roof of a 100 foot tall building is looking towards a skyscraper a few blocks away, wondering how tall it is. She measures the angle of declination from the roof of the building to the base of the skyscraper to be 20 degrees and the angle of inclination to the top of the skyscraper to be 42 degrees.
To approach this problem, it would be good to start with a picture. Although we are interested in the height, h, of the skyscraper, it can be helpful to also label other unknown quantities in the picture – in this case the horizontal distance x between the buildings and a, the height of the skyscraper above the person.
To start solving this problem, notice we have two right triangles. In the top triangle, we know one angle is 42 degrees, but we don’t know any of the sides of the triangle, so we don’t yet know enough to work with this triangle.
In the lower right triangle, we know one angle is 20 degrees, and we know the vertical height measurement of 100 ft. Since we know these two pieces of information, we can solve for the unknown distance x.
$$\tan (20{}^\circ )=\frac ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[7]/p[6]/span[1]/span[1], line 1, column 1 ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[7]/p[6]/span[1]/span[2], line 1, column 1 =\frac{100}{x}$$ Solving for x $x\tan (20{}^\circ )=100$ $x=\frac{100}{\tan (20{}^\circ )}$
Now that we have found the distance x, we know enough information to solve the top right triangle. $\tan (42{}^\circ )=\frac ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[7]/p[7]/span[1]/span[1], line 1, column 1 ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[7]/p[7]/span[1]/span[2], line 1, column 1 =\frac{a}{x} =\frac{a}{{\raise0.7ex\hbox{ 100 }\!\mathord{\left/{\vphantom{100 \tan (20{}^\circ )}}\right.\kern-\nulldelimiterspace}\!\lower0.7ex\hbox{ \tan (20{}^\circ ) }} }$ $\tan (42{}^\circ )=\frac{a\tan (20{}^\circ )}{100}$ $100\tan (42{}^\circ )=a\tan (20{}^\circ )$ $\frac{100\tan (42{}^\circ )}{\tan (20{}^\circ )} =a$
Approximating a value,
$$a=\frac{100\tan (42{}^\circ )}{\tan (20{}^\circ )} \approx 247.4$$ feet
Adding the height of the first building, we determine that the skyscraper is about 347 feet tall.
### Important Topics of This Section
• SOH CAH TOA
• Cofunction identities
• Applications with right triangles
Try it Now Answers $1. \sin (\alpha )=\frac{33}{65} \cos (\alpha )=\frac{56}{65} \sin (\beta )=\frac{56}{65} \cos (\beta )=\frac{33}{65}$
2. $$\cos \left(\frac{\pi }{3} \right)=\frac ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[8]/p[2]/span[1]/span[1], line 1, column 1 ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[8]/p[2]/span[1]/span[2], line 1, column 1 =\frac ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[8]/p[2]/span[1]/span[3], line 1, column 1 {20}$$ so, $${\rm adjacent}=20\cos \left(\frac{\pi }{3} \right)=20\left(\frac{1}{2} \right)=10$$
$$\sin \left(\frac{\pi }{3} \right)=\frac ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[8]/p[3]/span[1]/span[1], line 1, column 1 ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[8]/p[3]/span[1]/span[2], line 1, column 1 =\frac ParseError: invalid DekiScript (click for details) Callstack: at (TextMaps/Precalculus/Book:_Precalculus_-_An_Investigation_of_Functions_(Lippman_and_Rasmussen)/5:_Trigonometric_Functions_of_Angles/5.5:_Right_Triangle_Trigonometry), /content/body/div[8]/p[3]/span[1]/span[3], line 1, column 1 {20}$$ so, $${\rm opposite}=20\sin \left(\frac{\pi }{3} \right)=20\left(\frac{\sqrt{3} }{2} \right)=10\sqrt{3}$$
Missing angle = 180-90-60 = 30 degrees or $${\raise0.7ex\hbox{$$ $$}\!\mathord{\left/{\vphantom{\pi 6}}\right.\kern-\nulldelimiterspace}\!\lower0.7ex\hbox{$$ 6 $$}}$$. Section 5.5 Right Triangle Trigonometry 395
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Lecture Notes
Three Famous Angles
page 1
The exact value of trigonometric functions of most angles can not be determined by elementary techniques. At this point, we are to imagine mathematicians drawing right triangles and measuring sides to obtain approximate values. There are a few angles, however, that are exception to this; we can compute the exact values of trigonometric functions. Certain symmetries and the Pythagorean theorem enables us to do that, in case of 30 , 45 , and 60 .
Trigonometric Function Values of
30
and
60
We start by drawing an equilateral triangle with sides 2 units long. All three angles of this triangle measure 60 .
Let D be the midpoint of side AC . We connect points B and D. Because the triangle is isosceles, this line is perpendicular to the base AC and cuts the triangle into two identical parts. Consequently, the two angles created at point B both measure 30 .
Let us now focus on just one half of the picture, triangle ABD. This triangle has angles 30 , 60 , and 90 . We know two of its sides are 1 and 2 units long. Notice that the hypotenuse is 2 units long.
Lecture Notes
Three Famous Angles
page 2
We can easily compute the missing side using the Pythagorean theorem. If side BD is denoted by x, then x2 + 12 = 22 x2 + 1 = 4 x2 = 3 x = p 3 =) x = p 3
Now that we have the exact value of all three sides of the triangle, we can compute all trigonometric function values for 30 and 60 using this triangle.
hypotenuse 2 =p adjacent 3 p adjacent 3 p cot 30 = = = 3 opposite 1 sec 30 =
Trigonometric Function Values Of 60 : p 3 opposite sin 60 = = 2 hypotenuse cos 60 = 1 adjacent = 2 hypotenuse p 3 p opposite = 3 tan 60 = = 1 adjacent csc 60 = 2 hypotenuse =p opposite 3 2 hypotenuse = =2 1 adjacent adjacent 1 =p opposite 3
sec 60 =
cot 60 =
1 Note: Consider the exact value of tan 30 . The ratio of the sides is p . Sometimes this number is rationalized 3 p 3 and is presented as . Here is the computation: 3 p p 1 1 1 3 3 p = p 1= p p = 3 3 3 3 3
c copyright Hidegkuti, Powell, 2012 Last revised: August 3, 2012
Lecture Notes
Three Famous Angles
page 3
p 2 2 3 Similarly, csc 60 = p can be rationalized and presented as . While the rationalized form is considered 3 3 simplied , both forms have their own adventages, and it is a useful skill to know which form is better for us in dierent situations. For example, if we add several fractions of dierent denominators and tan 30 is one of them, the rationalized form is much better. On the other hand, if we need to square tan 30 , it is easier to work with 1 2 1 the other form: (tan 30 )2 = p = . 3 3
Trigonometric Function Values of
45
We start by drawing a square with sides 1 units long. We then draw the diagonal AC . This line cuts the square into two identical, isosceles right triangles. Consequently, the angles created at points A and C both measure 45 .
Lecture Notes
Three Famous Angles
page 4
To avoid confusion, we mark only one of the 45 angle on the picture below. Now that we know all three sides of the triangle, we can compute all trigonometric values of 45 .
Trigonometric Function Values Of 45 : opposite 1 sin 45 = =p hypotenuse 2 cos 45 = adjacent 1 =p hypotenuse 2 opposite 1 = =1 adjacent 1 p hypotenuse 2 p csc 45 = = = 2 opposite 1 p hypotenuse 2 p sec 45 = = = 2 adjacent 1 cot 45 = adjacent 1 = =1 opposite 1
tan 45 =
p p 1 1 1 1 2 2 Note: sin 45 = p can be rationalized: p = p 1 = p p = . Again, while the rationalized form is 2 2 2 2 2 2 considered simplied , both forms have their own adventages.
Note: There are some additional angles with algebraically approachable trigonometric function values. We can also compute the exact values for the trigonometric functions of 18 and 72 . These computations use the Pythagorean theorem and similar triangles and are very interesting.
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# 1.2. Exact Calculation, or Bound?#
All of the questions we asked in the previous section could be answered based on the information given. But sometimes the information at hand is not enough for us to be able to find the probability of an event of interest. Let’s look at an example of this and then see if there is anything useful we can say in situations where we can’t find an answer exactly.
As a reminder, here is the Pew Research Center graphic again. The experiment is to pick one teen at random from the population represented in the charts.
What is the chance that the selected teen used Facebook or Twitter most often?
This question has an exact answer based on the right hand chart. Each teen is in exactly one bar of that chart. The teens who used FB most often are an entirely distinct group from those who used Twitter most often.
Here is a Venn diagram representation of this. The number shown in each disc is the percent of teens in that category.
If you think of proportions as areas (a very good idea, as you will soon see), the green area is the proportion in question: the proportion of all teens who used FB or Twitter the most. The area is clearly 13% because there is no overlap between the two discs.
Because the discs are disjoint from each other, the answer for a randomly selected teen is
$P(\text{used Facebook or Twitter most often}) ~ = ~ 0.1 + 0.03 ~ = ~ 0.13$
What is the chance that the selected teen used Facebook or Twitter?
The left hand graph says that 51% of the teens used Facebook and 32% used Twitter. If we try to draw a Venn diagram that represents this, we run into a problem. Some teens might have used both Facebook and Twitter, and we don’t know the percent in this group.
So the first point to note is that there is not enough information to find an exact answer to the question.
That’s life. We usually have incomplete information and thus there are many questions that we can’t answer exactly.
But let’s not just throw up our hands. Let’s see if there is anything more that we can say about the chance other than, “We can’t find its exact value.”
## 1.2.3. Bounds#
When we can’t find the exact value of a chance, we can instead try to place some bounds on it.
One pair of bounds is easy: the chance is between 0 and 1 (or 0% and 100% if you prefer).
While this is doubtless true, it is not particularly useful. Can we do better?
For this we should examine where we got stuck.
• We want the percent of teens who used Facebook or Twitter.
• We know that 51% of teens used Facebook and 32% used Twitter.
• But we can’t draw a Venn diagram because we don’t know the extent of the overlap between the two groups. That is, we don’t know the percent of teens who used both platforms.
How big could the overlap be?
• The largest possible overlap would happen if every teen who used Twitter also used Facebook.
• The smallest possible overlap would happen if no teen used both Facebook and Twitter.
Both of these possibilities are unrealistic but they are theoretically possible. They are represented as Venn diagrams in the discussions below. Remember that in all the diagrams, the green region corresponds to teens who used either Facebook or Twitter.
## 1.2.4. Lower Bound on the Chance#
The percent of teens who used Facebook or Twitter must be at least 51% because the group includes all those who used Facebook. So the chance that the selected teen used Facebook or Twitter must be at least 51%.
Here is a Venn diagram of the extreme case when every teen who used Twitter also used Facebook.
In this diagram, the area of the green region is 51%. It is the same as the area of the Facebook region because the Twitter region is completely contained in it.
## 1.2.5. Upper Bound on the Chance#
The diagram below shows the other extreme, when the group that used Facebook is entirely distinct from the group that used Twitter.
