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How to find the common difference of an Arithmetic Progression whose sum is given?
• Last Updated : 25 Aug, 2021
A sequence is defined as the list of items or objects that are arranged in a sequential manner. It is also defined as the set of numbers in a defined order which follows some rule. Suppose if a1, a2, a3, a4… etc represents the terms of a given sequence, then the subscripts 1, 2, 3, 4… represent the position of the term of the sequence.
There are several types of sequences, some of which are:
• Arithmetic sequence
• Geometric sequence
• Harmonic sequence
The Arithmetic sequence is a sequence in which each term is created by the addition or subtraction of a constant number to the preceding number. For example, the series of whole numbers is 0, 1, 2, 3, 4,… is an arithmetic progression with a common difference of 1.
In an arithmetic progression (AP), there are three main terms. These are denoted as:
• Common difference (d)
• nth Term (an)
• Sum of the AP (Sn)
The difference between the two consecutive numbers is known as the common difference. Consider the numbers a1, a2, a3, a4an are in AP, the common difference is calculated by the expression:
d = a2 a1, a3 a2, a4 a3…, an an-1
The AP can also be written in the following form:
a, a+d, a+2d, a+3d,… … …, a+(n-1)d
where a is the first term of the progression, d is the common difference of the progression and a+(n-1)d is the nth term of AP.
The sum of the AP is given by
S = n/2[2a + (n – 1)d]
Steps to calculate the common difference when the sum is given:
The steps to calculate the common difference are given below:
• Substitute the values of sum, the number of terms, and the first term in the formula.
• Simplify the right-hand side.
• Solve for the value of d.
• Make sure that the calculation is correct.
Consider that the first term of an AP is 5, the number of terms in the AP is 9 and the sum of the AP is 189. Calculate the common difference of the above Arithmetic series.
Solution:
Given that,
The first term of the AP is 5,
The number of terms in the AP is 9 and
The sum of the AP is 189.
We know that the formula to calculate the sum of the AP is
Sn = n/2[2a + (n-1)d]
Substitute 9 for n, 5 for a and 189 for Sn into the formula.
189=9/2[2×5+(9-1)d]
42=10+8×d
8d=32
d=4
Hence, the common difference of the given series is 4.
Similar Questions
Question 1: Consider that the first term of an AP is 120, the number of terms in the AP is 5 and the sum of the AP is 650. Calculate the common difference of the above Arithmetic series.
Solution:
Given that,
The first term of the AP is 120,
The number of terms in the AP is 5 and
The sum of the AP is 650.
We know that the formula to calculate the sum of the AP is
Sn = n/2[2a + (n-1)d]
Substitute 5 for n, 120 for a and 650 for Sn into the formula.
650 = 5/2[2 × 120 + (5-1)d]
260 = 240 + 4 × d
4d = 20
d = 5
Hence, the common difference of the given series is 5.
Question 2: Consider that the first term of an AP is 5.5, the number of terms in the AP is 10 and the sum of the AP is 100. Calculate the common difference of the above Arithmetic series.
Solution:
Given that,
The first term of the AP is 5.5,
The number of terms in the AP is 10 and
The sum of the AP is 100.
We know that the formula to calculate the sum of the AP is
Sn = n/2[2a + (n-1)d]
Substitute 10 for n, 5.5 for a and 100 for Sn into the formula.
100 = 10/2[2 × 5.5 + (10-1)d]
20 = 11 + 9 × d
9d = 9
d = 1
Hence, the common difference of the given series is 1.
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# If $f(x) = \left( {\dfrac{{\sin 3x}}{{\sin x}}} \right),x \ne n\pi ,$ then the range of values of $f(x)$ for real values of $x$ is(a)$[1 - 3)$(b)$( - \infty , - 1]$(c)$(3, + \infty )$(d)$[ - 1,3]$
Last updated date: 03rd Mar 2024
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Hint:As we know that the above given question is related to trigonometric expression, sine and cosine are trigonometric ratios. Here we have to find the value of $f(x)$, so first of all we have to solve and simplify the value. We can convert the equation into basic trigonometric equations by applying the trigonometric identities.
Complete step by step solution:
As per the given question we have to solve the expression $f(x) = \dfrac{{\sin 3x}}{{\sin x}}$. We know that a trigonometric identity for $\sin 3x$ which is $3\sin x - 4{\sin ^3}x$, now by substituting this value we will expand and we get, $f(x) = \dfrac{{3\sin x - 4{{\sin }^3}x}}{{\sin x}}$. We can take the common $\sin x$ out and it gives,
$\dfrac{{\sin x(3 - 4{{\sin }^2}x)}}{{\sin x}} = 3 - 4{\sin ^2}x$. Now let us assume $3 - 4{\sin ^2}x = y$, so we get: ${\sin ^2}x = \dfrac{{3 - y}}{4}$.
Since we can say that $0 \leqslant \dfrac{{3 - y}}{4} \leqslant 1$. Now solving this by inequality :
$= 0 \leqslant 3 - y \leqslant 4 \Rightarrow - 3 \leqslant - y \leqslant 1$.
Now we can interchange the values but by keeping the signs not changed, $- 1 \leqslant y \leqslant 3$.
So we can write it as $y \in [ - 1,3]$.
Hence the correct option is (d) $[ - 1,3]$.
Note: Before solving this kind of question we should have the proper knowledge of all trigonometric ratios, identities and their formulas. To solve this trigonometric expression we should also have the proper knowledge of the inequality, and then we should solve it by avoiding mistakes and taking care of positive and negative signs. |
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Last updated date: 01st Dec 2023
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# In a class of $60$ children, $30\%$ are girls. How many boys are there?
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Hint: In the class, children comprise both the girls and boys. If the percentage of girls are given then subtracting it from the number hundred we get the percentage of boys. Simply find the percentage of boys for the total children in the class.
Given that: In a class of $60$ children, $30\%$ are girls
Class has both the girls and the boys.
Boys $\% = 100\% - 30\%$
Simplify finding the difference of the terms in the above expression –
Boys $\% = 70\%$
Now, the number of boys $= 70\%$ of the total children
Here “of” suggests the multiplication operator.
the number of boys $= 70\% \times 60$
percentage is expressed as the number upon hundred. So, place $70\% = \dfrac{{70}}{{100}}$ in the above expression –
The number of boys $= \dfrac{{70}}{{100}} \times 60$
Find the factors for the above expression –
The number of boys $= \dfrac{{7 \times 10 \times 6 \times 10}}{{10 \times 10}}$
Common factors from the numerator and the denominator cancel each other and therefore remove from the numerator and the denominator of the above expression.
The number of boys $= 7 \times 6$
Simplify finding the product of the terms in the above expression –
The number of boys $= 42$ Boys
Hence, there are $42$ boys in a class.
So, the correct answer is “42”.
Note: The above example can be used by finding the percentage given of girls in number of girls and then have to subtract the number of girls from the total number of children to get the number of boys since the class contains boys and girls. Always convert the given percentage in the form of the fraction where the numerator is hundred. |
# College:Product rule for differentiation
This page uses material from product rule for differentiation and Practical:Product rule for differentiation.
ORIGINAL FULL PAGE: Product rule for differentiation
STUDY THE TOPIC AT MULTIPLE LEVELS: Page for school students (first-time learners) | Page for college students (second-time learners) | Page for math majors and others passionate about math |
ALSO CHECK OUT: Practical tips on the topic |Quiz (multiple choice questions to test your understanding) |Pedagogy page (discussion of how this topic is or could be taught)|Page with videos on the topic, both embedded and linked to
## Name
This statement is called the product rule, product rule for differentiation, or Leibniz rule.
## Statement for two functions
### Statement in multiple versions
The product rule is stated in many versions:
Version type Statement
specific point, named functions Suppose $f$ and $g$ are functions of one variable, both of which are differentiable at a real number $x = x_0$. Then, the product function $f \cdot g$, defined as $x \mapsto f(x)g(x)$ is also differentiable at $x = x_0$, and the derivative at $x_0$ is given as follows:
$\! \frac{d}{dx} [f(x)g(x)]|_{x = x_0} = f'(x_0)g(x_0) + f(x_0)g'(x_0)$
or equivalently:
$\! \frac{d}{dx} [f(x)g(x)]|_{x = x_0} = \frac{d(f(x))}{dx}|_{x=x_0} \cdot g(x_0) + f(x_0)\cdot \frac{d(g(x))}{dx}|_{x = x_0}$
generic point, named functions, point notation Suppose $f$ and $g$ are functions of one variable. Then the following is true wherever the right side expression makes sense (see concept of equality conditional to existence of one side):
$\! \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$
generic point, named functions, point-free notation Suppose $f$ and $g$ are functions of one variable. Then, we have the following equality of functions on the domain where the right side expression makes sense (see concept of equality conditional to existence of one side):
$\! (f \cdot g)' = (f'\cdot g) + (f \cdot g')$
We could also write this more briefly as:
$\! (fg)' = f'g + fg'$
Note that the domain of $(fg)'$ may be strictly larger than the intersection of the domains of $f'$ and $g'$, so the equality need not hold in the sense of equality as functions if we care about the domains of definition.
Pure Leibniz notation using dependent and independent variables Suppose $u,v$ are variables both of which are functionally dependent on $x$. Then:
$\! \frac{d(uv)}{dx} = \left(\frac{du}{dx}\right) v + u \frac{dv}{dx}$
In terms of differentials Suppose $u,v$ are both variables functionally dependent on $x$. Then,
$\! d(uv) = v (du) + u (dv)$.
MORE ON THE WAY THIS DEFINITION OR FACT IS PRESENTED: We first present the version that deals with a specific point (typically with a $\{ \}_0$ subscript) in the domain of the relevant functions, and then discuss the version that deals with a point that is free to move in the domain, by dropping the subscript. Why do we do this?
The purpose of the specific point version is to emphasize that the point is fixed for the duration of the definition, i.e., it does not move around while we are defining the construct or applying the fact. However, the definition or fact applies not just for a single point but for all points satisfying certain criteria, and thus we can get further interesting perspectives on it by varying the point we are considering. This is the purpose of the second, generic point version.
### One-sided version
The product rule for differentiation has analogues for one-sided derivatives. More explicitly, we can replace all occurrences of derivatives with left hand derivatives and the statements are true. Alternately, we can replace all occurrences of derivatives with right hand derivatives and the statements are true.
### Partial differentiation
For further information, refer: product rule for partial differentiation
The product rule is also valid if we consider functions of more than one variable and replace the ordinary derivative by the partial derivative, directional derivative, or gradient vector.
## Statement for multiple functions
Below, we formulate the many versions of this product rule:
Version type Statement
specific point, named functions Suppose $f_1, f_2, \dots, f_n$ are functions defined and differentiable at a point $x_0$. Then the product $f_1 \cdot f_2 \cdot \dots \cdot f_n$ is also differentiable at $x_0$, and we have:
$\! \frac{d}{dx}[f_1(x)f_2(x)\dots f_n(x)]|_{x = x_0} = f_1'(x_0)f_2(x_0) \dots f_n(x_0) + f_1(x_0)f_2'(x_0) \dots f_n(x_0) + \dots + f_1(x_0)f_2(x_0) \dots f_{n-1}(x_0)f_n'(x_0)$
generic point, named functions, point notation Suppose $f_1, f_2, \dots, f_n$ are functions. Then the product $f_1 \cdot f_2 \cdot \dots \cdot f_n$ satisfies:
$\! (f_1 \cdot f_2 \cdot \dots \cdot f_n)'(x) = f_1'(x)f_2(x) \dots f_n(x) + f_1(x)f_2'(x) \dots f_n(x) + \dots + f_1(x)f_2(x) \dots f_{n-1}(x)f_n'(x)$ wherever the right side makes sense.
generic points, named functions, point-free notation Suppose $f_1, f_2, \dots, f_n$ are functions. Then the product $f_1 \cdot f_2 \cdot \dots \cdot f_n$ satisfies:
$\! (f_1 \cdot f_2 \cdot \dots \cdot f_n)' = f_1' \cdot f_2 \cdot \dots f_n + f_1 \cdot f_2' \cdot \dots \cdot f_n + \dots + f_1 \cdot f_2 \cdot \dots \cdot f_{n-1} \cdot f_n'$ wherever the right side makes sense. We could also write this more briefly as
$\! (f_1 f_2 \dots f_n)' = f_1' f_2 \dots f_n + f_1 f_2' \dots f_n + \dots + f_1 f_2 \dots \cdot f_{n-1}f_n'$
Pure Leibniz notation using dependent and independent variables Suppose $u_1,u_2,\dots,u_n$ are variables functionally dependent on $x$. Then $\frac{d(u_1u_2\dots u_n)}{dx} = \left(\frac{du_1}{dx}\right)(u_2u_3 \dots u_n) + u_1 \left(\frac{du_2}{dx}\right) (u_3 \dots u_n) + \dots + u_1u_2 \dots u_{n-1} \left(\frac{du_n}{dx}\right)$ wherever the right side make sense.
In terms of differentials Suppose $u_1,u_2,\dots,u_n$ are variables functionally dependent on $x$. Then $d(u_1u_2\dots u_n) = u_2u_3 \dots u_n(du_1) + u_1u_3 \dots u_n(du_2) + \dots + u_1u_2 \dots u_{n-1}(du_n)$
For instance, using the generic point, named functions notation for $n = 3$, we get:
$\! (f_1 \cdot f_2 \cdot f_3)'(x) = f_1'(x)f_2(x)f_3(x) + f_1(x)f_2'(x)f_3(x) + f_1(x)f_2(x)f_3'(x)$
## Significance
### Qualitative and existential significance
Each of the versions has its own qualitative significance:
Version type Significance
specific point, named functions This tells us that if $f$ and $g$ are both differentiable at a point, so is $f \cdot g$. The one-sided versions allow us to make similar statement for left and right differentiability.
generic point, named functions, point notation This tells us that if both $f$ and $g$ are differentiable on an open interval, then so is $f \cdot g$. The one-sided versions allow us to make similar statements for closed intervals where we require the appropriate one-sided differentiability at the endpoints.
generic point, point-free notation This can be used to deduce more, namely that the nature of $(f \cdot g)'$ depends strongly on the nature of $f$ and that of $g$. In particular, if $f$ and $g$ are both continuously differentiable functions on an interval (i.e., $f'$ and $g'$ are both continuous on that interval), then $(f \cdot g)$ is also continuously differentiable on that interval. This uses the sum theorem for continuity and product theorem for continuity.
### Computational feasibility significance
Each of the versions has its own computational feasibility significance:
Version type Significance
specific point, named functions This tells us that knowledge of the values (in the sense of numerical values) $\! f(x_0), g(x_0), f'(x_0), g'(x_0)$ at a specific point $x_0$ is sufficient to compute the value of $(f \cdot g)'(x_0)$. For instance, if we are given that $f(1) = 5, g(1) = 11, f'(1) = 4, g'(1) = 13$, we obtain that $(f \cdot g)'(1) = 4 \cdot 11 + 5 \cdot 13 = 44 + 65 = 109$.
A note on contrast with the (false) freshman product rule: [SHOW MORE]
generic point, named functions This tells us that knowledge of the general expressions for $f$ and $g$ and the derivatives of $f$ and $g$ is sufficient to compute the general expression for the derivative of $f \cdot g$. See the #Examples section of this page for more examples.
### Computational results significance
Each of the versions has its own computational results significance:
Shorthand Significance What would happen if the freshman product rule were true instead of the product rule?
significance of derivative being zero If $\! f'(x_0)$ and $\! g'(x_0)$ are both equal to 0, then so is $(f \cdot g)'(x_0)$. In other words, if the tangents to the graphs of $f,g$ are both horizontal at the point $x = x_0$, so is the tangent to the graph of $f \cdot g$. This result would still hold, but so would a stronger result: namely that if either $f'(x_0)$ or $g'(x_0)$ is zero, so is $(f \cdot g)'(x_0)$.
significance of sign of derivative $\! f'(x_0)$ and $\! g'(x_0)$ both being positive is not sufficient to ensure that $(f \cdot g)'(x_0)$ is positive. However, if all four of $\! f(x_0), g(x_0), f'(x_0), g'(x_0)$ are positive, then $(f \cdot g)'(x_0)$ is positive. This is related to the fact that a product of increasing functions need not be increasing. In that case, it would be true that $\! f'(x_0)$ and $\! g'(x_0)$ both being positive is sufficient to ensure that $(f \cdot g)'(x_0)$ is positive.
significance of uniform bounds $\! f',g'$ both being uniformly bounded is not sufficient to ensure that $(f \cdot g)'$ is uniformly bounded. However, if all four functions $\! f,g,f',g'$ are uniformly bounded, then indeed $(f \cdot g)'$ is uniformly bounded. In that case, it would be true that $\! f'(x_0)$ and $\! g'(x_0)$ both uniformly being bounded is sufficient to ensure that $(f \cdot g)'(x_0)$ is uniformly bounded.
## Examples
For practical tips and explanations on how to apply the product rule in practice, check out Practical:Product rule for differentiation
### Sanity checks
We first consider examples where the product rule for differentiation confirms something we already knew through other means. In all examples, we assume that both $f$ and $g$ are differentiable functions:
Case The derivative of $x \mapsto f(x)g(x)$ Direct justification (without use of product rule) Justification using product rule, i.e., computing it as $\! f'(x)g(x) + f(x)g'(x)$
$g$ is the zero function. zero function $\! f(x)g(x) = 0$ for all $x$, so its derivative is also zero . Both $\! g(x)$ and $\! g'(x)$ are zero functions, so $\! f'(x)g(x) + f(x)g'(x)$ is everywhere zero.
$g$ is a constant nonzero function with value $\lambda$. $\! \lambda f'(x)$ The function is $x \mapsto \lambda f(x)$, and the derivative is $\! \lambda f'(x)$, because the constant can be pulled out of the differentiation process. $\! f'(x)g(x)$ simplifies to $\! \lambda f'(x)$. Since $g$ is constant, $g'(x)$ is the zero function, hence so is $\! f(x)g'(x)$. The sum is thus $\! \lambda f'(x)$.
$f = g$ $\! 2f(x)f'(x)$ The derivative is $\! 2f(x)f'(x)$ by the chain rule for differentiation: we are composing the square function and $f$. We get $\! f'(x)f(x) + f(x)f'(x) = 2f(x)f'(x)$.
$g = 1/f$ zero function The product is $1$, which is a constant function, so its derivative is zero. We get $f(x)(1/f)'(x) + f'(x)/f(x)$. By the chain rule, $(1/f)'(x) = -f'(x)/(f(x))^2$, so plugging in, we get $-f(x)f'(x)/(f(x))^2 + f'(x)/f(x)$, which simplifies to zero.
### Nontrivial examples where simple alternate methods exist
Here is a simple trigonometric example:
$\! f(x) := \sin x \cos x$.
### Nontrivial examples where simple alternate methods do not exist
Consider a product of the form:
$\! f(x) := x \sin x$
Using the product rule, we get:
$f'(x) = \frac{dx}{dx} \sin x + x \frac{d}{dx}(\sin x) = 1(\sin x) + x \cos x = \sin x + x \cos x$
## Procedure to apply the product rule for differentiation
The product rule for differentiation is useful as a technique for differentiating functions that are expressed in the form of products of simpler functions.
### Most explicit procedure
The explicit procedure is outlined below:
1. Identify the two functions whose product is the given function. In other words, explicitly decompose the function as a product of two functions. We will here call the functions $f$ and $g$, though you may choose to give them different names.
2. Calculate the derivatives of $f$ and $g$ separately, on the side.
3. Plug into the product rule formula the expressions for the functions and their derivatives.
4. Simplify the expression thus obtained (this is optional).
Here is an example of a differentiation problem where we use this explicit procedure:
Differentiate the function $p(x) := (x^2 + 1)(x^3 + 2)$ with respect to $x$
We proceed step by step:
1. Identify the two functions: Define $f(x) = x^2 + 1$ and $g(x) = x^3 + 2$. Then, $p(x)= f(x)g(x)$ by definition.
2. Calculate the derivatives: The derivative of $f(x)$ is $f'(x) = 2x$ and the derivative of $g(x)$ is $g'(x) = 3x^2$.
3. Plug into the product rule formula: We get $p'(x) = f'(x)g(x) + f(x)g'(x) = (2x)(x^3 + 2) + (x^2 + 1)(3x^2)$.
4. Simplify the expression obtained: We get $p'(x) = 2x^4 + 4x + 3x^4 + 3x^2 = 5x^4 + 3x^2 + 4x$.
Here is another example of a differentiation problem where we use this explicit procedure:
Differentiate the function $p(x) := e^x \sin x$
1. Identify the two functions: Define $f(x) = e^x$ and $g(x) = \sin x$
2. Calculate the derivatives: The derivative of $f(x)$ is $f'(x) = e^x$ and the derivative of $g(x)$ is $g'(x) = \cos x$.
3. Plug into the product rule formula: We get $p'(x) = f'(x)g(x) + f(x)g'(x) = (e^x)(\sin x) + (e^x)(\cos x)$.
4. Simplify the expression obtained: The expression is already simplified. If we wish to collect terms, we can rewrite as $e^x(\sin x + \cos x)$.
### More inline procedure using Leibniz notation
Although the explicit procedure above is fairly clear, Step (2) of the procedure can be a waste of time in the sense of having to do the derivative calculations separately. If you are more experienced with doing differentiation quickly, you can combine Steps (2) and (3) by calculating the derivatives while plugging into the formula, rather than doing the calculations separately prior to plugging into the formula. Further, we do not need to explicitly name the functions if we use the Leibniz notation to compute the derivatives inline.
The shorter procedure is outlined below:
1. Identify the two functions being multiplied (but you don't have to give them names).
2. Plug into the formula for the product rule, using the Leibniz notation for derivatives that have not yet been computed.
3. Compute derivatives and simplify
For instance, consider the problem:
Differentiate the function $p(x) := (2x^3 + 3x - \sin x)(x^2 + \cos x - 3)$
The procedure is:
1. Identify the two functions: The functions are $x \mapsto 2x^3 + 3x - \sin x$ and $x \mapsto x^2 + \cos x - 3$.
2. Plug into the formula for the product rule: We get:
$\frac{d(2x^3 + 3x - \sin x)}{dx}(x^2 + \cos x - 3) + (2x^3 + 3x - \sin x)\frac{d(x^2 + \cos x - 3)}{dx}$
3. Compute derivatives and simplify: we get:
$(6x^2 + 3 - \cos x)(x^2 + \cos x - 3) + (2x^3 + 3x - \sin x)(2x - \sin x)$. The expression can be expanded and simplified if desired.
### Shortest inline procedure
If you are really experienced with doing derivatives in your head, you can shorten the procedure even further by combining Steps (2) and (3) in the previous procedure. The procedure has two steps:
1. Identify the two functions being multiplied (but you don't have to give them names).
2. Use the formula for the product rule, computing the derivatives of the functions while plugging them into the formula
For instance, going back to the example used in the beginning:
Differentiate the function $p(x) := (x^2 + 1)(x^3 + 2)$ with respect to $x$
This can be done quickly:
1. Identify the two functions: They are $x \mapsto x^2 + 1$ and $x \mapsto x^3 + 2$.
2. Use the formula for the product rule, computing the derivatives of the functions while plugging them into the formula: We get $(2x)(x^3 + 2) + (x^2 + 1)(3x^2) = 5x^4 + 3x^2 + 4x$.
### Choosing between procedures
The procedures are not fundamentally different, but they differ in the degree of explicitness of the steps. Generally speaking, the following are recommended:
• If the functions being multiplied are fairly easy to differentiate mentally, use the shortest inline procedure -- this is fast and reliable.
• If the functions being multiplied are somewhat more difficult to differentiate, then choose between the other two more explicit procedures, based on whether you are more comfortable with writing large inline expressions or with doing separate work on the side.
## Error types
### Incorrect formula
A common mistake in differentiating products of functions is the freshman product rule, i.e., the false rule that the derivative of the product is the product of the derivatives. The good news is that, generally speaking, it is easy to avoid this rule once you have enough experience with the actual product rule.
### Writing only one piece of the product rule
This is an error of the incomplete task form and is harder to avoid. What happens here is that you forget to write one of the two pieces being added for the product rule, so perhaps you end up doing:
$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) \qquad \mbox{WRONG! Forgot one half of the product rule}$
or
$\frac{d}{dx}[f(x)g(x)] = f(x)g'(x) \qquad\mbox{WRONG! Forgot one half of the product rule}$
A slight variant involves forgetting one of the factors being multiplied:
$\frac{d}{dx}[f(x)g(x)] = f'(x) + f(x)g'(x) \qquad \mbox{WRONG! Forgot one of the pieces being multiplied}$
Why this error occurs: Usually, this error is common if you are trying to use the shortest inline procedure, i.e., differentiating the functions and applying the product rule simultaneously, and one of the functions being differentiated is rather tricky to differentiate, requiring a product rule or chain rule for differentiation in and of itself.
How to avoid this error:
• When the functions being differentiated are tricky to differentiate, use either the fully explicit procedure or the inline procedure with Leibniz notation. Do not try to simultaneously differentiate the pieces and use the product rule.
• After finishing a product rule problem, ask the following sanity check question: did I get a sum of two products after the application of the product rule? If the answer is no, then check your work.
## How to remember the formula
### Different versions of the formula
Recall that we stated the product rule as:
$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$
The right side could be written in any of eight equivalent ways:
1. $f'(x)g(x) + f(x)g'(x)$
2. $g(x)f'(x) + f(x)g'(x)$
3. $g(x)f'(x) + g'(x)f(x)$
4. $f'(x)g(x) + g'(x)f(x)$
5. $f(x)g'(x) + f'(x)g(x)$
6. $g'(x)f(x) + f'(x)g(x)$
7. $g'(x)f(x) + g(x)f'(x)$
8. $f(x)g'(x) + g(x)f'(x)$
All of these are equivalent, so in some sense it does not matter which of the versions you choose to remember. However, the first version is somewhat preferable to the others for a number of reasons mentioned below:
• The order of multiplication is the same as in the expression being differentiated: the $f$-part is on the left and the $g$-part is on the right. This is true only of versions (1) and (5). This is particularly important when we consider generalization to non-commutative situations such as product rule for differentiation of cross product.
• The differentiation symbol (the prime) hops from left to right: Remembering the rule this way makes it easier to generalize to differentiating products of more than two functions (see product rule for differentiation#Statement for multiple functions).
### Quick rationalizations for the formula
If you are a little shaky about the product rule for differentiation, how do you do a reality check on the formula? A quick, intuitive version of the proof of product rule for differentiation using chain rule for partial differentiation will help. Basically, what it says is that to determine how the product changes, we need to count the contributions of each factor being multiplied, keeping the other constant. The contribution of $f$ keeping $g$ constant is $f'(x)g(x)$ and the contribution of $g$ keeping $f$ constant is $f(x)g'(x)$. |
# How do you graph f(x)=(x-2)/(x-4) using holes, vertical and horizontal asymptotes, x and y intercepts?
Jan 25, 2018
Undefined at $x = 4$
See explanation for the rest
#### Explanation:
$\textcolor{b l u e}{\text{Hole - undefined}}$
Hole is where the denominator becomes 0. The function becomes undefined at that point. So for this condition we have $x = 4$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Lets consider the behaviour close to $x = 4$
$\textcolor{b l u e}{\text{Vertical Asymptot } \to + \infty}$
If $x = 4 + \delta 4$ where $\delta 4 > 0$ and very small then we have
$\frac{4 + \delta 4 - 2}{\cancel{4} + \delta 4 - \cancel{4}} \to \frac{2}{\delta 4} + 1$
${\lim}_{\delta 4 \to {\textcolor{w h i t e}{}}^{+} 0} \frac{2}{\delta 4} + 1 \textcolor{w h i t e}{\text{dd")->color(white)("dd")kcolor(white)("dd")->color(white)("dd}} + \infty + 1 = + \infty$
$\textcolor{b l u e}{\text{Vertical Asymptot } \to - \infty}$
If $x = 4 - \delta 4$ where $\delta 4 > 0$ and very small then we have
$\frac{4 - \delta 4 - 2}{\cancel{4} - \delta 4 - \cancel{4}} \to \frac{2 - \delta 4}{- \delta 4} = - \frac{2}{\delta 4} + 1$
${\lim}_{\delta 4 \to {\textcolor{w h i t e}{}}^{+} 0} - \frac{2}{\delta 4} + 1 \textcolor{w h i t e}{\text{dd") ->color(white)("dd")kcolor(white)("dd")->color(white)("dd}} - \infty + 1 = - \infty$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Horizontal Asymptot } \to + \infty}$
$\frac{x - 2}{x - 4}$
As $x > 0$ becomes increasing greater and greater the influences of the -2 and -4 become less and less significant. This continues until we have
${\lim}_{x \to + \infty} \frac{x - 2}{x - 4} \textcolor{w h i t e}{\text{dd")->color(white)("dd")kcolor(white)("dd") ->color(white)("dd}} \frac{\infty}{\infty} = + 1$
As $x < 0$ becomes increasing less and less the influences of the -2 and -4 become less and less significant. This continues until we have
${\lim}_{x \to - \infty} \frac{x - 2}{x - 4} \textcolor{w h i t e}{\text{dd")->color(white)("dd")kcolor(white)("dd") ->color(white)("dd}} \frac{- \infty}{- \infty} = + 1$ |
A symbol for the set of rational numbers
The rational numbers (${\displaystyle \mathbb {Q} }$ are included in the real numbers (${\displaystyle \mathbb {R} }$), while themselves including the integers (${\displaystyle \mathbb {Z} }$), which in turn include the natural numbers (${\displaystyle \mathbb {N} }$)
In mathematics, a rational number is a number that can be expressed as the quotient or fraction ${\displaystyle {\tfrac {p}{q))}$ of two integers, a numerator p and a non-zero denominator q.[1] For example, ${\displaystyle {\tfrac {-3}{7))}$ is a rational number, as is every integer (e.g. 5 = 5/1). The set of all rational numbers, also referred to as "the rationals",[2] the field of rationals[3] or the field of rational numbers is usually denoted by boldface Q,[4] or blackboard bold ${\displaystyle \mathbb {Q} .}$[5]
A rational number is a real number. The real numbers that are rational are those whose decimal expansion either terminates after a finite number of digits (example: 3/4 = 0.75), or eventually begins to repeat the same finite sequence of digits over and over (example: 9/44 = 0.20454545...).[6] This statement is true not only in base 10, but also in every other integer base, such as the binary and hexadecimal ones (see Repeating decimal § Extension to other bases).
A real number that is not rational is called irrational.[7] Irrational numbers include square root of 2 (${\displaystyle {\sqrt {2))}$), π, e, and the golden ratio (φ). Since the set of rational numbers is countable, and the set of real numbers is uncountable, almost all real numbers are irrational.[1]
Rational numbers can be formally defined as equivalence classes of pairs of integers (p, q) with q ≠ 0, using the equivalence relation defined as follows:
${\displaystyle (p_{1},q_{1})\sim (p_{2},q_{2})\iff p_{1}q_{2}=p_{2}q_{1}.}$
The fraction ${\displaystyle {\tfrac {p}{q))}$ then denotes the equivalence class of (p, q).[8]
Rational numbers together with addition and multiplication form a field which contains the integers, and is contained in any field containing the integers. In other words, the field of rational numbers is a prime field, and a field has characteristic zero if and only if it contains the rational numbers as a subfield. Finite extensions of ${\displaystyle \mathbb {Q} }$ are called algebraic number fields, and the algebraic closure of ${\displaystyle \mathbb {Q} }$ is the field of algebraic numbers.[9]
In mathematical analysis, the rational numbers form a dense subset of the real numbers. The real numbers can be constructed from the rational numbers by completion, using Cauchy sequences, Dedekind cuts, or infinite decimals (see Construction of the real numbers).
## Terminology
The term rational in reference to the set ${\displaystyle \mathbb {Q} }$ refers to the fact that a rational number represents a ratio of two integers. In mathematics, "rational" is often used as a noun abbreviating "rational number". The adjective rational sometimes means that the coefficients are rational numbers. For example, a rational point is a point with rational coordinates (i.e., a point whose coordinates are rational numbers); a rational matrix is a matrix of rational numbers; a rational polynomial may be a polynomial with rational coefficients, although the term "polynomial over the rationals" is generally preferred, to avoid confusion between "rational expression" and "rational function" (a polynomial is a rational expression and defines a rational function, even if its coefficients are not rational numbers). However, a rational curve is not a curve defined over the rationals, but a curve which can be parameterized by rational functions.
### Etymology
Although nowadays rational numbers are defined in terms of ratios, the term rational is not a derivation of ratio. On the opposite, it is ratio that is derived from rational: the first use of ratio with its modern meaning was attested in English about 1660,[10] while the use of rational for qualifying numbers appeared almost a century earlier, in 1570.[11] This meaning of rational came from the mathematical meaning of irrational, which was first used in 1551, and it was used in "translations of Euclid (following his peculiar use of ἄλογος)".[12][13]
This unusual history originated in the fact that ancient Greeks "avoided heresy by forbidding themselves from thinking of those [irrational] lengths as numbers".[14] So such lengths were irrational, in the sense of illogical, that is "not to be spoken about" (ἄλογος in Greek).[15]
This etymology is similar to that of imaginary numbers and real numbers.
## Arithmetic
### Irreducible fraction
Every rational number may be expressed in a unique way as an irreducible fraction ${\displaystyle {\tfrac {a}{b)),}$ where a and b are coprime integers and b > 0. This is often called the canonical form of the rational number.
Starting from a rational number ${\displaystyle {\tfrac {a}{b)),}$ its canonical form may be obtained by dividing a and b by their greatest common divisor, and, if b < 0, changing the sign of the resulting numerator and denominator.
### Embedding of integers
Any integer n can be expressed as the rational number ${\displaystyle {\tfrac {n}{1)),}$ which is its canonical form as a rational number.