In this diagram the area of the green region is 51% + 32% = 83%.
That’s the biggest that the chance can be, because the other possible Venn diagrams look something like the one below and thus have a smaller green region.
We have shown that the chance that the randomly selected teen used Facebook or Twitter is at least 51% and at most 83%.
Though we don’t have an exact value for the chance, we have done much better than just saying that the chance is between 0% and 100%.
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# Finding Eigenvalues of a 3x3 Matrix (7.12-17)
Please check my work in finding eigenvalues for the following problem. I am working out of the textbook Advanced Engineering Mathematics by Erwin Kreyszig, 1988, John Wiley & Sons.
For reference the following identity is given because some textbooks reverse the formula having $\lambda$ subtract the diagonal elements instead of subtracting $\lambda$ from the diagonal elements:
$$det(A - \lambda I) = \begin{vmatrix} 13-\lambda & 0 & -15 \\ -3 & 4-\lambda & 9 \\ 5 & 0 & -7-\lambda \\ \end{vmatrix} = 0$$ Taking the center column we have: $$(4-\lambda) \begin{vmatrix} 13-\lambda & -15 \\ 5 & -7-\lambda \\ \end{vmatrix} \\ = (4-\lambda)[(13-\lambda)(-7-\lambda) + 5(15)] = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda - 91 + 75) = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda -21) = 0 \\ = -\lambda^3 + 10\lambda^2 - 24\lambda + 21\lambda - 84 = 0 \\ = \lambda^3 - 10\lambda^2 + 3\lambda + 84 = 0 \\$$ Using an online calculator the characteristic equation factors into: $$\lambda^3 - 10\lambda^2 + 3\lambda + 84 = 0 \\ (\lambda - 4)(\lambda^2 - 6\lambda - 21) = 0 \\$$ But answer in text is \begin{align*} \lambda_1 = 8 \qquad \lambda_2 = 4 \qquad \lambda_3 = -2 \\ \end{align*} Question: Although I can get $\lambda = 4$ out of the factored equation there is no way to get the other two eigenvalues. I suspect my algebra. Where did I go wrong?
• Your error is in = $$(4-\lambda)(\lambda^2 - 6\lambda - 91 + 75) = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda \color{red}{-21}) = 0$$. It should be $\color{blue}{-16}$ and then it factors well to give you the required answerr – Shailesh Jan 24 '16 at 0:11
• And then instead of multplying out $(4-\lambda)(\lambda^2-6\lambda -16)$, just factor $(4-\lambda)(\lambda-8)(\lambda +2)$. – Tim Raczkowski Jan 24 '16 at 0:13
Your error is in the step $$(4-\lambda)(\lambda^2 - 6\lambda - 91 + 75) = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda \color{red}{-21}) = 0$$ It should be $\color{blue}{-16}$ and then it factors well to give you the required answer
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Table of 114
Created by: Team Maths - Examples.com, Last Updated: June 10, 2024
Table of 114
The multiplication table of 114 lists the products of 114 with integers from 1 to 10. Starting with 114, it progresses through multiples: 228, 342, 456, 570, 684, 798, 912, 1026, and 1140. This table is a helpful tool for quick reference in arithmetic Operations and understanding the multiplication concept.
What is the Multiplication Table of 114?
The multiplication table of 114 outlines the sequential addition of 114 to itself for each step from 1 to 10. Starting with 114, the next value is found by adding another 114, resulting in 228. Adding 114 again to 228 gives 342. This pattern continues with each step increasing by 114: 456 follows, then 570, and so forth.
As the sequence progresses, 114 is added repeatedly: reaching 684, then moving to 798, next to 912, and then 1026. Finally, the addition sequence concludes at 1140 after adding 114 to 1026. This step-by-step addition method showcases how the table of 114 is built without directly using multiplication, highlighting the cumulative nature of repeated addition.
114ᵗʰ Table
This table helps visualize and understand the process of multiplying by 114, providing a quick reference for arithmetic operations involving this number.
Tips for 114ᵗʰ Table
1. Understand the Pattern: Recognize that each number in the table is 114 more than the previous one. This can be useful for mental calculations.
2. Use Addition: Instead of multiplying, add 114 repeatedly. For example, 114, 228 (114 + 114), 342 (228 + 114), and so on.
3. Break Down the Numbers: For complex multiplications, break them into simpler steps. For example, 114 x 4 can be broken into (100 x 4) + (10 x 4) + (4 x 4).
4. Practice Regularly: Write down the table and recite it daily. Repetition will help reinforce memory.
5. Utilize Multiples: Note that 114 is a multiple of smaller numbers (e.g., 2, 3, 6), which can help you relate it to easier tables you already know.
6. Flashcards: Create flashcards with each multiplication fact (e.g., 114 x 5) on one side and the answer (570) on the other. This can help with quick recall.
7. Real-World Examples: Apply the table to real-world situations, like calculating totals in shopping or measurements, to make it more relevant and easier to remember.
Simplest Way To Memorize Table 114
• One time 114 is 114.
• Two times 114 is 228.
• Three times 114 is 342.
• Four times 114 is 456.
• Five times 114 is 570.
• Six times 114 is 684.
• Seven times 114 is 798.
• Eight times 114 is 912.
• Nine times 114 is 1026.
• Ten times 114 is 1140.
To read the 114th table, start with 114 and sequentially add 114 for each subsequent value: 114, 228, 342, 456, 570, 684, 798, 912, 1026, and 1140.
Example 1: 114×1
Calculation: 114 × 1 = 114
Derivation: Multiplying any number by 1 gives the number itself.114 × 1 = 114
Example 2: 114 × 2
Calculation: 114 × 2 = 228
Derivation: 114 × 2 = 114 + 114 = 228
Example 3: 114 × 3
Calculation: 114 × 3 = 342
Derivation: 114 × 3 =114 + 114 + 114 = 342
Example 4: 114 × 4
Calculation: 114 × 4 = 4561
Derivation: 114 × 4 = 114 + 114 + 114 + 114 =456
Example 5: 114 × 5
Calculation: 114×5=570114×5=570
Derivation: 114 × 5 = 114 × (4+1) = (114×4) + 114 =456 + 114=570
Example 6: 114 × 6
Calculation: 114×6=684114×6=684
Derivation: 114 × 6 = 114 × (5+1) = (114×5) + 114 = 570+114 = 684
Example 7: 114 × 7
Calculation: 114 × 7 = 798
Derivation: 114 × 7 = 114 × (6+1) = (114×6) + 114 = 684 + 114 = 798
Example 8: 114 × 8
Calculation: 114 × 8 = 912
Derivation: 114 × 8 = 114 × (7+1) = (114×7) + 114 = 798 + 114 = 912
Example 9: 114 × 9
Calculation: 114 × 9 = 1026
Derivation: 114 × 9 = 114 × (8+1) = (114×8) + 114 = 912 + 114 = 1026
Example 10: 114 × 10
Calculation: 114 × 10 = 1140
Derivation: 114 × 10 = 114 × (5+5) = (114×5) + (114×5) = 570 + 570 = 1140
Text prompt
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## How do you solve 42 divided by 8?
The answer to the query: What is 42 divided by 8 is as follows:
1. 42 / 8 = 5.250. Instead of claiming 42 divided by 8 equals 5.250, you could simply use the department image, which is a slash, as we did above.
2. 42 ÷ 8 = 5.250.
3. 42 over 8 = 5.250.
4. 42⁄8 = 5.250.
## What is the rest of 42?
42 is the dividend, 10 is the divisor (modulo), 4 is the quotient explained beneath, and a pair of is named the remaining. The division remainder of 42 by 10 equals 2, and the price of the quotient is 4. Proof: 42 = (10×4) + 2.
What is the rest of 42 divided by 7?
The quotient (integer division) of 42/7 equals 6; the remainder (“left over”) is 0. 42 is the dividend, and seven is the divisor.
Related Posts
What can you divide with 42?
Division Method to Find Factors of 42: We see that once 42 is divided by 1, 2, 3, 6, 7, 14, 21, and 42, it leaves no remainder.
### How do you solve 42 divided by 3?
Multiply the latest quotient digit (4) by the divisor 3 . Subtract 12 from 12 . The results of department of 423 is 14 .
### What is 42 divided by Nine in a fragment?
Using a calculator, if you typed in 42 divided by 9, you’d get 4.6667. You may additionally specific 42/9 as a blended fraction: 4 6/9.
What is the remainder of 42 divided by 3?
The quotient (integer division) of 42/3 equals 14; the rest (“left over”) is 0. 42 is the dividend, and 3 is the divisor.
What is the quotient of 42 and seven?
(*8*)
14
The quotient of 42 and the variation of a number and 7 equals 14.
#### How do you show your work 42 divided by 7?
Using a calculator, if you typed in 42 divided by 7, you’d get 6. You may just additionally express 42/7 as a blended fraction: 6 0/7.
#### What are the high components for 42?
So, the top elements of 42 are 2 × 3 × 7, the place 2, Three and seven are top numbers.
What is 42 divided by 3 long?
The results of department of 423 is 14 .
What is the quotient of 42 divided by 3?
## What is 42 9 as a combined quantity?
42/Nine as a combined quantity is 4 2/3.
## How do you provide an explanation for 42 divided by 3?
The resolution to the question: What is 42 divided by 3 is as follows:
1. 42 / 3 = 14. Instead of saying 42 divided by Three equals 14, you could just use the department symbol, which is a slash, as we did above.
2. 42 ÷ 3 = 14.
3. 42 over 3 = 14.
4. 42⁄3 = 14.
What is the quotient of 42 and six?
The quotient (integer department) of 42/6 equals 7; the remaining (“left over”) is 0. 42 is the dividend, and 6 is the divisor.
What is the quotient of 42?
Explanation: A quotient is the answer to a division. A distinction is the answer to a subtraction. When we divide 42 by that distinction, the answer will be 14.
### What are the common elements of 30 and 42?
What is the GCF of 30 and 42? The GCF of 30 and 42 is 6. To calculate the best commonplace issue of 30 and 42, we want to factor each and every quantity (elements of 30 = 1, 2, 3, 5, 6, 10, 15, 30; factors of 42 = 1, 2, 3, 6, 7, 14, 21, 42) and choose the best issue that precisely divides both 30 and 42, i.e., 6.
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# Finding Factors
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## Objective
SWBAT determine all the factors of a number less than or equal to 100
#### Big Idea
Every positive integer (except for 1) can be represented as the product of at least 2 factors.
7 minutes
Students work in pairs to complete the Think About It problem. As they create their rectangles and multiplication sentences, I let students know that they don't need to display each rectangle twice. What I express to them, without using numbers, is that a 1x24 rectangle is the same as a 24x1 rectangle - I don't want to give away one of the factor pairs! I hold up a piece of paper and let students know that holding it horizontally or vertically doesn't change that it is one rectangle.
The most common student error here will be that students don't organize their work, and therefore forget one or more factor pair.