### Equality
${\displaystyle {\frac {a}{b))={\frac {c}{d))}$ if and only if ${\displaystyle ad=bc}$
If both fractions are in canonical form, then:
${\displaystyle {\frac {a}{b))={\frac {c}{d))}$ if and only if ${\displaystyle a=c}$ and ${\displaystyle b=d}$[8]
### Ordering
If both denominators are positive (particularly if both fractions are in canonical form):
${\displaystyle {\frac {a}{b))<{\frac {c}{d))}$ if and only if ${\displaystyle ad
On the other hand, if either denominator is negative, then each fraction with a negative denominator must first be converted into an equivalent form with a positive denominator—by changing the signs of both its numerator and denominator.[8]
Two fractions are added as follows:
${\displaystyle {\frac {a}{b))+{\frac {c}{d))={\frac {ad+bc}{bd)).}$
If both fractions are in canonical form, the result is in canonical form if and only if b, d are coprime integers.[8][16]
### Subtraction
${\displaystyle {\frac {a}{b))-{\frac {c}{d))={\frac {ad-bc}{bd)).}$
If both fractions are in canonical form, the result is in canonical form if and only if b, d are coprime integers.[16]
### Multiplication
The rule for multiplication is:
${\displaystyle {\frac {a}{b))\cdot {\frac {c}{d))={\frac {ac}{bd)).}$
where the result may be a reducible fraction—even if both original fractions are in canonical form.[8][16]
### Inverse
Every rational number ${\displaystyle {\tfrac {a}{b))}$ has an additive inverse, often called its opposite,
${\displaystyle -\left({\frac {a}{b))\right)={\frac {-a}{b)).}$
If ${\displaystyle {\tfrac {a}{b))}$ is in canonical form, the same is true for its opposite.
A nonzero rational number ${\displaystyle {\tfrac {a}{b))}$ has a multiplicative inverse, also called its reciprocal,
${\displaystyle \left({\frac {a}{b))\right)^{-1}={\frac {b}{a)).}$
If ${\displaystyle {\tfrac {a}{b))}$ is in canonical form, then the canonical form of its reciprocal is either ${\displaystyle {\tfrac {b}{a))}$ or ${\displaystyle {\tfrac {-b}{-a)),}$ depending on the sign of a.
### Division
If b, c, d are nonzero, the division rule is
${\displaystyle {\frac {\,{\dfrac {a}{b))\,}{\dfrac {c}{d))}={\frac {ad}{bc)).}$
Thus, dividing ${\displaystyle {\tfrac {a}{b))}$ by ${\displaystyle {\tfrac {c}{d))}$ is equivalent to multiplying ${\displaystyle {\tfrac {a}{b))}$ by the reciprocal of ${\displaystyle {\tfrac {c}{d)):}$[16]
${\displaystyle {\frac {ad}{bc))={\frac {a}{b))\cdot {\frac {d}{c)).}$
### Exponentiation to integer power
If n is a non-negative integer, then
${\displaystyle \left({\frac {a}{b))\right)^{n}={\frac {a^{n)){b^{n))}.}$
The result is in canonical form if the same is true for ${\displaystyle {\tfrac {a}{b)).}$ In particular,
${\displaystyle \left({\frac {a}{b))\right)^{0}=1.}$
If a ≠ 0, then
${\displaystyle \left({\frac {a}{b))\right)^{-n}={\frac {b^{n)){a^{n))}.}$
If ${\displaystyle {\tfrac {a}{b))}$ is in canonical form, the canonical form of the result is ${\displaystyle {\tfrac {b^{n)){a^{n))))$ if a > 0 or n is even. Otherwise, the canonical form of the result is ${\displaystyle {\tfrac {-b^{n)){-a^{n))}.}$
## Continued fraction representation
Main article: Continued fraction
A finite continued fraction is an expression such as
${\displaystyle a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{\ddots +{\cfrac {1}{a_{n))))))))},}$
where an are integers. Every rational number ${\displaystyle {\tfrac {a}{b))}$ can be represented as a finite continued fraction, whose coefficients an can be determined by applying the Euclidean algorithm to (a, b).
## Other representations
• common fraction: ${\displaystyle {\tfrac {8}{3))}$
• mixed numeral: ${\displaystyle 2{\tfrac {2}{3))}$
• repeating decimal using a vinculum: ${\displaystyle 2.{\overline {6))}$
• repeating decimal using parentheses: ${\displaystyle 2.(6)}$
• continued fraction using traditional typography: ${\displaystyle 2+{\tfrac {1}{1))+{\tfrac {1}{2))}$
• continued fraction in abbreviated notation: ${\displaystyle [2;1,2]}$
• Egyptian fraction: ${\displaystyle 2+{\tfrac {1}{2))+{\tfrac {1}{6))}$
• prime power decomposition: ${\displaystyle 2^{3}\times 3^{-1))$
• quote notation: ${\displaystyle 3'6}$
are different ways to represent the same rational value.
## Formal construction
A diagram showing a representation of the equivalent classes of pairs of integers
The rational numbers may be built as equivalence classes of ordered pairs of integers.[8][16]
More precisely, let ${\displaystyle (\mathbb {Z} \times (\mathbb {Z} \setminus \{0\}))}$ be the set of the pairs (m, n) of integers such n ≠ 0. An equivalence relation is defined on this set by
${\displaystyle (m_{1},n_{1})\sim (m_{2},n_{2})\iff m_{1}n_{2}=m_{2}n_{1}.}$[8][16]
Addition and multiplication can be defined by the following rules:
${\displaystyle (m_{1},n_{1})+(m_{2},n_{2})\equiv (m_{1}n_{2}+n_{1}m_{2},n_{1}n_{2}),}$
${\displaystyle (m_{1},n_{1})\times (m_{2},n_{2})\equiv (m_{1}m_{2},n_{1}n_{2}).}$[8]
This equivalence relation is a congruence relation, which means that it is compatible with the addition and multiplication defined above; the set of rational numbers ${\displaystyle \mathbb {Q} }$ is the defined as the quotient set by this equivalence relation, ${\displaystyle (\mathbb {Z} \times (\mathbb {Z} \ \{0\}))/\sim ,}$ equipped with the addition and the multiplication induced by the above operations. (This construction can be carried out with any integral domain and produces its field of fractions.)[8]
The equivalence class of a pair (m, n) is denoted ${\displaystyle {\tfrac {m}{n)).}$ Two pairs (m1, n1) and (m2, n2) belong to the same equivalence class (that is are equivalent) if and only if
${\displaystyle m_{1}n_{2}=m_{2}n_{1}.}$
This means that
${\displaystyle {\frac {m_{1)){n_{1))}={\frac {m_{2)){n_{2))))$
if and only if[8][16]
${\displaystyle m_{1}n_{2}=m_{2}n_{1}.}$
Every equivalence class ${\displaystyle {\tfrac {m}{n))}$ may be represented by infinitely many pairs, since
${\displaystyle \cdots ={\frac {-2m}{-2n))={\frac {-m}{-n))={\frac {m}{n))={\frac {2m}{2n))=\cdots .}$
Each equivalence class contains a unique canonical representative element. The canonical representative is the unique pair (m, n) in the equivalence class such that m and n are coprime, and n > 0. It is called the representation in lowest terms of the rational number.
The integers may be considered to be rational numbers identifying the integer n with the rational number ${\displaystyle {\tfrac {n}{1)).}$
A total order may be defined on the rational numbers, that extends the natural order of the integers. One has
${\displaystyle {\frac {m_{1)){n_{1))}\leq {\frac {m_{2)){n_{2))))$
If
{\displaystyle {\begin{aligned}&(n_{1}n_{2}>0\quad {\text{and))\quad m_{1}n_{2}\leq n_{1}m_{2})\\&\qquad {\text{or))\\&(n_{1}n_{2}<0\quad {\text{and))\quad m_{1}n_{2}\geq n_{1}m_{2}).\end{aligned))}
## Properties
The set ${\displaystyle \mathbb {Q} }$ of all rational numbers, together with the addition and multiplication operations shown above, forms a field.[8]
${\displaystyle \mathbb {Q} }$ has no field automorphism other than the identity. (A field automorphism must fix 0 and 1; as it must fix the sum and the difference of two fixed elements, it must fix every integer; as it must fix the quotient of two fixed elements, it must fix every rational number, and is thus the identity.)
${\displaystyle \mathbb {Q} }$ is a prime field, which is a field that has no subfield other than itself.[17] The rationals are the smallest field with characteristic zero. Every field of characteristic zero contains a unique subfield isomorphic to ${\displaystyle \mathbb {Q} .}$
With the order defined above, ${\displaystyle \mathbb {Q} }$ is an ordered field[16] that has no subfield other than itself, and is the smallest ordered field, in the sense that every ordered field contains a unique subfield isomorphic to ${\displaystyle \mathbb {Q} .}$
${\displaystyle \mathbb {Q} }$ is the field of fractions of the integers ${\displaystyle \mathbb {Z} .}$[18] The algebraic closure of ${\displaystyle \mathbb {Q} ,}$ i.e. the field of roots of rational polynomials, is the field of algebraic numbers.
The rationals are a densely ordered set: between any two rationals, there sits another one, and, therefore, infinitely many other ones.[8] For example, for any two fractions such that
${\displaystyle {\frac {a}{b))<{\frac {c}{d))}$
(where ${\displaystyle b,d}$ are positive), we have
${\displaystyle {\frac {a}{b))<{\frac {a+c}{b+d))<{\frac {c}{d)).}$
Any totally ordered set which is countable, dense (in the above sense), and has no least or greatest element is order isomorphic to the rational numbers.[19]
### Countability
Illustration of the countability of the positive rationals
The set of all rational numbers is countable, as is illustrated in the figure to the right. As a rational number can be expressed as a ratio of two integers, it is possible to assign two integers to any point on a square lattice as in a Cartesian coordinate system, such that any grid point corresponds to a rational number. This method, however, exhibits a form of redundancy, as several different grid points will correspond to the same rational number; these are highlighted in red on the provided graphic. An obvious example can be seen in the line going diagonally towards the bottom right; such ratios will always equal 1, as any non-zero number divided by itself will always equal one.
It is possible to generate all of the rational numbers without such redundancies: examples include the Calkin–Wilf tree and Stern–Brocot tree.
As the set of all rational numbers is countable, and the set of all real numbers (as well as the set of irrational numbers) is uncountable, the set of rational numbers is a null set, that is, almost all real numbers are irrational, in the sense of Lebesgue measure.
## Real numbers and topological properties
The rationals are a dense subset of the real numbers; every real number has rational numbers arbitrarily close to it.[8] A related property is that rational numbers are the only numbers with finite expansions as regular continued fractions.[20]
In the usual topology of the real numbers, the rationals are neither an open set nor a closed set.[21]
By virtue of their order, the rationals carry an order topology. The rational numbers, as a subspace of the real numbers, also carry a subspace topology. The rational numbers form a metric space by using the absolute difference metric ${\displaystyle d(x,y)=|x-y|,}$ and this yields a third topology on ${\displaystyle \mathbb {Q} .}$ All three topologies coincide and turn the rationals into a topological field. The rational numbers are an important example of a space which is not locally compact. The rationals are characterized topologically as the unique countable metrizable space without isolated points. The space is also totally disconnected. The rational numbers do not form a complete metric space, and the real numbers are the completion of ${\displaystyle \mathbb {Q} }$ under the metric ${\displaystyle d(x,y)=|x-y|}$ above.[16]
In addition to the absolute value metric mentioned above, there are other metrics which turn ${\displaystyle \mathbb {Q} }$ into a topological field:
Let p be a prime number and for any non-zero integer a, let ${\displaystyle |a|_{p}=p_{-n},}$ where pn is the highest power of p dividing a.
In addition set ${\displaystyle |0|_{p}=0.}$ For any rational number ${\displaystyle {\frac {a}{b)),}$ we set
${\displaystyle \left|{\frac {a}{b))\right|_{p}={\frac {|a|_{p)){|b|_{p))}.}$
Then
${\displaystyle d_{p}(x,y)=|x-y|_{p))$
defines a metric on ${\displaystyle \mathbb {Q} .}$[22]
The metric space ${\displaystyle (\mathbb {Q} ,d_{p})}$ is not complete, and its completion is the p-adic number field ${\displaystyle \mathbb {Q} _{p}.}$ Ostrowski's theorem states that any non-trivial absolute value on the rational numbers ${\displaystyle \mathbb {Q} }$ is equivalent to either the usual real absolute value or a p-adic absolute value.
Number systems
Complex ${\displaystyle :\;\mathbb {C} }$
Real ${\displaystyle :\;\mathbb {R} }$
Rational ${\displaystyle :\;\mathbb {Q} }$
Integer ${\displaystyle :\;\mathbb {Z} }$
Natural ${\displaystyle :\;\mathbb {N} }$
Zero: 0 One: 1 Prime numbers Composite numbers
Negative integers
Imaginary
## References
1. ^ a b Rosen, Kenneth (2007). Discrete Mathematics and its Applications (6th ed.). New York, NY: McGraw-Hill. pp. 105, 158–160. ISBN 978-0-07-288008-3.
2. ^ Lass, Harry (2009). Elements of Pure and Applied Mathematics (illustrated ed.). Courier Corporation. p. 382. ISBN 978-0-486-47186-0. Extract of page 382
3. ^ Robinson, Julia (1996). The Collected Works of Julia Robinson. American Mathematical Soc. p. 104. ISBN 978-0-8218-0575-6. Extract of page 104
4. ^ It was thus denoted in 1895 by Giuseppe Peano after quoziente, Italian for "quotient",[citation needed]
5. ^ It first appeared in Bourbaki's Algèbre.
6. ^ "Rational number". Encyclopedia Britannica. Retrieved 2020-08-11.
7. ^ Weisstein, Eric W. "Rational Number". mathworld.wolfram.com. Retrieved 2020-08-11.
8. Biggs, Norman L. (2002). Discrete Mathematics. India: Oxford University Press. pp. 75–78. ISBN 978-0-19-871369-2.
9. ^ Gilbert, Jimmie; Linda, Gilbert (2005). Elements of Modern Algebra (6th ed.). Belmont, CA: Thomson Brooks/Cole. pp. 243–244. ISBN 0-534-40264-X.
10. ^ Oxford English Dictionary (2nd ed.). Oxford University Press. 1989. Entry ratio, n., sense 2.a.
11. ^ Oxford English Dictionary (2nd ed.). Oxford University Press. 1989. Entry rational, a. (adv.) and n.1, sense 5.a.
12. ^ Oxford English Dictionary (2nd ed.). Oxford University Press. 1989. Entry irrational, a. and n., sense 3.
13. ^ Shor, Peter (2017-05-09). "Does rational come from ratio or ratio come from rational". Stack Exchange. Retrieved 2021-03-19.
14. ^ Coolman, Robert (2016-01-29). "How a Mathematical Superstition Stultified Algebra for Over a Thousand Years". Retrieved 2021-03-20.
15. ^ Kramer, Edna (1983). The Nature and Growth of Modern Mathematics. Princeton University Press. p. 28.
16. "Fraction - Encyclopedia of Mathematics". encyclopediaofmath.org. Retrieved 2021-08-17.
17. ^ Sūgakkai, Nihon (1993). Encyclopedic Dictionary of Mathematics, Volume 1. London, England: MIT Press. p. 578. ISBN 0-2625-9020-4.
18. ^ Bourbaki, N. (2003). Algebra II: Chapters 4 - 7. Springer Science & Business Media. p. A.VII.5.
19. ^ Giese, Martin; Schönegge, Arno (December 1995). Any two countable densely ordered sets without endpoints are isomorphic - a formal proof with KIV (PDF) (Technical report). Retrieved 17 August 2021.
20. ^ Anthony Vazzana; David Garth (2015). Introduction to Number Theory (2nd, revised ed.). CRC Press. p. 1. ISBN 978-1-4987-1752-6. Extract of page 1
21. ^ Richard A. Holmgren (2012). A First Course in Discrete Dynamical Systems (2nd, illustrated ed.). Springer Science & Business Media. p. 26. ISBN 978-1-4419-8732-7. Extract of page 26
22. ^ Weisstein, Eric W. "p-adic Number". mathworld.wolfram.com. Retrieved 2021-08-17. |
# T21 - Editorial
[Contest] T21
Author: Nandishwar Garg
SIMPLE
None
### PROBLEM:
You are given two large numbers m and n. You have to multiply m and n then product is needed to be divided by 3. What will be the remainder after division?
### EXPLANATION:
To solve the problem, just go through these 3 steps:
Step 1: The given numbers m and n are needed to be multiplied.
Step 2: The product which you get after multiplying the numbers is needed to be divided by 3.
Step 3: Print the remainder you get.
### AUTHOR’S SOLUTION:
Author’s solution can be found here
Editorial tag shoud be removed from this post. In my opinion this is not an editorial. Its just a duplicate post repeating question again.
### Misleading in the name of editorial.
No soln. No proper explanation. Nothing.
### Oh really !!
well i don't know how to multiply 2 numbers......XD:)
I think they are asking for the solution ! under the editorial tag… XD lol Xd :
I guess the concept here is that the program should cope when the numbers are too big to multiply. OR perhaps even hold as numbers. Python mostly laughs at those sort of restrictions, but not all are so blessed.
So: if the numbers are inconveniently large for your environment, it’s still easy to find the amswer with modular arithmetic. We can simplify by a number of stages:
Firstly, if we find the remainder of the two input numbers after division by 3, we can multiply those two results together instead of the original numbers, and the answer will have the same remainder. Much easier.
Why does this work? Well, if our two given numbers are a and b, and they have remainders respectively of c and d, this means that a = 3v + c and b = 3w + d. We don’t need to know what v and w are. Then multiplying we get ab = (3v + c)(3w + d) = 9vw + 3vd + 3wc + cd and since 9vw + 3vd + 3wc is clearly divisible by 3, ab and cd have the same remainders.
Now finding out what c and d are is our second stage of simplification. If you know the divisibility test for 3, you know that you can just add up the digits to find out whether a number is divisible by 3 or not. This works for much the same reason above, plus the fact that the remainder of 10 when divided by 3 is 1. This means that k and 10k have the same remainder on division by 3, and likewise 100k, 1000k, 10000k etc. So every digit in the number, no matter how many powers of ten it has been multiplied by to get to its place value, has the same remainder on division by 3 as just that digit alone. You can check this; 40 has the same remainder as 4, 500 has the same remainder as 5, 2000 has the same remainder as 2, etc.
So you can just sum the digits in each of a and b, find their remainders on division by 3, and that’s your c and d above.
If preferred you can bite off larger digestible pieces to evaluate interim remainders, which still works for the same reason - the larger pieces have the same remainder as the sum of their digits.
A more interesting challenge is to find the same but with remainders after division by 7, assuming that the numbers are inconveniently large. The trick then is to find the right size pieces to bite off, and in the right place. Working with single digits is no longer the right approach.
1 Like
exactly…it is not an editorial anyway …XD |
# Notes on Multivariate Data Analysis via Matrix Decomposition
### 2019-09-30
These are the notes taken on my master course Multivariate Data Analysis via Matrix Decomposition. If you’re confused with the course name, you can think of this as a statistical course on unsupervised learning. $\newcommand{1}[1]{\unicode{x1D7D9}_{\{#1\}}}\newcommand{Corr}{\text{Corr}}\newcommand{E}{\text{E}}\newcommand{Cov}{\text{Cov}}\newcommand{Var}{\text{Var}}\newcommand{span}{\text{span}}\newcommand{bs}{\boldsymbol}\newcommand{R}{\mathbb{R}}\newcommand{rank}{\text{rank}}\newcommand{\norm}[1]{\left\lVert#1\right\rVert}\newcommand{diag}{\text{diag}}\newcommand{tr}{\text{tr}}\newcommand{braket}[1]{\left\langle#1\right\rangle}\newcommand{C}{\mathbb{C}}$
# Notations
First let’s give some standard notations used in this course. Let’s assume no prior knowledge in linear algebra and start from matrix multiplication.
## Matrix Multiplication
We denote a matrix $\mathbf{A}\in\R^{m\times n}$, with its entries defined as $[a_{ij}]_{i,j=1}^{m,n}$. Similarly, we define $\mathbf{B}=[b_{jk}]_{j,k=1}^{n,p}$ and thus the multiplication is defined as $\mathbf{AB} = [\sum_{j=1}^n a_{ij}b_{jk}]_{i,k=1}^{n,p}$, which can also be represented in three other ways:
• vector form, using $\mathbf{a}$ and $\mathbf{b}$
• a matrix of products of $A$ and $\mathbf{b}$
• a matrix of products of $\mathbf{a}$ and $\mathbf{B}$
A special example of such representation: let’s assume
$$\mathbf{A}=[\mathbf{a}_1,\mathbf{a}_2,\ldots,\mathbf{a}_n]\in\R^{m\times n}\text{ and } \mathbf{D} = \diag(d_1,d_2,\ldots,d_n) \in\R^{n\times n},$$
then we have right away $\mathbf{AD}=[\mathbf{a}_id_i]_{i=1}^n$.
Exercise With multiplication we care ranks of matrices. There is a quick conclusion: If $\mathbf{x}\neq \bs{0}, \mathbf{y}\neq \bs{0}$, then $\rank(\bs{yx'})=1$. Conversely, if $\rank(\mathbf{A})=1$, then $\exists\ \mathbf{x}\neq \bs{0}, \mathbf{y}\neq \bs{0}$ s.t. $\bs{xy'}=\mathbf{A}$. Prove it.
## Norms
There are two types of norms in this course we consider:
• (Euclidean) We define the $l^1$-norm as $\norm{\mathbf{x}}_1 = \sum_{i=1}^n |x_i|$, define $l^2$-norm as $\norm{\mathbf{x}}_2 = \sqrt{\mathbf{x'x}}$, define $l^{\infty}$-norm as $\norm{\mathbf{x}}_{\infty} = \max_{1\le i \le n}\{|x_i|\}$, and define the Mahalanobis norm as $\norm{\mathbf{x}}_A = \sqrt{\mathbf{x'Ax}}$.
• (Frobenius) We define the Frobenius norm of a matrix as $\norm{\mathbf{A}}_F=\sqrt{\sum_{i=1}^m\sum_{j=1}^n a_{ij}^2}$. The spectral 2-norm of a matrix is defined as $\norm{\mathbf{A}}_2=\max_{\mathbf{x}\neq \bs{0}} \norm{\mathbf{Ax}}_2 / \norm{\mathbf{x}}_2$.
Properties of these norms:
• $\norm{\mathbf{v}}=0$ iff. $\mathbf{v}=\bs{0}$.
• $\norm{\alpha \mathbf{v}} = |\alpha|\cdot\norm{\mathbf{v}}$ for any $\alpha\in\R$ and any $\mathbf{v}\in\mathcal{V}$.
• (Triangular Inequality) $\norm{\mathbf{u} + \mathbf{v}} \le \norm{\mathbf{u}} + \norm{\mathbf{v}}$ for any $\mathbf{u}, \mathbf{v}\in\mathcal{V}$.
• (Submultiplicative) $\norm{\mathbf{AB}}\le \norm{\mathbf{A}}\cdot \norm{\mathbf{B}}$ for every formable matrices $\mathbf{A}$ and $\mathbf{B}$.
Exercise Try to prove them for Euclidean 2-norm, Frobenius norm and spectral 2-norm.
## Inner Products
There are two types of inner products we consider:
• (Euclidean) We define the inner product of vectors $\mathbf{x},\mathbf{y}\in\R^n$ as $\mathbf{x'y}=\sum_{i=1}^n x_iy_i$.
• (Frobenius) We define the inner product of matrices $\mathbf{A},\mathbf{B}\in\R^{m\times n}$ as $\braket{\mathbf{A},\mathbf{B}}=\tr(\mathbf{A'B})=\sum_{i=1}^m\sum_{j=1}^n a_{ij}b_{ij}$.
A famous inequality related to these inner products is the Cauchy-Schwarz inequality, which states
• (Euclidean) $|\mathbf{x'y}|\le \norm{\mathbf{x}}_2\cdot\norm{\mathbf{y}}_2$ for any $\bs{x,y}\in\R^n$.
• (Frobenius) $|\braket{\mathbf{A},\mathbf{B}}|\le\norm{\mathbf{A}}_F\cdot\norm{\mathbf{B}}_F$ for any $\mathbf{A},\mathbf{B}\in\R^{m\times n}$.
# Eigenvalue Decomposition (EVD)
The first matrix decomposition we’re gonna talk about is the eigenvalue decomposition.
## Eigenvalues and Eigenvectors
For square matrix $\mathbf{A}\in\R^{n\times n}$, if $\bs{0}\neq \mathbf{x}\in\C^n$ and $\lambda\in\C$ is s.t. $\mathbf{Ax} = \lambda\mathbf{x}$, then $\lambda$ is called an engenvalue of $\mathbf{A}$ and $\mathbf{x}$ is called the $\lambda$-engenvector of $\mathbf{A}$.
Ideally, we want a matrix to have $n$ eigenvectors and $n$ corresponding eigenvectors, linearly independent to each other. This is not always true.
## Existence of EVD
Theorem $\mathbf{A}\in\R^{n\times n}$ have $n$ eigenvalues iff. there exists an invertible $\mathbf{X}\in\R^{n\times n}$ s.t. $\mathbf{X}^{-1}\mathbf{A}\mathbf{X}=\bs{\Lambda}$, i.e. $\mathbf{A}$ is diagonizable. This gives $\mathbf{A}=\mathbf{X}\bs{\Lambda}\mathbf{X}^{-1}$, which is called the eigenvalue decomposition (EVD).
Theorem (Spectral Theorem for Symmetric Matrices) For symmetric matrix $\mathbf{A}\in\R^{n\times n}$ there always exists an orthogonal matrix $\mathbf{Q}$, namely $\mathbf{Q}'\mathbf{Q}=\mathbf{I}$, that gives
$$\mathbf{A}=\mathbf{Q}\bs{\Lambda}\mathbf{Q}’ = \sum_{i=1}^n \lambda_i \mathbf{q}_i \mathbf{q}_i'$$
where $\mathbf{q}$ are column vectors of $\mathbf{Q}$. This is called the symmetric EVD, aka. $\mathbf{A}$ being orthogonally diagonalizable.
## Properties of EVD
We have several properties following the second theorem above. For all $i=1,2,\ldots, n$
• $\mathbf{A}\mathbf{q}_i = \lambda_i \mathbf{q}_i$ (can be proved using $\mathbf{Q}^{-1}=\mathbf{Q}'$)
• $\norm{\mathbf{q}_i}_2=1$ (can be proved using $\mathbf{QQ}'=\mathbf{I}$)
The second theorem above can also be represented as
Theorem If $\mathbf{A}=\mathbf{A}'$, then $\mathbf{A}$ has $n$ orthogonal eigenvectors.
# Singular Value Decomposition (SVD)
For general matrices, we have singular value decomposition.
## Definition
The most famous form of SVD is define as
$$\mathbf{A} = \mathbf{U} \bs{\Sigma} \mathbf{V}'$$
where $\mathbf{A}\in\R^{m\times n}$, $\mathbf{U}\in\R^{m\times m}$, $\bs{\Sigma}\in\R^{m\times n}$ and $\mathbf{V}\in\R^{n\times n}$. Specifically, both $\mathbf{U}$ and $\mathbf{V}$ are orthogonal (i.e. $\mathbf{U}'\mathbf{U}=\mathbf{I}$, same for $\mathbf{V}$) and $\bs{\Sigma}$ is diagonal. Usually, we choose the singular values to be non-decreasing, namely
$$\bs{\Sigma}=\diag(\sigma_1,\sigma_2,\ldots,\sigma_{\min{m,n}})\quad\text{where}\quad \sigma_1 \ge \sigma_2 \ge \cdots \ge \sigma_{\min{m,n}}.$$
## Terminology
Here we define a list of terms that’ll be used from time to time:
• (SVD) $\mathbf{A} = \mathbf{U} \bs{\Sigma} \mathbf{V}'$.
• (Left Singular Vectors) Columns of $\mathbf{U}$.
• (Right Singular Vectors) Columns of $\mathbf{V}$.
• (Singular Values) Diagonal entries of $\bs{\Sigma}$.
## Three Forms of SVD
Besides the regular SVD given above, we have the outer product SVD:
$$\mathbf{A} = \sum_{i=1}^{\min{m,n}}!!!\sigma_i \mathbf{u}_i \mathbf{v}_i'$$
and condensed SVD:
$$\mathbf{A} = \mathbf{U}_r\bs{\Sigma}_r\mathbf{V}_r'$$
where $r=\rank(\mathbf{A})$ is also the number of non-zero singular values. In this form, we have $\bs{\Sigma}_r\in\R^{r\times r}$ with proper chunked $\mathbf{U}_r$ and $\mathbf{V}_r$.
## Existence of SVD
Theorem (Existence of SVD) Let $\mathbf{A}\in\R^{m\times n}$ and $r=\rank(\mathbf{A})$. Then $\exists\ \mathbf{U}_r\in\R^{m\times r}$, $\mathbf{V}_r\in\R^{n\times r}$ and $\bs{\Sigma}_r\in\R^{r\times r}$ s.t. $\mathbf{A} = \mathbf{U}_r\bs{\Sigma}_r\mathbf{V}_r'$ where $\mathbf{U}_r$ and $\mathbf{V}_r$ are orthogonal and $\bs{\Sigma}_r$ is diagonal. This means condensed SVD exists and therefore the rest two forms.
Proof. Define symmetric $\mathbf{W}\in\R^{(m+n)\times(m+n)}$ as
$$\mathbf{W} = \begin{bmatrix} \bs{0} & \mathbf{A} \ \mathbf{A}’ & \bs{0} \end{bmatrix}$$
which has an orthogonal EVD as $\mathbf{W} = \mathbf{Z}\bs{\Lambda}\mathbf{Z}'$ where $\mathbf{Z}'\mathbf{Z}=\mathbf{I}$. Now, assume $\mathbf{z}\in\R^{m+n}$ is an eigenvector of $\mathbf{W}$ corresponding to $\lambda$, then $\mathbf{W}\mathbf{z} = \lambda \mathbf{z}$. Denote the first $m$ entries of $\mathbf{z}$ as $\mathbf{x}$ and the rest $\mathbf{y}$, which gives
$$\begin{bmatrix} \bs{0} & \mathbf{A}\ \mathbf{A}’ & \bs{0} \end{bmatrix} \begin{bmatrix} \mathbf{x} \ \mathbf{y} \end{bmatrix} = \lambda \begin{bmatrix} \mathbf{x} \ \mathbf{y} \end{bmatrix} \Rightarrow \begin{cases} \mathbf{Ay} = \lambda \mathbf{x},\ \mathbf{A}’\mathbf{x} = \lambda \mathbf{y}. \end{cases}$$
Using this results
$$\begin{bmatrix} \bs{0} & \mathbf{A}\ \mathbf{A}’ & \bs{0} \end{bmatrix} \begin{bmatrix} \mathbf{x} \ -\mathbf{y} \end{bmatrix} = \begin{bmatrix} -\mathbf{Ay} \ \mathbf{A}’\mathbf{y} \end{bmatrix} = \begin{bmatrix} -\lambda \mathbf{x}\ \lambda \mathbf{y} \end{bmatrix} = -\lambda\begin{bmatrix} \mathbf{x}\ -\mathbf{y} \end{bmatrix}$$
which means $-\lambda$ is also an engenvalue of $\mathbf{W}$. Hence, we know
\begin{align} \mathbf{W} &= \mathbf{Z}\bs{\Lambda}\mathbf{Z}’ = \mathbf{Z}_r\bs{\Lambda}_r\mathbf{Z}_r’\ &= \begin{bmatrix} \mathbf{X} & \mathbf{X}\ \mathbf{Y} & -\mathbf{Y} \end{bmatrix} \begin{bmatrix} \bs{\Sigma} & \bs{0}\ \bs{0} & -\bs{\Sigma} \end{bmatrix} \begin{bmatrix} \mathbf{X} & \mathbf{X}\ \mathbf{Y} & -\mathbf{Y} \end{bmatrix}’\ &= \begin{bmatrix} \bs{0} & \mathbf{X}\bs{\Sigma}\mathbf{Y}’\ \mathbf{Y}\bs{\Sigma}\mathbf{X}’ & \bs{0} \end{bmatrix}. \end{align}
Therefore, we conclude $\mathbf{A}=\mathbf{X}\bs{\Sigma}\mathbf{Y}'$ where now all we need to prove is the orthogonality of $\mathbf{X}$ and $\mathbf{Y}$. Let’s take a look at $\mathbf{z}=(\mathbf{x},\mathbf{y})$ we just defined. Let
$$\norm{\mathbf{z}}=\mathbf{z}’\mathbf{z}=\mathbf{x}’\mathbf{x} + \mathbf{y}’\mathbf{y} = 2.$$
From orthogonality of eigenvectors corresponding to different eigenvalues, we also know
$$\mathbf{z}’\bar{\mathbf{z}} = \mathbf{x}’\mathbf{x} - \mathbf{y}’\mathbf{y} = 0$$
which altogether gives $\norm{\mathbf{x}}=\norm{\mathbf{y}}=1$.Q.E.D.
## How to Calculate SVD
Three steps to calculate the SVD of $\mathbf{A}\in\R^{m\times n}$:
• $\bs{\Sigma}$: calculate eigenvalues of $\mathbf{AA}'$, then let $\bs{\Sigma}=\diag\{\sigma_1,\sigma_2,\ldots,\sigma_r\}\in\R^{m\times n}$.
• $\mathbf{U}$: calculate eigenvectors of $\mathbf{AA}'$, then normalize them to norm $1$, then $\mathbf{U}=(\bs{u_1},\bs{u_2},\ldots,\mathbf{u}_m)$.
• $\mathbf{V}$: calculate eigenvectors of $\mathbf{A}'\mathbf{A}$, then normalize them to norm $1$, then $\mathbf{V}=(\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n)$.
Remark: Alternatively you may use formula $\mathbf{U}=\mathbf{AV}\bs{\Sigma}^{-1}\Rightarrow \mathbf{u}_i=\mathbf{Av}_i/\sigma_i$.
## Properties of SVD
There are several characteristics we have about SVD:
• The $\mathbf{W}$ decomposition above.
• The left singular vector $\mathbf{v}$ given by $\mathbf{Au} = \sigma \mathbf{v}$, and the right singular vector $\mathbf{u}$ given by $\mathbf{A}'\mathbf{v} = \sigma \mathbf{u}$.
• Relationship with eigenvectors/eigenvalues…
• of $\mathbf{A}'\mathbf{A}$: $\mathbf{A}'\mathbf{A}\mathbf{u} = \sigma\mathbf{A}'\mathbf{v} = \sigma^2\mathbf{u}$.
• of $\mathbf{AA}'$: $\mathbf{AA}'\mathbf{v} = \sigma\mathbf{A}\mathbf{u} = \sigma^2\mathbf{v}$.
• Frobenius norms (eigenvalues cannot define a norm!):
• $\norm{\mathbf{A}}_F^2 = \sum_{i,j=1}^{m,n}a_{ij}^2=\sum_{i=1}^r\sigma_i^2$.
• $\norm{\mathbf{A}}_2 = \max_{\mathbf{x}\neq 0} \norm{\mathbf{Ax}}_2 / \norm{\mathbf{x}}_2 = \sigma_1$.
Exercise Show how to use SVD to calculate these two norms.
## Applications of SVD
1) Projections. One of the most importance usages of SVD is computing projections. $\mathbf{P}\in\R^{n\times n}$ is a projection matrix iff. $\mathbf{P}^2=\mathbf{P}$. More commonly, we consider orthogonal projection $\mathbf{P}$ that’s also symmetric. Now let’s consider dataset $\mathbf{A}\in\R^{m\times n}$ where we have $n$ observations, each with $m$ dimensions. Suppose we want to project this dataset onto $\mathbf{W}\subseteq\R^m$ that has $k$ dimensions, i.e.
$$\mathbf{W} = \span{\mathbf{q}_1,\mathbf{q}_2,\ldots,\mathbf{q}_k},\quad \mathbf{q}_i’\mathbf{q}_j=\1{i=j}$$
then the projection matrix would be $\mathbf{P}_{\mathbf{W}}=\mathbf{Q}_k\mathbf{Q}_k'$.
Nearest Orthogonal Matrix. The nearest orthogonal matrix of $\mathbf{A}\in\R^{p\times p}$ is given by
$$\min_{\mathbf{X}’\mathbf{X}=\mathbf{I}}\norm{\mathbf{A}-\mathbf{X}}_F$$
which solves if we have optima for
\begin{align} \min_{\mathbf{X}'\mathbf{X}=\mathbf{I}}\norm{\mathbf{A}-\mathbf{X}}_F^2 &= \min_{\mathbf{X}'\mathbf{X}=\mathbf{I}}\tr[(\mathbf{A}-\mathbf{X})'(\mathbf{A}-\mathbf{X})]\\&= \min_{\mathbf{X}'\mathbf{X}=\mathbf{I}}\tr[\mathbf{A}'\mathbf{A} - \mathbf{X}'\mathbf{A} - \mathbf{A}'\mathbf{X} + \mathbf{X}'\mathbf{X}]\\&= \min_{\mathbf{X}'\mathbf{X}=\mathbf{I}} \norm{\mathbf{A}}_F^2 - \tr(\mathbf{A}'\mathbf{X}) - \tr(\mathbf{X}'\mathbf{A}) + \tr(\mathbf{X}'\mathbf{X})\\&= \norm{\mathbf{A}}_F^2 + n - 2\max_{\mathbf{X}'\mathbf{X}=\mathbf{I}} \tr(\mathbf{A}'\mathbf{X}) \end{align}.
Now we try to solve
$$\max_{\mathbf{X}’\mathbf{X}=\mathbf{I}} \tr(\mathbf{A}’\mathbf{X})$$
and claim the solution is given by $\mathbf{X} = \mathbf{U}\mathbf{V}'$ where $\mathbf{U}$ and $\mathbf{V}$ are derived from SVD of $\mathbf{A}$, namely $\mathbf{A} = \mathbf{U}\bs{\Sigma} \mathbf{V}'$. Proof: We know
$$\tr(\mathbf{A}’\mathbf{X}) = \tr(\mathbf{V}\bs{\Sigma}’\mathbf{U}’\mathbf{X}) = \tr(\bs{\Sigma}’\mathbf{U}’\mathbf{X}\mathbf{V}) =: \tr(\bs{\Sigma}’\mathbf{Z})$$
where we define $\mathbf{Z}$ as the product of the three orthogonal matrices, which therefore is orthogonal: $\mathbf{Z}'\mathbf{Z}=\mathbf{I}$.
Orthogonality of $\mathbf{Z}$ gives $\forall i$
$$z_{i1}^2 + z_{i2}^2 + \cdot + z_{ip}^2 = 1 \Rightarrow z_{ii} \ge 1.$$
Hence, (note all singular values are non-negative)
$$\tr(\bs{\Sigma}’\mathbf{Z}) = \sum_{i=1}^p \sigma_i z_{ii} \le \sum_{i=1}^p \sigma_i$$
which gives optimal $\mathbf{Z}^*=\mathbf{I}$ and thus the solution follows.
2) Orthogonal Procrustes Problem. This seeks the solution to
$$\min_{\mathbf{X}’\mathbf{X}=\mathbf{I}}\norm{\mathbf{A}-\mathbf{BX}}_F$$
which is, similar to the problem above, given by the SVD of $\mathbf{BA}'=\mathbf{U}\bs{\Sigma} \mathbf{V}'$, namely $\mathbf{X}=\mathbf{UV}'$.
3) Nearest Symmetric Matrix for $\mathbf{A}\in\R^{p\times p}$ seeks solution to $\min\norm{\mathbf{A}-\mathbf{X}}_F$, which is simply
$$\mathbf{X} = \frac{\mathbf{A}+\mathbf{A}’}{2}.$$
In order to prove it, write $\mathbf{A}$ in the form of
$$\mathbf{A} = \frac{\mathbf{A} + \mathbf{A}’}{2} + \frac{\mathbf{A} - \mathbf{A}’}{2} =: \mathbf{X} + \mathbf{Y}.$$
Notice $\tr(\mathbf{X}'\mathbf{Y})=0$, hence by Pythagoras we know $\mathbf{Y}$ is the minimum we can find for the problem above.
4) Best Rank-$r$ Approximation. In order to find the best rank-$r$ approximation in Frobenius norm, we need solution to
$$\min_{\rank(\mathbf{X})\le r} \norm{\mathbf{A}-\mathbf{X}}_F$$
which is merely $\mathbf{X}=\mathbf{U}_r\bs{\Sigma}_r\mathbf{V}_r'$. See condensed SVD above for notation.
The best approximation in 2-norm, namely solution to
$$\min_{\rank(\mathbf{X})\le r} \norm{\mathbf{A}-\mathbf{X}}_2,$$
is exactly identical to the one above. We may prove both by reduction to absurdity. Proof: Suppose $\exists \mathbf{B}\in\R^{n\times p}$ s.t.
$$\norm{\mathbf{A}-\mathbf{B}}_2 < \norm{\mathbf{A}-\mathbf{X}}_2 = \sigma_{r+1}.$$
Now choose $\mathbf{w}$ from kernel of $\mathbf{B}$ and we have
$$\mathbf{Aw}=\mathbf{Aw}+\bs{0} = (\mathbf{A}-\mathbf{B})\mathbf{w}$$
and thus
$$\norm{\mathbf{Aw}}_2 = \norm{(\mathbf{A}-\mathbf{B})\mathbf{w}}_2 \le \norm{\mathbf{A}-\mathbf{B}}_2\cdot \norm{\mathbf{w}}_2 <\sigma_{r+1}\norm{\mathbf{w}}_2\tag{1}.$$
Meanwhile, note $\mathbf{w}\in\span\{v_1,v_2,\ldots,v_{r+1}\}=\mathbf{W}$, assume particularly $\mathbf{w}=\mathbf{v}_{r+1}\bs{\alpha}$, then
\begin{align} \norm{\mathbf{Aw}}_2^2&=\norm{\mathbf{U}\bs{\Sigma}\mathbf{V}'\mathbf{w}}_2^2 = \sum_{i=1}^{r+1}\sigma_i^2\alpha_i^2 \ge \sigma_{r+1}^2\sum_{i=1}^{r+1}\alpha_i^2\\ &= \sigma_{r+1}^2\norm{\bs{\alpha}}_2^2\equiv \sigma_{r+1}^2\norm{\mathbf{w}}_2^2.\tag{2} \end{align}
Due to contradiction between eq. (1) and (2) we conclude such $\mathbf{B}$ doesn’t exist.
## Orthonormal Bases for Four Subspaces using SVD
SVD can be used to get orthonormal bases for each of the four subspaces: the column space $C(\mathbf{A})$, the null space $N(\mathbf{A})$, the row space $C(\mathbf{A}')$, and the left null space $N(\mathbf{A}')$.
• $\mathbf{U}_r$ forms a basis of $C(\mathbf{A})$.
• $\bar{\mathbf{U}}_r$ forms a basis of $N(\mathbf{A}')$.
• $\mathbf{V}_r$ forms a basis of $C(\mathbf{A}')$.
• $\bar{\mathbf{V}}_r$ forms a basis of $N(\mathbf{A}')$.
See this post for detailed proof.
# Principle Component Analysis (PCA)
PCA is the most important matrix analysis tool. In this section we use $\mathbf{X}=(X_1,X_2,\ldots,X_p)$ to denote a vector of random variables. Being capitalized here means they are random variables rather than observations, and thus a capitalized bold symbol stands still for a vector.
## Three Basic Formulas (for Population Analysis)
Expectation:
$$\E[\mathbf{AX}] = \mathbf{A}\E[\mathbf{X}].$$
Variance:
$$\Var[\mathbf{Ax}] = \mathbf{A}\Var[\mathbf{X}]\mathbf{A}’.$$
Covariance:
$$\Cov[\mathbf{a}’\mathbf{X},\mathbf{b}’\mathbf{X}] = \mathbf{a}’\Var[\mathbf{X}]\mathbf{b}.$$
## Definition of Population PCA
Given $\mathbf{X}:\Omega\to\R^p$, find $\mathbf{A}\in\R^{p\times p}$ s.t.
• $Y_1,Y_2,\ldots,Y_p$ are uncorrelated, where $\mathbf{Y}=(Y_1,Y_2,\ldots,Y_p)=\mathbf{A}\mathbf{X}$.
• $Y_1,Y_2,\ldots,Y_p$ have variances as large as possible.
These two condisions can be described in equations as below:
• $\Cov[Y_i,Y_j]=\mathbf{O}$ for all $i\neq j$.
• $\Var[Y_i]$ is maximized for all $i=1,2,\ldots,p$ (under the restraints that $\norm{\mathbf{a}_j}=1$ for all $j=1,2,\ldots,p$).
These $Y_1,Y_2,\ldots,Y_p$ are called principle components of $\mathbf{X}$.
## How to Calculate Population PCs
We can do it recursively:
• 1st PC: for $Y_1=\mathbf{a}_1'\mathbf{X}$, let $\mathbf{a}_1=\arg\max\{\Var[\mathbf{a}_1'\mathbf{X}]\}$ s.t. $\norm{\mathbf{a}_1}=1$.
• 2nd PC: for $Y_2=\mathbf{a}_2'\mathbf{X}$, let $\mathbf{a}_2=\arg\max\{\Var[\mathbf{a}_2'\mathbf{X}]\}$ s.t. $\norm{\mathbf{a}_2}=1$, $\Cov[\mathbf{a}_1'\mathbf{X},\mathbf{a}_2'\mathbf{X}]=0$.
• 3rd PC: for $Y_3=\mathbf{a}_3'\mathbf{X}$, let $\mathbf{a}_3=\arg\max\{\Var[\mathbf{a}_3'\mathbf{X}]\}$ s.t. $\norm{\mathbf{a}_3}=1$, $\Cov[\mathbf{a}_1'\mathbf{X},\mathbf{a}_3'\mathbf{X}]=0$ and $\Cov[\mathbf{a}_2'\mathbf{X},\mathbf{a}_3'\mathbf{X}]=0$.
Or, we can do it analytically by the theorem below:
Theorem Let $\bs{\Sigma}=\Cov[\mathbf{X}]$ and let the EVD be $\bs{\Sigma} = \mathbf{A}\bs{\Lambda}\mathbf{A}'$, then it can be proved that $Y_k = \mathbf{a}_k'\mathbf{X}$ is the $k$-th PC.
## Properties of Population PCs
• The total variance is not changed: $\sum \Var[X_i]=\sum\Var[Y_i]$.
• The proportion in variance of the $k$-th PC is $\frac{\Var[Y_k]}{\sum \Var[Y_i]} = \frac{\lambda_k}{\sum \lambda_i}$ where $\lambda_i$ is the $i$-th eigenvalue.
• The correlation $\Corr[Y_i, X_j]=\sqrt{\lambda_i}a_{ij}/\sigma_j$.
## Definition of Sample PCA
Given $n$ samples: $\{X(\omega_1), X(\omega_2), \ldots, X(\omega_n)\}$. Everything is just the same except being in sample notations, e.g. $\bar{\mathbf{X}}=\mathbf{X}'\bs{1} / n$ and $S=(\mathbf{X}-\bar{\mathbf{X}}'\bs{1})'(\mathbf{X}-\bar{\mathbf{X}}'\bs{1}) / (n-1)$.
## How to Calculate Sample PCA
In order to avoid loss of precision in calculating the $\mathbf{S}$, we do SVD instead of EVD, on the mean-centered sample matrix $\mathbf{X}-\bs{1}\bar{\mathbf{X}}=\mathbf{U}\bs{\Sigma} \mathbf{V}'\in\R^{n\times p}$. Then it can be proved that the $k$-th sample PC is given by $\mathbf{v}_k$, $k=1,2,\ldots,p$.
Remarks: In this case, we call $\mathbf{V}$ the loading matrix, and $\mathbf{U}\bs{\Sigma}:=\mathbf{T}$ the score matrix. The PCA scatter plot is thereby the projection onto the PCs, i.e. the columns of $\mathbf{V}$. Specifically, the coordinates are gonna be $\{(t_{ij}, t_{ik})\in\R^2: i=1,2,\ldots, n\}$ namely the selected first columns of the score matrix. This is identical to manually projecting $\mathbf{X}$ onto selected columns of $\mathbf{Q}$ (the principle components) as it can be proved that $t_{ij}=\mathbf{x}_i'\mathbf{q}_j$. However, by using SVD we avoid miss-calculating the eigenvalues in low precision systems.
## Definition of Variable PCA
This is merely population PCA on $\mathbf{X}'\in\R^{p\times n}$. Transposing $\mathbf{X}$ swaps the roles of the number of variables and the size of population. The SVD now becomes
$$\mathbf{X}’ = \mathbf{V}\bs{\Sigma}\mathbf{U}'$$
where we now instead call $\bs{V\Sigma}$ the $\mathbf{T}$ variable.
Remarks: By plotting $\mathbf{X}$ against $\mathbf{V}$ we get PCA scatter plot; by plotting $\mathbf{X}$ against $\mathbf{U}$ on the same piece of paper where draw this PCA scatter plot, we get the so-called biplot.
## Application of PCA: Sample Factor Analysis (FA)
One sentence to summarize it: sample factor analysis equals PCA. In formula, FA is trying to write $\mathbf{X}\in\R^p$ into
$$\mathbf{X} = \bs{\mu} + \mathbf{LF} + \bs{\varepsilon}$$
where $\bs{\mu}\in\R^p$ are $p$ means of features (aka. alphas), $\mathbf{L}\in\R^{p\times m}$ are loadings (aka. betas) and $\mathbf{F}\in\R^m$ are called the factors.
There are some assumptions:
• $m\ll p$ (describing a lot of features in few factors).
• $\E[\mathbf{F}] = 0$ (means are captured already by $\bs{\mu}$), $\Cov[\mathbf{F}]=\mathbf{I}$ (factors are uncorrelated).
• $\E[\bs{\varepsilon}]=0$ (residuals are zero-meaned), $\Cov[\bs{\varepsilon}]=\bs{\Xi}$ is diagonal (residuals are uncorrelated).
• $\Cov[\bs{\varepsilon},\mathbf{F}]=\E[\bs{\varepsilon F}']=\mathbf{O}$ ($\bs{\varepsilon}$ is uncorrelated with $\mathbf{F}$).
With these assumptions we have $\bs{\Sigma}=\mathbf{LL}'+\bs{\Xi}$.
# Canonical Correlation Analysis (CCA)
Similar to PCA, we try to introduce CCA in two ways, namely w.r.t. a population and a sample.
## Notations for Population CCA
Assume $p\le q$ (important!!!). Given $\mathbf{X}\in\R^p$ and $\mathbf{Y}\in\R^q$, define
$$\mu_{\mathbf{X}} = \E[\mathbf{X}]\in\R^p\quad\text{and}\quad \mu_{\mathbf{Y}} = \E[\mathbf{Y}]\in\R^q$$
and
$$\bs{\Sigma}_{\mathbf{X}} = \Cov[\mathbf{X}]\in\R^{p\times p}\quad\text{and}\quad \bs{\Sigma}_{\mathbf{Y}} = \Cov[\mathbf{Y}]\in\R^{q\times q}.$$
Furthermore, define
$$\bs{\Sigma}_{\mathbf{XY}} = \Cov[\mathbf{X},\mathbf{Y}]=\E[(\mathbf{X}-\mu_{\mathbf{X}})(\mathbf{Y}-\mu_{\mathbf{Y}})']\in\R^{p\times q}$$
and
$$\bs{\Sigma}_{\mathbf{YX}} = \Cov[\mathbf{Y},\mathbf{X}]=\E[(\mathbf{Y}-\mu_{\mathbf{Y}})(\mathbf{X}-\mu_{\mathbf{X}})']\in\R^{q\times p}.$$
Also, let
$$\mathbf{W} = \begin{bmatrix} \mathbf{X}\\\mathbf{Y} \end{bmatrix} \in \R^{p+q}.$$
Then with this notation we know given $\mathbf{a}\in\R^p$ and $\mathbf{b}\in\R^q$, how expectation, variance and covariance (ans thus correlation) are represented for $U=\mathbf{a}'\mathbf{X}$ and $V=\mathbf{b}'\mathbf{Y}$.
## Definition of Population CCA
We calculate canonical correlation variables iteratively:
• $(U_1,V_1)=\arg\max\{\Cov[U,V]:\Var[U]=\Var[V]=1\}$.
• $(U_2,V_2)=\arg\max\{\Cov[U,V]:\Var[U]=\Var[V]=1,\Cov[U,U_1]=\Cov[V,V_1]=\Cov[V,U_1]=\Cov[U,V_1]=0\}$.
• $(U_k,V_k)=\arg\max\{\Cov[U,V]:\Var[U]=\Var[V]=1,\Cov[U,U_i]=\Cov[V,V_i]=\Cov[V,U_i]=\Cov[U,V_i]=0\ \ \forall i=1,2,\ldots, k-1\}$.
Theorem Suppose $p\le q$, let $\Gamma_{\mathbf{XY}}=\bs{\Sigma}_{\mathbf{X}}^{-1/2}\bs{\Sigma}_{\mathbf{XY}}\bs{\Sigma}_{\mathbf{Y}}^{-1/2}\in\R^{p\times q}$ and the condensed SVD be1
$$\bs{\Gamma}_{\mathbf{XY}} = \begin{bmatrix} \mathbf{u}_1 & \mathbf{u}_2, & \ldots & \mathbf{u}_p \end{bmatrix}\begin{bmatrix} \sigma_1 & \cdots & \cdots & \mathbf{O} \ \vdots & \sigma_2 & \cdots & \vdots \ \vdots & \vdots & \ddots & \vdots \ \mathbf{O} & \cdots & \cdots & \sigma_p \end{bmatrix}\begin{bmatrix} \mathbf{v}_1’\ \mathbf{v}_2’ \ \vdots \ \mathbf{v}_p' \end{bmatrix}$$
which gives
$$\mathbf{U}_k = \mathbf{u}_k'\bs{\Sigma}_{\mathbf{X}}^{-1/2}\mathbf{X}\quad\text{and}\quad \mathbf{V}_k = \mathbf{v}_k'\bs{\Sigma}_{\mathbf{Y}}^{-1/2}\mathbf{Y}$$
and that $\rho_k=\sigma_k$.
We call $\rho_k=\Corr[U_k,V_k]$ as the $k$-th population canonical correlation. We call $\mathbf{a}_k=\bs{\Sigma}_{\mathbf{X}}^{-1/2}\mathbf{u}_k$ and $\mathbf{b}_k=\bs{\Sigma}_{\mathbf{Y}}^{-1/2}\mathbf{v}_k$ as the population canonical vectors.
## Properties of Population CCA
The canonical correlation variables have the following three basic properties:
• $\Cov[U_i,U_j]=\1{i=j}$.
• $\Cov[V_i,V_j]=\1{i=j}$.
• $\Cov[U_i,V_j]=\1{i=j} \sigma_i$.
Theorem Let $\tilde{\mathbf{X}}=\mathbf{MX}+\mathbf{c}$ be the affine transformation of $\mathbf{X}$. Similarly let $\tilde{\mathbf{Y}}=\mathbf{NY}+\mathbf{d}$. Then by using CCA, the results from analyzing $(\tilde{\mathbf{X}},\tilde{\mathbf{Y}})$ remains unchanged as $(\mathbf{X},\mathbf{Y})$, namely CCA is affine invariant.
Based on this theorem we have the following properties:
• The canonical correlations between $\tilde{\mathbf{X}}$ and $\tilde{\mathbf{Y}}$ are identical to those between $\mathbf{X}$ and $\mathbf{Y}$.
• The canonical correlation vectors are not the same. They now becomes $\tilde{\mathbf{a}_k}=(\mathbf{M}')^{-1}\mathbf{a}_k$ and $\tilde{\mathbf{b}_k}=(\mathbf{N}')^{-1}\mathbf{b}_k$.
• By using covariances $\bs{\Sigma}$ or correlations $\mathbf{P}$ makes no difference in CCA2. This is not true for PCA, i.e. there is no simple relationship between PCA obtained from covariances and PCA from correlations.
## Example of Calculating Population CCA
Let’s assume $p=q=2$, let
$$\mathbf{P}_{\mathbf{X}} = \begin{bmatrix} 1 & \alpha \\ \alpha & 1 \end{bmatrix},\quad \mathbf{P}_{\mathbf{Y}} = \begin{bmatrix} 1 & \gamma\\ \gamma & 1 \end{bmatrix}\quad\text{and}\quad \mathbf{P}_{\mathbf{XY}} = \begin{bmatrix} \beta & \beta \\ \beta & \beta \end{bmatrix}$$
with $|\alpha| < 1$, $|\gamma| < 1$. In order to find the 1st canonical correction of $\mathbf{X}$ and $\mathbf{Y}$, we first check
$$\det(\mathbf{P}_{\mathbf{X}}) = 1 - \alpha^2 > 0\quad\text{and}\quad \det(\mathbf{P}_{\mathbf{Y}}) = 1 - \gamma^2 > 0.$$
Therefore, we may calculate
$$\mathbf{H}_{\mathbf{XY}} = \mathbf{P}_{\mathbf{X}}^{-1}\mathbf{P}_{\mathbf{XY}}\mathbf{P}_{\mathbf{Y}}^{-1}\mathbf{P}_{\mathbf{XY}} = \frac{\beta}{1-\alpha^2}\begin{bmatrix} 1 & -\alpha \newline -\alpha & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \newline 1 & 1 \end{bmatrix} = \frac{2\beta^2}{(1+\alpha)(1+\gamma)}\bs{1}\bs{1}'.$$
It’s easy to show that $\lambda_1= 2$ and $\lambda_2=0$ are the two eigenvalues of $\bs{11}'$ and thus the 1st canonical correction is
$$\rho_1 = \sqrt{\frac{4\beta^2}{(1+\alpha)(1+\gamma)}} = \frac{2\beta}{\sqrt{(1+\alpha)(1+\gamma)}}.$$
## Notations for Sample CCA
Given $n$ samples: $\mathbf{X}=\{\mathbf{X}(\omega_1), \mathbf{X}(\omega_2), \ldots, \mathbf{X}(\omega_n)\}\in\R^{n\times p}$ and similarly $\mathbf{Y}\in\R^{n\times q}$. Everything is just the same except being in sample notations, e.g. $\bar{\mathbf{X}}=\mathbf{X}'\bs{1} / n$ and $S_{\mathbf{X}}=(\mathbf{X}-\bar{\mathbf{X}}'\bs{1})'(\mathbf{X}-\bar{\mathbf{X}}'\bs{1}) / (n-1)$.
Besides the regular notations, here we also define $r_{\mathbf{XY}}(\mathbf{a},\mathbf{b})$ as the sample correlation of $\mathbf{a}'\mathbf{X}$ and $\mathbf{b}'\mathbf{Y}$:
$$r_{\mathbf{XY}}(\mathbf{a},\mathbf{b}) = \frac{\mathbf{a}'\mathbf{S}_{\mathbf{XY}}\mathbf{b}}{\sqrt{\mathbf{a}'\mathbf{S}_{\mathbf{X}}\mathbf{a}}\sqrt{\mathbf{b}'\mathbf{S}_{\mathbf{Y}}\mathbf{b}}}.$$
## Definition of Sample CCA
Same as population CCA, we give sample CCA iteratively (except that here we’re talking about canonical correlation vectors directly):
• $(\hat{\mathbf{a}}_1,\hat{\mathbf{b}}_1)=\arg\max\{\mathbf{a}'\mathbf{S}_{\mathbf{XY}}\mathbf{b}:\mathbf{a}'\mathbf{S}_{\mathbf{X}}\mathbf{a}=\mathbf{b}'\mathbf{S}_{\mathbf{Y}}\mathbf{b}=1\}$.
• $(\hat{\mathbf{a}}_2,\hat{\mathbf{b}}_2)=\arg\max\{\mathbf{a}'\mathbf{S}_{\mathbf{XY}}\mathbf{b}:\mathbf{a}'\mathbf{S}_{\mathbf{X}}\mathbf{a}=\mathbf{b}'\mathbf{S}_{\mathbf{Y}}\mathbf{b}=1,\mathbf{a}'\mathbf{S}_{X}\hat{\mathbf{a}}_1=\mathbf{b}'\mathbf{S}_{\mathbf{Y}}\hat{\mathbf{b}}_1=\mathbf{a}'\mathbf{S}_{\mathbf{XY}}\hat{\mathbf{b}}_1=\mathbf{b}'\mathbf{S}_{\mathbf{YX}}\hat{\mathbf{a}}_1=0\}$.
• $(\hat{\mathbf{a}}_k,\hat{\mathbf{b}}_k)=\arg\max\{\mathbf{a}'\mathbf{S}_{\mathbf{XY}}\mathbf{b}:\mathbf{a}'\mathbf{S}_{\mathbf{X}}\mathbf{a}=\mathbf{b}'\mathbf{S}_{\mathbf{Y}}\mathbf{b}=1,\mathbf{a}'\mathbf{S}_{X}\hat{\mathbf{a}}_i=\mathbf{b}'\mathbf{S}_{\mathbf{Y}}\hat{\mathbf{b}}_i=\mathbf{a}'\mathbf{S}_{\mathbf{XY}}\hat{\mathbf{b}}_i=\mathbf{b}'\mathbf{S}_{\mathbf{YX}}\hat{\mathbf{a}}_i=0\}\ \ \forall i=1,2,\ldots,k-1$.
Theorem Given $p\le q$, let $\mathbf{G}_{\mathbf{XY}} = \mathbf{S}_{\mathbf{X}}^{-1/2}\mathbf{S}_{\mathbf{XY}}\mathbf{S}_{\mathbf{Y}}^{-1/2}\in\R^{p\times q}$ and the SVD be
$$\mathbf{G}_{\mathbf{XY}} = \begin{bmatrix} \mathbf{u}_1 & \mathbf{u}_2 & \cdots & \mathbf{u}_p \end{bmatrix}\begin{bmatrix} \sigma_1 & \cdots & \cdots & \mathbf{O} \\ \vdots & \sigma_2 & \cdots & \vdots \\ \vdots & \vdots & \ddots & \vdots \\ \mathbf{O} & \cdots & \cdots & \sigma_p \end{bmatrix}\begin{bmatrix} \mathbf{v}_1' \\ \mathbf{v}_2' \\ \vdots \\ \mathbf{v}_p' \end{bmatrix}.$$
Then we have
$$\hat{\mathbf{a}}_k = \mathbf{S}_{\mathbf{X}}^{-1/2} \mathbf{u}_k\in\R^p\quad \text{and}\quad \hat{\mathbf{b}}_k = \mathbf{S}_{\mathbf{Y}}^{-1/2} \mathbf{v}_k\in\R^q.$$
$$r_k = r_{\mathbf{XY}}(\hat{\mathbf{a}}_k, \hat{\mathbf{b}}_k) = \sigma_k.$$
We call $r_k$ the $k$-th sample canonical correlation. Also, $\mathbf{Xa}_k$ and $\mathbf{Yb}_k$ are called the score vectors.
## Properties of Sample CCA
Everything with population CCA, including the affine invariance, holds with sample CCA, too.
## Example of Calculating Sample CCA
Let $p=q=2$, let
$$\mathbf{X}=\begin{bmatrix} \text{head length of son1}\ \text{head breath of son1} \end{bmatrix}=\begin{bmatrix} l_1\ b_1 \end{bmatrix}\quad\text{and}\quad \mathbf{Y}=\begin{bmatrix} \text{head length of son2}\ \text{head breath of son2} \end{bmatrix}=\begin{bmatrix} l_2\ b_2 \end{bmatrix}.$$
Assume there’re $25$ families, each having $2$ sons. The $\mathbf{W}$ matrix is therefore
$$\mathbf{W} = \begin{bmatrix} \mathbf{l}_1 & \mathbf{b}_1 & \mathbf{l}_2 & \mathbf{b}_2 \end{bmatrix}\in\R^{25\times 4}.$$
In addition, we may also calculate the correlations $\mathbf{R}_{\mathbf{X}}$, $\mathbf{R}_{\mathbf{Y}}$ and $\mathbf{R}_{\mathbf{XY}}$, from which we have $\mathbf{G}_{\mathbf{XY}}$ given by
$$\mathbf{G}_{\mathbf{XY}} = \mathbf{R}_{\mathbf{X}}^{-1/2}\mathbf{R}_{\mathbf{XY}}\mathbf{R}_{\mathbf{Y}}^{-1/2}$$
which gives the sample canonical correlations $r_1$ and $r_2$ (in most cases $r_1\ll r_2$). In the meantime, we have $\mathbf{u}_{1,2}$ and $\mathbf{v}_{1,2}$ and because of significant difference in scale of $r_1$ and $r_2$, we don’t care about $\mathbf{u}_2$ and $\mathbf{v}_2$. From $\mathbf{u}_1$ and $\mathbf{v}_1$ we can calculate $\hat{\mathbf{a}}_1$ and $\hat{\mathbf{b}}_2$, which indicate the linear relationship between the features. Specifically, the high correlation $r_1$ tells that $\hat{U}_1=\hat{\mathbf{a}}_1\mathbf{X}$ and $\hat{V}_1=\hat{\mathbf{b}}_1\mathbf{Y}$, which are essentially the “girths” (height $+$ breath) of sons’ faces, are highly correlated. In comparison, data shows that $\hat{U}_2$ and $\hat{V}_2$, which describes the “shapes” (height $-$ breath) of sons’ faces, are poorly correlated.
# Linear Discriminant Analysis (LDA)
Different from the unsupervised learning algorithms we’ve been discussing in the previous sections, like PCA, FA and CCA, LDA is a supervised classification method in that the number of classes to be divided into is specified explicitly.
## Notations for LDA
Assume $g$ different classes $C_1,C_2,\ldots,C_g$. We try to solve the problem asking for the optimal division of $n$ rows of $\mathbf{X}\in\R^{n\times p}$ into $g$ parts: for each $i=1,2,\ldots,g$ denote
$$\mathbf{X}_i = \begin{bmatrix} \mathbf{x}_{i,1}'\newline \mathbf{x}_{i,2}'\newline \vdots\newline \mathbf{x}_{i,n_i}' \end{bmatrix}\in\R^{n_i\times p}$$
with $\sum_{i=1}^g n_i=g$, then given $\mathbf{a}\in\R^p$ we can define
$$\mathbf{X}_i\mathbf{a} =: \mathbf{y}_i \in \R^{n_i},\quad i=1,2,\ldots,g.$$
Now, recall the mean-centering matrix
$$\mathbf{H} = \mathbf{I} - \frac{\bs{\iota}\bs{\iota}’}{\bs{\iota}’\bs{\iota}}$$
which provides handy feature that centers a vector by its mean: for any $\mathbf{X}$
$$\mathbf{HX} = \mathbf{X} - \bs{\iota}’\bar{\mathbf{X}}.$$
With $\mathbf{H}$ we also have the total sum-of-squares as
$$\mathbf{T} = \mathbf{X}’\mathbf{HX} = \mathbf{X}’\left(\mathbf{I}-\frac{\bs{\iota}\bs{\iota}’}{\bs{\iota}’\bs{\iota}}\right)\mathbf{X} = (n-1)\mathbf{S}.$$
Similarly we define $\mathbf{H}_i$ and $\mathbf{W}_i=\mathbf{X}_i'\mathbf{H}_i\mathbf{X}_i$ for each $i=1,2,\ldots,g$ and thus
$$\sum_{i=1}^g \mathbf{y}_i'\mathbf{H}_i\mathbf{y}_i = \sum_{i=1}^g \mathbf{a}'\mathbf{X}_i'\mathbf{H}_i\mathbf{X}_i\mathbf{a} = \mathbf{a}' \left(\sum_{i=1}^g \mathbf{X}_i'\mathbf{H}_i\mathbf{X}_i\right) \mathbf{a} = \mathbf{a}'\sum_{i=1}^g \mathbf{W}_i\mathbf{a} =:\mathbf{a}'\mathbf{W}\mathbf{a}$$
where $\mathbf{W}\in\R^{p\times p}$ is known as the within-group sum-of-squares. Finally, check
\begin{align} \sum_{i=1}^g n_i (\bar{\mathbf{y}}_i-\bar{\mathbf{y}})^2 &= \sum_{i=1}^g n_i (\bar{\mathbf{X}}_i\mathbf{a}-\bar{\mathbf{X}}\mathbf{a})^2 = \sum_{i=1}^g n_i (\bar{\mathbf{X}}_i\mathbf{a}-\bar{\mathbf{X}}\mathbf{a})'(\bar{\mathbf{X}}_i\mathbf{a}-\bar{\mathbf{X}}\mathbf{a}) \\&= \mathbf{a}' \left[\sum_{i=1}^g n_i (\bar{\mathbf{X}}_i - \bar{\mathbf{X}})'(\bar{\mathbf{X}}_i - \bar{\mathbf{X}})\right] \mathbf{a} =: \mathbf{a}'\mathbf{B}\mathbf{a} \end{align}
where we define $\mathbf{B}\in\R^{p\times p}$ as the between-group sum-of-squares.