I have students share out the rectangles that they were able to create, and then ask students to raise a hand if they forgot a factor pair on their lists. I frame this lesson by telling students they'll have the chance to practice their division fluency and learn a way to organize their work that will help them to capture all of the factor pairs for a given number.
## Intro to New Material
10 minutes
Students worked with factors and multiples in 4th grade. This lesson is a review of previously learned materials, and it's designed to set students up for our work finding the greatest common factors of numbers and solving application problems involving GCFs.
The Intro to New Material section in this lesson is a short one. The goal is to get kids using the academic vocabulary that goes along with this lesson, and then also have them work systematically using t-charts to organize their factor pairs. Below are the steps that students will follow.
Steps for Finding Factors of a Number
1) Create a small T-Chart and write the number on top.
2) On the left side of the chart start with the first factor, 1. On the right side of the chart across from the 1, next to the 1 write what number you multiply 1 by to get the original number, which will be that original number.
3) Continuing in numerical order, try each number mentally (or by dividing off to the side), and write each factor pair.
4) Stop when a factor pair repeats itself.
After we go through the process of creating a t-chart for the factor pairs of 36, I guide students through filling in the definitions at the bottom of the page. These terms are not new for students. I want my students to use these terms throughout the lessons in this unit.
Factor – one of two or more whole numbers that are multiplied to get a product.
Factor pair – two whole numbers that are multiplied to get a product.
Divisible – a number is divisible by another number when it can be divided by that number and the remainder is zero.
Relatively prime numbers - two numbers are relatively prime if they have no common factors other than 1.
Prime number – a number with only 2 factors, 1 and itself. Or, a number with one factor pair.
Composite number – a number with more than 2 factors. Or, a number with more than one factor pair.
## Partner Practice
20 minutes
Students work in pairs on the Partner Practice problem set. As students work, I circulate around the room and check in with each group. I am looking for:
• Are students talking about their thinking with their partners?
• Are students identifying the prime numbers?
• Are students working systematically to find all of the factors?
• Explain you determined the factors.
• How did you begin looking for the factors?
• How did you know you found all of the factors?
• What numbers is ___ divisible by? How do you know? Is it divisible by ______? How do you know it isn’t?
In the t-chart section, 3 of the 6 problems are prime numbers. This was intentional, as I want students to encounter prime numbers and become fluent in recognizing them. This will be important in our upcoming work, with prime factorization and GCF.
Before students work on the Check for Understanding, I use the partner practice problems to quickly review vocabulary. I display one student's t-charts on the document camera. I ask students which numbers had only 2 factors, and then ask students for the word for numbers that have only 2 factors. I then ask for the term we use for numbers that have more than one factor. Students then complete the CFU questions independently.
## Independent Practice
15 minutes
Students work on the Independent Practice problem set.
In this problem set, there is less of a reliance on t-charts. Students will get more practice with finding all of the factors of numbers as we work with greatest common factors. Students have the opportunity to justify their responses when answering the 'how do you know' questions.
Problem 16 gives students a preview of the work to come. This problem may be difficult for kids, but it is a good chance to have them build perseverance in problem solving.
## Closing and Exit Ticket
8 minutes
After independent work time, we come together as a class to discuss Problem 17. I ask if anyone has come up with what they think is the mystery number. If there are responses, I record them on the write board and then ask students to test each of the responses against the clues with their partners. After 2-3 minutes of work time (depending on the number of responses), I then have students share out their thinking about each possible answer.
If students do not have an answer for Problem 17, then we'll talk about each clue as a class. I'll ask students what each clue tells us, and we'll come up with possible answers ourselves, until we get to 64.
Students then complete the Exit Ticket independently to close the lesson.
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# When the value of angle A is ${{90}^{\circ }}$ and B is ${{0}^{\circ }}$ then find the value of ${{\sin }^{2}}A-{{\sin }^{2}}B.$$\left( a \right)0$$\left( b \right)\dfrac{1}{2}$$\left( c \right)1$$\left( d \right)2$
Last updated date: 19th Sep 2024
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. We will first of all assume the variables for sin A and sin B and then by calculating the value of variables of sin A and sin B. We will calculate ${{\sin }^{2}}A$ by using ${{\sin }^{2}}A=\left( \sin A \right)\left( \sin A \right)$ and similarly ${{\sin }^{2}}B$ by using ${{\sin }^{2}}B=\left( \sin B \right)\left( \sin B \right).$ And finally subtract them to get the result.
Complete step-by-step solution
We are given to find the value of the expression
${{\sin }^{2}}A-{{\sin }^{2}}B......\left( i \right)$
Let a = sin A and b = sin B. We are given $A={{90}^{\circ }}$ and $B={{0}^{\circ }}.$ We know that the value of $\sin {{90}^{\circ }}=1$ and the value of $\sin {{0}^{\circ }}=0.$
$\Rightarrow \sin A=\sin {{90}^{\circ }}=1$
$\Rightarrow \sin B=\sin {{0}^{\circ }}=0$
Then the value of ${{\sin }^{2}}A$ can be obtained by using ${{\sin }^{2}}A=\left( \sin A \right)\left( \sin A \right).$
$\Rightarrow {{\sin }^{2}}A={{\sin }^{2}}{{90}^{\circ }}=\left( \sin {{90}^{\circ }} \right)\left( \sin {{90}^{\circ }} \right)$
$\Rightarrow {{\sin }^{2}}A={{\sin }^{2}}{{90}^{\circ }}=1\times 1$
$\Rightarrow {{\sin }^{2}}A={{\sin }^{2}}{{90}^{\circ }}=1$
$\Rightarrow {{\sin }^{2}}A=1$
And
${{\sin }^{2}}B={{\sin }^{2}}{{0}^{\circ }}=\left( \sin \left( {{0}^{\circ }} \right) \right)\sin {{0}^{\circ }}$
$\Rightarrow {{\sin }^{2}}B={{\sin }^{2}}{{0}^{\circ }}=0\times 0$
$\Rightarrow {{\sin }^{2}}B={{\sin }^{2}}{{0}^{\circ }}=0$
Then the value of ${{\sin }^{2}}A-{{\sin }^{2}}B$ will be
${{\sin }^{2}}A-{{\sin }^{2}}B={{\sin }^{2}}{{90}^{\circ }}-{{\sin }^{2}}{{0}^{\circ }}$
$\Rightarrow {{\sin }^{2}}A-{{\sin }^{2}}B=1-0$
$\Rightarrow {{\sin }^{2}}A-{{\sin }^{2}}B=1$
Therefore, the correct option is (a).
Note: The possibility of mistake can be when the angles are $A={{45}^{\circ }}$ or $B={{45}^{\circ }}$ then $\sin {{45}^{\circ }}$ would be $\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ and ${{\sin }^{2}}{{45}^{\circ }}=\left( \sin {{45}^{\circ }} \right)\left( \sin {{45}^{\circ }} \right)=\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{\sqrt{2}} \right)=\dfrac{1}{2}.$ So, there can be a difference between ${{\sin }^{2}}{{45}^{\circ }}$ and $\sin {{45}^{\circ }}.$ Here it was $A={{90}^{\circ }}$ and $\sin {{90}^{\circ }}=1={{\sin }^{2}}{{90}^{\circ }},$ so that is the same here.
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## Judy is baking 5 cheesecakes. A recipe for one of the cheesecakes requires 1 1 2teaspoons of vanilla. Judy’s bottle of vanilla contains 12 t
Question
Judy is baking 5 cheesecakes. A recipe for one of the cheesecakes requires 1 1 2teaspoons of vanilla. Judy’s bottle of vanilla contains 12 teaspoons of vanilla. After Judy bakes the 5 cheesecakes, how many teaspoons of vanilla will be left in the bottle?
in progress 0
5 months 2021-08-26T12:47:42+00:00 1 Answers 14 views 0
1. Answer:4 1/2 Teaspoons of vanilla will be left in the bottle
Step-by-step explanation:
Step 1
Recipe for 1 cheesecake = 1 1/2 teaspoons of vanilla
Since she s making 5 cheesecakes, she will requite = 5 x 1 1/2 teaspoons
= 5 x 3/2= 15/2 = 7 1/2 teaspoons of vanilla.
Step 2
If a bottle of vanilla contains = 12 teaspoons of vanilla
After baking 5 cheesecakes which consumed = 7 1/2 teaspoons of vanilla
Teaspoons left of vanilla = Total teaspoons of vanilla-teaspoons of vanilla used for 5 cheesecakes= 12 – 7 1/2 =4 1/2
Therefore 4 1/2 Teaspoons of vanilla will be left in the bottle
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Before understanding Numerical Integration, we will first take a brief look on definite integration which forms as essential base for understanding numerical integration.
A definite integral is the area under a curve or a function $$f(x)$$, called the integrand, between two points $$[a,b]$$ in the real line. Mathematically, it is defined as
$$I=\int^{b}_{a}f(x)dx$$
The symbol $$dx$$, is called the differential of the variable $$x$$, indicates that the variable of integration is $$x$$. It means a very small change in $$x$$ as in differentiation. Integration is the process of computing the value of a definite integral. The first fundamental theorem of calculus says that the derivative of an integral is the integrand. So, we sometimes call the integral the anti-derivative of the function. Mathematically,
$$\displaystyle f(x)=\frac{d}{dx}\int_{0}^{x}f(t)dt$$
Reimann gave the first practical definition of an integral. The basic idea of the Reimann integral is to partition the area into many rectangles to approximate the area and then to sum them up, just loke integrating them and hence the word integration was used. By taking better and better approximations, which are infinitesimally small, then in the limit, we get exactly the same area under the curve.
Consider a partition of an interval $$[a,b]$$ is a finite sequence of points, then mathematically
$$a=x_0 <x_1<\hdots x_n=b$$
Now, here there is a distinguished point $$t_i\epsilon[x_i,x_{i+1}]$$. The Reimann sum of $$f$$ with respect to the partition is defined as the sum of all teh rectangles.
$$S_n=\sum^{n-1}_{i=0}f(t_i)(x_{i+1}-x_1)$$
When we take the limit that $$n$$ goes to infinity, the Reimann sum becomes the integral value,
$$I=\lim_{n\rightarrow\infty}S_n$$
Numerical Integration is a technique that approximates a definite integral using a similar concept like in a definite integral, which is using a weighted average approximation of the limited sample values of the integrand to replace the value of the function. Numerical integration is an important tool because as it is often not possible to find the antiderivatives in their closed form analytically. Even if the antiderivatives exist, they may not necessarily be easy to compute as they cannot be composed of elementary functions. And moreover it is comparatively easier to compute a numerical approximation than a antiderivative in terms of special functions or the infinite series. Moreover, the integrand may be known only at certain points, such as when obtained by sampling. Hence some of the embedded systems and some other computer applications may need numerical integration for the same.