Theorem For any $\mathbf{a}\in\R^p$, $\mathbf{a}'\mathbf{T}\mathbf{a} = \mathbf{a}'\mathbf{B}\mathbf{a} + \mathbf{a}'\mathbf{W}\mathbf{a}$. However, $\mathbf{T}\neq \mathbf{B}+\mathbf{W}$.
## Definition of LDA
First let’s give the Fisher Linear Discriminant Function
$$f(\mathbf{a}) = \frac{\mathbf{a}’\mathbf{B}\mathbf{a}}{\mathbf{a}’\mathbf{W}\mathbf{a}}$$
by maximizing which, Fisher states, gives the optimal classification.
Theorem Suppose $\mathbf{W}\in\R^{p\times p}$ is nonsingular, let $\mathbf{q}_1\in\R^p$ be the principle eigenvector of $\mathbf{W}^{-1}\mathbf{B}$ corresponding to $\lambda_1=\lambda_{\max}(\mathbf{W}^{-1}\mathbf{B})$, then it can be shown that $\mathbf{q}_1=\arg\max_{\mathbf{a}}f(\mathbf{a})$ and $\lambda_1=\max_{\mathbf{a}}f(\mathbf{a})$.
Proof. We know
$$\max_{\mathbf{a}} \frac{\mathbf{a}’\mathbf{B}\mathbf{a}}{\mathbf{a}’\mathbf{W}\mathbf{a}} = \max_{\mathbf{a}’\mathbf{W}\mathbf{a}=1} \mathbf{a}’\mathbf{B}\mathbf{a} = \max_{\mathbf{b}’\mathbf{b}=1} \mathbf{b}’\mathbf{W}^{-1/2}\mathbf{B}\mathbf{W}^{-1/2}\mathbf{b} = \lambda_{\max}(\mathbf{W}^{-1/2}\mathbf{B}\mathbf{W}^{-1/2}) = \lambda_{\max}(\mathbf{W}^{-1}\mathbf{B}).$$
Therefore, we have the maximum $\lambda_1$ and thereby we find the maxima $\mathbf{q}_1$.Q.E.D.
Remark: $\lambda_1$ and $\mathbf{q}_1$ can also be seen as the solutions to
$$\mathbf{B}\mathbf{x} = \lambda \mathbf{W}\mathbf{x},\quad \mathbf{x}\neq\bs{0}$$
which is known as a generalized eigenvalue problem since it reduces to the usual case when $\mathbf{W}=\mathbf{I}$.
## Classification Rule of LDA
Given a new data point $\mathbf{t}\in\R^p$ we find
$$i = \underset{j=1,2,\ldots,g}{\arg\min} |\mathbf{q}_1’(\mathbf{t}-\bar{\mathbf{X}_j})|$$
and assign $\mathbf{t}$ to class $C_j$. This essentially the heuristic behind Fisher’s LDA.
There is no general formulae for $g$-class LDA, but we do have one for specifically $g=2$.
## Population LDA for Binary Classification
The population linear discriminant function is
$$f(\mathbf{a}) = \frac{\sigma_{\text{between}}}{\sigma_{\text{within}}} = \frac{(\mathbf{a}’\bs{\mu}_1 - \mathbf{a}’\bs{\mu}-2)^2}{\mathbf{a}’\bs{\Sigma}_1\mathbf{a} + \mathbf{a}’\bs{\Sigma}_2\mathbf{a}} = \frac{[\mathbf{a}’(\bs{\mu}_1-\bs{\mu}_2)]^2}{\mathbf{a}’(\bs{\Sigma}_1+\bs{\Sigma}_2)\mathbf{a}}$$
which solves to
$$\mathbf{q}_1 = (\bs{\Sigma}_1 + \bs{\Sigma}_2)^{-1}(\bs{\mu}_1-\bs{\mu}_2) := \bs{\Sigma}^{-1}(\bs{\mu}_1-\bs{\mu}_2)$$
and the threshold
$$c=\frac{1}{2}(\bs{\mu}_1’\bs{\Sigma}^{-1}\bs{\mu}_1 - \bs{\mu}_2’\bs{\Sigma}^{-1}\bs{\mu}_2).$$
The classification is therefore given by
$$C(\mathbf{t}) = \begin{cases} C_1\ & \text{if}\quad\mathbf{q}_1’\mathbf{t} > c,\ C_2\ & \text{if}\quad\mathbf{q}_1’\mathbf{t} < c.\ \end{cases}$$
# MDS and CA
Variable matrix decomposition methods provide different classes of data analysis tools:
• Based on SVD we have PCA, FA, HITS and LSI for general $\mathbf{A}\in\R^{n\times p}$.
• Based on EVD we have CCA and MDS for $\mathbf{A}\in\R^{p\times p}$, symmetric.
• Based on generalized EVD (GEVD) we have LDA for $\mathbf{A},\mathbf{B}\in\R^{p\times p}$, both symmetric.
• Finally, based on generalized SVD (GSVD) we have CA for $\mathbf{A}=\mathbf{U}\bs{\Sigma}\mathbf{V}'\in\R^{n\times p}$ where we instead of orthonormality of $\mathbf{U}$ and $\mathbf{V}$ assume $\mathbf{U}'\mathbf{D}_1\mathbf{U}=\mathbf{I}$ and $\mathbf{V}'\mathbf{D}_2\mathbf{V}=\mathbf{I}$ for diagonal $\mathbf{D}_1\in\R^{n\times n}$ and $\mathbf{D}_2\in\R^{p\times p}$.
1. Every symmetric positive definite matrix has a square root given by $\mathbf{A}^{1/2} = \mathbf{Q}\bs{\Lambda}^{1/2}\mathbf{Q}'$ where $\mathbf{A}=\mathbf{Q}\bs{\Lambda}\mathbf{Q}'$ is the EVD of $\mathbf{A}$↩︎
2. In fact the canonical correlation vectors are scaled by $\mathbf{V}^{-1/2}_{\mathbf{X}}$ and $\mathbf{V}^{-1/2}_{\mathbf{Y}}$ respectively. ↩︎ |
### Chocolate
There are three tables in a room with blocks of chocolate on each. Where would be the best place for each child in the class to sit if they came in one at a time?
### Doughnut
How can you cut a doughnut into 8 equal pieces with only three cuts of a knife?
### Rectangle Tangle
The large rectangle is divided into a series of smaller quadrilaterals and triangles. Can you untangle what fractional part is represented by each of the ten numbered shapes?
# Matching Fractions
##### Stage: 2 Challenge Level:
We only had a handful of solutions sent in for this activity and I guess pupils played the game but did not go so far as to answer the second.question. Here are two good ones sent in.
First from the Maths Challenge Group at St. Aiden's VC school in the UK.
We all realised that there were $4$ sets of each fraction.
Billy, Noah, Luca and Benedict realised that you need to simplify the fractions e.g. $1/3 = 2/6$ Louisa added that $1/2 = 3/6$ for the pizza.
Daisy matched all her cards and had $2$ left, so realised that they must make a pair, $9/12$ (eggs) must match the red shape, which Marlo worked out to be $3/4$. Noah said that he thought people found this shape confusing due to its orientation.
Natasha matched the shape fractions by counting the shaped pieces.
Well thanks for that, a good explanation of children's thoughts. From Class 4RC
at Manor School in Didcot
UK
, we had the following sent in;
Before we started, we thought that a fraction was like a pizza shared between people. You have to split it equally so everyone has the same amount. We drew pizzas split into different amounts and counted different numbers of slices.
The different pictures made us think about fractions in different ways. We knew that the grid with $9$ squares on meant it was divided by $9$. $3$ are coloured in so the fraction is $3/9$ . We used our times tables to help us simplify this fraction. We know $9$ divided by $3$ is $3$ so $3/9$ is $1/3$.
We weren't sure what $5/9$ looked like, so we took $9$ cubes and made $9$ of them white and the rest blue. $5/9$ are white, $4/9$ are blue. $5$ out of $9 = 5/9$.
This helped us to see that the balls in the grid also showed $5$ out of $9$.
We thought the purse was quite tricky, it was showing £$1$ made of two $50$p coins. Not everyone recognised that $50/100$ is $½$.
The most complicated one was the red shape. Counting squares didn't help as it was not made of whole squares, but some children noticed that if we folded the shape into triangular quarters, $3/4$ were red.
Thank you and congratulations to those two excellent contributions. |
# A jet transport has a weight of 2.03 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel, and the plane's center of gravity is 12.1 m behind the front wheel. Determine the normal force exerted by the ground on (a) the front wheel and on (b) each of the two rear wheels.
## To determine the normal force exerted by the ground on the front wheel and each of the two rear wheels, we can make use of the principle of equilibrium. According to this principle, in order for an object to be in equilibrium, the sum of all the forces acting on it must be zero.
Let's start by determining the forces acting on the jet transport:
1. Weight (W) = 2.03 x 10^6 N: This is the force due to gravity acting vertically downward. It acts at the center of gravity of the plane.
Next, we need to determine the distances involved:
1. Distance between the front wheel and the rear wheels (L) = 16.0 m.
2. Distance between the front wheel and the center of gravity (d) = 12.1 m.
Now, let's solve the problem step by step:
(a) Normal force on the front wheel:
Since the plane is at rest on the runway, the vertical forces acting on it must balance out. The normal force exerted by the ground on the front wheel (Nf) must equal the weight of the plane (W).
Therefore, Nf = W = 2.03 x 10^6 N.
(b) Normal force on each of the two rear wheels:
Since the plane is at rest and in equilibrium, the sum of the moments about any point must be zero. We can choose the front wheel as the reference point.
The clockwise moment (or torque) due to the weight of the plane (W) acting at the center of gravity (12.1 m from the front wheel) must be balanced by the anticlockwise moment due to the normal force acting on the rear wheels.
The moments can be calculated using the formula: Moment = Force x Distance.
The moment due to the weight of the plane is given by: Moment(W) = W x d.
The moment due to the normal force on each rear wheel is given by: Moment(Nr) = 2 x Nr x L, as there are two rear wheels.
Since the plane is at rest, the sum of the moments must be zero.
Therefore, Moment(W) = Moment(Nr1) + Moment(Nr2), where Nr1 and Nr2 are the normal forces on each of the two rear wheels.
W x d = 2 x Nr x L.
We can rearrange this equation to solve for Nr:
Nr = (W x d) / (2 x L).
Substituting the given values, we have:
Nr = (2.03 x 10^6 N x 12.1 m) / (2 x 16.0 m).
Calculating this equation gives us:
Nr = 1.224 x 10^6 N.
Hence, the normal force exerted by the ground on each of the two rear wheels is 1.224 x 10^6 N.
## To determine the normal force exerted by the ground on the front wheel and each of the two rear wheels of the jet transport, we can use the principle of moments.
(a) Normal force on the front wheel:
Step 1: Calculate the weight distribution of the jet transport.
The weight of the jet transport is given as 2.03 x 10^6 N. Since the plane is at rest, the weight is balanced by the normal forces on the front and rear wheels. Therefore, the weight is distributed between these two wheels.
Let Wf be the normal force on the front wheel, and Wr be the normal force on each rear wheel.
According to the principle of moments, the net moment about the rear wheels must be zero for the system to be in rotational equilibrium. Therefore, we can use the equation:
(Wf × 16.0 m) - (Wr × 12.1 m) = 0
Step 2: Calculate the weight distribution.
The weight distribution is given by the equation:
Wf + Wr + Wr = 2.03 x 10^6 N
Rearranging the equation, we get:
Wf + 2Wr = 2.03 x 10^6 N
Step 3: Substitute the value of Wr from Step 1 into the weight distribution equation.
Substituting (Wr × 12.1 m)/16.0 m for Wf in the weight distribution equation, we get:
(Wr × 12.1 m)/16.0 m + 2Wr = 2.03 x 10^6 N
Step 4: Solve the equation to find Wr.
Multiply through by 16.0 m to eliminate the denominator:
(Wr × 12.1 m) + 2Wr × 16.0 m = (2.03 x 10^6 N) × 16.0 m
Simplifying:
12.1Wr + 32.0Wr = 32.48 x 10^6 N
44.1Wr = 32.48 x 10^6 N
Wr = (32.48 x 10^6 N) / 44.1
Wr ≈ 737,810.20 N
Step 5: Calculate the normal force on the front wheel.
Using the equation from Step 3:
Wf ≈ (737,810.20 N × 12.1 m) / 16.0 m
Wf ≈ 556,381.36 N
Therefore, the normal force exerted by the ground on the front wheel is approximately 556,381.36 N.
(b) Normal force on each rear wheel:
The normal force on each rear wheel is equal to Wr, which we calculated in Step 4.
Therefore, the normal force exerted by the ground on each of the two rear wheels is approximately 737,810.20 N. |
## Prime factorization
### Six consecutive numbers
B3, 2011
Prove that there are infinitely many perfect squares that can be written as a sum of six consecutive natural numbers. Find the smallest such square.
Solution
The sum of six consecutive numbers is of the form:
$\frac{(n+6)(n+7)}{2} - \frac{n(n+1)}{2} \text{ for some } n$
The above expression simplifies to $$3(2n+7)$$. This number is a perfect square whenever $$2n+7=3k^2$$. Any odd number greater than one can be substituted for $$k$$. The smallest value of $$k$$ is 3, so the smallest such square is 81.
$81 = 11+12+\cdots+16$
### Two variables one equation
C1, 2011
Show that there are exactly 16 pairs of integers $$(x,y)$$ such that $$11x+8y+17=xy$$.
Solution
An equation of the form $$xy=ax+by+c$$ can be written as:
$(x-a)(y-b) = c+ab$
So the given equation is:
$(x-8)(y-13) = 105$
The number of solutions to the above equation is the number of divisors of 105. We can factorize 105 as $$1\cdot 3\cdot 5\cdot 7$$. Every subset of these four numbers corresponds to a divisor, so there are $$2^4=16$$ solutions.
### Number of divisors
A1, 2019
For a natural number $$m,$$ define $$\Phi_{1}(m)$$ to be the number of divisors of $$m$$ and for $$k \geq 2$$ define $$\Phi_{k}(m):=\Phi_{1}\left(\Phi_{k-1}(m)\right) .$$ For example, $$\Phi_{2}(12)=\Phi_{1}(6)=4 .$$ Find the minimum $$k$$ such that
$\Phi_{k}\left(2019^{2019}\right)=2$
Solution
A number $$n$$ that has prime factorizaton $$p_1^{k_1} p_2^{k_2}\ldots p_t^{k_t}$$ has $$(k_1+1)(k_2+1)\ldots(k_t+1)$$ divisors.
IterationNumberFactorizationResult
1$$2019^{2019}$$$$3^{2019}\cdot 673^{2019}$$$$2020^2$$
2$$2020^2$$$$2^4\cdot 5^2\cdot 101^2$$$$45$$
3$$45$$$$3^2\cdot 5$$$$6$$
4$$6$$$$3\cdot 2$$$$4$$
5$$4$$$$2^2$$$$3$$
6$$3$$$$3$$$$2$$
Comment. A bit tedious task for a first problem. Is there a simpler way?
### Number of divisors II
A8, 2020
Find the number of positive integers $$n\leq 60$$ having $$6$$ divisors.
Solution
Suppose the prime factorization of $$n$$ is $$p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$$, then $$n$$ has $$(a_1+1)(a_2+1)\cdots (a_k+1)$$ factors. If $$n$$ has six factors, its prime factorizaton must be of the form $$p_1p_2^{2}$$ or $$p_1^5$$. There are nine numbers less than 60 that satisfy this condition:
$2\cdot3^2, 2\cdot 5^2, 3\cdot2^2, 5\cdot 2^2, 5\cdot 3^2, 7\cdot 2^2, 11\cdot 2^2, 13\cdot 2^2 \mbox{ and } 2^{5}$
### Prime factorization and perfect squares again
B3, 2013
A positive integer $$N$$ has its first, third and fifth digits equal and its second, fourth and sixth digits equal. In other words, when written in the usual decimal system it has the form $$x y x y x y,$$ where $$x$$ and $$y$$ are the digits. Show that $$N$$ cannot be a perfect power, i.e., $$N$$ cannot equal $$a^{b},$$ where $$a$$ and $$b$$ are positive integers with $$b>1$$.
Solution
We have: $N=\left(10^{5}+10^{3}+10\right) x+\left(10^{4}+10^{2}+1\right) y=10101(10 x+y)= 3 \times 7 \times 13 \times 37 \times(10 x+y)$
Therefore for $$N$$ to be a perfect power, the primes 3,7,13,37 must all occur as factors in the prime factorization of $$10 x+y .$$ In particular $$10 x+y \geq 10101$$. But since $$x$$ and $$y$$ are digits, each is between 0 and $$9,$$ so $$10 x+y \leq 99$$. So $$N$$ cannot be a perfect power.
### Prime factorization and divisibility
A6, 2014
Find the smallest $$n$$ for which $$\displaystyle \frac{50!}{24^n}$$ is not an integer.
Solution
The expression will not be an integer if and only if the numerator has a prime power lesser than the corresponding power in the denominator.
We have $$24=2^3\cdot 3$$. Let us find the the powers of 2 and 3 in $$50!$$.
The power of 2 is given by: $\lfloor 50/2 \rfloor + \lfloor 50/4 \rfloor + \lfloor 50/8 \rfloor + \lfloor 50/16 \rfloor + \lfloor 50/32 \rfloor = 47$
The power of 3 is given by:
$\lfloor 50/3 \rfloor + \lfloor 50/9 \rfloor + \lfloor 50/27 \rfloor = 22$
The numerator is of the form: $$2^{47}3^{22}\cdot k$$. If $$n=16$$, the denominator has $$2^{48}$$ as a factor which is sufficient to make the number a non-integer.
### When is $$a^2-a$$ divisible by 10000?
B3, 2015
(a) Show that there are exactly 2 numbers $$a$$ in $$\{2,3, \ldots, 9999\}$$ for which $$a^{2}-a$$ is divisible by $$10000 .$$ Find these values of $$a$$.
(b) Let $$n$$ be a positive integer. For how many numbers $$a$$ in $$\left\{2,3, \ldots, n^{2}-1\right\}$$ is $$a^{2}-a$$ divisible by $$n^{2} ?$$ State your answer suitably in terms of $$n$$ and justify.
Solution
(a) We have $$10000=16 \times 625$$ as product of prime powers. Recall the notation $$a \mid b,$$ meaning $$b$$ is divisible by a. We have $$10000 \mid a^{2}-a$$ if and only if $$(625 \mid a(a-1)$$ and $$16 \mid a(a-1)) .$$ Because $$a$$ and $$a-1$$ cannot share a factor, in turn this is equivalent to having both the conditions $$(1) 625 \mid a$$ or $$625 \mid a-1$$ AND (2) $$16 \mid a$$ or $$16 \mid a-1 .$$ Now if the coprime integers 16 and 625 both divide the same natural number (in our case $$a$$ or $$a-1),$$ their product 10000 will also divide this number. In our case this would force $$a=0,1,$$ or $$\geq 10000,$$ all of which are not allowed. Thus the given requirement on $$a$$ is equivalent to having either (1) $$16 \mid a$$ and $$625 \mid a-1$$ OR (2) $$16 \mid a-1$$ and $$625 \mid$$ a. Each case has a unique solution, respectively $$a=9376$$ and $$a=625$$ (e.g. use modular arithmetic: in case $$1,$$ we have $$a=625 k+1,$$ which is $$k+1$$ mod $$16,$$ forcing $$k=15$$ because $$16 \mid a$$ and $$a \in\{2,3, \ldots, 9999\})$$
(b) Let $$n=p_{1}{ }^{e_{1}} \ldots p_{k}{ }^{e_{k}}$$ be the factorization of $$n$$ into powers of distinct primes. The analysis in part (a) tells us that required values of $$a$$ are obtained as follows: write $$n^{2}=x y$$ as a product of two coprime integers and find values of $$a$$ in $$\left\{2,3, \ldots, n^{2}-1\right\}$$ that are simultaneously 0 mod $$x$$ and 1 mod $$y$$. These are precisely the values of $$a$$ that we want. This is because each $$p_{i}^{2 e_{i}}$$ must divide $$a$$ or $$a-1,$$ as $$a$$ and $$a-1$$ are coprime.
Now the Chinese remainder theorem tells us that there is always an $$a$$ that is 0 mod $$x$$ and 1 mod $$y$$. It is also unique modulo $$x y=n^{2}$$ because difference between any two solutions would be divisible by $$x y$$.
The total number of ways to write $$n^{2}=x y$$ as a product of coprime integers is exactly $$2^{k}$$ as it amounts to choosing which of the $$k$$ distinct primes to include in $$x$$ and then the rest go into $$y$$. (Notice that $$x$$ and $$y$$ are not interchangeable.) However, we have to delete the two cases $$x=1, y=n^{2}$$ and $$y=1, x=n^{2},$$ as these will respectively lead to solutions $$a=1$$ and $$a=0$$ or $$n^{2},$$ which are not in $$\left\{2,3, \ldots, n^{2}-1\right\} .$$ Finally it is easy to see that different choices of $$x$$ lead to different values of $$a$$. This is because, of the primes $$p_{1}, \ldots, p_{k}$$ in the factorization of $$n,$$ precisely the ones dividing $$x$$ will divide $$a$$ and the remaining primes will not, because they divide $$a-1$$.
Thus the final answer is $$2^{k}-2 .$$ Note that this matches with the special case in part (a). Finally, note that there was nothing special about taking a square: instead of $$n^{2}$$ it could be any positive integer $$m$$ and we would proceed the same way to find requisite integers $$a$$ in $$\{2,3 \ldots, m-1\}$$ based on prime factorization of $$m$$
### Primes in an algebraic equation ＊
B6, 2016
Find all pairs $$(p, n)$$ of positive integers where $$p$$ is a prime number and $$p^{3}-p=n^{7}-n^{3}$$
Solution
The given equation is $$p(p-1)(p+1)=n^{3}\left(n^{2}+1\right)(n+1)(n-1) .$$ As the factor $$p$$ on the LHS is a prime, it must divide one of the factors $$n-1, n, n+1, n^{2}+1$$ on the RHS.
Key lemma. $$p>n^{2}$$
Proof The LHS $$=p^{3}-p$$ is an increasing function of $$p$$ for $$p \geq 1,$$ e.g. because the derivative $$3 p^{2}-1$$ is positive. So for any given $$n \geq 1$$, there is exactly one real value of $$p$$ for which LHS=RHS. Trying $$p=n^{2}$$ gives LHS$$=n^{6}-n^{2}< n^{7}-n^{3}=$$RHS. This is because $$n^{7}-n^{3}-\left(n^{6}-n^{2}\right)=\left(n^{6}-n^{2}\right)(n-1)>0$$. $$\,\square$$
As the prime $$p$$ is greater than $$n^{2},$$ it cannot divide any of $$n-1, n, n+1 .$$ So $$p$$ must divide $$n^{2}+1$$ and therefore we must have $$p=n^{2}+1,$$ again because $$p>n^{2} .$$ Substituting this in the given equation, we get $$\left(n^{2}+1\right) n^{2}\left(n^{2}+2\right)=n^{3}\left(n^{2}+1\right)(n+1)(n-1)$$. Canceling common factors gives $$n^{2}+2=n^{3}-n,$$ i.e. $$2=n^{3}-n^{2}-n .$$ This has a unique integer solution $$n=2,$$ e.g. because the factor $$n$$ on the RHS must divide 2 and now one checks that $$n=2$$ works. So $$n=2$$ and the prime $$p=n^{2}+1=5$$ give a unique solution to the given equation. |
# Word Problem With Multiple Decimal Operations: Problem Type 2 Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to Word Problem With Multiple Decimal Operations: Problem Type 2. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Moshe pays for his new car in 36 monthly payments. If his car costs $20,975.64, and he makes a down payment of$1000, how much will Moshe pay each month?
A - $555.88 B -$554.89
C - $554.88 D -$554.98
### Explanation
Step 1:
Cost of car = $20,975.64; Down payment =$1000
Number of months = 36
Step 2:
Each monthly instalment = ($20,975.64 −$1000)/36
= 19, 975.64/36 = $554.88 Q 2 - Sasha paid$30 to buy 7 hot dogs and got back change of $4.17. How much did each hot dog cost? A -$3.69
B - $3.79 C -$3.68
Step 2:
A - 6 classes
B - 7 classes
C - 5 classes
D - 8 classes
### Explanation
Step 1:
Amount charged for classes = $225 −$45 = $180 Amount charged per class =$30
Step 2:
Number of classes = 180/30 = 6
Q 6 - If the division of p by 13.2 gives a quotient of 9.8 and remainder 8, find p.
A - 138.36
B - 137.46
C - 137.36
D - 137.63
### Explanation
Step 1:
p/13.2 = 9.8 R 8
Step 2:
p = 9.8 × 13.2 + 8 = 129.36 + 8 = 137.36
Q 7 - A papaya is 0.5 Kg heavier than 1/3th the weight of a water melon. If the watermelon has a weight of 5.1 kg, what is the weight of the papaya?
A - 2.1 Kg
B - 2.2 Kg
C - 2.3 Kg
D - 2.02 Kg
### Explanation
Step 1:
Weight of water melon = 5.1 Kg
Step 2:
Weight of papaya = 1/3 × 5.1 + 0.5 = 1.7 + 0.5 = 2.2 Kg
Q 8 - Jack bought 9.75 kg of sugar. He poured the sugar equally into 5 bottles. There was 0.45 kg of sugar left over. What was the weight of sugar in 1 bottle?
A - 1.85 Kg
B - 1.96Kg
C - 1.87 Kg
D - 1.86 Kg
### Explanation
Step 1:
Amount of sugar in 5 bottles = 9.75 – 0.45 = 9.3 Kg
Step 2:
Weight of one sugar bottle = 9.3/5 = 1.86 Kg
Q 9 - A carton containing 12 tins of powdered milk weighs 8.7 kg including its own weight of 0.3 kg. What is the weight of each tin of powdered milk?
A - 0.701 Kg
B - 0.600 Kg
C - 0.700 Kg
D - 0.800 Kg
### Explanation
Step 1:
Weight of the 12 tins of milk powder = 8.7 – 0.3 = 8.4 Kg
Step 2:
Weight of 1 tin of milk powder = 8.4/12 = 0.700 Kg
Q 10 - Mike bought 5 CDs that were each the same price including tax. He paid a total of $63.75. Each CD had a tax of$0.30. What was the price of each CD before tax?
A - $12.45 B -$12.55
C - $12.65 D -$13.45
### Explanation
Step 1:
Cost of 5 CDs including tax = $63.75 Cost of 1 CD including tax = 63.75/5 =$12.75
Tax on each CD = $0.30 Step 2: Cost of 1 CD before tax =$12.75 − $0.30 =$12.45
word_problem_with_multiple_decimal_operations_problem_type2.htm |
# Solve domain and range
This can help the student to understand the problem and how to Solve domain and range. We will give you answers to homework.
## Solving domain and range
These can be very helpful when you're stuck on a problem and don't know how to Solve domain and range. Here, let's say that the UV solution of each pixel in rasterize is actually done with the barycentric coordinates. The barycentric coordinates represent a weight, that is, with the barycentric coordinates, the properties of a certain point on the triangle can be obtained by the weighted average of three vertices. The third model of the triangular pyramid receiving ball is to determine the star to construct a right triangle. This model usually appears in the center of a curved regular triangle in the triangular pyramid of a postal triangle.
Since we regard the state to be solved as an arbitrary state, the final differential equation actually replaces the whole by studying the tiny local area. We think that as long as the local area is solved, it can be directly extended to the whole or the whole through recursion, so recursion is the essential idea of global solution. In practical application, many problems are quite complex, and the differential equations constructed are also extremely complex. It is impossible to get the expression of y = f (x).
In addition, it should be noted that the number of calculation questions this year is still quite large, and on the admission ticket for the postgraduate examination of Tsinghua University, the unified requirements for the preliminary examination are not allowed to use calculators. Therefore, when doing examples, it is best to exercise your written arithmetic ability. Of course, we don't need to worry too much. The problems requiring written calculation only involve some addition and subtraction methods and simple integration. For exponential and logarithmic operations, you can express them without calculating the specific values.
May we grow up with children and achieve each other! There are often two voices for dealing with emotions first or solving problems first. One voice is: we must solve the problem first, because bad emotions are caused by problems. If the problem is not solved, the mood will not be better. Another voice is: deal with emotions first, because good emotions are the prerequisite for solving problems.
## We solve all types of math troubles
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Xavienna Moore
Great app I love using this to help me understand any mistakes I've made in a math problem and how the problem SHOULD be done. I would really like it if the "why" button was free but overall, I think it's great for anyone who is struggling in math or simply wants to check their answers.
Gabrielle Long |
"
Mental methods are about trying to get to the correct answer in the quickest and easiest way! There are lots of ways to add and subtract numbers in your head. By the time you have looked through the factsheets in this module you will have used 5 different methods for mental addition and subtraction! Try to remember them by using the word "Shared".
• • •
When you've read about them, practise the methods you like and can remember most easily. Ask other people about the methods they use - and share your methods too! Keep practising, and HAVE FUN!
Here are three reminders to help you with mental addition and subtraction. A reminder about place value Have a look at the number 623.
• • •
6 is the hundreds digit. 2 is the tens digit. 3 is the units digit.
Here you can see that the same numbers added together in a different order will give the same answer. Addition and subtraction are opposites. You can check the answer to a subtraction sum by turning the numbers around and adding them up. Have a look below.
Here you've done the sum 10 - 25 and got the answer 15. To check the answer, turn the sum around to 15 + 10 and see if you get 25.
Splitting up numbers
Splitting up numbers is a good method to use for both addition and subtraction. It is sometimes called partitioning. Addition Take a look at this addition sum: 80 + 49 To make it easier, split the 49 into 40 + 9. This makes the sum: 80 + 40 + 9 = 129 First, add the first two numbers: 80 + 40 = 120 Then add the result of that sum to the third number to get the answer: 120 + 9 = 129 Subtraction Take a look at this subtraction sum: 150 - 34 To make it easier, split the - 34 into 30 - 4. This makes the sum: 150 - 30 - 4 First, subtract the 30 from 150: 150 - 30 = 120 Then, subtract the 4 from the 120 to get the answer: 120 - 4 = 116
Hundreds, tens and ones
To make addition sums easier, you can separate the hundreds, tens and units and add them up separately. Have a look at how separating works for this sum: 31 + 22
Using this method you can work out that 31 + 22 = 53. Now look at this sum with hundreds as well as tens and units: 125 + 100 + 235 + 132
Using this method you can work out that 125 + 100 + 235 + 132 = 592.
Rounding is a method for mental addition which is useful in many different situations. Imagine you are in a shop and you have to quickly work out an amount. £3.70 + £1.00 = £4.70. Taking away 10p gives £4.60 Have a look at how this addition sum can be solved with rounding: 75 + 19 First, round the 19 up to 20 and work out the sum: 75 + 20 = 95
As 20 is 1 more than 19, you then need to subtract 1 from the total. 95 - 1 = 94 Then you can see that: 75 + 19 = 94 Further reading. These tables have some more information to help you with addition by rounding. The first shows methods and examples for adding a number between 11 and 14 to another number.
This table shows methods and examples for adding a number between 15 and 19 to another number.
Rounding - subtraction
Rounding is a method for mental subtraction which is useful in many different situations. Imagine you are shopping and need to work out an amount quickly. £3.70 - £1.00 = £2.70. Adding 10p gives £2.80. Have a look at how this subtraction sum can be solved with rounding: 64 - 17 First, round the 17 up to 20 and work out the sum: 64 - 20 = 44 As 20 is 3 more than 17, you have taken 3 too many from the total. So you need to add 3: 44 + 3 = 47 So you can see that: 64 - 17 = 47 Further reading. These tables have some more information to help you with subtraction by rounding. The first shows you a methods and examples for subtracting a number between 11 and 14 from another number.
This table shows methods and examples for subtracting a number between 15 and 19 to another number.
Empty number line for counting on
Counting on using an empty number line is a good method for subtracting numbers mentally. Use this method to find the difference between 37 and 50. This is the same as the sum 50 - 37.
When you have pictured that line, count on from 37 to 40, which makes 3. Keep that 3 in your head. Then, count from 40 to 50, which is 10. Have a look below to see how this works.
Now all you need to do is add the 3 to the 10. This makes 13. So: The difference between 37 and 50 is 13. Or 50 - 37 = 13
Doubling
If you are adding together two numbers that are nearly the same, you can double one of them and then adjust the difference. Imagine you are adding together 38 and 35.
Here are some of the words which will crop up when doing addition sums.
Have a look below to see how they can be used in the simple sum 3 + 4 = 7. Add 3 add 4 is 7 Altogether Altogether, 3 and 4 make 7. Increase If you increase 3 by 4 you get 7. More 7 is 3 more than 4. Plus 3 plus 4 is 7. Sum The sum of 3 and 4 is 7. Total The total of 3 and 4 is 7.
Key words for mental subtraction
Here are some of the words which will crop up when doing subtraction sums.
Have a look below to see how they can be used in the simple sum 8 - 5 = 3. Decrease If you decrease 8 by 5 you get 3.
Difference The difference between 8 and 5 is 3. Fewer than 3 is 5 fewer than 8. Less than 3 is 5 less than 8. Minus 8 minus 5 is 3. Reduce If you reduce 8 by 5 you get 3. Subtract 8 subtract 5 is 3. Take away 8 take away 5 is 3.
Mental multiplication methods
Mental methods are about trying to get to the correct answer in the quickest and easiest way! Here are some of the mental methods you can use. When you've read about them, practise the methods you like and can remember most easily. Multiplying the tens then the units, then adding them together.
Rounding up one number to the nearest 10 and adjusting the answer.
Doubling one number, then halving the answer.
Tip: Have you ever worried that if it's called 'mental methods' you have to do it in your head? Well, you don't have to! It can really help to jot down some figures which make the sum easier for you.
Some more mental multiplication methods
Mental methods are about trying to get to the correct answer in the quickest and easiest way! Here are two more mental methods you can use. When you've read about them, practise the methods you like and can remember most easily. Changing the order to make the numbers easier to work with
Numbers can be split into factors to make multiplying simpler.
Tip: Remember! Using mental methods is about choosing the method that works for you and for the numbers you're working with.
Multiplication Glossary
Here are some of the words which will crop up when doing multiplication sums.
Have a look below to see how they can be used in the simple sums 2 x 2 = 4. Factors 2 is a factor of 4. One number is a factor of another number if it divides, or goes into it exactly. Divisible 6 is exactly divisible by 3. 7 is not exactly divisible by 3. Groups of 2 groups of 2 make 4. Lots of 2 lots of 2 make 4. Multiple 4 is a multiple of 2. Multiply If you multiply 2 by 2 you get 4. Product The product of 2 and 2 is 4. Sets of 2 sets of 2 make 4. Times 2 times 2 is 4.