The numerical integration formulas are called as numerical quadrature formulas. There are many methods to do numerical integration. They are differed by how many partitions they divide the function into, whether the partitions are equally spaced, or how we choose the distinguished points, whether it is open or closed or how we do extrapolation.
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# Find the value of $\tan \left[ \tan \left( \pi -{{\tan }^{-1}}z \right) \right]$?A. $2z$B. $-z$C. $z$D. None of these
Last updated date: 25th Jul 2024
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Hint: We first assume the conversion of ${{\tan }^{-1}}z=\alpha$ which gives $\tan \alpha =z$. We use the associative angle formula to apply the associative angle. We use the identity formula of $\tan \left[ -x \right]=-\tan x$.
We assume that ${{\tan }^{-1}}z=\alpha$. Taking the inverse form, we get $\tan \alpha =z$.
For general form of $\tan \left( x \right)$, we need to convert the value of x into the closest multiple of $\dfrac{\pi }{2}$ and add or subtract a certain value $\alpha$ from that multiple of $\dfrac{\pi }{2}$ to make it equal to x.
Let’s assume $x=k\times \dfrac{\pi }{2}+\alpha$, $k\in \mathbb{Z}$. Here we took the addition of $\alpha$. We also need to remember that $\left| \alpha \right|\le \dfrac{\pi }{2}$.
The final form becomes $\tan \left( \pi -\alpha \right)=\tan \left( 2\times \dfrac{\pi }{2}-\alpha \right)=-\tan \alpha =-z$.
So, $\tan \left[ \tan \left( \pi -{{\tan }^{-1}}z \right) \right]=\tan \left[ -z \right]=-\tan z$.
Note: There are two tan ratio operations which will give values instead of angles. The identity $\tan \left[ -x \right]=-\tan x$ works in the principal domain of ratio tan.
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# interval
Loosely speaking, an interval is a part of the real numbers that start at one number and stops at another number. For instance, all numbers greater that $1$ and smaller than $2$ form in interval. Another interval is formed by numbers greater or equal to $1$ and smaller than $2$. Thus, when talking about intervals, it is necessary to specify whether the endpoints are part of the interval or not. There are then four types of intervals with three different names: open, closed and half-open. Let us next define these precisely.
1. 1.
The open interval contains neither of the endpoints. If $a$ and $b$ are real numbers, then the open interval of numbers between $a$ and $b$ is written as $(a,b)$ and
$(a,b)=\{x\in\mathbb{R}\mid a
2. 2.
The closed interval contains both endpoints. If $a$ and $b$ are real numbers, then the closed interval is written as $[a,b]$ and
$[a,b]=\{x\in\mathbb{R}\mid a\leq x\leq b\}.$
3. 3.
A half-open interval contains only one of the endpoints. If $a$ and $b$ are real numbers, the half-open intervals $(a,b]$ and $[a,b)$ are defined as
$\displaystyle(a,b]$ $\displaystyle=$ $\displaystyle\{x\in\mathbb{R}\mid a $\displaystyle\,\![a,b)$ $\displaystyle=$ $\displaystyle\{x\in\mathbb{R}\mid a\leq x
Note that this definition includes the empty set as an interval by, for example, taking the interval $(a,a)$ for any $a$.
An interval is a subset $S$ of a totally ordered set $T$ with the property that whenever $x$ and $y$ are in $S$ and $x then $z$ is in $S$. Applied to the real numbers, this encompasses open, closed, half-open, half-infinite, infinite, empty, and one-point intervals. All the various different types of interval in $\mathbb{R}$ have this in common. Intervals in $\mathbb{R}$ are connected under the usual topology.
There is a standard way of graphically representing intervals on the real line using filled and empty circles. This is illustrated in the below figures:
The logic is here that a empty circle represent a point not belonging to the interval, while a filled circle represents a point belonging to the interval. For example, the first interval is an open interval.
## Infinite intervals
If we allow either (or both) of $a$ and $b$ to be infinite, then we define
$\displaystyle(a,\infty)$ $\displaystyle=$ $\displaystyle\{x\in\mathbb{R}\mid x>a\},$ $\displaystyle\,\![a,\infty)$ $\displaystyle=$ $\displaystyle\{x\in\mathbb{R}\mid x\geq a\},$ $\displaystyle(-\infty,a)$ $\displaystyle=$ $\displaystyle\{x\in\mathbb{R}\mid x $\displaystyle(-\infty,a]$ $\displaystyle=$ $\displaystyle\{x\in\mathbb{R}\mid x\leq a\},$ $\displaystyle(-\infty,\infty)$ $\displaystyle=$ $\displaystyle\mathbb{R}.$
The graphical representation of infinite intervals is as follows:
## Note on naming and notation
In [1, 2], an open interval is always called a segment, and a closed interval is called simply an interval. However, the above naming with open, closed, and half-open interval seems to be more widely adopted. See e.g. [3, 4, 5]. To distinguish between $[a,b)$ and $(a,b]$, the former is sometimes called a right half-open interval and the latter a left half-open interval [6]. The notation $(a,b)$, $[a,b)$, $(a,b]$, $[a,b]$ seems to be standard. However, some authors (especially from the French school) use notation $]a,b[$, $[a,b[$, $]a,b]$, $[a,b]$ instead of the above (in the same ). Bourbaki, for example, uses this notation.
This entry contains content adapted from the Wikipedia article http://en.wikipedia.org/wiki/Interval_(mathematics)Interval (mathematics) as of November 10, 2006.
## References
• 1 W. Rudin, Principles of Mathematical Analysis, McGraw-Hill Inc., 1976.
• 2 W. Rudin, Real and complex analysis, 3rd ed., McGraw-Hill Inc., 1987.
• 3 R. Adams, Calculus, a complete course, Addison-Wesley Publishers Ltd., 3rd ed., 1995.
• 4 L. Råde, B. Westergren, Mathematics Handbook for Science and Engineering, Studentlitteratur, 1995.
• 5 R.A. Silverman, Introductory Complex Analysis, Dover Publications, 1972.
• 6 S. Igari, Real analysis - With an introduction to Wavelet Theory, American Mathematical Society, 1998.
The metapost code for the figures can be found http://aux.planetmath.org/files/objects/4446/here.
Title interval Canonical name Interval Date of creation 2013-03-22 13:44:58 Last modified on 2013-03-22 13:44:58 Owner PrimeFan (13766) Last modified by PrimeFan (13766) Numerical id 16 Author PrimeFan (13766) Entry type Definition Classification msc 12D99 Classification msc 26-00 Classification msc 54C30 Related topic OpenSetsInMathbbRnContainsAnOpenRectangle Related topic LineSegment Related topic CircularSegment Defines open interval Defines closed interval Defines half-open interval Defines right half-open interval Defines left-half-open interval Defines segment
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### Introduction to Linear Algebra
In the previous quiz, we started looking at an algorithm for solving systems of linear equations, called Gaussian elimination. In this quiz, we’ll take a deeper look at this algorithm, why it works, and how we can speed it up.
# The Full Story of Gaussian Elimination
To save ourselves some effort, let’s introduce some new notation. Instead of having to write $x, y, z$ (or whatever the variables are) multiple times, we can simplify things using matrix notation:
\begin{aligned} x + 2y + 3z &= 18 \\ 3x + 4y + z &= 12 \implies \begin{pmatrix}1&2&3\\3&4&1\\2&-1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}18\\12\\6\end{pmatrix}. \\ 2x - y + z &= 6 \end{aligned}
Here, the first block is the coefficients of the equations, one equation per row. The second block is the variables, one per row and listed in the same order as the coefficients. Finally, the last block is the RHS (short for righthand side) of each equation, one per row. This lets us represent the whole system in just one equation!
Which column of values makes the above matrix equation true?
Note: All options contain the same numbers, but the order in which those numbers are listed is different and important!
# The Full Story of Gaussian Elimination
As with systems, some matrix equations are equivalent, which is important because some matrix equations are easier to solve than others. As a result, we can often transform a difficult matrix system into one that is much easier to solve. Over the next three problems, we’ll see the three fundamental ways to do this, using elementary matrix operations.
The first type of equivalence is fairly simple: since each row of a matrix system represents an equation, nothing changes if we swap the rows around.
Which of the following represents the same system as this matrix equation: $\begin{pmatrix}1&-1&1\\2&3&-2\\-3&3&3\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\-1\\12\end{pmatrix}?$
A. $\quad \begin{pmatrix}2&3&-2\\1&-1&1\\-3&3&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\-1\\12\end{pmatrix}$
B. $\quad \begin{pmatrix}-3&3&3\\2&3&-2\\1&-1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\-1\\12\end{pmatrix}$
C. $\quad \begin{pmatrix}1&-1&1\\-3&3&3\\2&3&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\-1\\12\end{pmatrix}$
D. $\quad \begin{pmatrix}2&3&-2\\-3&3&3\\1&-1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}-1\\12\\2\end{pmatrix}$
# The Full Story of Gaussian Elimination
Just as we can multiply both sides of an equation by the same amount, we can multiply any of the rows of the matrices by constants (as long as we do it to both sides).
Which of the following represents the same system as this matrix equation: $\begin{pmatrix}1&-1&1\\2&3&-3\\-3&3&3\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\-1\\12\end{pmatrix}?$
A. $\quad \begin{pmatrix}2&-2&2\\2&3&-3\\-3&3&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\-1\\12\end{pmatrix}$
B. $\quad \begin{pmatrix}1&-1&1\\4&6&-6\\-3&3&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\-1\\12\end{pmatrix}$
C. $\quad \begin{pmatrix}1&-1&1\\2&3&-3\\-6&6&6\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\-1\\12\end{pmatrix}$
D. $\quad \begin{pmatrix}2&-2&2\\4&6&-6\\-6&6&6\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}4\\-2\\24\end{pmatrix}$
# The Full Story of Gaussian Elimination
Finally, just as we can add two equations together, we can add one row of a matrix to another. As always, we have to be careful to add the corresponding RHS!
Which of the following represents the same system as this matrix equation: $\begin{pmatrix}1&-1&1\\2&3&-3\\-3&3&3\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\-1\\12\end{pmatrix}?$
A. $\quad \begin{pmatrix}1&-1&1\\3&2&-2\\-3&3&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\-1\\14\end{pmatrix}$
B. $\quad \begin{pmatrix}1&-1&1\\2&3&-3\\-2&2&4\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\1\\12\end{pmatrix}$
C. $\quad \begin{pmatrix}1&-1&1\\2&3&-3\\-1&6&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\-1\\11\end{pmatrix}$
D. $\quad \begin{pmatrix}-2&2&4\\2&3&-3\\-3&3&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\-1\\12\end{pmatrix}$
# The Full Story of Gaussian Elimination
Now we can finally restate Gaussian elimination in matrix form. First let’s simplify the notation even further, dropping the variables completely: \begin{aligned} x + 2y + 3z &= 6 \\ 3x + 4y + z &= 8 \implies \begin{pmatrix}1&2&3&6\\3&4&1&8\\2&-1&1&2\end{pmatrix}, \\ 2x - y + z &= 2 \end{aligned} where the last column represents the RHS of each equation. Now we follow this algorithm:
• Determine what's called a “pivot,” i.e. the top leftmost nonzero number that isn’t in a row/column already used. (If necessary, scale this whole row to make the pivot entry 1). Here we have the pivots in blue.
\begin{aligned} \begin{pmatrix} \color{#3D99F6}{1}&2&3&6\\3&4&1&8\\2&-1&1&2 \end{pmatrix} \end{aligned}
• For each row $r$ below the pivot row $p,$ multiply the pivot row by the leading entry and subtract: $r \mapsto r- (\text{leading entry of}\ r) \cdot p$
\begin{aligned} r_{2} &\rightarrow r_{2}-3 r_{1} \\ r_{3} &\rightarrow r_{3}-2 r_{1} \end{aligned} \begin{aligned} \begin{pmatrix}\color{#3D99F6}{1}&2&3&6\\0&-2&-8&-10\\0&-5&-5&-10\end{pmatrix} \end{aligned}
Each row below the pivot row will be left with zeros in that column.