Mental division tips
A division sum can be shown in several different ways.
Estimating When you divide any numbers, it is a good idea to estimate a rough answer first. Your estimate can then be checked against your actual answer. 92 ÷ 3 is approximately 90 ÷ 3 which is 30 143 ÷ 7 is approximately 140 ÷ 7 which is 20 994 ÷ 5 is approximately 1 000 ÷ 5 which is 200 Check by multiplying Multiplication and division are inverses (opposites). Division sums can be checked by multiplying, like this: 81 ÷ 3 = 27 27 x 3 = 81 Jot it down Have you ever worried that if it's called 'mental methods' you have to do it in your head? Well, you don't have to! It can really help to jot down some figures which make the sum easier for you.
Some mental division methods
Mental methods are about trying to get to the correct answer in the quickest and easiest way. Here are two of the mental methods you can use. When you've read about them, practise the methods you like and can remember most easily. Splitting the number you're dividing into, to make it simpler.
Numbers can be split into factors to make dividing simpler.
Some more mental division
Mental methods are about trying to get to the correct answer in the quickest and easiest way. Here are two of the mental methods you can use. When you've read about them, practise the methods you like and can remember most easily. Spacesaver division This is long division without all the written bits! Let's look at the sum 22 972 ÷ 4.
1. 4 into 2 won't go - so carry 2 2. 4 into 22 (5 x 4 = 20) - so carry 2 3. 4 into 29 (7 x 4 = 28) - so carry 1 4. 4 into 17 (4 x 4 = 16) - so carry 1 5. 4 into 12, that will be 3 exactly With this method you're doing a division sum, but all the thinking is multiplication and subtraction! Dividing with even numbers 120 ÷ 40 is the same as: (keep halving both numbers) 60 ÷ 20 30 ÷ 10 15 ÷ 5 which is 3 Tip: Remember! Using mental methods is about choosing the method that works for you and for the numbers you're working with. If you're in a group you could vote for the most popular methods, then let us know the result!
Division Glossary
Here are some of the words which will crop up when doing division sums.
Have a look below to see how they can be used in the simple sums 6 ÷ 3 = 2 and 7 ÷ 3. Divide If you divide 6 by 3 you get 2. Divisible 6 is exactly divisible by 3. 7 is not exactly divisible by 3. Groups There are 3 groups of 2 in 6. Left over If you divide 7 by 3 the answer is 2 with 1 left over. Remainder If you divide 7 by 3 the answer is 2 with 1 remainder. Share If you share 6 toffees between 3 people, each person gets 2.
What is ratio?
Ratio is a way of comparing amounts of something. It shows how much bigger one thing is than another. For example:
• • •
Use 1 measure screen wash to 10 measures water Use 1 shovel of cement to 3 shovels of sand Use 3 parts blue paint to 1 part white
Ratio is the number of parts to a mix. The paint mix is 4 parts, with 3 parts blue and 1 part white. The order in which a ratio is stated is important. For example, the ratio of screenwash to water is 1:10. This means for every 1 measure of screenwash there are 10 measures of water. Mixing paint in the ratio 3:1 (3 parts blue paint to 1 part white paint) means 3 + 1 = 4 parts in all.
3 parts blue paint to 1 part white paint = is ¾ blue paint to ¼ white paint. If the mix is in the right proportions, we can say that it is in the correct ratio.
Understanding direct proportion
Two quantities are in direct proportion when they increase or decrease in the same ratio. For example you could increase something by doubling it or decrease it by halving. If we look at the example of mixing paint the ratio is 3 pots blue to 1 pot white, or 3:1.
But this amount of paint will only decorate two walls of a room. What if you wanted to decorate the whole room, four walls? You have to double the amount of paint and increase it in the same ratio. If we double the amount of blue paint we need 6 pots. If we double the amount of white paint we need 2 pots.
The amount of blue and white paint we need increase in direct proportion to each other. Look at the table to see how as you use more blue paint you need more white paint: Pots of blue paint Pots of white paint 3 1 6 2 9 3 12 4
Have a look at this graph:
Two quantities which are in direct proportion will always produce a graph where all the points can be joined to form a straight line.
Using direct proportion
Understanding proportion can help in making all kinds of calculations. It helps you work out the value or amount of quantities either bigger or smaller than the one about which you have information. Here are some examples: Example 1: If you know the cost of 3 packets of batteries is £6.00, can you work out the cost of 5 packets? To solve this problem we need to know the cost of 1 packet. If three packets cost £6.00, then you divide £6.00 by 3 to find the price of 1 packet. (6 ÷ 3 = 2) Now you know that they cost £2.00 each, to work out the cost of 5 packets you multiply £2.00 by 5. (2 x 5 = 10) So, 5 packets of batteries cost £10.00 Example 2: You've invited friends round for a pizza supper. You already have the toppings, so just need to make the pizza base. Looking in the recipe book you notice that the quantities given in the recipe are for 2 people and you need to cook for 5! Pizza base - to serve 2 people: 100 g flour 60 ml water 4 g yeast 20 ml milk pinch of salt The trick here is to divide all the amounts by 2 to give you the quantities for 1 serving. Then multiply the amounts by the number stated in the question, 5. For 1 serving, divide by 2: 100 g ÷ 2 = 50 g 60 ml ÷ 2 = 30 ml 4 g ÷ 2 = 2 g 20 ml ÷ 2 = 10 ml For 5 servings, multiply by 5: 50 g x 5 = 250 g 30 ml x 5 = 150 ml 2 g x 5 = 10 g 10 ml x 5 = 50 ml The pinch of salt is up to you!
Simplifying ratios
We can often make the numbers in ratios smaller so that they are easier to compare. You do this by dividing each side of the ratio by the same number, the highest common factor. This is called simplifying. Example: In a club the ratio of female to male members is 12:18 Both 12 and 18 can be divided by 2. 12 ÷ 2 = 6 18 ÷ 2 = 9 So a simpler way of saying 12:18 is 6:9. To make the ratio simpler again, we can divide both 6 and 9 by 3 6÷3=2 9÷3=3 So a simplest way of saying 12:18 is 2:3. These are all equivalent ratios, they are in the same proportion. All these ratios mean that for every 2 female members in the club there are 3 males: 12:18 6:9 2:3 2:3 is easier to understand than 12:18!
Tips for ratio and proportion sums
Ratio can be used to solve many different problems, for example recipes, scale drawing and map work. Changing a ratio A common test question will ask you to change a ratio - the reverse of cancelling down. Example: A map scale is 1 : 25 000. On the map the distance between two shopping centres is 4 cm. What is the actual distance between the shopping centres? Give your answer in km. A scale of 1 : 25 000 means that everything in real life is 25 000 times bigger than on the map. So 4 cm on the map is the same as 4 x 25 000 = 100 000 cm in real life. (Reminder 1 m = 100 cm and 1 km = 1 000 m) Now change the real life distance of 100 000 cm to metres 100 000 ÷ 100 = 1 000 m And 1 000 m is the same as 1 km. So the shopping centres are 1 km apart.
Keeping things in order When working with ratios keep both the words and the numbers in the same order as they are given in the question. Example: Share a prize of £20.00 between Dave and Adam in the ratio 3:2. The trick with this type of question is to add together the numbers in the ratio to find how many parts there are, divide by the number of parts to find the value of 1 part, then multiply by the number of parts you want to calculate.
• • • • •
First add together the number of parts in the ratio: 3 + 2 = 5 Divide to find out how much 1 part will be: £20.00 ÷ 5 = £4.00 To find Dave's share multiply £4.00 x 3 = £12.00 Adam's share is £4.00 x 2 = £8.00 Dave's £12.00 is of 5 parts). of £20.00 (3 of 5 parts). Adam's £8.00 is of £20.00 (2
You can check that you have worked out the ratio correctly by adding the shares together. In this sum Dave's and Adam's shares should equal £20.00 Let's check: £12.00 + £8.00= £20.00 Correct! Use the same units Always check that the things you are comparing are measured in the same units. Example: Jenna has 75 pence. Hayley has £1.50 What is the ratio of Jenna's money to Hayley's?. In this problem one amount is in pence, the other in pounds. Before you calculate the ratio you have to make sure they are the same units. We have to convert Hayley's amount into pence first. There are 100 pence to a pound Hayley's £1.50 = 150 pence So the ratio is 75 : 150 You can simplify this ratio as both numbers are divisble by 75. The ratio is 1:2.
Key words for ratio and proportion
Ratio is a way in which quantities can be divided or shared. Example: Share £20 between two people in a ratio of 3:1. A ratio of 3 + 1 = 4 parts, the money needs to be divided into 4 parts. 20 ÷ 4 = £5. If one person is getting three parts they will have 3 x 5 = £15
The other person will have one part, £5.
Simplest form. Ratios can be simplfied by finding common factors. Direct proportion. Ratios are in direct proportion when they increase or decrease in the same ratio. Equivalent ratios. When both sides of a ratio can be multiplied or divided by the same number to give an equivalent ratio. Example: In a group there are 15 males and 12 females. What is the ratio of males to females? Give your example in its simplest form. So the ratio of males to females is 15:12. However, both sides of the ratio can be divided by 3. Dividing 15 and 12 by 3 gives 5:4. 5:4 is the ratio in its simplest form. 5:4 and 15:12 are equivalent ratios.
Factor The factors of a number are those numbers which divide into it exactly. Example: 1 x 12 = 12 2 x 6 = 12 3 x 4 = 12 So the factors of 12 are 1, 2, 3, 4, 6 and 12.
What is rounding?
Rounding is a way of simplifying numbers. If the driveway of a house is 5 metres and 7 cm long we would usually just say it is 5 m long. Saying it's 5 m long will be close enough most of the time. Here is another example. The picture shows a stick of rock next to a ruler. The ruler has only got the 10 cm points marked on it.
We can't see exactly how long the rock is. But we can see to the nearest 10 cm. The end of the rock is close to the 20 cm mark. So we say that the rock is 20 cm long to the nearest 10 cm. What about this longer stick. How long is it to the nearest 10 cm?
It is closer to 30 cm than 20 cm. So we say it is 30 cm long to the nearest 10 cm. Rounding numbers to the nearest 10 means finding which 10 they are nearest to. Rounding a number to the nearest hundred or to the nearest thousand can be done in the same kind of way. There is more about this in the other factsheets.
Rules for rounding
Example A stick of rock is 27 cm long. How long is it to the nearest 10 cm? Answer 27 cm is between 20 cm and 30 cm. So 27 cm will get rounded to either 20 cm or 30 cm. To get the right answer we need to decide whether 27 is nearer to 20 or 30.
You can see from the picture that it is closer to 30. So 27 cm is rounded up to 30 cm. So the stick of rock is 30 cm long when we measure to the nearest 10 cm. For the same reasons 26, 27, 28 and 29 all get rounded up to 30. And 21, 22, 23 and 24 all get rounded down to 20 What about 25? It's exactly half way between 20 and 30. It has to be rounded one way or the other. The rule that everyone usually follows is that 25 gets rounded up to 30. The Rules In this way we get the rules about rounding up and down. 1, 2, 3 and 4 get rounded down 5, 6, 7, 8 and 9 get rounded up These rules work for all numbers, whether you are using tens, hundreds or thousands (or anything else). There is more about these rules in the other factsheets.
Rounding tens, hundreds and thousands
Rounding a number is another way of writing a number approximately. We often don't need to write all the figures in a number, as an approximate one will do. For a population of 27 653 the number is large and will change daily. It is better to round up and say 28 000.
Rounding to the nearest ten
To round a number to the nearest 10, you have to decide if the number is nearest to 10, 20 30 etc. To do this you follow a rule.
Is 37 nearer to 30 or to 40?
As the unit figure is 7, you round up to 40. Rounding to the nearest 10 can help you estimate the cost of your shopping.
Rounding to the nearest hundred
To round a number to the nearest 100, you have to decide if the number is nearest to 100, 200, 300 etc. The rule is the same as for rounding to the nearest 10, but this time look at the tens figure.
Is 236 nearer to 200 or to 300?
As the tens figure is 3, you round down to 200. Rounding to the nearest 100 can help you estimate your yearly spending on rent or mortgage.
Rounding to the nearest thousand
To round a number to the nearest 1 000, you have to decide if the number is nearest to 1 000, 2 000, 3 000 etc. Follow the rules as above now looking at the hundreds figure.
Is 8 572 nearer to 8 000 or to 9 000?
As the hundreds figure is 5, follow the rule and round up to 9 000. When a figure is halfway between two hundreds, the rule is to round up. Rounding to the nearest 1 000 can help you estimate the number of people who attended a pop concert or football match.
Example
If 43 715 tickets were sold for a football match, that number could be rounded to the nearest ten, hundred or thousand:
• • •
rounding 43 715 to the nearest 10 would give 43 720 rounding 43 715 to the nearest 100 would give 43 700 rounding 43 715 to the nearest 1,000 would give 44 000.
Estimating using rounding
We can use rounding numbers to get a rough idea or an estimate. An estimate might be a little more or a little less than the actual amount. By carrying out an estimate we can check that the answers to problems are sensible. If you were buying 9 identical shirts for the school's sports team that cost £7.80 each, to get a rough idea of the total cost you could round up £7.80 to £8.00. You could also round up 9 shirts to 10 shirts. Your calculation would then be 10 x £8.00 = £80.00 The actual cost would be 9 x £7.80 = £70.20 Notice that the actual cost of £70.20 is a little less than our £80.00 estimate. This is because we rounded up. When using a calculator it is a good idea to estimate the answer first in case you make keying errors.
Example
To estimate the cost of 11 pens at 95p each, you could round down 11 to 10 pens and round up 95p to £1.00 The estimated cost would then be 10 x £1.00 = £10.00
Key words for rounding and estimating
Here are some words that you'll come across when rounding and estimating. For example with the sum: 197 – 50 = 147 Rounding To write a number to a given amount of accuracy. Rounding 197 to the nearest 100 would be 200. Estimate To give a rough answer that may be a little less or a little more than the actual result. To estimate 197 – 50 you may instead work out 200 – 50 to give an estimate of 150. Approximate An answer that is not exactly correct but is close enough to be useful in working out a sum. To approximate 197 – 50 you may instead work out 200 – 50 to give an estimate of 150. To the nearest.. rounding off.. A guide to how accurate your rounding needs to be. Rounding 197 to the nearest 100 would be 200. Rounding off 147 to the nearest 10 would be 150. Actual The correct answer to a sum. The estimate of 197 – 50 is 150. The actual answer is 147.
Writing big numbers
We come across large numbers in our everyday life so it is important to be able to read them. To help with numbers that have more than five figures, which might be difficult to read, we use place value. Place value is the idea that a figure has a different value when used in different places.
Below is a place value table with the numbers 7 853 and 5 387. Note: Each column can only contain one figure from 0 to 9.
In the number 7 853 (seven thousand eight hundred and fifty three) the 7 has the value 7 thousand. This number is 7 000 + 800 + 50 + 3. In the number 5 387 (five thousand three hundred and eighty seven) the 7 has the value 7 units. This number is 5 000 + 300 + 80 + 7. In these two numbers the 7 stands for different values when it is in different places.
Writing numbers up to a million
Using the place value table can help you to write large numbers. Look at the following numbers: Numbers in figures 10 100 1 000 10 000 100 000 1 000 000 Numbers in words Ten Hundred Thousand Ten thousand Hundred thousand Million
You will notice that the numbers are grouped in three figures. There is a space between each group of three figures (counting from right to left). You will sometimes see a comma used to separate the three figures. (If there is no comma in a large number and you have problem saying it, try putting in the comma.) This grouping can help you to say the number 405 000. The first group of three figures is four hundred and five and the last three figures
show thousands (since there are three zeros in a thousand). The number is four hundred and five thousand. This example shows how important zero is. In 405 000 the zero between the 4 and 5 keeps the place for the missing tens of thousands. Without the zero, the number is forty five thousand (45 000).
Writing numbers in words in figures
There are times when you may need to write down a large number in figures that someone has told you in words. Newspaper stories often have large numbers written in figures that may be difficult to make sense of unless you can say them in words. Examples 1. Write five thousand, three hundred and six in figures. Put the 5 in the thousands column and the 3 in the hundreds column. The 6 should go in the units column so make sure you fill the tens column with a 0 to show no tens. Th 5 H 3 T 0 U 6
The number is 5 306. 2. Write twenty six thousand, seven hundred and fifty in figures. For this number we would start with the tens of thousands column. Any number larger than 9 999 would have 5 figures. Start with the 2 in the tens of thousands column and continue by putting the 6 in the thousands column, the 7 in the hundreds column and the 5 in the tens column. The units column must have a 0 to show no units. TTh 2 Th 6 H 7 T 5 U 0
The number is 26 750. 3. Write 58 432 in words. TTh 5 Th 8 H 4 T 3 U 2
Again by writing in the letters TTh, Th, H, T and U above the number can help with writing it in words. Grouping the numbers in threes from right to left will let you know that the number must be fifty eight thousand and something. The last three figures can be read on their own as four hundred and thirty two. The number is fifty eight thousand, four hundred and thirty two. 4. Write 1 200 in words. Th 1 H 2 T 0 U 0
Start by writing the letters Th, H, T and U above the number you have been given. You can then see that the 1 is in the thousands column and the 2 is in the hundreds column. The number is one thousand, two hundred. Note: Another way of writing this number would be twelve hundred. Although this is not incorrect if you think writing numbers such as these as twelve hundred might confuse you, stick to writing them in terms of thousands.
Ordering large numbers
When you have a series of large numbers, which are not in number order, it is sometimes difficult to make sense of them. Here is a table showing the daily profits of a supermarket written in order of days.
If these numbers were put into a place value table, it would be easier to arrange them in order.
Look at each column in turn. The figures for Friday and Saturday will be the largest as these have figures in the tens of thousands column. Looking at the thousands column shows that since there is a 4 in the thousand column for Saturday and a 0 in the thousands column for Friday that Saturday has the largest number. Carry on for each of the other numbers.
Writing figures in words
0 1 2 3 4 5 6 7 8 9 zero one two three four five six seven eight nine 10 11 12 13 14 15 16 17 18 19 ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen 20 30 40 50 60 70 80 90 100 1 000 1 000 000 twenty thirty forty fifty sixty seventy eighty ninety hundred thousand million
Big numbers glossary
Here are some of the words you may come across to do with big numbers. Place value A figure has a different value when used in different places. For example, in these three numbers, the 4 stands for a different value: 45 The number 4 has a value of 40 (4 tens) 405 The number 4 has a value of 400 (4 hundreds) 54 The number 4 has a value of 4 (4 units) Digit A figure or a number. 45 is a two-digit number whereas 405 is a three-digit number. Billion When we talk about a billion we mean a thousand million or 1 000 000 000. If you see a billion in a news story it is referring to a thousand million. Such big numbers can be difficult to imagine.
Numerical order The order that you would write numbers if you were counting from the lowest up. 405, 406 and 407 are in numerical order. Unit The word unit means one. It is the smallest number and is always on the righthand side of a whole number: 5 This number has 5 units 72 This number has 2 units 591 This number has 1 unit
Multiples
Multiples of a number can be made by multiplying the number by any whole number. The first four multiples of 2 are 2, 4, 6 and 8. You get them by doing 2 x 1, 2 x 2, 2 x 3 and 2 x 4 The numbers you find in the 2-times table are all multiples of 2.
Reminder: when you do multiplication you can write the numbers in any order and get the same answer. 6 x 2 is the same as 2 x 6. Here is how to make multiples of 10. Just multiply 10 by a whole number each time. 1 x 10 = 10, 2 x 10 = 20, 3 x 10 = 30, 4 x 10 = 40, 5 x 10 = 50, 6 x 10 = 60, and so on ... The first six multiples of 10 are 10, 20, 30, 40, 50 and 60. Example 1 Is 12 a multiple of 3? If you multiply 3 by 4 you get 12, so 12 is a multiple of 3. Example 2 20 is a multiple of 5 because 4 x 5 = 20. 20 is a multiple of 4 too, because 5 x 4 = 20. Example 3 Is 15 a multiple of 3? 3 x 5 = 15. So 15 is a multiple of 3, (and also of 5). Example 4 Is 21 a multiple of 6? 21 is not a multiple of 6 because you can't make 21 by multiplying 6 by any whole number. 6 x 3 = 18 and 6 x 4 = 24 but there is no whole number between 3 and 4 that could give us an answer of 21. Example 5 Is 30 a multiple of 15? 30 = 2 x 15, so 30 is a multiple of 15. You can also see that 2 x 3 x 5 = 30 so 30 is a multiple of 2, 3 and 5. And 30 = 3 x 10 so 30 is a multiple of 10. Also 30 = 5 x 6 so 30 is a multiple of 6 too.
Factors
In arithmetic, a factor is a whole number that divides exactly into another whole number. For example, what are the factors of 12? Try making 12 in different ways. Your answer should look like this: 6 x 2 = 12 12 x 1 = 12 4 x 3 = 12 Remember that you can write your numbers in any order you like for a multiplication so: 2 x 6 is the same as 6 x 2 1 x 12 is the same as 12 x 1 3 x 4 is the same as 4 x 3.
The full list of factors of 12 is 1, 2, 3, 4, 6, and 12. Some numbers have many factors, so it is a good idea to work in an organised way or you may miss some. Don't forget to include 1 and the number itself in your list. Here is one way to find the factors of 48. Start with 1 and pair off your numbers. 1 x 48, 2 x 24, 3 x 16, 4 x 12 and 6 x 8 all make 48. Write the list in order: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Here is another way: Write your first pair of factors with a reasonable space between them, then move on to the next pair until you have them all. (You don't need to put in the lines.) This way, when you get to the 6,8 pair, you can stop because 7 is not a factor and you already have 8 in your list.
Sequences
A sequence is a set of numbers arranged in order according to a rule. Each number in a sequence is called a term. Multiplication tables give good examples of sequences. For example the 2-times table gives you the sequence 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ... and so on Each term comes from the 2-times table. The rule for this sequence is 'add 2' each time. The first three terms of the four times table are 4, 8, 12. You can see that each term in the sequence increases by four. If you carried on with this sequence you would eventually reach 92 (try it!). What is the next term after 92? Using the rule of adding four each time gives you the next term, 96 (because 92 + 4 = 96). Example What is the next term in the sequence 35, 32, 29, 26, ...? This time each term is three less than the one before it. Using this rule (take three away each time) gives the fifth term as 23, because 26 - 3 = 23. Example A sequence begins 64, 32, 16. What are the next two terms? The numbers are decreasing, but not by equal amounts. The rule for this sequence is 'Divide by two'. The next term will be 8, because 16 ÷ 2 = 8. The term after that will be 4, because 8 ÷ 2 = 4. Example What are the next two terms in the sequence 1, 2, 4, 8, ? The rule is 'multiply by 2 each time'. The next two terms are 8 x 2 = 16 and then 16 x 2 = 32. Example Look at this sequence: 3, 5, 8, 12, ... It doesn't follow any of the rules above. But if you look at the differences between
each pair of terms, you can see that they are 2, 3 and 4. The next difference will be 5 and so the fifth term is 17, because 12 + 5 = 17.
Number patterns
Some sequences can be shown as number patterns, for example:
The difference between one term and the next is 2. So the next term in the sequence will be 10. (We don't need the pattern to work it out.) Example Here the patterns are made from circles.
The differences between the terms are 2, 3 and 4. The next difference will be 5, so the fifth term is 15. Square Numbers You square a number by multiplying it by itself. For example 5 squared is 5 x 5 = 25 and not 5 x 2 = 10. It is an easy mistake to make! The first four terms of the sequence of square numbers are 1, 4, 9 and 16. They are worked out by squaring the numbers 1, 2, 3 and 4 like this:
These can be written 1², 2², 3² and 4². We say this as 'one squared', 'two squared' and so on. The tenth term will be 'ten squared' which is written 10². That is 10 x 10 = 100
What are digits?
Numbers are made from combinations of the digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 For example 816 is a three digit number. It has 3 digits. You might also call it a three figure number.
Some examples The number 3 538 has four digits 3 5 3 and 8 The number 276 has three digits 2 7 and 6 The number 41 has two digits 4 and 1 The number 5 has only one digit 5
Multiplying by 10, 100 and 1 000
Multiplying by 10 When you multiply a decimal number by 10 you move all the digits one place to the left. The number becomes 10 times bigger. Example 2.63 x 10 = 26.3 You can see that the digits move along to the left. Units move to Tens, and the others follow like this:
Multiplying by 100 When you multiply a decimal number by 100 you move all the digits two places to the left. The number becomes 100 times bigger. Example 2.63 x 100 = 263
Multiplying by 1 000 When you multiply a number by 1 000 you move all the digits three places to the left. The number becomes 1 000 times bigger. Example 2.63 x 1 000 = 2 630
Dividing by 10, 100 and 1 000
Dividing by 10 When you divide a decimal number by 10 you move all the digits one place to the right. The number becomes 10 times smaller. Example 3 502 ÷ 10 = 350.2 You can see that the digits move along to the right. Thousands move to Hundreds, and the others follow like this:
Dividing by 100 When you divide a decimal number by 100 you move all the digits two places to the right. The number becomes 100 times smaller. Example 3 502 ÷ 100 = 35.02
Dividing by 1 000 When you divide a decimal number by 1 000 you move all the digits three places to the right. The number becomes 1 000 times smaller. Example 3 502 ÷ 1 000 = 3.502
Shortcuts
When multiplying by 10, 100 and 1 000 there's a pattern that can help you get the right answer very quickly. This method moves the decimal point rather than the digits. Multiplying number 10 100 1 000 10 000 Number of places to move the decimal point 1 2 3 4
The zeros in the multiplying number tell you how many places to move the decimal point. Example Multiply 2.341 by 100 100 has two zeros. Make 2.34 bigger by moving the decimal point two places.
It's the same when you divide except you have to remember to move the decimal point the other way. Remember when you divide you are making the number smaller. Example Divide 761.2 by 10 10 has one zero. Make 761.2 smaller by moving the decimal point 1 place.
When multiplying (or dividing) by 10, 100, 1 000, etc count the zeros to find out how much bigger (or smaller) your number must be. Make sure you move the digits (or the decimal point) in the correct direction!
Working with metric and decimal units
Being able to multiply or divide by 10, 100 and 1 000 is useful when you want to convert between units. Here are some rules and examples, starting with pounds and pence . £ 1 = 100 p Multiply by 100 to change pounds into pence. Divide by 100 to change pence into pounds. Examples £2 = 2 x 100 = 200 p £15.38 = 15.38 x 100 = 1 538 p 139 p = 139 ÷ 100 = £1.39 225 p = 225 ÷ 100 = £2.25 1 kg = 1 000 g Multiply by 1 000 to change kilograms into grams. Divide by 1 000 to change grams into kilograms. Examples 3 kg = 3 x 1 000 = 3 000 g 2.5 kg = 2.5 x 1 000 = 2 500 g 4 000 g = 4 000 ÷ 1 000 = 4 kg 1 500 g = 1 500 ÷ 1 000 = 1.500 kg = 1.5 kg
1 litre = 1 000 millilitres Multiply by 1 000 to change litres into millilitres. Divide by 1 000 to change millilitres into litres. Examples 6 l = 6 x 1 000 = 6 000 ml 3.25 l = 3.25 x 1 000 = 3 250 ml 10 000 ml = 10 000 ÷ 1 000 = 10 l 750 ml = 750 ÷ 1 000 = 0.750 l = 0.75 l 1 km = 1 000 m Multiply by 1 000 to change kilometres into metres. Divide by 1 000 to change metres into kilometres. Examples 7 km = 7 x 1 000 = 7 000 m 3.3 km = 3.3 x 1 000 = 3 300 m 1 250 m = 1 250 ÷ 1 000 = 1.250 km = 1.25 km 750 m = 750 ÷ 1 000 = 0.750 km = 0.75 km 1 m = 100 cm Multiply by 100 to change metres into centimetres. Divide by 100 to change centimetres into metres. Examples 3 m = 3 x 100 = 300 cm 2.51 m = 2.51 x 100 = 251 cm 345 cm = 345 ÷ 100 = 3.45 m 902 cm = 902 ÷ 100 = 9.02 m 1 cm = 10 mm Multiply by 10 to change centimetres into millimetres. Divide by 10 to change millimetres into centimetres. Examples 4 cm = 4 x 10 = 40 mm 31 cm = 31 x 10 = 310 mm 105 mm = 105 ÷ 10 = 10.5 cm 50 mm = 50 ÷ 10 = 5 cm
Place value
All numbers use one or more of these ten digits 1 2 3 4 5 6 7 8 9 0 For example 816 is a three digit number. It has 3 digits. You might also call it a three figure number.
Digits can be used on their own to give us small numbers like 2 and 4. They can be used together to make bigger numbers, like 27, 431 and 2 146 Question Is the digit 4 always worth 4? Answer No. For example 4 is worth a different amount in each of these numbers: 4, 40, 400, 4 000 Because we only have ten digits, the same ten have to be used in such a way that we always know whether a 4 stands for four, forty, four hundred or four thousand. Place value helps with this. Understanding place value tells us whether we have been given a bill for four pounds, forty pounds or four hundred pounds: £4, £40, £400 Place value is vital. It means putting digits into columns. These columns are always in the same order. thousands hundreds tens units (ones)
The value of a digit depends on which column it is in - units, hundreds, tens, thousands, etc. Let's look at the number 4 444. There are four 4s, but each 4 means something different. How much is each 4 worth? thousands 4 hundreds 4 tens 4 units (ones) 4
4 444 is worth four thousand four hundred and forty four.
Place holders
Look at the number 4 040. How much is it worth? thousands 4 hundreds 0 tens 4 units (ones) 0
This number is four thousand and forty. We must use zeros to keep the digits in the correct columns. If we missed out the zeros from the number above we would have 44 and that is a very different number from 4 040. 44 = four tens and four units thousands hundreds tens 4 units (ones) 4
4 040 is much bigger than 44. 4 040 = four thousands, no hundreds, four tens and no units thousands 4 hundreds 0 tens 4 units (ones) 0
Both the fours and the zeros are important in this number. Zero is called a place holder. It is not worth anything on its own, but it changes the value of other digits. In this case zeros change the number 44 to the much larger number 4 040. The digit on the right of any number must always go into the units column. If there are no units there will be a zero. For example in the number 20 there is a zero in the units place. thousands 0 hundreds 0 tens 2 units (ones) 0
Common mistakes
Whenever we work with numbers we must always remember to use place value. If we don't our answers will be wrong. Have a go at working out this sum 142 + 56 Did you put the digits into columns? Did you put the digits on the right into the units column?
If you don't put the digits into the correct columns you will get a wrong answer. Forgetting to use place holders is another common error. How would you write the number six thousand, three hundred and nine? It should look lke this thousands 6 hundreds 3 tens 0 units 9
That's 6 309. If you missed out the place holder (the zero) you would have written 639. That would be six hundred and thirty-nine, which would be a completely different number. Another common error is mistaking big digits for big numbers. For example 1 111 may look smaller than 999 because it is made up of small digits, but put them into columns to see that 1 111 is bigger. thousands 1 hundreds 1 9 tens 1 9 units 1 9
Inequalities - more than and less than
You will use a lot of different symbols when you are working with numbers. You probably know some of them already. + - x ÷ = These are all mathematical symbols. There are also symbols to show 'less than' and 'more than'. What symbol would you use to could show 1p is less than £1 ? 1p ? £1 The symbol to show 'less than' is <. It looks like an equals sign squashed to a point at one end. The two ends are not equal to show that the numbers are not equal.
So to say that 1p is less than £1 we write
1p < £1
What symbol could show that £1 is more than 1p ? £1 ? 1p We can use the < sign if we turn it around. This is the greater than sign > Then we can use it like this £1 > 1p The widest part of the arrow is always next to the largest amount. The pointed end is on the small side.
Seven digit numbers
How do you write one million pounds in numbers? Like this £1 000 000 The number 1 000 000 has seven digits. These are the seven columns
How would you write one million five hundred thousand?
That is 1 500 000. How would you write one million, three hundred and twenty thousand and fifty four? Put place holder zeros into the empty columns like this
That number is 1 320 054. It would be very different if it was 1 302 054
That is one million, three hundred and two thousand and fifty four, which is smaller than 1 320 054 1 302 054 < 1 320 054 |
# Fall Semester, SY Mrs. Abejero John Burroughs Middle School
## Presentation on theme: "Fall Semester, SY Mrs. Abejero John Burroughs Middle School"— Presentation transcript:
Fall Semester, SY 2009-2010 Mrs. Abejero John Burroughs Middle School
Algebra Readiness Fall Semester, SY Mrs. Abejero John Burroughs Middle School
Course syllabus Unit 1 – Scope and Sequence
Number Sense, Integers, Fractions, & Algebraic Thinking. Students work with whole numbers, integers and their operations. Next they study the properties of fractions. Recommended Focus Standards 7 NS 1.2 Add, subtract, multiply and divide rational numbers (integers, fractions, and terminating decimals) and take positive rational numbers to whole number powers. 7 NS 2.1 Understand negative whole-number exponents. Multiply and divide expressions involving exponents with a common base. 7 AF 1.3 Simplify numerical expression by applying properties of rational numbers (e.g. identity, inverse, distributive, associative and commutative) and justify the process used. 7 AF 2.1 Simplify and evaluate expressions that include exponents (including positive whole number powers and negative whole number powers) Students will be give an addendum to these topics if the teacher deems it necessary to make any modifications.
Unit 2 –Scope and Sequence
Fractions, Decimals, Ratios and Proportions. Students transition to operations on fractions and mixed numbers and then move to work on decimals and operations with them then study ratios, rates &proportions. Recommended Focus Standards 7 NS 1.2 Add, subtract, multiply and divide rational numbers (integers, fractions, and terminating decimals) and take positive rational numbers to whole number powers. 7 NS 1.3 Convert fractions to decimals and percents and use these representations in estimations, computations and applications. 7 NS 1.5 Know that every rational number is either a terminating or repeating decimal and be able to convert terminating decimals into reduced fractions. 7 AF 4.2 Solve multistep problems involving rate, average speed, distance and time or a direct variation. 7 MG 1.3 Use measures expressed as rates (e.g., speed, density) and measures expressed as products (e.g., person-days) to solve problems, check the units of the solution; and use dimensional analysis to check the reasonableness of the answer.
teacher deems it necessary to make any modifications.