• Repeat until there are no more pivots. At this point, the matrix will have all 0s in the lower left triangle.
\begin{aligned} \ \ r_{2} & \rightarrow -\frac{1}{2} r_{2} \\ \ r_{3} & \rightarrow r_{3} + 5 r_{2} \end{aligned}
\begin{aligned} \begin{pmatrix}\color{#3D99F6}{1}&2&3&6\\0&\color{#3D99F6}{1}&4&5\\0&0&15&15\end{pmatrix} \end{aligned}
Now we need pivot in $r_3$:
\begin{aligned} r_{3} \rightarrow \frac{1}{15} r_{3} \end{aligned}
\begin{aligned} \begin{pmatrix}\color{#3D99F6}{1}&2&3&6\\0&\color{#3D99F6}{1}&4&5\\0&0&\color{#3D99F6}{1}&1\end{pmatrix} \end{aligned}
Now let’s look at that last row--the equation it represents is $z = 1$. Then we can find $y$ and $x$ directly.
# The Full Story of Gaussian Elimination
The “simple” matrices we saw in the last pane have a special name: row-echelon form. They are defined as matrices with the following two properties:
1. Every row containing a nonzero number is above all the rows that only have zeros.
2. Every row’s pivot (i.e. leftmost nonzero number) is to the right of the pivot in the row above it.
For example, $\begin{pmatrix}1&3&2&4&0\\0&0&8&2&1\\0&0&0&3&-1\end{pmatrix}$ Is in row-echelon form. As we saw, these are important because they are easy to solve, and Gaussian elimination is useful because it produces precisely these matrices.
Which of the following is a row-echelon form of this matrix: $\begin{pmatrix}1&5&4&3\\2&6&3&4\\1&2&6&2\end{pmatrix}?$
A. $\quad \begin{pmatrix}1&5&4&3\\0&-4&-5&-2\\0&-3&2&-1\end{pmatrix}$
B. $\quad \begin{pmatrix}1&5&4&3\\0&-4&-5&-2\\0&0&23&2\end{pmatrix}$
C. $\quad \begin{pmatrix}1&5&4&3\\0&-4&-5&-2\\0&0&0&12\end{pmatrix}$
D. $\quad \begin{pmatrix}1&5&4&3\\0&0&3&4\\0&0&12&2\end{pmatrix}$
# The Full Story of Gaussian Elimination
We’ve now seen how Gaussian elimination provides solutions to matrix equations of the form $Ax = b,$ where $A$ is the matrix of coefficients, $x$ is the matrix of variables, and $b$ is the matrix of the right hand side (RHS).
As we know, not all systems have solutions. Is there a matrix for $b$ that guarantees there is a solution for $x$, regardless of what $A$ is?
# The Full Story of Gaussian Elimination
In the previous problem, we saw that $Ax=0$ always has at least one solution, namely $x=0$. But in some cases, there might be more solutions as well. For instance, if $A = \begin{pmatrix}1&2&-3\\2&-2&-3\end{pmatrix},$ then $x = \begin{pmatrix}4\\1\\2\end{pmatrix}$ is a solution as well (try it!).
We'll use this example to see something very important about general solutions of linear systems. First, let's consider a system similar to the one above: $A x = \begin{pmatrix}1&2&-3\\2&-2&-3\end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix}3\\0\end{pmatrix} = b.$
Which of the following is also a solution?
# The Full Story of Gaussian Elimination
In this chapter, we saw how Gaussian elimination translates naturally to the language of matrices, which provides a natural starting point into their exploration. Indeed, linear algebra is all about solving the equation $Ax = b$.
We also saw the special importance of the case where $b = 0$, which naturally leads into the topic of vector spaces--as we will see in the next chapter!
# The Full Story of Gaussian Elimination
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# Is a point in a triangle?
The calculator determines if an arbitrary point lies inside 2D triangle. The triangle is defined by 3 cartesian coordinate pairs.
The calculator below determines if a given point is inside a 2D triangle. The calculator uses a simple algorithm based on vector cross product features. The algorithm detailed description is right behind the calculator.
#### Point in triangle
Digits after the decimal point: 2
Is point in triangle
AB x AP
BC x BP
CA x CP
## A point inside a triangle. The algorithm description
3-dimensional vectors a and b cross product result for the right orthonormal basis is defined as:
$\mathbf {a} \times \mathbf {b} =(a_{y}b_{z}-a_{z}b_{y},\;a_{z}b_{x}-a_{x}b_{z},\;a_{x}b_{y}-a_{y}b_{x})$ [1].
Cross product is anticommutative:
$\mathbf {a} \times \mathbf {b} =-\mathbf {b} \times \mathbf {a}$
It is important feature to solve the point and triangle problem.
To determine if a point P is inside a triangle ABC we compute 3 cross products: ABxAP, BCxBP and CAxCP. Since we are in 2D space, third vector coordinate (z) equals to 0. According to formula [1] we don't need to calculate first two coordinates, since the result x,y coordinate is always 0, because z coordinate is 0 (the result vector is perpendicular to ABC plane). The sign of remained (z) coordinate depends on the relative position of the vectors. If the first vector appears on the right of the second one, the z coordinate is positive, if the first vector appears on the left of the second one, the z coordinate is negative, otherwise (both vectors point to the same direction) the result is 0.
The remaining step to solve our problem is to analyse signs of the z coordinates. If we have both positive and negative results then the point is outside the triangle, if we have only positive or only negative signs - the point is inside the triangle.
The following table illustrates all possible solutions
Drawing Description
All results are negative or positive.
The point is inside the triangle.
One result is zero, the remaining ones is all positive or all negative.
The point lies on a side of the triangle.
If two results are zero, then the point coincides with a triangle vertex.
We have both positive and negative results.
The point is outside the triangle.
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PLANETCALC, Is a point in a triangle?
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# Test: Rectangles And Squares- 1
## 15 Questions MCQ Test Quantitative Aptitude for GMAT | Test: Rectangles And Squares- 1
Description
Attempt Test: Rectangles And Squares- 1 | 15 questions in 30 minutes | Mock test for GMAT preparation | Free important questions MCQ to study Quantitative Aptitude for GMAT for GMAT Exam | Download free PDF with solutions
QUESTION: 1
Solution:
QUESTION: 2
Solution:
QUESTION: 3
### Area of square ABCD is equal to the area of rectangle PQRS. If the length of rectangle PQRS is 21% more than its breadth, then what is the ratio of perimeter of ABCD to perimeter of PQRS?
Solution:
QUESTION: 4
What is the area of rectangle ABCD?
1. Length of each diagonal of rectangle ABCD = 13cm
2. Perimeter of rectangle ABCD = 34cm
Solution:
QUESTION: 5
A border of uniform width of ‘w’ inches is placed around a rectangular photograph. What is the area of the border, in square inches?
1. w = 2 inches
2. Perimeter of rectangular photograph = 20 inches
Solution:
QUESTION: 6
In the given trapezium, what is the length of the diagonal shown by the dashed line?
Solution:
Step 1: Question statement and Inferences
Let’s label the trapezium as ABCD, where Angles A and B are right angles.
To find the length of the diagonal, focus on the right triangle ABC. If we can find the length of side AB, we will be able to find the length of diagonal AC by applying the Pythagoras Theorem.
Step 2: Finding required values
Let us drop a perpendicular DE on the base BC.
So, BEDA is a rectangle. Therefore,
And, AB = DE
Now, BC = BE + EC
4 = 3 + EC
Therefore, EC = 1 cm
Let the length of side DE be h cm.
By applying Pythagoras Theorem in right triangle DEC, we get:
12 + h2 = 22
1 + h2 = 4
h2 = 3
h = √3 cm
Step 3: Calculating the final answer
AB = DE = √3 cm
Let the length of AC be d cm
Therefore, in right triangle ABC, by applying Pythagoras Theorem, we get:
(√3)2 + 42 = d2
3 + 16 = d2
19 = d2
d = √19
QUESTION: 7
The length of a rectangle is 9 cm more than its width, and the area of the rectangle is 486 cm2. What is the length of the diagonal (in cm) of the rectangle?
Solution:
To find – Length of Diagonal of the given rectangle.
Assumptions – Let the length of this rectangle be a cm and width be b cm.
Given
• Area of rectangle = 486 cm2
• a = b + 9
Solution
• We know the relationship between the diagonal and the length and width of the rectangle.
• Diagonal =
• So we find the values of length and width, we will be able to calculate the diagonal.
• From the given information of the rectangle – area and the relationship between the length and the width, we should be able to find the values of length and width.
Area of rectangle – length x width
• a * b = 486
• Substituting a = b + 9 above we get:
• (b + 9) * (b) = 486
• b2 + 9b – 486 = 0
• (b+27)(b-18) = 0
• b = -27 and 18
Since b is a measure of length, it cannot be negative. Thus b = 18. This implies a = b + 9 = 27
Thus, diagonal can be calculated as follows:
Diagonal =
=
=
=
QUESTION: 8
A jogger runs along the track PABCQRCDASP. How many times will he need to run around this track to cover 5 miles? Note that ABCD is a square.
(1) ∠PSD = ∠QRD = 90° and PA = AB = CQ
(2) PS = QR = BD = 0.35 miles (rounded to nearest tenth)
Solution:
Steps 1 and 2 – Understand the question and draw inferences
To find – Number of rounds that the jogger will run around the path to cover 5 miles.
Given
• ∠PAB = ∠BCQ = 90°
Approach
Since we are given the total distance as 5 miles, to find the number of rounds taken by the jogger, we need to find the distance that the jogger will run in one round. Thus, we need to find the perimeter of the path.
To find the perimeter of the path we need the measure of each individual side that make up the path. Thus, we should look for information in the statements that help us arrive at the individual lengths of each side of the jogging path
Step 3 – Analyze Statement 1
Statement 1 indicates that ∠PSD = ∠QRD = 90° and PA = AB = CQ.