Algebra Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root. They understand and use the rules of exponents. Students will be give an addendum to these topics if the teacher deems it necessary to make any modifications. Unit 3 –Scope and Sequence Percents and Algebraic Problem Solving. The unit opens with the study of percents, including percent increase and decrease. Students make the transition to algebraic problem solving including word problems including those involving average speed, distance and time. They solve simple (one and two step) equations and inequalities. Recommended Focus Standards 7 NS 1.3 Convert fractions to decimals and percents and use these representations in estimations, computations and applications.
teacher deems it necessary to make any modifications.
7 NS 2.1 Understand negative whole-number exponents. Multiply and divide expressions involving exponents with a common base. 7 AF 1.1 Use variables and appropriate operations to write an expression, an equation, an inequality, or a system of equations or inequalities that represents a verbal description (e.g., three less than a number, half as large as area A) 7 AF 1.3 Simplify numerical expression by applying properties of rational numbers (e.g. identity, inverse, distributive, associative and commutative) and justify the process used. 7 AF 4.1 Solve two-step linear equations and inequalities in one variable over the rational numbers, interpret the solution or solutions in the context from which they arose, and verify the reasonableness of the results. 7 AF 4.2 Solve multistep problems involving rate, average speed, distance and time or a direct variation. Algebra Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to fractional power. They understand and use the rules of exponents (excluding fractional powers). Students will be give an addendum to these topics if the teacher deems it necessary to make any modifications.
Unit 4 – Scope and Sequence
Linear Functions and their Graphs, Linear Equations, and Pythagorean Theorem. Students use the Cartesian coordinate system to graph points and lines and work with proportional relationships and linear functions using the slope to graph linear functions. Additionally students continue algebraic problem solving including multi-step problems. Note: As students in this course will take the General Mathematics CST examination, in Unit 4 they study additional topics in Statistics, Data Analysis and Probability, Measurement and Geometry and the Real Number System. Recommended Focus Standards 7 AF 3.3 Graph linear functions, noting that the vertical change (change in y-value) per unit of horizontal change (change in x-value) is always the same and know that the ratio (“rise over run”) is called the slope of a graph.
teacher deems it necessary to make any modifications.
7 AF 3.4 Plot the values of quantities whose ratios are always the same (e.g., cost tot the number of an item, feet to inches, circumference to diameter of a circle). Fit a line to the plot and understand that the slope of the line equals the quantities. 7 MG 3.3 Know and understand the Pythagorean Theorem and its converse and use it to find the length of the missing side of a right triangle and the lengths of other line segments and, in some situations, empirically verify the Pythagorean theorem by direct measurement. Algebra Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root. They understand and use the rules of exponents. Algebra Students simplify expressions before solving linear equations in one variable, such as 3(2x – 5) + (4(x – 2) = 12. Algebra Students solve multi-step problems, including word problems, involving linear equations in one variable and provide justification for each step. Students will be give an addendum to these topics if the teacher deems it necessary to make any modifications.
Grade = Assignments Average + Quizzes Average 2 or (Grade = 50% Assignments Ave + 50% Quizzes Ave) (Assignments include Class Work, Homework, Dispatch) Grade Scale: Note: Poor Grades will be due to the 90% - 100% A following reasons: 80% - 89% B Missing assignments 70% - 79% C Poor test grade 60% - 69% D Lack of effort below 60% Failed Not paying attention Not taking notes in class Not studying at home & at school Not bringing the required materials in everyday
Excellent (E) Satisfactory (S) Complete assignments 85% of assignments completed [accomplished with honesty and integrity] Active participation in Satisfactory participation in [class discussions and activities] 95%Attendance/No tardy marks 90% Attendance/Two excused tardy marks Unsatisfactory(U) 50% of assignments accomplished. Little participation in class discussions and activities. Irregular attendance Not bringing the required materials in class in everyday
Excellent Citizenship (E) Follows the Student Behavior Code of Conduct, Student Uniform Policy and Classroom Rules at all times. Satisfactory Citizenship (S) Follows the Student Behavior Code of Conduct, Student Uniform Policy and Classroom Rules but needs to be reminded. Two teacher warnings and two warning marks. Poor Citizenship (U) Disregards School’s Code of Conduct, Student Uniform Policy and Classroom Rules. More than two teacher warnings and warning marks Consequences If you choose to break a rule. Maintains school’s program of “Progressive Discipline”. Severe Disruption: Student sent immediately to office.
Bring the following materials in class each day.
Requirements: Bring the following materials in class each day. a blue or black pen, a red pen, pencil, pencil eraser loose leaf notebook paper a folder a notebook and/or binder Student Workbook Textbook when asked to bring Grades will be lowered for the following reasons. Breaks a classroom rule or Student Behavior Code of Conduct Excessive Unexcused absences and tardiness Excessive Talking Disruptive Behavior Being unprepared. (including not bringing materials/workbook/ textbook in class)
Classroom Rules Classroom Rules
Eyes Watching (I will raise my hand to ask permission to speak.) Ears Listening (I will listen and follow directions the first time they are given.) Hands Still (I will keep hands, feet, and other objects to myself.) Brain Thinking (I will pay attention at all times.) Heart Caring (I will use appropriate language. I will follow the Student Behavior Code of Conduct and Student Appearance.) Basic Student Responsibilities Keep track of your own assignments. (Homework/Classwork/Tests/Warm-up/Projects). Start work on time and turn-in assignments on time. Do you own work and ask for help when you need it. Bring all materials/workbooks/textbook in class everyday. Accept responsibilities for grades and other assignments.
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## WILL GIVE BRAINLIEST!! If the length of a rectangle is decreased by 4 cm and the width is increased by 5 cm, the result will be a squa
Question
WILL GIVE BRAINLIEST!!
If the length of a rectangle is decreased by 4 cm and the width is increased by 5 cm, the result will be a square, the area of which will be 40 cm2 greater than the area of the rectangle. Find the area of the rectangle.
in progress 0
5 months 2021-08-25T17:30:09+00:00 2 Answers 3 views 0
The rectangle is 15 cm by 24 cm, and its area is 360 cm^2.
hope this helps 🙂 plz brainliest
Step-by-step explanation:
Step-by-step explanation:
Let,
The length of rectangle = x
The width of rectangle = y
The length of square = x – 4
The width of square = y + 5
Area of rectangle = xy
Area of square = xy + 40
Condition No. 1:
Area of Square = Length of square * Width of square
xy + 40 = (x – 4)(y – 5)
xy + 40 = xy – 5x – 4y + 20
– 5x – 4y = 40 – 20
– 5x – 4y = 20
5x + 4y = -20 ——————(1)
Condition No. 2:
We know that:
Length of square = Width of square
x – 4 = y + 5
x = y + 5 + 4
x = y + 9 —————————(2)
Solution:
Put Eq. (2) in (1)
5 (y + 9) + 4y = -20
5y + 45 + 4y = -20
9y + 45 = -20
9y = -20-45
y = -65 / 9
Now, Put the value of y in Eq. (2)
x = (-65 / 9) + 9
x = (-65 + 81) / 9
x = 16 / 9 cm
Now,
Area of rectangle = xy
Area = ( -65 / 9 ) * ( 16 / 9 )
Area = (-65*16) / (9*9)
Area = -1040/81
Area = 12.8 cm² (Neglecting -ve sign)
Hope this helped! |
In this video, we are going to look at how to simplify numbers raised to negative exponents. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.
## Example of Negative Exponents
To simplify the following:
$4^{-2}$
we have to move the 4 to the denominator and make the exponent positive.
$\dfrac{1}{4^2}$
Then, multiply the 4 by itself, 2 times which would simplify to:
$4 \times 4=16$
so the final answer is
$\dfrac{1}{16}$
To simplify the following:
$x^{-3}$
we have to move the x to the denominator and make the exponent positive.
$\dfrac{1}{x^3}$
If there is something more complex such as $5x^{-2}$ then only the x will go to the denominator since it is the one with the negative exponent. This will be simplified to
$\dfrac{5}{x^2}$
Other examples:
$x^{-b}= \dfrac{1}{x^b}$
$8^{-3}= \dfrac{1}{8^3}$
## Video-Lesson Transcript
Let’s have a quick look at negative exponents.
Here we have $4^{-2}$
If this is $4^2$, it’s just going to be
$4^2 = 4 \times 4 = 16$
But here we have a negative exponent.
This has a different meaning.
We’re still going to do $4^2$ but the negative is going to change where we do that.
So whenever we have a number raised to a negative exponent, we’re going to move it to the denominator and make the exponent positive.
In this case, we’ll have
$4^{-2} = \dfrac{1}{4^2} = \dfrac{1}{16}$
Notice that our answer is not negative. The negative in the exponent has nothing to do with our final answer.
It just shows where we’re going to do $4^2$. We’re going to move it to the denominator.
Let’s look at another one.
$3^{-2}$
So, we’ll just move it to the denominator.
$3^{-2} = \dfrac{1}{3^2}$
If you’re wondering where our numerator came from, it’s a regular thing. Since $3^{-2}$ is the same as $\dfrac{3^{-2}}{1}$.
So let’s continue solving.
$3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}$
Let’s have another one.
$5^{-3}$
So we’ll do the same thing. we’ll move it to the denominator and change the exponent to be positive.
$\dfrac{5^{-3}}{1} = \dfrac{1}{5^3} = \dfrac{1}{125}$
And the same is true for variables.
We have $x^{-3}$,
this is simply $= \dfrac{1}{x^3}$
The reason why I’m teaching it as numerator-denominator style is that it may become more complex.
If we have
$5x^{-2}$
the thing here is that the $x$ is the only one raised to a negative exponent. Not $5$.
So when we solve this, $5$ stays on the numerator and the $x^{-2}$ will be on the denominator and change the exponent to positive.
$5x^{-2} = \dfrac{5}{x^2}$
Let’s have another one.
$3x^{-2} y^3$
Here, we’ll have
$\dfrac{3x^{-2} y^3}{1} = \dfrac{3y^3}{x^2}$
The only variable to change to be a denominator is $x^{-2}$, the rest stays where they are.
Let’s have another example.
$\dfrac{2}{x^-2}$
In all the other cases, we have a negative exponent but all are in the numerator.
In this case, we have a negative exponent on the denominator.
So what we’ll do is to do the opposite.
We’re going to place the negative exponent to the numerator then change the exponent to positive.
$\dfrac{2}{x^-2} = 2x^2$
So wherever the term with a negative exponent is written, put it to the opposite location.
If it’s on the numerator, put it to the denominator and make the exponent positive.
If it’s on the denominator, put it to the numerator and make the exponent positive.
Keep in mind that the negative exponent does not affect the sign of the answer. |
# Answer in Statistics and Probability for napie #250967
In a survey taken 10 years ago, it was found that 10% of customers of a supermarket brought along their own shopping bags. A recent survey aimed to prove that the current percentage of customers bringing along their own shopping bags is different from 10%. In the survey, it was found that 92 of the 1 000 customers surveyed brought along their own shopping bags. We want to test the claim that the current percentage is not 10%, at the 5% significance level
(a) State the appropriate null and alternative hypothesis.
(b) State and calculate the appropriate test statistic.
(c) Determine the critical value of the test or the p–value of the test.
(d) State whether or not you reject the null hypothesis, giving the reason.
(e) Draw an appropriate conclusion
To answer the question we should test hypothesis about the probability of the event
We got the relative frequency “w = {frac {92} {100}}”, and want to test if it’s significantly different from “p{scriptscriptstyle 0} = 0.1”
Null hypothesis: p = 0.1
Alternative hypothesis: “p not =0.1”
Since we want to test the claim that current percenatge is not 10%, then if we reject the null hypothesis it will mean that this claim is correct
The test statistic: “U={frac {(w-p{scriptscriptstyle 0})sqrt{n}} {sqrt{p{scriptscriptstyle 0}*(1-p{scriptscriptstyle 0})}}}”
In our case: “U={frac {(0.092-0.1)sqrt{1000}} {sqrt{0.1*0.9}}}= -0.84”
Due to the form of the alternative hypothesis, two-tailed test is required
The critical interval is “(-infty;-u{scriptscriptstyle 0}) u222a(u{scriptscriptstyle 0};+infty)”, where “P(N(0,1)>u{scriptscriptstyle 0}) = {frac alpha 2} = 0.025to u{scriptscriptstyle 0} = 1.96”
So, the critical interval is “(-infty;-1.96) u222a(1.96;+infty)”
Our U-value does not belong to the critical interval, so we fail to reject the null hypothesis, which means there are no statistically significant evidence that current percentage is different from 10%
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# PSAT Math : Rectangles
## Example Questions
### Example Question #1 : How To Find The Length Of The Diagonal Of A Rectangle
What is the length of the diagonal of a rectangle that is 3 feet long and 4 feet wide?
Explanation:
The diagonal of the rectangle is equivalent to finding the length of the hypotenuse of a right triangle with sides 3 and 4. Using the Pythagorean Theorem:
Therefore the diagonal of the rectangle is 5 feet.
### Example Question #1 : How To Find The Length Of The Diagonal Of A Rectangle
The length and width of a rectangle are in the ratio of 3:4. If the rectangle has an area of 108 square centimeters, what is the length of the diagonal?
12 centimeters
15 centimeters
18 centimeters
9 centimeters
24 centimeters
15 centimeters
Explanation:
The length and width of the rectangle are in a ratio of 3:4, so the sides can be written as 3x and 4x.
We also know the area, so we write an equation and solve for x:
(3x)(4x) = 12x= 108.
x2 = 9
x = 3
Now we can recalculate the length and the width:
length = 3x = 3(3) = 9 centimeters
width = 4x = 4(3) = 12 centimeters
Using the Pythagorean Theorem we can find the diagonal, c:
length2 + width2 = c2
92 + 12= c2
81 + 144 = c2
225 = c2
= 15 centimeters
### Example Question #24 : Quadrilaterals
The two rectangles shown below are similar. What is the length of EF?
10
8
5
6
10
Explanation:
When two polygons are similar, the lengths of their corresponding sides are proportional to each other. In this diagram, AC and EG are corresponding sides and AB and EF are corresponding sides.
To solve this question, you can therefore write a proportion:
AC/EG = AB/EF ≥ 3/6 = 5/EF
From this proportion, we know that side EF is equal to 10.
### Example Question #2 : How To Find The Length Of The Side Of A Rectangle
A rectangle is x inches long and 3x inches wide. If the area of the rectangle is 108, what is the value of x?
12
8
3
4
6
6
Explanation:
Solve for x
Area of a rectangle A = lw = x(3x) = 3x2 = 108
x2 = 36
x = 6
### Example Question #1 : How To Find The Length Of The Side Of A Rectangle
If the area of rectangle is 52 meters squared and the perimeter of the same rectangle is 34 meters. What is the length of the larger side of the rectangle if the sides are integers?
14 meters
13 meters
15 meters
12 meters
16 meters
13 meters
Explanation:
Area of a rectangle is = lw
Perimeter = 2(l+w)
We are given 34 = 2(l+w) or 17 = (l+w)
possible combinations of l + w
are 1+16, 2+15, 3+14, 4+13... ect
We are also given the area of the rectangle is 52 meters squared.
Do any of the above combinations when multiplied together= 52 meters squared? yes 4x13 = 52
Therefore the longest side of the rectangle is 13 meters
### Example Question #1 : How To Find If Rectangles Are Similar
Note: Figure NOT drawn to scale.
In the above figure,
.
.
Give the perimeter of .
Explanation:
We can use the Pythagorean Theorem to find :
The similarity ratio of to is
so multiplied by the length of a side of is the length of the corresponding side of . We can subsequently multiply the perimeter of the former by to get that of the latter:
### Example Question #2 : How To Find If Rectangles Are Similar
Note: Figure NOT drawn to scale.
In the above figure,
.
.
Give the area of .
Insufficient information is given to determine the area.
Explanation:
Corresponding sidelengths of similar polygons are in proportion, so
, so
We can use the Pythagorean Theorem to find :
The area of is
### Example Question #1 : How To Find If Rectangles Are Similar
Note: Figure NOT drawn to scale.
In the above figure,
.
.
Give the area of Polygon .
Explanation:
Polygon can be seen as a composite of right and , so we calculate the individual areas and add them.
The area of is half the product of legs and :
Now we find the area of . We can do this by first finding using the Pythagorean Theorem:
The similarity of to implies
so
The area of is the product of and :
Now add: , the correct response.
### Example Question #1 : How To Find If Rectangles Are Similar
Note: Figure NOT drawn to scale.
Refer to the above figure.
and .
What percent of has been shaded brown ?
Insufficient information is given to answer the problem.
Explanation:
and , so the similarity ratio of to is 10 to 7. The ratio of the areas is the square of this, or
or
Therefore, comprises of , and the remainder of the rectangle - the brown region - is 51% of .
### Example Question #313 : Plane Geometry
Note: figure NOT drawn to scale.
Refer to the above figure.
.
Give the area of .
. |
# Mean and Third Proportional
We will learn how to find the mean and third proportional of the set of three numbers.
If x, y and z are in continued proportion then y is called the mean proportional (or geometric mean) of x and z.
If y is the mean proportional of x and z, y^2 = xz, i.e., y = +$$\sqrt{xz}$$.
For example, the mean proportion of 4 and 16 = +$$\sqrt{4 × 16}$$ = +$$\sqrt{64}$$ = 8
If x, y and z are in continued proportion then z is called the third proportional.
For example, the third proportional of 4, 8 is 16.
Solved examples on understanding mean and third proportional
1. Find the third proportional to 2.5 g and 3.5 g.
Solution:
Therefore, 2.5, 3.5 and x are in continuous proportion.
$$\frac{2.5}{3.5}$$ = $$\frac{3.5}{x}$$
⟹ 2.5x = 3.5 × 3.5
⟹ x = $$\frac{3.5 × 3.5}{2.5}$$
⟹ x = 4.9 g
2. Find the mean proportional of 3 and 27.
Solution:
The mean proportional of 3 and 27 = +$$\sqrt{3 × 27}$$ = +$$\sqrt{81}$$ = 9.
3. Find the mean between 6 and 0.54.
Solution:
The mean proportional of 6 and 0.54 = +$$\sqrt{6 × 0.54}$$ = +$$\sqrt{3.24}$$ = 1.8
4. If two extreme terms of three continued proportional numbers be pqr, $$\frac{pr}{q}$$; what is the mean proportional?
Solution:
Let the middle term be x
Therefore, $$\frac{pqr}{x}$$ = $$\frac{x}{\frac{pr}{q}}$$
⟹ x$$^{2}$$ = pqr × $$\frac{pr}{q}$$ = p$$^{2}$$r$$^{2}$$
⟹ x = $$\sqrt{p^{2}r^{2}}$$ = pr
Therefore, the mean proportional is pr.
5. Find the third proportional of 36 and 12.
Solution:
If x is the third proportional then 36, 12 and x are continued proportion.
Therefore, $$\frac{36}{12}$$ = $$\frac{12}{x}$$
⟹ 36x = 12 × 12
⟹ 36x = 144
⟹ x = $$\frac{144}{36}$$
⟹ x = 4.
6. Find the mean between 7$$\frac{1}{5}$$and 125.
Solution:
The mean proportional of 7$$\frac{1}{5}$$and 125 = +$$\sqrt{\frac{36}{5}\times 125} = +\sqrt{36\times 25}$$ = 30
7. If a ≠ b and the duplicate proportion of a + c and b + c is a : b then prove that the mean proportional of a and b is c.
Solution:
The duplicate proportional of (a + c) and (b + c) is (a + c)^2 : (b + c)^2.
Therefore, $$\frac{(a + c)^{2}}{(b + c)^{2}} = \frac{a}{b}$$
⟹ b(a + c)$$^{2}$$ = a(b + c)$$^{2}$$
⟹ b (a$$^{2}$$ + c$$^{2}$$ + 2ac) = a(b$$^{2}$$ + c$$^{2}$$ + 2bc)
⟹ b (a$$^{2}$$ + c$$^{2}$$) = a(b$$^{2}$$ + c$$^{2}$$)
⟹ ba$$^{2}$$ + bc$$^{2}$$ = ab$$^{2}$$ + ac$$^{2}$$
⟹ ba$$^{2}$$ - ab$$^{2}$$ = ac$$^{2}$$ - bc$$^{2}$$
⟹ ab(a - b) = c$$^{2}$$(a - b)
⟹ ab = c$$^{2}$$, [Since, a ≠ b, cancelling a - b]
Therefore, c is mean proportional of a and b.
8. Find the third proportional of 2x^2, 3xy
Solution:
Let the third proportional be k
Therefore, 2x^2, 3xy and k are in continued proportion
Therefore,
\frac{2x^{2}}{3xy} = \frac{3xy}{k}
⟹ 2x$$^{2}$$k = 9x$$^{2}$$y$$^{2}$$
⟹ 2k = 9y$$^{2}$$
⟹ k = $$\frac{9y^{2}}{2}$$
Therefore, the third proportional is $$\frac{9y^{2}}{2}$$.
`
● Ratio and proportion |
TRIG.SEQ.T Solving Tangent Equations
Consider the function $f(x) = A\tan(Bx+C)$.
A standard form of a trigonometric equation with variable $x$ is an equation of the form $f(x) = A\tan(Bx+C) = D$ with $A, B \ne 0$.
Trigonometry courses spend a considerable amount of time solving this type of equation.
When $B = 1$ and $C=0$ this equation has at least one solution $x = \arctan ( \frac DA)$ and others determined by the fact that $\tan(x+\pi) =\tan(x)$.
When $B\ne 1$ , $B \ne 0$ and $C=0$ the equation can be solved so that the key solution is $x =\frac {\arctan(\frac D A)}B$
The following examples illustrate these solutions with both mapping diagrams and graphs..
Example SEQ.T.1 : Suppose $10 \tan(x)= 5$. Find $x$.
Example SEQ.T.2 :Suppose $10\tan(4x)= 5$. Find $x$.
Now consider solving equations of the form $f(x) = A\cdot trig(Bx-C) +k = D$ with $A,B \ne 0$ and $trig = \sin, \cos$ or $\tan$.
Example SEQ.T.3 : Suppose $4* \tan(2x-1) +1 = 9$. Find $x$.
This example shows an important visual connection between a mapping diagram for a trigonometric function as a composition with core linear functions and the algebra used in solving a trigonometric equation.
The included GeoGebra mapping diagram can be used to visualize the algebra for solving equations of the form $f(x) = A\cdot trig(Bx-C) +k = D$ with $A,B \ne 0$ and $trig = \sin, \cos$ or $\tan$.
You can use this next dynamic example to solve trigonometric equations like those in Examples SEQ.T.1 and SEQ.T.2 visually with a mapping diagram of $f$ and the lines in the graph of $f$.
Example TRIG.DSEQ.T.0 Dynamic Views for solving an equation $f(x) = A\tan(Bx) = D$ on Graphs and Mapping Diagrams |
Trees Problem 4
Solution
Let's define n=pq to be the area of the chocolate bar. The claim we need to prove is that n-1 breaks are required to divide the bar into individual squares.
Proof by induction on n.
Base case: At n=1, the bar already consists of a single square. So we don't need to break it up further. That is, we need 0, i.e. n-1 breaks to divide it up. So the claim is true.
Inductive hypothesis: any chocolate bar of area n can be divided into individual squares using n-1 breaks, for n from 1 up through k-1.
Rest of inductive step: suppose that B is a chocolate bar of area k, where k is larger than 1. Let's break B along any grid line, creating two bars X and Y.
Let x and y be the areas of X and Y. Since we are mathematicians and broke the chocolate bar perfectly without leaving any flakes, we haven't lost any area. So x+y=k.
By the inductive hypothesis, we can reduce X to individual squares using x-1 breaks. Similarly, we can reduce Y to individual squares using y-1 breaks.
Therefore, to reduce B to individual squares, we use our initial break, then break up X and Y using x-1 and y-1 breaks. So the total number of breaks required to divide up B is $$1 + (x-1) + (y-1) = x + y -1 = k-1$$. So breaking up B requires k-1 breaks, which is what we needed to prove.
Self-check
Notice that the inductive step chooses some arbitrary line for making the initial break. If you make some specific choice (e.g. pick the top horizontal line first), then you'll only prove that that specific strategy requires n-1 breaks. Perhaps some other strategy would have required more breaks or fewer breaks?
The claim was about chocolate bars, so make sure that your proof is talking about chocolate bars. Your proof may also involve various equations, but they need to be related to the main topic of chocolate bars. |
# 5+ Missing Number 2nd Grade Worksheets
How do you explain a missing number?
Missing numbers are the numbers that got missed in the given series of numbers with similar differences among them. The process of writing the missing numbers is termed as finding similar changes between those numbers and filling their missing values in their specific series and places.
How do you teach the missing number in Year 2?
How do I find a missing number for kids?
## How do you find the missing number in a pattern?
• Identity, if the order of number given is ascending ( smaller to larger number) or descending ( larger to smaller number)
• Calculate the differences between those that are next to each other.
• Estimate the difference between numbers to calculate the missing number.
• ## How do you teach subtracting missing numbers?
If the missing number is immediately after the equals sign: Subtract the second number from the first. If the missing number is being subtracted from a larger number: Subtract the given answer from the larger number. If the missing number is immediately before the subtraction sign: Add the two other numbers together.
## What is missing activity for kids?
Put three or four objects on a plate and tell your child to look at them. Ask them to close their eyes (or put a pillow over them) and take one item away. When your child opens their eyes, ask them to guess what item is missing. You can focus on letters, numbers, shapes, or kitchen items.
## How do you teach using a number line?
• Find the dot marked on a number (and name the dot)
• Place a dot at the desired location.
• Find the desired number below the number line's tick mark*
• Learn the numbers to the right are higher than numbers to left.
• Determine which number is greater than, less than or equals.
## What does addends mean in math?
: a number to be added to another.
In math, an addend can be defined as the numbers or terms added together to form the sum. Here, the numbers 7 and 8 are addends.
## How do I find a missing character?
Go through the given solved examples missing character to understand the concept better. Explanation: The pattern is : 62+28/10 = 9; 18+52/10 = 7. Explanation: The pattern is: (12×4)-(5×3)=33,(15×7)-(8×6)=57. ∴ Missing number =(17×9)-(9×7)=90.
## What number is missing 234 23 Tip It's not 4?
The missing no.is 4 .
## What is the answer to the Triangle riddle?
The answer is 25! There are 24 in the entire shape and the 25th triangle is in the artist's signature. Luckily, mathematician Martin Silvertant created a diagram that easily explains where all 24 triangles are located in the figure.
## What is the missing pattern?
A missing data pattern is said to be univariate if there is only one variable with missing data. A missing data pattern is said to be monotone if the variables Yj can be ordered such that if Yj is missing then all variables Yk with k>j are also missing. This occurs, for example, in longitudinal studies with drop-out.
## How do you teach missing Addends?
Write a missing addend math problem on the board. Lay out counters to match the known addend and the given sum. Match each counter from the known addend group to a counter in the sum group. The unmatched counters in the sum group equal the number needed for the missing addend.
## How do you play what's missing game?
• Arrange them in front of your child.
• Name the things to make sure he/she knows the names of all the things.
• Ask the child to close their eyes or turn around.
• Child can open eyes and name the missing item.
• Take turns and have fun finding the missing item.
• Color What's Missing Game.
• ## What zoom games are missing?
Need a game to play with your students over Zoom or Google Meets? What's Missing is an observation game. Students will look at various pictures and study the objects. Students will then look at the same picture, but has to figure out what object is missing.
## What grade do kids learn number line?
Using Number Lines to Learn About Numbers and Operations
Number lines are a key concept that begins to be used in Grade 2. The length of a line segment with one end placed at zero and the other end of the segment placed on a number represents the magnitude of that number.
## What is a number line to 20?
A Number Line to 20 is a series of numbers from 0-20 that are placed on a line. Number lines are great visual resources for visual learners and show children each number in that sequence. Count forwards and backwards with a number line to really heighten learning.
## What is an addend of 10?
If you add 10 to 15, then 10 is the addend. You can also call an addend a summand.
## What is addend and Augend?
In addition, an augend and an addend are added to find a sum. In the following equation, 6 is the augend, 3 is the addend, and 9 is the sum: 6 + 3 = 9. NOTE: Sometimes both the augend and addend are called addends. Sometimes the sum is called the total.
## What is minuend and subtrahend?
Minuend: a quantity or number from which another is to be subtracted. Subtrahend: a quantity or number to be subtracted from another. |
## Welcome to H3 Maths
Blog Support for Growing Mathematicians
### Lightning and Lotteries – What are the odds?
November23
The Mathematics of Probability is a fascinating one, and every day we take our chances in a variety of different situations (actually, some of our students take big chances every day, simply by running around corners without looking!). Impossible means a probability of 0; while certainty is a probability of 1.0 – that is, every possible outcome is a “favourable”one. For example, I was teaching Year 10 students (aka sophomores) the other day and I said that, “The probability that a student chosen from random in my room is a Year 10 student is 1.0“. My sample space would be all my students.
In the United States, an average of 80 people are killed by lightning each year. Considering being killed by lightning to be our ‘favourable outcome’, the sample space contains the entire population of the United States (about 250 million).
If we assume that all the people in our sample space are equally likely to be killed by lightning (so people who never go outside have the same chance of being killed by lightning as those who stand by flagpoles in large open fields during thunderstorms), the chance of being killed by lightning in the United States is equal to 80/250 million, or a probability of about .000032%.
Clearly, you are much more likely to die in a car accident than by being struck by lightning!
#### The Lottery
What if you want to know the probability of winning the lottery? Combination and permutation formulas are very useful for solving probability problems. Imagine a lottery where you pick six numbers from 1-49 and for a winning number, their order matters. Here you must use the formula for permutations to figure out the size of the sample space, which consists of the number of permutations of size k that can be taken from a set of n objects.
Since only one possible ordering of the six numbers can win the lottery, there is only one favourable outcome.
The sample space, however, is quite large because it is equal to 49! (Factorial 49) which is 49 x 48 x 47 x 46….. which is roughly 10 billion! This means that the probability of winning the lottery is about 1 in 10 billion. (source: Ask DrMath)
by posted under Uncategorized | Comments Off on Lightning and Lotteries – What are the odds?
#### Post Support
10 x 9 x 8 + (7 + 6) x 5 x 4 x (3 + 2) x 1 = 2020
NCEA Level 2 Algebra Problem. Using the information given, the shaded area = 9, that is:
y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?] |
# NCERT solution class 9 chapter 10 Circles exercise 10.6 mathematics
## EXERCISE 10.6
#### Question 1:
Prove that line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Let two circles having their centres as O and intersect each other at point A and B respectively. Let us join O.
In ΔAO and BO,
OA = OB (Radius of circle 1)
A = B (Radius of circle 2)
O = O (Common)
ΔAO ≅ ΔBO (By SSS congruence rule)
∠OA = ∠OB (By CPCT)
Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.
#### Question 2:
Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Draw OM ⊥ AB and ON ⊥ CD. Join OB and OD.
(Perpendicular from the centre bisects the chord)
Let ON be x. Therefore, OM will be 6− x.
In ΔMOB,
In ΔNOD,
We have OB = OD (Radii of the same circle)
Therefore, from equation (1) and (2),
From equation (2),
Therefore, the radius of the circle iscm.
#### Question 3:
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
Let AB and CD be two parallel chords in a circle centered at O. Join OB and OD.
Distance of smaller chord AB from the centre of the circle = 4 cm
OM = 4 cm
MB =
In ΔOMB,
In ΔOND,
Therefore, the distance of the bigger chord from the centre is 3 cm.
#### Question 4:
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
In ΔAOD and ΔCOE,
OA = OC (Radii of the same circle)
OD = OE (Radii of the same circle)
∴ ΔAOD ≅ ΔCOE (SSS congruence rule)
∠OAD = ∠OCE (By CPCT) … (1)
∠ODA = ∠OEC (By CPCT) … (2)
Also,
∠OAD = ∠ODA (As OA = OD) … (3)
From equations (1), (2), and (3), we obtain
∠OAD = ∠OCE = ∠ODA = ∠OEC
Let ∠OAD = ∠OCE = ∠ODA = ∠OEC = x
In Δ OAC,
OA = OC
∴ ∠OCA = ∠OAC (Let a)
In Δ ODE,
OD = OE
∠OED = ∠ODE (Let y)
∴ ∠CAD + ∠DEC = 180° (Opposite angles are supplementary)
x + a + x + y = 180°
2x + a + y = 180°
y = 180º − 2x − a … (4)
However, ∠DOE = 180º − 2y
And, ∠AOC = 180º − 2a
∠DOE − ∠AOC = 2a − 2= 2− 2 (180º − 2x − a)
= 4a + 4x − 360° … (5)
∠BAC + ∠CAD = 180º (Linear pair)
⇒ ∠BAC = 180º − ∠CAD = 180º − (a + x)
Similarly, ∠ACB = 180º − (a + x)
In ΔABC,
∠ABC + ∠BAC + ∠ACB = 180º (Angle sum property of a triangle)
∠ABC = 180º − ∠BAC − ∠ACB
= 180º − (180º − a − x) − (180º − a −x)
= 2a + 2− 180º
[4a + 4x − 360°]
∠ABC = [∠DOE − ∠ AOC] [Using equation (5)]
#### Question 4:
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
In ΔAOD and ΔCOE,
OA = OC (Radii of the same circle)
OD = OE (Radii of the same circle)
∴ ΔAOD ≅ ΔCOE (SSS congruence rule)
∠OAD = ∠OCE (By CPCT) … (1)
∠ODA = ∠OEC (By CPCT) … (2)
Also,
∠OAD = ∠ODA (As OA = OD) … (3)
From equations (1), (2), and (3), we obtain
∠OAD = ∠OCE = ∠ODA = ∠OEC
Let ∠OAD = ∠OCE = ∠ODA = ∠OEC = x
In Δ OAC,
OA = OC
∴ ∠OCA = ∠OAC (Let a)
In Δ ODE,
OD = OE
∠OED = ∠ODE (Let y)
∴ ∠CAD + ∠DEC = 180° (Opposite angles are supplementary)
x + a + x + y = 180°
2x + a + y = 180°
y = 180º − 2x − a … (4)
However, ∠DOE = 180º − 2y
And, ∠AOC = 180º − 2a
∠DOE − ∠AOC = 2a − 2= 2− 2 (180º − 2x − a)
= 4a + 4x − 360° … (5)
∠BAC + ∠CAD = 180º (Linear pair)
⇒ ∠BAC = 180º − ∠CAD = 180º − (a + x)
Similarly, ∠ACB = 180º − (a + x)
In ΔABC,
∠ABC + ∠BAC + ∠ACB = 180º (Angle sum property of a triangle)
∠ABC = 180º − ∠BAC − ∠ACB
= 180º − (180º − a − x) − (180º − a −x)
= 2a + 2− 180º
[4a + 4x − 360°]
∠ABC = [∠DOE − ∠ AOC] [Using equation (5)]
#### Question 6:
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
It can be observed that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is 180°.
∠AEC + ∠CBA = 180°
∠AEC + ∠AED = 180° (Linear pair)
∠AED = ∠CBA … (1)
For a parallelogram, opposite angles are equal.
From (1) and (2),
AD = AE (Angles opposite to equal sides of a triangle)
#### Question 7:
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
Let two chords AB and CD are intersecting each other at point O.
In ΔAOB and ΔCOD,
OA = OC (Given)
OB = OD (Given)
∠AOB = ∠COD (Vertically opposite angles)
ΔAOB ≅ ΔCOD (SAS congruence rule)
AB = CD (By CPCT)
Similarly, it can be proved that ΔAOD ≅ ΔCOB
∴ AD = CB (By CPCT)
Since in quadrilateral ACBD, opposite sides are equal in length, ACBD is a parallelogram.
We know that opposite angles of a parallelogram are equal.
∴ ∠A = ∠C
However, ∠A + ∠C = 180° (ABCD is a cyclic quadrilateral)
⇒ ∠A + ∠A = 180°
⇒ 2 ∠A = 180°
⇒ ∠A = 90°
As ACBD is a parallelogram and one of its interior angles is 90°, therefore, it is a rectangle.
∠A is the angle subtended by chord BD. And as ∠A = 90°, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.
#### Question 8:
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° .
It is given that BE is the bisector of ∠B.
∴ ∠ABE =
However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)
Similarly, ∠ACF = ∠ADF = (Angle in the same segment for chord AF)
Similarly, it can be proved that
#### Question 9:
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
AB is the common chord in both the congruent circles.
∴ ∠APB = ∠AQB
In ΔBPQ,
∠APB = ∠AQB
∴ BQ = BP (Angles opposite to equal sides of a triangle)
#### Question 10:
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC.
Let perpendicular bisector of side BC and angle bisector of ∠A meet at point D. Let the perpendicular bisector of side BC intersect it at E.
Perpendicular bisector of side BC will pass through circumcentre O of the circle. ∠BOC and ∠BAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively. We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠BOC = 2 ∠BAC = 2 ∠A … (1)
In ΔBOE and ΔCOE,
OE = OE (Common)
OB = OC (Radii of same circle)
∠OEB = ∠OEC (Each 90° as OD ⊥ BC)
∴ ΔBOE ≅ ∠COE (RHS congruence rule)
∠BOE = ∠COE (By CPCT) … (2)
However, ∠BOE + ∠COE = ∠BOC
⇒ ∠BOE +∠BOE = 2 ∠A [Using equations (1) and (2)]
⇒ 2 ∠BOE = 2 ∠A
⇒ ∠BOE = ∠A
∴ ∠BOE = ∠COE = ∠A
The perpendicular bisector of side BC and angle bisector of ∠A meet at point D.
∴ ∠BOD = ∠BOE = ∠A … (3)
Since AD is the bisector of angle ∠A, |
# Full-Length ALEKS Math Practice Test-Answers and Explanations
Did you take the ALEKS Math Practice Test? If so, then it’s time to review your results to see where you went wrong and what areas you need to improve.
## ALEKS Math Practice Test Answers and Explanations
$$7×8=56$$, Petrol use: $$8×1.5=12$$ liters
Petrol cost: $$12×1.5=18$$. Money earned: $$56-18=38$$
First, multiply the tenth place of 4.3 by 5.12. The result is 1.536. Next, multiply 4 by 5.12 which results in 20.48. The sum of these two numbers is: 1.536 + 20.48 = 22.016
Dividing 85 by $$25\%$$, which is equivalent to 0.25, gives 340. Therefore, $$25\%$$ of 340 is 85.
4- The answer is $$25\%$$
Use this formula: Percent of Change $$=\frac{New \space Value-Old \space Value}{Old \space Value} ×100\%$$
$$\frac{24,000-32,000}{32,000}×100\%=25\%$$ and $$\frac{18,000-24,000}{24,000}×100\%=25\%$$
Let x be the original price. If the price of the sofa is decreased by $$13\%$$ to $426.3, then: $$87\%$$ of $$x=426.3 ⇒ 0.87x=426.3 ⇒ x=426.3÷0.87=490$$ 6- The answer is $$15\%$$ The percent of girls playing tennis is: $$30\% × 50\% = 0.30 × 0.50 = 0.15= 15 \%$$ 7- The answer is $$200\%$$ Write the equation and solve for B: $$0.80 A = 0.40 B$$, divide both sides by 0.40, then: $$\frac{0.80}{0.40}A=B$$, therefore: $$B = 2A$$, and B is 2 times of A or it’s $$200\%$$ of A. 8- The answer is 6 hours and 30 minutes Use distance formula: Distance = Rate × time ⇒ 390 = 60 × T, divide both sides by 60⇒ $$\frac{390}{60}$$ = T ⇒ T = 6.5 hours. Change hours to minutes for the decimal part. 0.5 hours = 0.5 × 60 = 30 minutes. 9- The answer is $$120\%$$ Use percent formula: $$part = \frac{percent}{100}×whole$$ , $$84=\frac{percent}{100}×70 ⇒ 84=\frac{percent ×70}{100} ⇒$$, multiply both sides by 100 ⇒ 8400 = percent × 70, divide both sides by 70. 120 = percent 10- The answer is 6.832 To add decimal numbers, line them up and add from the right. $$1.78+3.045+2.007=6.832$$ ## The Absolute Best Book to Ace the ALEKS Math Test Original price was:$28.99.Current price is: $14.99. Satisfied 213 Students 11- The answer is 17 Use Pythagorean Theorem: $$a^2+b^2=c^2, 82 + 152 = c^2 ⇒ 64+225=c^2 ⇒ 289=c^2⇒c=17$$ 12- The answer is 40 The sum of supplement angles is 180. Let $$x$$ be that angle. Therefore, $$x = \frac{2}{9}×180, x = 40$$ 13- The answer is 6:7 The average speed of john is: $$210÷7=30$$ km The average speed of Alice is: $$175÷5=35$$ km Write the ratio and simplify. $$30: 35 ⇒ 6: 7$$ 14- The answer is$40
$$8.5×10=85$$, Petrol use: $$10×3=30$$ liters, Petrol cost: $$30×1.5=45$$
Money earned: $$85-45=40$$
Let $$x$$ be the number. Write the equation and solve for $$x$$.
$$(45 – x) ÷ x = 4$$
Multiply both sides by $$x$$.
$$(45 – x) = 4x$$, then add $$x$$ both sides. $$45 = 5x$$, now divide both sides by 5. $$x = 9$$
If the length of the box is 40, then the width of the box is one-fourth of it, 10, and the height of the box is 5 (one second of the width). The volume of the box is: V = lwh = (40) (10) (5) = 2000
To find the number of possible outfit combinations, multiply the number of options for each factor: 4 × 5 × 7 = 140
The area of the trapezoid is: $$Are=\frac{1}{2} h(b_1+b_2 )=\frac{1}{2}(x)(24+16)=300→20x=300→x=15$$⇒ $$z=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17$$
The perimeter of the trapezoid is: $$16+8+17+16+15=72$$
19- The answer is $$\frac{8}{25}$$
There are 25 integers from 5 to 30. Set of numbers that are not composite between 5 and 30 is: $${ 5,7,11,13,17,19,23,29}$$⇒ 8 integers are not composite. Probability of not selecting a composite number is:
$$Probability= \frac{number \space of \space desired \space outcomes}{number \space of \space total \space outcomes}= \frac{8}{25}$$
20- The answer is 26 miles
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem:
$$a^2+ b^2=c^2 , 24^2+ 10^2=c^2 ⇒ 576+100= c^2 ⇒ 676=c^2 ⇒ c=26$$
## Best ALEKS Math Prep Resource for 2022
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21- The answers are 11 and 13.5
You can find the possible values of and b in $$(ax+3)(bx-4)$$ by using the given equation $$a+b=5$$ and finding another equation that relates the variables a and b. Since $$(ax+3)(bx-4)=-6x^2+2cx+10$$, expand the left side of the equation to obtain
$$abx^2-4ax+3bx-12=-6x^2+2cx+10$$
Since ab is the coefficient of $$x^2$$ on the left side of the equation and 10 is the coefficient of $$x^2$$ on the right side of the equation, it must be true that $$ab=-6$$
The coefficient of $$x$$ on the left side is $$3b-4a$$ and the coefficient of $$x$$ in the right side is $$2c$$. Then: $$\frac{3b-4a}{2}=c, a+b=5$$, then: $$a=5-b$$
Now, plug in the value of $$a$$ in the equation $$ab=-6$$. Then: $$ab=-6→(5-b)b=-6→5b-b^2=-6$$⇒ Add $$-5b+b^2$$ both sides. Then: $$b^2-5b-6=0$$
Solve for $$b$$ using the factoring method. $$b^2-5b-6=0→(b-6)(b+1)=0$$
Thus, either $$b=6$$ and $$a =-1$$, or $$b =-1$$ and $$a =6$$. If $$b =6$$ and $$a =-1$$, then
$$c=\frac{3b-4a}{2}=\frac{3(6)-4(-1)}{2}=11$$. If $$b=-1$$ and $$a =6$$, then, $$c=\frac{3b-4a }{2}=\frac{3(-1)-4(6)}{2}=\frac{27}{2}=13.5→c=13.5$$. Therefore, the two possible values for $$c$$ are 11 and 13.5.
22- The answer is $$15x+2$$
If $$f(x)=7x+2(1-x)$$, then find $$f(3x)$$ by substituting $$3x$$ for every $$x$$ in the function. This gives: $$f(3x)=7(3x)+2(1-(3x))$$ ⇒ It simplifies to: $$f(3x)=7(3x)+2(1-(3x))=21x+2-6x=15x+2$$
The input value is 5. Then: $$x=-4⇒ f(x)=2x^2+4x+1→f(-4)=2(-4)^2+4(-4)+1=17$$
24- The answer is $$cosA=\frac{3}{5}$$
To solve for $$cosA$$ first identify what is known.
The question states that ∆ABC is a right triangle whose $$n∠B=90\circ$$ and $$sinC=\frac{3}{5}$$.
It is important to recall that any triangle has a sum of interior angles that equals 180 degrees. Therefore, to calculate $$cosA$$ use the complementary angles identify of a trigonometric function. $$cosA=cos(90-C)$$, Then: $$cosA=sinC$$
For complementary angles, the sin of one angle is equal to the $$cos$$ of the other angle. $$cosA=\frac{3}{5}$$
25- The answer is $$y=x^2-4x+1$$
To figure out what the equation of the graph is, first find the vertex. From the graph, we can determine that the vertex is at $$(2,-3)$$.
We can use vertex form to solve for the equation of this graph.
Recall vertex form, $$y=a(x-h)^2+k$$, where h is the x coordinate of the vertex, and k is the y coordinate of the vertex. Plugging in our values, you get $$y=a(x-2)^2-3$$
To solve for a, we need to pick a point on the graph and plug it into the equation.
Let’s pick $$(4,1)$$, $$1=a(4-2)^2-3$$, $$1=a(2)^2-3$$, $$1=4a-3$$, $$a=1$$⇒ Now the equation is : $$y=(x-2)^2-3$$, Let’s expand this, $$y=(x^2-4x+4)-3$$, $$y=x^2-4x+1$$
26- The answer is $$-\frac{2}{3}$$
Multiplying each side of $$\frac{7}{x}=\frac{21}{x+4} by x(x+4)$$ gives $$7(x+4)=21x$$, divide two side by 7.
$$x+4=3x$$ or $$x=2$$. Therefore, the value of $$-\frac{x}{3}=-\frac{2}{3}$$.
It is given that $$g(6)=8$$. Therefore, to find the value of $$f(g(6))$$, then $$f(g(6))=f(8)=12$$
28- The answer is $$49\sqrt{3}$$
The area of the triangle is: $$\frac{1}{2}$$ AD × BC and AD is perpendicular to BC. Triangle ADC is a $$30^\circ-60^\circ- 90^\circ$$ right triangle. The relationship among all sides of the right triangle $$30^\circ-60^\circ- 90^\circ$$ is provided in the following triangle: In this triangle, the opposite side of the 30° angle is half of the hypotenuse. And the opposite side of $$60^\circ$$ is opposite of $$30^circ × \sqrt{3}$$
CD = 7, then AD = $$7 × \sqrt{3}$$
Area of the triangle ABC is: $$\frac{1}{2}$$ AD×BC = $$\frac{1}{2}$$ $$7\sqrt{3}×14=49\sqrt{3}$$
It is given that $$g(4)=10$$. Therefore, to find the value of $$f(g(4))$$, substitute 10 for $$g(4)$$.
$$f(g(4))=f(10)=40$$.
30- The answer is $$\sqrt{17}$$
The equation of a circle with center (h, k) and radius r is $$(x-h)^2+(y-k)^2=r^2$$. To put the equation $$x^2+y^2-6x+4y=4$$ in this form, complete the square as follows:
$$x^2+y^2-6x+4y=4, (x^2-6x)+(y^2+4y)=4$$
$$(x^2-6x+9)-9+(y^2+4y+4)-4=4, (x-6)^2+(y+4)^2=17$$
Therefore, the radius of the circle is $$\sqrt{17}$$
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# GMAT Math Questions | Simple Interest Q15
#### GMAT Sample Questions | Simple Interest | Word Problem
The given question is a GMAT 550 level problem solving question from the topic Simple Interest. This GMAT sample question tests the concept of computing the number of years for which an investment was made.
Question 15: Braun invested a certain sum of money at 8% p.a. simple interest for 'n' years. At the end of 'n' years, Braun got back 4 times his original investment. What is the value of n?
1. 50 years
2. 25 years
3. 12 years 6 months
4. 37 years 6 months
5. 40 years
## GMAT Live Online Classes
### Explanatory Answer | GMAT Simple Interest
#### Step 1 : Assume initial investment (Principal) and compute final value (amount) after n years
Simple interest = $$frac{\text{Principal × number of years × rate of interest}}{\text{100}}$ . . . .$1)
For any assumed principal, the number of years is going to remain the same because the amount is expressed as ‘x’ times the principal i.e., 4 times in this case.
Let us assume the initial investment (principal) by Braun to be $100. Amount = 4 × Principal =$400
Amount = Principal + Simple Interest
Therefore, the Simple Interest earned = 400 - 100 = $300. #### Step 2 : Find the number of years n Substitute assumed value of principal and the corresponding interest earned, and rate of interest in equation 1. 300 = $$frac{\text{100 × n × 8}}{\text{100}}$ Or 8n = 300 Or n = 37.5 years Any amount, when invested for 37.5 years at 8% per annum simple interest would become 4 times the initial value. #### Alternative Method: Find the number of years required to double the initial amount$principal). When the initial investment doubles, the interest earned is the same as the initial investment (principal). So, if principal = 100, interest earned = 100 and r = 8%. 100 = $$frac{\text{100 × n × 8}}{\text{100}}$ So, n =$\frac{\text{100 }}{\text{8}}$= 12.5 years. Initial investment of$100 becomes $400 after earning an interest of$300. To earn $100 interest it took 12.5 years. Hence, it will take 3 × 12.5 = 37.5 years to earn$300 interest. #### Choice D is the correct answer. #### GMAT Online CourseTry it free! Register in 2 easy steps and Start learning in 5 minutes! #### Already have an Account? #### GMAT Live Online Classes Next Batch June 8, 2024 ## Additional Practice Questions in Ratio Proportion, Mixtures, Percents, Fractions, Profits, and Interest Work @ Wizako ##### How to reach Wizako? Mobile:$91) 95000 48484
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# Angle
something that is formed when two rays meet at a single or same point
(Redirected from Straight angle)
When two straight lines come together, they make an angle. The two lines are called the sides[1] of the angle, and they meet at a point. A flat surface (called a plane) also forms an angle when it meets another.
To represent an angle, Greek letters such as ${\displaystyle \alpha }$ (alpha), ${\displaystyle \beta }$ (beta), ${\displaystyle \gamma }$ (gamma) and ${\displaystyle \theta }$ (theta) are sometimes used.[2] An angle indicates the space between its sides, or the amount of rotation needed to make one side coincide the other.[3][4]
To measure the size of an angle, we use units called degrees. A degree is a standard unit and we use the symbol ° after a number to show that it is a number of degrees. We can use a decimal number or a fraction for part of a degree, but a degree can also be divided into 60 minutes (1° = 60'), and a minute can be divided into 60 seconds (1' = 60"). So 22.5°, 2212° and 22° 30' are all the same angle.
In mathematics, angles can also be (and often are) measured in radians instead of degrees, by using the conversion factor ${\displaystyle 2\pi {\mbox{ rad}}=360^{\circ }}$ (for example, ${\displaystyle 22.5^{\circ }={\tfrac {\pi }{8}}{\mbox{ rad}}}$). Yet another unit of angle is gradian,[4] with ${\displaystyle 100{\text{ grad}}=90^{\circ }}$.
Angles are studied in geometry, where an angle where edges meet is often called a vertex. For example, the three sides of a triangle are its edges and two of the edges meet at each vertex. Similarly, two of the six sides (or faces) of a cube meet at each of its twelve edges, and three edges meet at each of its eight corners (or vertices, which is the plural version of vertex).
## Types of angles
### Individual angles
• In a zero angle the lines lie one upon the other thus creating a 0° angle aka the zero angle.
• An angle measuring greater (wider) than 0° but less (narrower) than 90° is called an acute angle.
• An angle more than 90° but less than 180° is called an obtuse angle.
• An angle that measures 180° is called a straight angle.
• An angle wider than 180° and narrower than 360° is called a reflex angle.
• An angle that has a made or full circle/completed 360° is called a full or complete angle.[5]
### Special angle pairs
In geometry, there are pairs angles having a special relationship with each other, making them interesting and convenient.
There is a pair of angles called complementary angles to which the sum of their measure (wideness) is equal to one right angle (which is equal to ${\displaystyle {\tfrac {1}{4}}}$ turn, 90°, or ${\displaystyle {\tfrac {\pi }{2}}}$ radians). Supplementary angles are also two angles, this time their combined measure is a straight angle (${\displaystyle {\tfrac {1}{2}}}$ turn, 180°, or ${\displaystyle \pi }$ radians). Two angles that total to a full angle (${\displaystyle 1}$ turn, 360°, or ${\displaystyle 2\pi }$ radians) are referred to as explementary or conjugate angles.
## References
1. Campana, D. M. (2016-09-06). The Teacher of Geometrical Drawing - For High Schools, Manual Training Schools, Technical Schools, Etc. Read Books Ltd. ISBN 978-1-4733-5366-4.
2. "Compendium of Mathematical Symbols". Math Vault. 2020-03-01. Retrieved 2020-08-17.
3. "Definition of angle | Dictionary.com". www.dictionary.com. Retrieved 2020-08-17.
4. Weisstein, Eric W. "Angle". mathworld.wolfram.com. Retrieved 2020-08-17.
5. "Angles - Acute, Obtuse, Straight and Right". www.mathsisfun.com. Retrieved 2020-08-17. |
How Many Diagonals Does a Hexagon Have
# How Many Diagonals Does a Hexagon Have
Edited By Team Careers360 | Updated on Mar 24, 2023 04:42 PM IST
## Introduction
A closed figure made up of lines is called a polygon. We can classify the polygon on the basis of the number of sides. The minimum number of sides of the polygon is three. A polygon has sides, faces, edges, and vertices. All polygons with different numbers of sides have different properties.
## Hexagon
A two-dimensional closed figure with six sides is called a hexagon. The honeycomb is the next example of the nature of the hexagon shape. A hexagon has 6 sides, 6 vertices, 6 angles, and 6 faces. By adding six sides to the hexagon we can find the perimeter of the hexagon.
## Types of Hexagon
The hexagon can be classified based on sides and on the basis of angles.
Hexagon on the basis of sides;
• Irregular Hexagon
A hexagon with irregular sides is called an irregular hexagon. All angles and sides of irregular hexagons are not the same.
• Regular Hexagon
A hexagon with regular sides is called a regular hexagon. All angles and sides of a regular hexagon are the same. Each angle is 60°.
Hexagon based on the side;
• Convex Hexagon
A hexagon with vertices pointing in an outward direction is called a convex hexagon. All interior angles of the convex hexagon are less than 180°.
• Concave Hexagon
A hexagon with at least one vertex pointing inward is called a concave hexagon. At least one e of the concave hexagon is more than 180°.
## 3. Diagonals
The opposite of the polygon is connected by the line called the diagonal. We can also say diagonals connect two non-adjacent sides of the polygons.
Diagnosis is the root word for the diagonal. Diagnosis is a Greek word that means angle to angle.
There is a common formula for finding the diagonal of all types of the polygon.
The formula for finding the diagonal is;
(n2-3n)/2
n is the number of the sides of the polygon.
## Diagonal of the Hexagon
Hexagon has a six sides, so n=6
\begin{aligned}
& (n^{^{2}}-3\times n)\div 2
\\&=(6^{^{2}}-3\times 6)\div 2
\\&=9
\end{aligned}
1. How many edges does the hexagon have?
Hexagons have 6 edges.
2. What is the total sum of all the angles of the hexagon?
The sum of all the angles of the hexagon is 720°.
3. What is the line of symmetry of the regular hexagon?
Regular hexagon has 6 lines of symmetry.
4. Write the two properties of the pentagon.
The two properties of the hexagon;
1. The Pentagon has five sides.
2. The sum of all angles of the pentagon is 540°.
5. Name the polygon with eight sides.
Octagon has eight sides.
Get answers from students and experts |
# Question #96293
Jan 19, 2018
On the domain, $\left\{x \in R | 0 \le x \le 2 \pi\right\}$
$x = \frac{\pi}{6} , x = \frac{5 \pi}{6} , x = \frac{7 \pi}{6} , x = \frac{11 \pi}{6}$
Or, in degrees;
On the domain, $\left\{x \in R | {0}^{\circ} \le x \le {360}^{\circ}\right\}$
$x = {30}^{\circ} , x = {150}^{\circ} , x = {210}^{\circ} , x = {330}^{\circ}$
#### Explanation:
You are going to need to use the pythagorean identity (${\sin}^{2} x + {\cos}^{2} x = 1$) to solve this problem.
$7 {\sin}^{2} x + 3 {\cos}^{2} x = 4$
As, ${\sin}^{2} x + {\cos}^{2} x = 1$ then; ${\cos}^{2} x = 1 - {\sin}^{2} x$ which we can use to substitute into the given equation.
$7 {\sin}^{2} x + 3 \left(\textcolor{red}{1 - {\sin}^{2} x}\right) = 4$
$7 {\sin}^{2} x + 3 - 3 {\sin}^{2} x = 4$
Combining like terms;
$4 {\sin}^{2} x + 3 = 4$
Moving all of the information to the left of the equality yields;
$4 {\sin}^{2} x - 1 = 0$
The next clever piece of maths comes because you will need to factorise this expression. So what we do is substitute a variable in for the trigonometric ratio.
i.e. let $u = \sin x$
$\therefore 4 {\sin}^{2} x - 1 = 4 {u}^{2} - 1 = 0$
Which we can factorise using the difference of perfect squares.
$4 {u}^{2} - 1 = \left(2 u + 1\right) \left(2 u - 1\right) = 0$
By the null factor law we can conclude that $u = \pm \frac{1}{2}$
$\therefore \sin x = \pm \frac{1}{2}$
So, on the domain, $\left\{x \in R | 0 \le x \le 2 \pi\right\}$
$x = \frac{\pi}{6} , x = \frac{5 \pi}{6} , x = \frac{7 \pi}{6} , x = \frac{11 \pi}{6}$
Or, in degrees;
On the domain, $\left\{x \in R | {0}^{\circ} \le x \le {360}^{\circ}\right\}$
$x = {30}^{\circ} , x = {150}^{\circ} , x = {210}^{\circ} , x = {330}^{\circ}$
I hope that helps :) |
What is the antiderivative of 6/x?
Understand the Problem
The question is asking for the antiderivative (or indefinite integral) of the function 6/x. This involves finding a function whose derivative is equal to 6/x, which can be solved using integration techniques.
Answer
The antiderivative of $\frac{6}{x}$ is $6 \ln |x| + C$.
Answer for screen readers
The antiderivative of $\frac{6}{x}$ is $6 \ln |x| + C$.
Steps to Solve
1. Identify the Integral to Solve We need to solve the integral of the function $\frac{6}{x}$ with respect to $x$. This can be rewritten as: $$\int \frac{6}{x} , dx$$
2. Factor Out the Constant Since integration allows us to factor out constants, we can take the number 6 outside of the integral: $$6 \int \frac{1}{x} , dx$$
3. Integrate the Function The integral of $\frac{1}{x}$ is a standard result. We know that: $$\int \frac{1}{x} , dx = \ln |x| + C$$ where $C$ is the constant of integration.
4. Combine the Results Now we can combine our results from the previous steps: $$6 \int \frac{1}{x} , dx = 6 (\ln |x| + C)$$ This simplifies to: $$6 \ln |x| + C$$
The antiderivative of $\frac{6}{x}$ is $6 \ln |x| + C$.
More Information
The result $6 \ln |x| + C$ represents a family of functions whose derivative gives $\frac{6}{x}$. The $C$ is an arbitrary constant representing all possible vertical shifts of this function.
Tips
• Forgetting to include the constant of integration $C$, which is essential in indefinite integrals.
• Not recognizing that the integral of $\frac{1}{x}$ is not simply $x$, but rather a logarithmic function.
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# 6.2 Using the normal distribution (Page 4/25)
Page 4 / 25
## References
“Naegele’s rule.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Naegele's_rule (accessed May 14, 2013).
“403: NUMMI.” Chicago Public Media&Ira Glass, 2013. Available online at http://www.thisamericanlife.org/radio-archives/episode/403/nummi (accessed May 14, 2013).
“Scratch-Off Lottery Ticket Playing Tips.” WinAtTheLottery.com, 2013. Available online at http://www.winatthelottery.com/public/department40.cfm (accessed May 14, 2013).
“Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed May 14, 2013).
## Chapter review
The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean µ and the standard deviation σ . A special normal distribution, called the standard normal distribution is the distribution of z -scores. Its mean is zero, and its standard deviation is one.
## Formula review
Normal Distribution: X ~ N ( µ , σ ) where µ is the mean and σ is the standard deviation.
Standard Normal Distribution: Z ~ N (0, 1).
Calculator function for probability: normalcdf (lower x value of the area, upper x value of the area, mean, standard deviation)
Calculator function for the k th percentile: k = invNorm (area to the left of k , mean, standard deviation)
How would you represent the area to the left of one in a probability statement?
P ( x <1)
What is the area to the right of one?
Is P ( x <1) equal to P ( x ≤ 1)? Why?
Yes, because they are the same in a continuous distribution: P ( x = 1) = 0
How would you represent the area to the left of three in a probability statement?
What is the area to the right of three?
1 – P ( x <3) or P ( x >3)
If the area to the left of x in a normal distribution is 0.123, what is the area to the right of x ?
If the area to the right of x in a normal distribution is 0.543, what is the area to the left of x ?
1 – 0.543 = 0.457
Use the following information to answer the next four exercises:
X ~ N (54, 8)
Find the probability that x >56.
Find the probability that x <30.
0.0013
Find the 80 th percentile.
Find the 60 th percentile.
56.03
X ~ N (6, 2)
Find the probability that x is between three and nine.
X ~ N (–3, 4)
Find the probability that x is between one and four.
0.1186
X ~ N (4, 5)
Find the maximum of x in the bottom quartile.
Use the following information to answer the next three exercise: The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will break down during the guarantee period.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
2. P (0< x <____________) = ___________ (Use zero for the minimum value of x .)
1. Check student’s solution.
2. 3, 0.1979
Find the probability that a CD player will last between 2.8 and six years.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
2. P (__________< x <__________) = __________
Find the 70 th percentile of the distribution for the time a CD player lasts.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the lower 70%.
2. P ( x < k ) = __________ Therefore, k = _________
1. Check student’s solution.
2. 0.70, 4.78 years
what is statistics
can anyone explain it better for me
frequency distribution
noun STATISTICS a mathematical function showing the number of instances in which a variable takes each of its possible values.
Robin
ok
Common language-- taking a bunch of information and seeing if it is related or not to other info
Mandy
Does standard deviation have measuring unit?
Mohamed
yes, the measuring unit of the data you are looking at, for example centimetres for height.
Emma
thanks
Mohamed
is that easy to plot a graph between three axis?
Mohamed
yes we can but we do not have that much effective tools. If the graph is normal or less complicated then it is plotted effectively otherwise it will give you nightmare.
umair
whats the difference between discrete and contineous data
umar
Discrete variables are variables that can assume finite number of values. Continuous variables are variables that can assume infinite number of values
Mike
i will give you an example: {0,4,84} it contains discrete or limited values like it can also contain boolean values{true,false} or {0,1} and continuous are like {1,2,3,4,5......} , {0,0.1,0.2,0.3,0.4...........}
umair
a no. of values which are countable are called discrete variables on the other hand, a no. of values which are not countable are called continuous variables
Aliya
Yup, I would like to support Mr.Umair's argument by saying that it can only apply if we have a 3-D graph,otherwise a plane graph will not apply at all
festus
Aliya and Mike thnks to both of you ❤❤
umar
what's variance
what's case control study?
Shakilla
hi
Noman
?
Sulaiman
what is covariance
In probability theory and statistics, covariance is a measure of the joint variability of two random variables.[1] If the greater values of one variable mainly correspond with the greater values of the other variable, and the same holds for the lesser values, (i.e., the variables tend to show simila
Robin
Economics department, faculty of social sciences, NOUN. You are required to calculate: the covariance and State whether the covariance is positive or negative. (11½ marks) Observation E D 1 15 17.24 2 16 15.00 3 8 14.91 4 6 4.50 5 15 18.00 6 12 6.29 7 12 19.23 8 18 18.69 9 12 7.21 10 20 4
Florence
In probability theory and statistics, covariance is a measure of the joint variability of two random variables.
Robin
what is the purpose of statistics and why it is important that statistics to be a solo and one complete field?
to organize,analyze and interpret information in order to make decision
Berema
what is noun?
so simple. the name of any person,place or thing.
Edu-info
Using the Chi-square test, two coins were flipped a hundred times. What will be the chances of getting a head and getting a tale? Given observed values is 62 heads and 38 tails. Expected value is 50 heads, 50 tails. Is the difference due to chance or a significant error? a. Draw your hypothesis
how can I win
what is difference between the blocking and confounding
how do you get 2/50 ?
can you explained it for me
korankye
an easier definition of inferential statistics
Inferential statistics makes inferences and predictions about a population based on a sample of data taken from the population in question.
Rukhsana
Inferential statistics helps you to extract insights from a random sample data which then helps you to use specific predictive Modeling/machine learning technic to predict or forecast.
Manish
what is stemplot? can anyone explain?
Javokhirbek
what is statistics
what is collection of data
ernest
no collection data was provided just the mean =14
Leticia
sd=14 describe the position of score to the mean how many points below or above z=1.00 z=1.50
Leticia
I have this sample score 14 18 12 22 14 22 21 20 13 26 13 26 16 21 they want me to.compute the z- score of x= 15 ×=40 and x=9?
Leticia
how do you understand that it is the mean?
Kenedy
fact and figure
hira
factors to consider when using secondary data
define binomial distribution
the distribution in which the outcome is of dichotomous
bimal
can you tell me Standar division is =14 what is the position of the score relative to the mean how many point above/below the mean?
Leticia
What do you call a measure of central tendency (i.e., average) appropriate for data measured on the continuous scale
arithmetic mean
bimal |
# 2.16: Integer Division and Modulus
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## Overview
In integer division and modulus, the dividend is divided by the divisor into an integer quotient and a remainder. The integer quotient operation is referred to as integer division, and the integer remainder operation is the modulus.[1][2]
## Discussion
By the time we reach adulthood, we normally think of division as resulting in an answer that might have a fractional part (a floating-point data type). This type of division is known as floating-point division. However, division, when both operands are of the integer data type, may act differently, depending on the programming language, and is called: integer division. Consider:
11 / 4
Because both operands are of the integer data type the evaluation of the expression (or answer) would be 2 with no fractional part (it gets thrown away). Again, this type of division is called integer division and it is what you learned in grade school the first time you learned about division.
Integer division as learned in grade school.
In the real world of data manipulation there are some things that are always handled in whole units or numbers (integer data type). Fractions just don’t exist. To illustrate our example: I have 11 dollar coins to distribute equally to my 4 children. How many do they each get? The answer is 2, with me still having 3 left over (or with 3 still remaining in my hand). The answer is not 2 ¾ each or 2.75 for each child. The dollar coins are not divisible into fractional pieces. Don’t try thinking out of the box and pretend you’re a pirate. Using an axe and chopping the 3 remaining coins into pieces of eight. Then, giving each child 2 coins and 6 pieces of eight or 2 6/8 or 2 ¾ or 2.75. If you do think this way, I will change my example to cans of tomato soup. I dare you to try and chop up three cans of soup and give each kid ¾ of a can. Better yet, living things like puppy dogs. After you divide them up with an axe, most children will not want the ¾ of a dog.
What is modulus? It’s the other part of the answer for integer division. It’s the remainder. Remember in grade school you would say, “Eleven divided by four is two remainder three.” In many programming languages, the symbol for the modulus operator is the percent sign (%).
11 % 4
Thus, the answer or value of this expression is 3 or the remainder part of integer division.
Many compilers require that you have integer operands on both sides of the modulus operator or you will get a compiler error. In other words, it does not make sense to use the modulus operator with floating-point operands.
Don’t let the following items confuse you.
6 / 24 which is different from 6 % 24
How many times can you divide 24 into 6? Six divided by 24 is zero. This is different from: What is the remainder of 6 divided by 24? Six, the remainder part is given by modulus.