So, we just need to find the value of a to calculate the perimeter of the path.
In triangle ABD, as ∠BAD = 900, using the Pythagoras theorem, we can write, 2a2 = 0.352
Thus, we can determine the value of a. With this we have all the individual lengths along the jogging path. Thus, we can determine the perimeter of the jogging path.
Thus, Statements (1) and (2) together yield a unique answer.
QUESTION: 9
If the shaded area represents a kids’ playground in a rectangular park, what is the perimeter of the rectangular park?
(1) At a rate of \$80 per meter, it costs \$1920 to fence the two smaller sides of the kids’ playground.
(2) The area of the kids’ playground is 64 m2.
Solution:
Steps 1 and 2 – Understand the question statement and draw inferences
To find – perimeter of rectangular park = 2 (a+b)
Given -The shaded area is as shown with the measurements in the figure.
• Base of triangular shaded area = a/2
• Height of triangular shaded area = b/2
Step 3 – Analyze Statement 1
At a rate of \$80 per meter, it costs \$1920 to fence the two smaller sides of the kids’ playground
Cost to fence = \$1920
Fencing rate = \$80/m
Therefore, length of the two smaller sides = 1920/ 80 = 24
So, we can write the equation:
• We know the sum of a and b
• Thus, we can find the value of perimeter of the rectangle as 2(a+b)
Statement 1 provides sufficient information to arrive at a unique answer.
Step 4 – Analyze Statement 2
The area of the kids’ playground is 64 m2
or ab = 512
• This relationship gives us the product of a and b.
• It does not give us the individual values of a and b or the sum of a and b
• Thus, we cannot find the value of perimeter of the rectangle.
Statement 2 does not provide sufficient information to arrive at a unique answer.
Step 5 – Analyze Statement 1 and 2 together
Since statement 1 already provides sufficient information to arrive at a unique answer, we do not need to do step 5.
QUESTION: 10
What is the cost to fence a rectangular plot at the rate of \$20/meter?
(1) If length of the plot is increased by 10 meters, the area of the plot increases by 100 sq. m.
(2) If length of the plot is increased by 10% and width of the plot is decreased by 10%, the perimeter decreases by 5%.
Solution:
Steps 1 and 2 – Understand the question and draw inferences
To find – Cost to fence a rectangular plot at \$20/meter
Given
• Rectangular plot
• Rate of fencing -\$20/meter
Approach
• Fencing a rectangular plot implies going around its perimeter.
• So we need to find the perimeter of the rectangular plot.
• For this we need the value of the sum of the length and the width.
Step 3 – Analyze Statement (1)
Statement (1): If length of the plot is increased by 10 meters, the area of the plot increases by 100 sq. m.
Let’s assume the length and the width of the plot as ‘a’ and ‘b’ meters.
New Length = a + 10
New Area = (a+10) * b
Original Area = a * b
According to Statement (1):
(a + 10) * b = a * b + 100
Simplifying this further we get:
• ab + 10b = ab + 100
• 10b = 100
• b = 10
Even though we know the value of b, we cannot find the value of a since no other information is provided about the rectangular plot.
Thus, this statement does not provide any information about the sum of the length and the width and hence the information is not sufficient to arrive at a unique answer.
Step 4 – Analyze Statement (2)
Statement II: If length of the plot is increased by 10% and width of the plot is decreased by 10%, the perimeter decreases by 5%.
Let’s assume the length and width of the plot as ‘a’ and ‘b’ meters.
Original perimeter = 2(a + b)
New Length = a + 0.1a = 1.1a
New Width = b -0.1b = 0.9b
New perimeter = 2(1.1a + 0.9b)
According to Statement (2):
2(1.1a + 0.9b) = 0.95 (2(a + b))
• 1.1a + 0.9b = 0.95a + 0.95b
• 0.15a = 0.05b
• 3a = b
Even though we know the relationship between a and b, we cannot find the value of sum of a and b since no other information is provided.
Thus, this statement does not provide any information about the sum of length and width and hence the information is not sufficient to arrive at a unique answer.
Step 5 – Analyze both Statements Together
• Per Statement (1): b = 10
• Per Statement (2): 3a = b
So combining these two statements, we can find individual values of a and b and hence we can find the sum of a and b.
Thus implies that we can find a unique answer to the question about determining the cost of fencing at the given rate.
Thus, Statements (1) and (2) together yield a unique answer.
Correct Answer Choice – Choice C
QUESTION: 11
ABCD is a rectangle whose longer side is AB. E, F, G and H are the mid-points of the sides AB, BC, CD and DA respectively. The line segment joining E and G intersects with the line segment joining F and H at point P. The perimeter of the rectangle EBFP is 28 units and the area of the triangle AEP is 24 square units. What is the length of each of the smaller sides, AD and BC, of the rectangle ABCD?
Solution:
Step 1: Question statement and Inferences
Since EG is joining the midpoints of the lengths of the rectangle, it is parallel to AD and BC. (i.e., EG is perpendicular to AB and CD).
Similarly, FH is parallel to AB and CD. (FH is perpendicular to AD and BC).
Both EG and FH bisect each other at the point P.
Let the length of AE and EP be x and y respectively.
Area = 24
½ * x * y= 24
xy = 48 ------equation (1)
Note that AE = EB = HP = PF = DG = GC = x
Similarly, AH = HD = EP = PG = BF = FC = y
Perimeter of EBFP = 28
2*(x + y) = 28
(x + y) = 14 ------Equation (2)
Step 2:Finding required values
Upon substituting x = 48/y from equation (2) in equation (1),
Multiplying with y on both sides of the equation, we have
y2 – 14y + 48 = 0
y2 – 8y – 6y + 48 = 0
(y – 8)(y – 6) = 0
y = 8 or y = 6
Step 3: Calculating the final answer
Substituting the value of y in x = 48/y, to obtain the value of x.
If y = 8, then x = 6 and if y = 6, then x = 8.
The longer dimension is 8 units while the smaller dimension is 6 units.
Therefore BF = 6
The smaller of the sides of the rectangle = BC = 2 * 6 = 12 units
QUESTION: 12
A rectangular photograph shown by the white region in the figure is enclosed by a frame of uniform width. The shaded region in the figure represents the frame. What is the ratio of AB to BC?
(1) The ratio of the longer dimension to the smaller dimension of the photograph is 3:2.
(2) If a frame of twice the width was used then the ratio of AB to BC would have been 5:4.
Solution:
Steps 1 & 2: Understand Question and Draw Inferences
Notice that the frame has the same width throughout the border.
Let us say that the longer dimension of the photograph is L, the smaller dimension of the photograph is B and the width of the frame is x.
Then, AB = L + 2x
And, BC = B + 2x
We need to find the ratio of AB:BC. To do so, we ought to know the values of L, B, and x.
Step 3: Analyze Statement 1
The ratio of the longer dimension to the smaller dimension of the photograph is 3:2
Let’s say that for the white area the length of the longer dimension and the length of the shorter dimension are 3z and 2z where z is a positive number.
Thus,
With different values of z and x, we shall obtain different values of the required ratio.
For example, if z =1, x = 1, then AB:BC = 5:4. Whereas, if z = 1, x = 2, then AB:BC = 7:6.
INSUFFICIENT.
Step 4: Analyze Statement 2
If a frame of twice the width was used then the ratio of AB to BC would have been 5:4
In this case, AB = L + 2(2x) = L + 4x
And, BC = B + 2(2x) = B + 4x
We are given that AB:BC = 5 : 4
We need to find the ratio of
Without the values of L, B, and x, it is impossible to determine its value.
INSUFFICIENT.
Step 5: Analyze Both Statements Together (if needed)
Combining statements (1) and (2),
We know that L:B = 3:2, but we do not know the values of L and B.
Let’s assume the values of L and B to be 3z and 2z and the original width is x.
Then,
Required ratio
The ratio has been determined.
SUFFICIENT.
QUESTION: 13
A triangle ABC is placed inside a rectangle PQRS such that the minimum distance of any vertex of the triangle from the rectangle is 2 inches. What is the area of the rectangle PQRS?
(1) Triangle ABC is an equilateral triangle
(2) The area of Triangle ABC is 4√3 square inches and the length of the side AB of the triangle is 50% of the length of side PQ of the rectangle.
Solution:
Steps 1 and 2 – Understand the question statement and draw inferences
To find – Area of rectangle PQRS given the conditions about triangle ABC.
Given – Minimum distance from any vertex of the triangle to the frame = 2”.
The Area of the Rectangle = PQ*PS . . . (1)
So, in order to find the area, we need to know the values of PQ and PS.
Now, from the given diagram, it’s clear that PQ = 2 + AB +2
That is, PQ = AB + 4 . . . (2)
Similarly, let’s try to express PS in terms of a dimension of the triangle ABC. (Note: We are expressing the dimensions of the rectangle in terms of the dimensions of the triangle because both St. 1 and 2 give us information regarding the triangle only)
Let’s drop a perpendicular CD from C on the side AB.
By doing so, we get: PS = 2 + CD + 2
That is, PS = CD + 4 . . . (3)
Substituting (2) and (3) in (1), we get:
Area of rectangle = (AB+4)(CD+4) . . . (4)
Thus, in order to find the area of rectangle PQRS, we need to find the values of AB and CD.
With this understanding, let’s analyze the given statements.
Step 3 – Analyse Statement (1)
Per statement (1), Triangle ABC is an equilateral triangle.
This means, AB = BC = AC
And, CD = √3/2 * AB
However, since this statement doesn’t provide us the exact value of AB, we will not be able to find the area of rectangle PQRS.
Therefore, statement 1 does not provide sufficient information to arrive at a unique answer.
Step 4 – Analyse Statement (2)
Statement 2 provides two pieces of information
• Area of Triangle ABC = 4√3 in2.
• AB = 50% of PQ.
Area of triangle ABC = ½ * AB * CD = 4√3
• AB * CD = 8√3 ….(5)
Also,
• AB = 0.5 * PQ
• 2AB = AB + 4
• AB = 4 . . . (6)
Substituting (6) in (5), we get:
4CD = 8√3
So, CD = 2√3 . . .(7)
Substituting (6) and (7) in (4), we can find the area of rectangle PQRS.
Thus, St. 2 is sufficient to determine the area of the rectangle.
Step 5 – Analyze both Statements together
This step is not required since statement 2 helps us arrive at a unique answer.
QUESTION: 14
The floor of a company's storage room has an area, of 20,000 square feet. If the floor is in the shape of a square, approximately how many feet long is each side?
Solution:
Factorize 20000 to its prime we get
20000=2∗2∗2∗2∗2∗5∗5∗5∗520000=2∗2∗2∗2∗2∗5∗5∗5∗5
or, 20000=24∗2∗5420000=24∗2∗54
Taking square root of this we get 22∗52∗222∗52∗2
which is equal to 100√2
QUESTION: 15
squares are placed one inside another, leaving a strip of uniform width around each square. If
PQ = 2√2 units, QR = 2 units and RS = 10 units, what is the area of the shaded region?
Solution:
Width of strip is uniform around each square. PQ = 2√2 units, QR = 2 units and RS = 10 units.
irst of all we will draw a perpendicular PU from vertex P on side QT.
So in Triangle PUQ:-
Let PU = QU = x units. (width of the shaded region)
By applying Pythagoras theorem,
(PQ)2 = (PU)2 + (QU)2
(2√2)2 = x2 + x2
2x2 = 8
x2 = 4
x = 2 units.
As, RS =10 units.
So, QT = RS – QR = 10 – 2 = 8 units.
And, PV = QT – 2(PU) = 8-2(2) =4 units.
Area of Shaded region = (Area of square with side QT) – (Area of square with side PV) = (8 × 8) – (4 × 4)= 48 sq units.
|
# What Is the Least Common Multiple of 10 and 12?
Author Lee Cosi
Posted Aug 28, 2022
The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both of those numbers. The LCM of 10 and 12 is 60.
Multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, ... Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, ...
The LCM of 10 and 12 is 60 because 60 is the smallest number that is a multiple of both 10 and 12. 60 is a multiple of 10 because 60 ÷ 10 = 6. 60 is a multiple of 12 because 60 ÷ 12 = 5.
## What is the greatest common factor of 10 and 12?
There are several methods that can be used to find the greatest common factor of 10 and 12. One method is to list the factors of each number and then find the largest number that is common to both lists. The factors of 10 are 1, 2, 5, and 10. The factors of 12 are 1, 2, 3, 4, 6, and 12. The greatest common factor of 10 and 12 is 2.
Another method is to use the prime factorization of each number. The prime factorization of 10 is 2 times 5. The prime factorization of 12 is 2 times 2 times 3. The greatest common factor of 10 and 12 is 2.
Yet another method is to use the greatest common divisor (GCD) formula. The GCD of two numbers is the largest number that will divide both numbers evenly. The GCD of 10 and 12 is 2.
The greatest common factor of 10 and 12 is 2.
## How do you find the least common multiple of two numbers?
To find the least common multiple (LCM) of two numbers, you first need to understand what a multiple is. A multiple is simply a number that is a product of two other numbers. So, the LCM of two numbers is the smallest multiple that is a product of those two numbers.
There are a few different ways that you can find the LCM of two numbers. One way is to list out the multiples of each number until you find a common multiple. For example, if you were finding the LCM of 6 and 8, you would list out the multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120. Then, you would list out the multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120. As you can see, the LCM of 6 and 8 is 24.
Another way to find the LCM of two numbers is to use the greatest common divisor (GCD) method. To do this, you would first find the GCD of the two numbers. The GCD is the largest number that divides evenly into both numbers. For example, the GCD of 6 and 8 is 2. Then, you would take the product of the two numbers and divide it by the GCD. So, in this example, you would take 6 x 8 and divide it by 2, which would give you 24.
You can also use a prime factorization method to find the LCM of two numbers. To do this, you would write out the prime factorization of each number. For example, the prime factorization of 6 is 2 x 3 and the prime factorization of 8 is 2 x 2 x 2. To find the LCM, you would take the product of each unique factor. So, in this example, you would take 2 x 3 x 2 x 2, which would give you 48.
No matter which method you use, finding the LCM of two numbers is a relatively simple process. By understanding what a multiple is and using one of the methods described above, you should be able to find the LCM of any two numbers.
## What is the LCM of 10 and 12?
What is the LCM of 10 and 12?
The LCM of 10 and 12 is 30. The LCM is the smallest positive integer that is divisible by both 10 and 12.
In mathematics, the least common multiple, or lowest common multiple, of two integers a and b, usually denoted by LCM(a, b), is the smallest positive integer that is evenly divisible by both a and b. For example, LCM(10,12) = 60.
The LCM of 10 and 12 is 30.
## What's the smallest number that is a multiple of both 10 and 12?
What's the smallest number that is a multiple of both 10 and 12?
The smallest number that is a multiple of both 10 and 12 is 120.
120 is a multiple of 10 because it is evenly divisible by 10. That is, 10 goes into 120 12 times with no remainder.
120 is also a multiple of 12 because it is evenly divisible by 12. That is, 12 goes into 120 10 times with no remainder.
Since 120 is a multiple of both 10 and 12, it is the smallest number that is a multiple of both 10 and 12.
### What are the first few multiples of 10 and 12?
What is the LCM of 10 and 12? The LCM of 10 and 12 is 24.
### What are the multiples of 8 10 and 12?
8, 10 and 12 are the LCM of 120.
### What does LCM mean in math?
The Least Common Multiple is the smallest positive integer that is evenly divisible by both a and b.
### What is the LCM of 12 and 30?
The LCM of 12 and 30 is 60.
### What is the LCM of 10 12 15 15 75?
LCM = 300
Featured Images: pexels.com
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Methods of Solving Simultaneous Equations
# Methods of Solving Simultaneous Equations | The Complete SAT Course - Class 10 PDF Download
## What are Simultaneous Equations?
Simultaneous equations, also known as a system of equations, are a set of equations that involve multiple variables and are related to each other. The solution to a set of simultaneous equations is the common solution that satisfies all the given equations.
• Simultaneous equations can be classified into different types based on the type of equations involved. The most common type of simultaneous equations is linear equations, which involve variables raised to the first power and can be represented on a graph as a straight line. Non-linear equations, on the other hand, involve variables raised to a power greater than one and cannot be represented as a straight line.
• To solve simultaneous equations, we need to find the values of the variables that satisfy all the given equations. There are different methods to solve simultaneous equations, such as substitution method, elimination method, and graphical method. These methods involve manipulating the equations to eliminate one or more variables and then solving for the remaining variables.
The general form of simultaneous linear equations is given as:
ax +by = c
dx + ey = f
## Methods for Solving Simultaneous Equations
The simultaneous linear equations can be solved using various methods. There are three different approaches to solve the simultaneous equations such as substitution, elimination, and augmented matrix method. Among these three methods, the two simplest methods will effectively solve the simultaneous equations to get accurate solutions. Here we are going to discuss these two important methods, namely,
1. Elimination Method: In the elimination method, we eliminate one of the variables from the equations by adding or subtracting the equations from each other. The aim is to get one of the variables to have the same coefficient in both equations so that we can add or subtract them easily.
Steps to solve simultaneous equations using the elimination method:
• Identify which variable to eliminate and write the equations in standard form.
• Multiply one or both of the equations by suitable constants to get the same coefficient for one of the variables.
• Add or subtract the equations to eliminate one of the variables.
• Solve the remaining equation to find the value of one of the variables.
• Substitute the value of the variable into either of the original equations and solve for the other variable.
• Check the solution by substituting the values back into both equations.
2. Substitution Method: In the substitution method, we solve one of the equations for one of the variables in terms of the other variable and then substitute that expression into the other equation.
Steps to solve simultaneous equations using the substitution method:
• Solve one of the equations for one of the variables in terms of the other variable.
• Substitute the expression obtained in step 1 into the other equation to get an equation in one variable.
• Solve the equation for the variable.
• Substitute the value of the variable into either of the original equations and solve for the other variable.
• Check the solution by substituting the values back into both equations.
### Simultaneous Equation Example
Solving Simultaneous Linear Equations Using Elimination Method: Go through the solved example given below to understand the method of solving simultaneous equations by the elimination method along with steps.
Example: Solve the following simultaneous equations using the elimination method.
4a + 5b = 12,
3a – 5b = 9
To solve the simultaneous equations using the elimination method, we need to eliminate one of the variables by adding or subtracting the equations. Here, we can eliminate b by adding the equations together because the coefficients of b are opposites:
4a + 5b = 12
3a – 5b = 9
7a = 21
Dividing both sides by 7, we get:
a = 3
Now, we can substitute the value of a into one of the equations and solve for b. Let's use the first equation:
4a + 5b = 12
4(3) + 5b = 12
12 + 5b = 12
5b = 0
b = 0
Therefore, the solution to the simultaneous equations is:
a = 3, b = 0.
Solving Simultaneous Linear Equations Using Substitution Method: Below is the solved example with steps to understand the solution of simultaneous linear equations using the substitution method in a better way.
Example: Solve the following simultaneous equations using the substitution method.
b= a + 2
a + b = 4.
Using the substitution method, we can substitute the first equation into the second equation:
b = a + 2
a + (a + 2) = 4
Simplifying the second equation:
2a + 2 = 4
2a = 2
a = 1
Now substituting the value of a in the first equation:
b = 1 + 2
b = 3
Therefore, the solution to the given simultaneous equations is a = 1 and b = 3.
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# Free Math Games for 7th Graders
Does your 7th grader need some extra practice in math? Review the activities below and consider using them to provide your child with some interactive math practice at home.
## What Will My 7th Grader Learn in Math?
Your 7th grade child will be focusing his or her math studies on three main areas: ratio and proportional relationships, the number system and geometry. In the ratio and proportions focus area, your child will be applying his or her knowledge of ratios to solve multi-step word problems. These word problems will involve determining the ratios of items, such as length, area and percents. Students will also be using the four basic math operations to solve problems containing rational numbers. Finally, in geometry, they will determine the area, volume and surface area of 2-dimensional and 3-dimensional objects.
### How Much Did I Make?
Have your child plan a pretend bake sale to raise money. After your child determines which items will be baked for the bake sale, have him or her calculate the cost of the necessary grocery items. This may be easier for your child if you take him or her on a trip to the grocery store to price items. Once your child has calculated the amount of money needed for the ingredients, he or she should determine the price for each item. Ask your child to calculate the percent markup for the baked items and figure how much money he or she will profit if all items are sold.
### What Is the Capacity?
For this activity, you will need to gather paper of different sizes, such as computer paper, construction paper and an index card. Have your child roll each piece of paper to make a cylinder shape and tape the ends together. Ask your child to arrange the cylinders from least capacity to greatest capacity. Your child will then need to measure out dry beans and pour them into the created cylinders. Have your child record the amount of beans each cylinder held and use the record to check his or her predictions.
### Math Connect Four
Before beginning this activity, draw out a grid on a piece of poster paper. Write a variety of equations containing rational numbers on two differently colored index cards. Assign a color to each player. Players will take turns drawing one of their cards and solving the equation. If it is solved correctly, the player will place the card on the board inside the square of his or her choice. The first player to connect four cards in a row wins the game!
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# numerical3j
## {14} • Numerical
{14} • Numerical Methods For another example, when we approximated 2 above, it took 4 iterations to find 2 to 1 decimal place. It takes a further 7 iterations to find 2 to 2 decimal places! Exercise 9 Continue the example above approximating to 2, up to [a 11 ,b 11 ]. Verify that it takes 11 iterations to approximate 2 to two decimal places. Although the bisection method doubles its accuracy at each stage, its accuracy in terms of decimal places is much more haphazard! We’ll see later, in the Links Forward section, that there is a nice way to predict, based on the desired accuracy of our solution, just how many iterations of the bisection algorithm we need to calculate. On the other hand, we’ll also see that, if we want a desired number of decimal places of accuracy, then it’s harder to predict how many iterations of the bisection algorithm we will need. Convergence of the bisection method The best thing about the bisection method is that it is guaranteed to work. Provided you’re using the method appropriately, you have a failsafe guarantee that the method works. That guarantee is the best possible type of guarantee: a mathematical theorem. That is, if you’re trying to solve f (x) = 0 in [a,b], for a continuous function f , where f (a) and f (b) have opposite signs, then the bisection method is guaranteed to give you an arbitrarily good approximation to a solution. “Arbitrarily good" means “as good as you want". You can say how good you want your approximation to be, and then, by applying the bisection method enough times, you can get the approximation you want. More formally, suppose you say that you want your approximation to be accurate to within a number ɛ (the Greek letter epsilon). That is, you want to arrive at an interval [a n ,b n ] where b n − a n < ɛ. Then, by applying the bisection method enough times, you can obtain such an interval [a n ,b n ] with b n − a n < ɛ. We can summarise the above formally as a theorem. Theorem Let f : [a,b] → R be a continuous function and suppose f (a), f (b) are both nonzero and have opposite signs. Let ɛ be a positive number. Then the bisection method, after a finite number N of iterations, will produce an interval [a N ,b N ] such that b N − a N < ɛ. That is, after some finite number N of iterations, we have a solution to f (x) = 0 to within
A guide for teachers – Years 11 and 12 • {15} an accuracy of ɛ. Why is theorem true? The idea is simply that the lengths of the intervals [a n ,b n ] halve at each stage. When you take a number and continually halve it, the number gets very small: as small as you like, smaller than any ɛ you prefer! We’ll see more details in the Links Forward section. There we will also see that there is a formula for N , the number of iterations of the bisection method required to obtain the desired accuracy. The best thing about the bisection method may be that it is guaranteed to work, but the worst thing about the bisection method is that it is slow. We’ll now see a method that is often much quicker in finding a solution — but not guaranteed to work every time. Newton’s method Truth is much too complicated to allow anything but approximations. – John Von Neumann Finding a solution with geometry Newton’s method for solving equations is another numerical method for solving an equation f (x) = 0. It is based on the geometry of a curve, using the tangent lines to a curve. As such, it requires calculus, in particular differentiation. Roughly, the idea of Newton’s method is as follows. We seek a solution to f (x) = 0. That is, we want to find the red dotted point in the picture below. We start with an initial guess x 1 . We calculate f (x 1 ). If f (x 1 ) = 0, we are very lucky, and have a solution. But most likely f (x 1 ) is not zero. Let f (x 1 ) = y 1 , as shown.
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# Square of Trinomial
In this chapter we will learn formula for square of trinomial along with solved examples at the end.
## Trinomial Square Formula
If a , b & c are the given entities then square of sum of a + b + c is given by following formula.
\mathtt{( a+b+c)^{2} =\ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}
Hence, square of sum (a + b + c) is equal to square of individual terms and addition of 2ab+ 2bc+ 2ca.
### Deriving Square of Trinomial Formula
The given expression is: \mathtt{(a+b+c)^{2}}
Let a + b = x
Rewriting the equation;
\mathtt{\Longrightarrow \ ( x\ +\ c)^{2}}
Now using the square of sum formula;
\mathtt{\Longrightarrow \ ( x\ +\ c)^{2} \ }\\\ \\ \mathtt{\Longrightarrow \ \ x^{2} +2xc+c^{2}}
Putting the value of x = a + b in equation.
\mathtt{\Longrightarrow \ ( a+b)^{2} +2( a+b) c\ +c^{2}}
Again using sum of square formula for \mathtt{( a+b)^{2}}
\mathtt{\Longrightarrow \ a^{2} +b^{2} +2ab\ +2( a+b) c\ +c^{2}}\\\ \\ \mathtt{\Longrightarrow \ a^{2} +b^{2} +2ab\ ++2ac+2bc+c^{2}}\\\ \\ \mathtt{\Longrightarrow \ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}
Hence we get the formula;
\mathtt{( a+b+c)^{2} =\ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}
### Proof of Square of Trinomial formula
Consider the number \mathtt{( 3+4+2)^{2}}
Finding value of number using simple calculation.
\mathtt{\Longrightarrow ( 3+4+2)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 9)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 81}
Hence, the value of expression is 81.
Now find the value using Square of Trinomial Formula
Using the formula;
\mathtt{( a+b+c)^{2} =\ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}
Putting the values;
\mathtt{\Longrightarrow \ ( 3+4+2)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 3^{2} +4^{2} +2^{2} +2.3.4+2.4.2+2.2.3}\\\ \\ \mathtt{\Longrightarrow \ 9\ +\ 16\ +\ 4\ +\ 24+\ 16\ +\ 12}\\\ \\ \mathtt{\Longrightarrow \ 81}
Using the formula we get the same value 81.
Hence the above formula is valid.
## Square of Trinomial – Solved problems
Example 01
Expand \mathtt{( 2x+4y+5z)^{2}}
Solution
The expression is in form of square of trinomial.
We will use the formula;
\mathtt{( a+b+c)^{2} =\ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}
Putting the values;
\mathtt{\Longrightarrow \ ( 2x+4y+5z)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 2x)^{2} +( 4y)^{2} +( 5z)^{2} + 2( 2x)( 4y) +2( 4y)( 5z) +2( 5z)( 2x)}\\\ \\ \mathtt{\Longrightarrow \ 4x^{2} +\ 16y^{2} +25z^{2} +16xy+40yz+20zx}
Hence, the above expression is the expanded form of given trinomial square.
Example 02
Expand \mathtt{( x-7y+2z)^{2}}
Solution
Using the formula;
\mathtt{( a+b+c)^{2} =\ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}
Putting the values;
\mathtt{\Longrightarrow \ ( x-7y+2z)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( x)^{2} +( -7y)^{2} +( 2z)^{2} +2( x)( -7y) +2( -7y)( 2z) +2( 2z)( x)}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +\ 49y^{2} +4z^{2} -14xy-28yz+4zx}
Hence the above expression is expanded form of trinomial square.
Example 03
Expand \mathtt{( 5x-y-3z)^{2}}
Solution
Using the formula;
\mathtt{( a+b+c)^{2} =\ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}
Putting the values;
\mathtt{\Longrightarrow \ ( 5x-y-3z)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 5x)^{2} +( -y)^{2} +( -3z)^{2} +2( 5x)( -y) +2( -y)( -3z) +2( -3z)( 5x)}\\\ \\ \mathtt{\Longrightarrow \ 25x^{2} +\ y^{2} +9z^{2} -10xy+6yz-30zx}
Hence, the above expression is the expanded term.
Example 04
Expand \mathtt{\ ( -6x-10y-3z)^{2}}
Solution
Using the square of trinomial formula;
\mathtt{( a+b+c)^{2} =\ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}
Putting the values;
\mathtt{\Longrightarrow \ ( -6x-10y-3z)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( -6x)^{2} +( -10y)^{2} +( -3z)^{2} +2( -6x)( -10y) +2( -10y)( -3z) +2( -3z)( -6x)}\\\ \\ \mathtt{\Longrightarrow \ 6x^{2} +100y^{2} +9z^{2} +120xy+60yz+36zx}
Hence, the above expression is the solution.
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# Ones – Definition with Examples
Home » Math Vocabulary » Ones – Definition with Examples
## What is Ones?
In elementary school, students learn to count using numbers, and the first number they learn is “1” or “one.”
The value “one” represents any quantity that is singular in number. Every other number is composed of this “one” in the form of combinations or groups. For example, if we had more than one number of objects like in the given image
We could call this “one and one more,” but if the number of objects increased further, it would become very cumbersome to keep writing in the form of “one and one and….” or we could simply compose another number and call this as “2 ones” or simply “2”.
The numbers 3 to 9 are all composed similarly.
However, as we go further and the numbers become bigger, it becomes difficult to represent numbers by just using the number “one.” What if we have a lot of objects? Like in this image:
There are more than 9 objects here, but if we only keep using “ones,” it would become cumbersome every time we wanted to tell the quantity of an object to someone and started drawing so many of them. This is where “tens” comes to our rescue. We combine “9 ones” and one more “ones” to make a “10” or “1 ten”.
Now, we can represent the same quantity as “1 ten and 2 ones” or the number “12”.
As the number of objects increases, we keep grouping and making tens.
For example, the number 37 means “3 tens and 7 ones”.
Thus we form bigger numbers and use “tens” and “ones” as our basic units to make counting easier.
These units are called place values.
Place Values
All two-digit numbers are composed of tens and ones.
For example, in the number 56, 5 is at the tens place, and 6 is at the ones place.
That is, the number 56 means 5 tens and 6 ones and would represent these many objects:
Here, each rod represents 1 ten and is comprised of 10 ones.
Factoid 1: The numbers 10, 20, 30, 40… can be written as 1 ten, 2 tens, 3 tens, 4 tens, and so on.
Factoid 2: The largest one-digit number that we can make using ones is 9 and the largest two-digit number is 99, which means “9 tens and 9 ones”.
## Solved Examples
Example 1. Write the number of objects given below in the form of tens and ones.
Answer: 1 ten and 4 ones.
Example 2. Which digit is at the ones place in 68?
Example 3. What is the place of the underlined digit in 95?
Example 4. Identify the number represented in the given picture.
Answer: There are 4 tens and 4 ones, so the number is 44.
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How do you find the first and second derivative of ln(ln x)?
Sep 11, 2016
$\frac{1}{x \ln x}$
Explanation:
First derviative: $\frac{d}{\mathrm{dx}} \ln \left(\ln \left(x\right)\right)$
Chain rule: $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$
Let g=ln(x)
$f \left(g\right) = \ln \left(g\right) , g \left(x\right) = \ln \left(x\right)$
$f ' \left(g\right) = \frac{1}{g} , g ' \left(x\right) = \frac{1}{x}$
$f ' \left(x\right) = \frac{1}{\ln} \left(a\right)$
$f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right) = \left(\frac{1}{\ln} \left(x\right) \cdot \frac{1}{x}\right)$
$\frac{1}{x \ln x}$
Second Derivative: $\frac{d}{\mathrm{dx}} \left(\frac{1}{\ln} \left(x\right) \cdot \frac{1}{x}\right)$
Product rule: $f \left(x\right) g ' \left(x\right) + f ' \left(x\right) g \left(x\right)$
$f \left(x\right) = \frac{1}{\ln} x , g \left(x\right) = \frac{1}{x}$
$\frac{1}{\ln} x \frac{d}{\mathrm{dx}} \left({x}^{-} 1\right) + \frac{d}{\mathrm{dx}} {\left(\ln x\right)}^{-} 1 \left(\frac{1}{x}\right)$
-1/(x^2lnx)-1/(x^2ln^2(x)
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