Evaluate the following division expressions:
1. 14 / 4
2. 5 / 13
3. 7 / 2.0
Evaluate the following modulus expressions:
1. 14 % 4
2. 5 % 13
3. 7 % 2.0
## Key Terms
integer division
Division with no fractional parts.
modulus
The remainder part of integer division.
## References
1. Wikipedia: Division (mathematics)
2. Wikipedia: Modulo operation
2.16: Integer Division and Modulus is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. |
AP State Syllabus AP Board 6th Class Maths Solutions Chapter 3 HCF and LCM Ex 3.5 Textbook Questions and Answers.
## AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM Ex 3.5
Question 1.
Find the HCF of the following by prime – factorization and by continued division method.
i) 48, 64
ii) 126, 216
iii) 40, 60, 56
iv) 10, 35, 40
i) 48,64
Factorization method:
48 = 2 x 2 x 2 x 2 x 3
64 = 2 x 2 x 2 x 2 x 2 x 2
HCF = 2 x 2 x 2 x 2 = 16
∴ HCF of 48 and 64 is 16.
Continued division method:
HCF = 16
ii) 126, 216
Factorization method:
126 = 2 x 3 x 3 x 7
216 = 2 x 2 x 2 x 3 x 3 x 3
HCF = 2 x 3 x 3 = 18
∴ HCF of 126 .and 216 is 18.
Continued division method:
HCF = 18
iii) 40, 60, 56
Factorization method:
40 = 2 x 2 x 2 x 5
60 = 2 x 2 x 3 x 5
∴ HCF of 40, 56 and 60 is 4.
Continued division method:
First find the HCF of the smallest and the biggest of given numbers i.e., 40 and 60
Now, find the HCF of 20 and the remaining number 56.
HCF = 4
iv) 10, 35,40
Factorization method
10 = 2 x 5
35 = 5 x 7 40 = 2 x 2 x 2 x 5
HCF = 5
Continued division method:
First find the HCF of the smallest and biggest numbers of the given numbers, i.e., 10 and 40.
HCF of 10 and 40 is 10.
10 is a factor of 40.
So, HCF (10, 40) is 10.
Now, find HCF of 10 and the remaining number i.e., 10 and 35.
HCF of 10 and 35 is 5.
∴ HCF of 10, 35 and 40 is 5.
Question 2.
Two milk cans have 60 and 165 liters of milk. Find a can of maximum capacity which can exactly measure the milk in two cans.
Capacity of first can = 60 l
Capacity of second can = 165 l
Capacity of required can = HCF (60,165)
HCF of 60 and 165 is 15.
Therefore maximum capacity of the required can is 15 liters.
Question 3.
Three different containers contain different quantities of milk whose measures are 403 lit, 465 lit, 527 liters. What biggest measure must be’ there to measure all different quantities in exact number of times? |
# Unlocking the Secret to Calculating the Number of Sides in a Polygon with a Given Interior Angle
## Introduction to Calculating the Number of Sides in a Polygon Using its Interior Angle
Polygons are 2-dimensional shapes which have 3 or more straight sides. The internal angles of a polygon add up to a certain value, depending on the number of sides. In this blog we will explain how to use the interior angles of a polygon to calculate the number of sides it has.
To begin with, letâs start by talking about what an interior angle is. An interior angle is formed when two sides of a shape meet at one point. Interior angles are also referred to as vertex angles or corner angles because they measure how much one side turns relative to another when two lines intersect.
The Interior Angle Sum Theorem states that for any simple polygon, the sum of all its interior angles will always equal (n-2)â˘180Ë, where n represents the number of sides in the shape. This formula works regardless of whether you are dealing with a triangle, quadrilateral, pentagon, and so on – so it can be applied no matter what kind of inner angle calculation you’re trying to make!
Now that we know the math equation behind calculating an interior angle, let’s look at how this translates into actionable steps for finding out your own geometry problem! First off, decide if your shape is regular or irregular. A regular polygon is one whose sides are equal and its angles are congruent; meaning all six sides and six angles measure exactly the same size. An irregular polygon is just like it sounds â any type or combination of side lengths with different sizes or angles from each other.
Now determine how many interior angles there will be (for example if youâre looking at a triangle there should be three total since it has three distinct corners). Next take these totals (and remembering our formula from above) subtract two from whatever total you get and then multiple that result by 180° to get your sum total for all the inner angles combined together within that particular figure! So if we had three individual inner corners here then 3 -2 = 1 and 1 * 180° = 180°; giving us our total sum answer!
Finally divide this sum by each individual inner corner (remembering the rule that no matter what kind of shape it is, all its internal angels have identical values) – so in this case we would simply divide 180 by 3 resulting in 60° per corner securing success in solving our initial query â âhow do we uniquely identify internal corners using already known information?â In other words: If a shape has x inner corners then each must measure 360/x degrees (\$frac{360}{x}\$). VoilĂ ! There you have it: an easy yet reliable way use information gathered from inside a geometric figure itself when attempting not only calculate but also uniquely identify each separate adjacent inside corner measurement along those same edges simultaneouslyâ
## Step-by-Step Instructions on How to Calculate the Number of Sides
Calculating the number of sides in a shape can be tricky to do, especially if you are not familiar with basic geometry. That said, it isn’t as hard as it seems! Here are step-by-step instructions on how to calculate the number of sides for any polygon or figure:
Step 1: Identify What Kind of Figure You Are Dealing With
The first step is to identify what kind of figure you are dealing with; this will help determine which methods and formulas you should use. Youâll need to consider whether your shape is convex (open) or concave (closed). Common examples include squares, rectangles, pentagons, hexagons, heptagons and octagons.
Step 2: Using Line Counting To Determine The Number Of Sides
If your figure can be closed into an enclosed shape without crisscrossing lines forming additional triangles or other shapes inside the original form, then you can simply count the number of lines that separate each space within the perimeter. This method will give you an accurate measurement of the number of sides for that particular shape.
Step 3: Using Angles To Find The Number Of Sides
In some cases â when a Polygon has internal angles rather than straight lines separating its pieces â counting may not be possible. In these cases, use a Protractor to measure each angle surrounding the points in order to get a good estimate. Then utilize a mathematical formula to convert angles into side numbers; it typically takes fewer acute angles (smaller than 90 degrees) divided by 180 degrees to equal one side per angle.
Step 4: Apply These Numbers To A Formula To Calculate The Number Of Sides
Once you have all your necessary components recorded, itâs time to apply them together using a formula that gives you the final number of total sides in the shape before you. Depending on the type of shape (convex or concave), this formula might look different for each case but should still generally involve adding together each angle and dividing by 180 degrees multiplied by two times two plus two times three minus two etc., where applicable. Thatâs it! Now that you know how to calculate the number of sides in any given figure â from circles and squares all way up through more complex polygons â go ahead and set out on your own geometric exploration!
## Advanced Techniques for Finding the Number of Sides with More Complex Math
When it comes to finding the number of sides on a polygon or other shape, simple math can often come up short. For more complex calculations, advanced techniques may be required. Here are some of the most useful methods for uncovering the number of sides when faced with a more difficult problem.
First and foremost, visualizing the shape can make all the difference. By carefully sketching out each line and vertex point, it is much easier to understand how many points and lines comprise it – as well as make sense of any measurements or given values present in the question. Often times, this method can make solving an otherwise convoluted problem much simpler.
For more intricate shapes, calculating angles comes into play. Knowing that angles around any point total to 360° makes turning angles into sides much easier to do — simply divide 360 by each angle measurement and youâll figure out how many sides each angle connects with! Additionally, determining perimeter in such cases allows for a simple side index calculation by learning ratios between two pieces: if two known side lengths join at a single vertex they must always add up to 180°; use this ratio along with any available information to deduce other missing side lengths.
Finally, if these techniques donât work â try breaking long equations down into smaller parts until understanding becomes a manageable task: separate curves from straight edges and combine smaller angles that may have been initially overlooked until one large answer presents itself! This method works particularly well for problems which involve circles intersecting polygons â which take lots of measuring to solve without making division mistakes!
No matter what kind of polygon is being attempted, always keep in mind that like numbers should always stay together during calculations: adding all interior angles together before attempting division or multiplication means accounting for all measurements alongside each other rather than singling out one value at a time â simplifying equations greatly! With these tricks under your belt next time you encounter an especially tricky number-of-sides problem you should have no fear knowing advanced math will pull through when geometry alone fails!
## FAQs About Calculating Polygon Sides from Interior Angles
Q: What is a polygon?
A: A polygon is a shape with three or more straight sides that are joined together. Polygons can be regular or irregular, convex or concave. Regular polygons have all equal side lengths and equal interior angles. Irregular polygons donât usually have equal sides or angles.
Q: What are the different types of polygons?
A: The two main categories of polygons are convex and concave. Convex polygons have all their angles pointing outward, while concave polygons have at least one angle pointing inward. Specific names for common types of polygons include triangle (3 sides), quadrilateral (4 sides), pentagon (5 sides) hexagon (6 sides), heptagon (7 sides), octagon (8 sides), nonagon (9 sides) and decagon (10 sides).
Q: How do you calculate the number of sides in a polygon from its interior angles?
A: Since each interior angle in a regular n-sided polygon has an measure of 180(n-2)/n there is a formula to calculate the number of sides when given the measure of each angle: n = 360/(180-[given angle]). For example, if the interior angle measure is 140° then the formula would give 5 result [360/[180 – 140] = 5].
## Top 5 Facts about Calculating Sides from Interior Angles
1. The interior angles of a triangle always add up to 180°. This is one of the main properties of a triangle, so itâs useful to keep in mind when calculating the side lengths from its interior angles.
2. The formula used to calculate the sides of an arbitrary triangle from its interior angles is known as âThe Law of Cosinesâ and reads: a² = b² + c² â 2bc cosA, where A is any given angle and a, b, and c are its opposite side lengths, respectively.
3. Furthermore, once either two of the side lengths or two of the interior angles are known in combination, you can use them with The Law Of Sines formula to determine the third unknown side or angle which reads: a/sinA = b/sinB = c}.
4. When available (which isn’t always the case), knowing all three sides (a,b,c) can also be used to calculate an unknown internal angle through another concept called The Law Of Cevians which utilizes Heron’s Formula for Area AtForTriangles built on using three sides and Semi-Perimeters: s= (a+b+c)/2 . From here you solve for an angle using tan(A)=[sqrt(s*(s-a)*(s-b)*(s-c))]/(ab).
5. Last but not least when calculating sides or angles based on Interior Angle measurements keep in mind that this method only applies to right triangles which have one 90° internal angle measurement – as stated earlier 180° is still applicable as long as thereâs one right angle present; so if you know only 180° out of three most likely your shape isn’t ‘triangular’ but rather âother polygonâ instead!
## Conclusion: How to Easily Find the Number of Sides with an Interior Angle Measurement
Finding the number of sides with an Interior Angle Measurement can seem tricky at first, but it is actually quite easy. All you have to do is start by determining the sum of interior angles in a polygon. The formula for this sum is (n â 2) x 180°, where ânâ represents the total number of sides on the polygon. To find the number of sides for a polygon with an interior angle measurement, you need to solve for ânâ in this expression by dividing both sides by 180 and subtracting 2 from either side, thus giving you your answer: n = (measured_angle/180) + 2.
For instance, if we have a shape with an interior angle measuring 120°, then all we have to do is plug that into our equation: n = (120°/180°) + 2 which simplifies to n = 1 + 2 which results in n = 3 , meaning our shape has 3 total sides!
In summary, finding the number of sides for any given shape is quite straightforward as long as you keep track of your measurements and adhere to the simple equation provided. Remember that for any polygon consisting of ‘n’ sides, its sum of interior angles will always equate to (n â 2) x 180°; from there, all you need to do is divide each side by 180 and add two back onto either side in order to obtain ânâ – or simply put – the total number of its sides! |
# Class 9 Maths Chapter 8 Quadrilaterals MCQs
Class 9 Maths Chapter 8 Quadrilaterals MCQs are available here with answers, online. The questions are prepared, as per the CBSE syllabus and NCERT curriculum. Students can prepare for exams with the help of these objective questions to score good marks. The answers are available with detailed explanations. Get chapter-wise MCQs at BYJU’S and also check Important Questions for Class 9 Maths.
## MCQs on Class 9 Maths Chapter 8 Quadrilaterals
Solve the questions provided below with four multiple options and choose the correct one.
1) The quadrilateral whose all its sides are equal and angles are equal to 90 degrees, it is called:
a. Rectangle
b. Square
c. Kite
d. Parallelogram
2) The sum of all the angles of a quadrilateral is equal to:
a. 180°
b. 270°
c. 360°
d. 90°
3) A trapezium has:
a. One pair of opposite sides parallel
b. Two pairs of opposite sides parallel to each other
c. All its sides are equal
d. All angles are equal
Explanation: A trapezium has only one pair of opposite sides parallel to each other, and the other two sides are non-parallel.
4) A rhombus can be a:
a. Parallelogram
b. Trapezium
c. Kite
d. Square
5) A diagonal of a parallelogram divides it into two congruent:
a. Square
b. Parallelogram
c. Triangles
d. Rectangle
6) In a parallelogram, opposite angles are:
a. Equal
b. Unequal
c. Cannot be determined
d. None of the above
7) The diagonals of a parallelogram:
a. Equal
b. Unequal
c. Bisect each other
d. Have no relation
8) Each angle of the rectangle is:
a. More than 90°
b. Less than 90°
c. Equal to 90°
d. Equal to 45°
Explanation: Let ABCD is a rectangle, and ∠A = 90°
AD || BC and AB is a transversal
∠ A + ∠ B = 180° (Interior angles on the same side of the transversal)
∠ A = 90°
So, ∠ B = 180° – ∠ A = 180° – 90° = 90°
Now, ∠ C = ∠ A and ∠ D = ∠ B (Opposite angles of the parallelogram)
So, ∠ C = 90° and ∠ D = 90°
Hence all sides are equals to 90°.
9) The angles of a quadrilateral are in the ratio 4: 5: 10: 11. The angles are:
a. 36°, 60°, 108°, 156°
b. 48°, 60°, 120°, 132°
c. 52°, 60°, 122°, 126°
d. 60°, 60°, 120°, 120°
Explanation: Let x be the common angle among all the four angles of a quadrilateral.
As per angle sum property, we know:
4x+5x+10x+11x = 360°
30x = 360°
x = 12°
Hence, angles are
4x = 4 (12) = 48°
5x = 5 (12) = 60°
10x = 10 (12) = 120°
11x = 11 (12) = 132°
10) If ABCD is a trapezium in which AB || CD and AD = BC, then:
a. ∠A = ∠B
b. ∠A > ∠B
c. ∠A < ∠B
d. None of the above
Explanation: Draw a line through C parallel to DA intersecting AB produced at E.
CE = AD (Opposite sides)
AD = BC (Given)
BC = CE
⇒ ∠CBE = ∠CEB
also,
∠A + ∠CBE = 180° (Angles on the same side of transversal and ∠CBE = ∠CEB)
∠B + ∠CBE = 180° ( As Linear pair)
⇒ ∠A = ∠B
11) Which of the following is not true for a parallelogram?
(a) Opposite sides are equal
(b) Opposite angles are equal
(c) Opposite angles are bisected by the diagonals
(d) Diagonals bisect each other.
Explanation: Opposite angles are bisected by the diagonals is not true for a parallelogram. Whereas opposite sides are equal, opposite angles are equals, diagonals bisect each other are the properties of a parallelogram.
12) Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is
(a) 90º
(b) 95º
(c) 105º
(d) 120º
Explanation: We know that the sum of angles of a quadrilateral is 360º.
Let the unknown angle be x.
Therefore, 75º+90º+75º+x = 360º
x = 360º – 240º = 120º.
13) ABCD is a rhombus such that ∠ACB = 40º. Then ∠ADB is
(a) 40º
(b) 45º
(c) 50º
(d) 60º
Explanation: We know that the diagonals of the rhombus bisect each other perpendicularly.
By using the alternate interior angles, and angle sum property of triangle, we can say:
From the triangle, BOC,
∠BOC + ∠OCB + ∠OBC = 180º
(where ∠BOC= 90º, ∠OCB = 40º)
90º+40º+ ∠OBC = 180º
∠OBC = 180º – 130º
∠OBC = 50º
∠OBC =∠DBC
Now, by using alternate angles, we can say
14) The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if
(a) PQRS is a rhombus
(b) PQRS is a parallelogram
(c) Diagonals of PQRS are perpendicular
(d) Diagonals of PQRS are equal.
Explanation: The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus if the diagonals of PQRS are equal.
15) A diagonal of a rectangle is inclined to one side of the rectangle at 25º. The acute angle between the diagonals is
(a) 25º
(b) 40º
(c) 50º
(d) 55º
Explanation: Consider the rectangle ABCD
In a triangle BOC,
∠OBC = ∠OCB (Opposite angles of isosceles triangle)
Therefore, ∠OBC + ∠OCB+∠BOC = 180º
25º+25º + ∠BOC = 180º
∠BOC = 180º- 50º
∠BOC = 130º.
By using the linear pair,
∠AOB + ∠BOC = 180º
∠AOB = 180º – 130º
∠AOB= 50º
Hence, the acute angle between the diagonals is 50º.
16) If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3:7:6:4, then ABCD is a
(a) Kite
(b) Rhombus
(c) Parallelogram
(d) Trapezium
Explanation:
Given that, the angles A, B, C and D of a quadrilateral are in the ratio 3:7:6:4.
We know that A+B+C+D = 360º
Hence, now we can assume, 3k+7k+6k+4k= 360º
20k = 360º
k=18º
Therefore, A = 3k = 54º
B= 7k = 126º
C = 6k = 108º
D = 4k= 72º
If we draw the quadrilateral with these angles, we get a trapezium.
17) The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if
(a) PQRS is a rectangle
(b) PQRS is a parallelogram
(c) Diagonals of PQRS are perpendicular
(d) Diagonals of PQRS are equal
Explanation: The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if the diagonals of PQRS are perpendicular.
18) The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is
(a) a square
(b) a rhombus
(c) a rectangle
(d) any parallelogram
Explanation: Let ABCD be a rhombus. P, Q, R, S be the midpoint of the sides AB, BC, CD and DA. If we join the midpoints, we will get the shape rectangle.
19) If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form
(a) Square
(b) Rectangle
(c) Rhombus
(d) any other parallelogram
Explanation:
Hence, the bisectors of the angles APQ, BPQ, CQP and PQD form the shape rectangle.
20) Which of the following is not a quadrilateral?
(a) Kite
(b) Square
(c) Triangle
(d) Rhombus
Explanation: square, kite and rhombus are quadrilaterals as it has four sides. Whereas a triangle is not a quadrilateral as it has only three sides.
Stay tuned with BYJU’S – The Learning App and download the app today to get more class-wise concepts.
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Volume of rectangular prism (Step by Step)
volume of Rectangular Prism
We calculate the different volume of a rectangular prism having a different dimension and different shapes. All calculations are in meters, feet or in centimeters.
After this reading, you are able to do every volume calculation of prism (rectangular). By applying the same strategies which we have done in these examples.
We divided the calculation into different examples. For your better understanding first goes for less difficult examples and then for difficult ones… Let’s jump into the example
1: Example
Let’s suppose we have a rectangular prism having a length 0.9 m, width 0.5 m, and having a height 0.5 m. Calculate the volume of the rectangular prism.
Length = 0.9 m
Width = 0.5 m
Height = 0.5 m
Solution:
The solution is divided into
Step – 1
First, you have to look at the dimensions (Length, Width, Height). So the given dimensions were in meters.
Step – 2
Calculate the volume. The formula is (L x W x H). By putting the given values in this formula
Volume 1 = L x W x H
= 0.9 m x 0.5 m x 0.5 m
= 0.225 meter cube Ans...
In the first we got the result of our rectangular prism is 0.225 meter cube. The same formula you can use in feet and centimeters dimensions by using the same strategy. Let’s jump into the next example.
2: Example:
Let’s suppose we have a difficult prism having different dimensions in different sections as shown in figure
It looks difficult but when you simplify it, it is very easy to calculate the volume. let’s do some math and simplify the given figure.
So we divide the rectangular pyramid into 3 rectangles. let’s do some math to calculate volume. (L x W x H)
Rectangle 1 = (0.8 m x 0.9 m x 0.5 m) = 0.36 meter cube
Rectangle 2 = (0.8 m x (0.9 – o.o5 – o.05) m x 0.5 m) = 0.32 meter cube
Rectangle 3 = (0.8 m x 0.7 m x 0.5 m) = 0.28 meter cube
By adding the rectangle 1,2,3 we get the total Volume of rectangular prism.
= 0.36 + 0.32 + 0.28
= 0.96 meter cube Ans…
So the whole prism have a 0.96 meter cube volume. In this way, the calculation was done. Let’s jump into the next example for more understanding.
3 Example
Let’s suppose we have a loading truck. Its backside is in rectangular and we have to calculate its Volume. (as shown in the figure.
The calculation is very easy and simple just like we do in the first example.
Length = 15 ft
Width = 7.5 ft
Height = 7.5 ft
Solution:
same first we note its dimension and then proceed its next calculation.
= (L x W x H)
= 15 x 7.5 x 7.5
= 843.75 cubic feet Ans…
So the truck storage volume is 843.75 cubic feet. In this way the calculation was done you may |
# Can You Pass Algebra?
EDUCATION
By: Jouviane Alexandre
4 Min Quiz
Image: Shutterstock
We've all been there! You might've gotten your first taste of algebra in elementary school and took your last course in high school or college. Test your knowledge and see if you could still pass the course!
# How would you write '10 decreased by X?'
When 10 is decreased by amount 'X,' you are subtracting X from the amount.
# Simplify: -8 - 1 =
-8-1 = (-8) + (-1) = -9
# Combine like terms: 3x^2 + 7x - 4x - 3x^2
The solutions breaks down to (-3x^2) cancelling out (3x^2) which leaves you with 7x - 4x = 3x
# Solve this equation: x + 8 = -5
When following the steps for this equation, subtracting 8 from both sides will leave you with the solution x=-13.
# Expand: (5x + 2)(3x - 4)
Multiplying the first two terms of each factor will yield 15x^2. Multiplying the last two terms of each factor will yield -8. Multiplying 5x and -4 while yield -20x while the product of 2 and 3x is 6x. When 6x is added to -20x, the sum is -14x.
# Factor: x^2 + 11x + 28
Although the product of 14 and 2 equals 28, the numbers used must be 7 and 4 because when added together, you will receive the middle term of 11x.
# Simplify: (11x/10) - (x/10)
When the numerator is handles, you get the difference to be 10x. When the numerator (10x) is divided by the denominator (10), the quotient is x.
# Find the solution: 2x-4y = 12
When (0,-3) is made to replace x and y, respectively, the equation becomes 0 + 12 = 12.
# The sum of two numbers is 22. The larger is two less than twice the smaller number. What is the larger number?
If x equals the smaller number and the larger number is two less than twice the smaller, the larger number can be written as 2x-2. When 2x - 2 + x = 22, the solution for x = 8. When 8 is substituted into the larger number (2x-2), the answer is 14.
# What is the square root of 28?
When solving radicals that aren't perfect squares, the first step is to split them up in a fashion where you can solve for a perfect square. Radical 28 can be simplified to radical 4 and radical 7. Since 4 is a perfect square, the radical 4 gets simplified to 2 and the whole answer is written as 2 radical 7.
# What is (3x)^3 in expanded form?
When expanding with exponents, you take the base and multiply it how ever many times the number of the exponent is. In this example, you would take 3x and multiply it 3 times.
# Multiply: (-3)(-3)
When two negative numbers are multiplied, the product is automatically positive. Thus, you can look at is as (3)(3) which is 9.
# Divide c^6/C^2
When dividing exponents that share the same base, subtract the denominator's exponent from the numerators exponent for the answer.
# 25 is 4 less than a number. What is the number?
If 25 is 4 less than a number, adding 25 and 4 together will create the sum of the number thus the answer is 29.
# Expand: (2x-7)^2
The first term is the product of (2x)^2 which is 4x^2. The last term is (-7)^2 which is 49. The middle term comes from the product of of 2[(2x)(-7)] which is -28.
# Factor: 25-x^2
The answer must have alternating signs in order for it to cancel out the middle term commonly seen. The signs must also be alternating to yield a negative x^2.
# What quadrant does the point (-2,1) fall in?
On a graph, the first quadrant (I) is made up of positive x's and y's. Quadrant II is made up of negative x's and positive y's. The quadrants continue in a counterclockwise fashion.
First radical 24 is split into two radicals (4 and 6). Since four is a perfect square, it can be solved to 2. This number is then multiplied by the three in front of the radical leaving the solution as 6 radical 6.
# If x=2, what is (3x)^3
If x=2, the product inside the parentheses is 6. Then hold 6 to the third power (6x6x6) and the product is 216.
# Simplify: -6+5-2+1-9
Because of the PEMDAS rule, addition and subtraction can occur at the same time. This problem can be easily solved by going in order.
# What is the product of: x(x^3)(x^4)
The rule for multiplying exponents with the same base is to add the exponents. While you might be tempted to think the answer is x^7, the first term 'x' has an invisible exponent - 1. Thus the addition for the exponents is 1+3+4 which will give you the final exponent of 8.
# Solve this inequality: 3-x < 4
For this equation, adding 'x' to both sides and subtracting '4' from both sides of the inequality will give you the answer that x > -1.
# Multiply: x(3x-4)
When x is multiplied along the terms in the parentheses, you get the answer 3x^2 - 4x
# Factor: 5x^2 - 11x + 2
In order to get the first term as 5x^2, the first two terms of the factors needs to yield a product of 5x^2 which can be achieved by multiplying x by 5x. In order to get a positive 2 as the last term, the last two terms of the factors need to both carry positive signs or both carry negative signs as in answer 3.
# What is the slope of the equation: 3x + 4y = 12
In order to solve for slope, 3x would have to be subtracted from both sides of the equation. Both sides of the equation would be divided by 4 leaving us with the equation y = -3/4x + 3 where the slope is -3/4.
# Write using exponents: (a)(a)(a) + (b)(b)(b)(b) - (c)(c)
(A) multiplied twice yields (a)^2. How ever many times you see the letter appear is the number of its exponent.
# What is the answer to 14/0?
When a number is divided by 0, the answer is never 0. It is always undefined.
# What kind of line is the equation x=2?
If the equation is x=2, no matter what the y-value is, x will always be 2. This means that the line will intersect the x-axis at x=2 and will continue to move vertically.
# Combine like terms: 2x + y - 5 - 5x -2y + 1
When 2x and (-5x) are combined, the difference is -3x. When y and (-2y) are combined, the answer is -y. when 1 is added to (-5) the answer is -4. This gives us the solution -3x - y - 4
# The sum of three consecutive numbers is 108. What is the middle number?
If x = the smallest number, the other two numbers are x + 1 and x + 2 because they are consecutive. You'd then set up the equation x + x + 1 + x + 2 = 108. Simplifying this equation will give you 3x + 3 = 108. Simplifying this further would create the equation 3x = 105 where x = 35. In order to find the middle/second number, you would replace x into the equation for the second number which is x + 1 giving you the answer of 36.
# Solve for x: 3x^2 - 3x = 0
For this equation, you want two numbers that when replaced as x, they will yield 0. Zero is an immediate answer because the product of any number with 0 is 0. The value -1 would make the first term 3 would introducing it to the second term would not properly cancel it. The other value is 1. When added to the first term (3x^2), this will yield 3. When introduced to the second term, (-3x), it will yield -3 which will cancel the first term and make the solution 0.
# What is the y-intercept of the equation 3x + 4y = 12.
In order to find the y-intercept of the equation, pretend x = 0 and solve for y. This will give you the y-intercept of (0,3).
# What is the name of this: c^2 = a^2 + b^2
The Pythagorean Theorem is a common formula used in algebra when dealing with a right triangle.
# In this equation, which letter represents the slope : y = mx + b.
In this equation, which is known as slope-intercept form, m is used to represent the slope while b is the y-intercept.
# a(bc) = (ab)c is known as what property?
This is an example of the associative property of multiplication. It states that the numbers can be multiplied regardless of the way you group them (i.e. whether the parentheses are placed around the a and b or the b and c). |
# FIND MISSING COORDINATE USING DISTANCE FORMULA
## About "Find missing coordinate using distance formula"
Find missing coordinate using distance formula :
Here we are going to see some examples to find missing coordinate using distance formula.
Example 1 :
Find the value of a if the distance between the points at (7, 5) and (a, -3) is 10 units.
Solution :
d = (x2 - x1)2 + (x2 - x1)2
Let x2 = a, x1 = 7, y2 = -3, y1 = 5
Here d = 10
10 = (a - 7)2 + (-3-5)2
10 = (a - 7)2 + (-8)2
10 = (a - 7)2 + 64
10 (a2 - 2 a (7) + 72) + 64
10 (a2 - 14 a + 113)
Taking squares on both sides
102 = ((a2 - 14 a + 113)2
a2 - 14 a + 113 - 100 = 0
a2 - 14 a + 13 = 0
(a - 1) (a - 13) = 0
a = 1 (or) a = 13
Hence the missing coordinates are 1 or 13.
Example 2 :
Find the value of a if the distance between the points at (3, -1) and (a, 7) is 10 units.
Solution :
d = (x2 - x1)2 + (x2 - x1)2
Let x2 = a, x1 = 3, y2 = 7, y1 = -1
Here d = 10
10 = (a - 3)2 + (7 - (-1))2
10 = (a - 3)2 + (7 + 1)2
10 = (a - 3)2 + (8)2
10 (a2 - 2 a (3) + 32) + 64
10 (a2 - 6a + 9 + 64)
10 (a2 - 6a + 73)
Taking squares on both sides
102 = ((a2 - 6 a + 73)2
a2 - 6 a + 73 - 100 = 0
a2 - 6 a - 27 = 0
(a - 9) (a + 3) = 0
a = 9 (or) a = -3
Hence the missing coordinates are 9 or -3.
Example 3 :
Find the value of a if the distance between the points at (10, a) and (1, -6) is 145 units.
Solution :
d = (x2 - x1)2 + (x2 - x1)2
Let x2 = 1, x1 = 10, y2 = -6, y1 = a
Here d = 145
145 = (1 - 10)2 + (-6 - a)2
145 = (- 9)2 + (-6 - a)2
145 = √81 + (6 + a)2
145 = √81 + (62 + 2(6) a + a2)
145 = √(81 + 36 + 12 a + a2)
145 = a2 + 12a + 117
Taking squares on both sides
(√145)2 = (a2 + 12a + 117)2
a2 + 12 a + 117 - 145 = 0
a2 + 12 a - 28 = 0
(a + 14) (a - 2) = 0
a = -14 (or) a = 2
Hence the missing coordinates are -14 or 2.
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Ayushi Sharma
1
Simple, substitute n=1 in 3^n to get 3
n = 2 in 3^n to get 9
n = 3 in 3^n to get 27
n = 4 in 3^n to get 81
n = 5 in 3^n to get 243
3,9,27,81,243 are the first 5 terms
Sangeetha Pulapaka
0
STEP 1: Recall what is a sequence
https://www.qalaxia.com/viewDiscussion?messageId=5d1f5ee65bbeee0fdf682c41
STEP 2: Recall what is a term in a sequence
https://virtualnerd.com/algebra-1/relations-functions/arithmetic-sequences/arithmetic-sequence-definitions/sequence-terms-definition
STEP 3: Plug in n= 1 to n = 5 in the given term to get a sequence
Plugging in n = 1
3^{n} = 3^{1} = 3
Plugging in n =2
3^{n} = 3^{2} = 9
Plugging in n = 3
3^{n} = 3^{3} = 27
Plugging in n = 4
3^{n} = 3^{4} = 81
Plugging in n = 5
3^{n} = 3^{5} = 243
So, the first five terms are 3, 9, 27, 81, 243 |
# Lesson 11: Inference for One Mean: Sigma Unknown
(Redirected from Lesson 11)
These optional videos discuss the contents of this lesson.
## 1 Lesson Outcomes
By the end of this lesson, you should be able to:
• Confidence Intervals for a single mean with σ unknown:
• Calculate and interpret a confidence interval for a population mean given a confidence level.
• Identify a point estimate and margin of error for the confidence interval.
• Show the appropriate connections between the numerical and graphical summaries that support the confidence interval.
• Check the requirements the confidence interval.
• Hypothesis Testing for a single mean with σ unknown:
• State the null and alternative hypothesis.
• Calculate the test-statistic, degrees of freedom and p-value of the hypothesis test.
• Assess the statistical significance by comparing the p-value to the α-level.
• Check the requirements for the hypothesis test.
• Show the appropriate connections between the numerical and graphical summaries that support the hypothesis test.
• Draw a correct conclusion for the hypothesis test.
## 2 What If We Don't Know $\sigma$?
Historical Note:
William Sealy Gosset (1876-1937) was a British industrial scientist and statistician best known for his discovery of the $t$-distribution.
In practice, we almost never know the population standard deviation, $\sigma$. So, it is generally not appropriate to use the formula $$z = \frac{ \bar x - \mu }{ \sigma / \sqrt{n} }$$ In 1908, William Sealy Gosset published a solution to this problem . He found a way to appropriately compute the confidence interval for the mean when $\sigma$ is not known. The basic idea is to use the sample standard deviation, $s$ in the place of the true population standard deviation, $\sigma$. If $\sigma$ is not known, we cannot base the calculations on the standard normal distribution, and we cannot use the formula above to conduct hypothesis tests.
In a remarkable piece of work, Gosset found the appropriate distribution to use when $\sigma$ is unknown. At the time of this discovery, Gosset worked for the Guinness brewery. To avoid problems with industrial espionage, Guinness prohibited employees from publishing any research results. Knowing his work provided a significant contribution to Statistics, Gosset chose convinced his employer to allow him to publish under a pseudonum. He chose the pseudonym "Student". Gosset's test statistic was denoted by the letter $t$, this distribution has come to be known as Student's t-distribution.
$$t = \frac{ \bar x - \mu }{ s / \sqrt{n} }$$
The $t$-distribution is bell-shaped and symmetrical. The $t$-distribution has a mean of 0, but it has more area in the tails than the standard normal distribution. The exact shape of the $t$-distribution depends on a parameter called the degrees of freedom (abbreviated $df$). The degrees of freedom is related to the sample size. As the sample size goes up, the degrees of freedom increase accordingly. For the procedures discussed in this lesson, the degrees of freedom equal the sample size minus one: $df = n-1$.
Click here to explore how the shape of the $t$-distribution changes with the degrees of freedom. Notice that as the degrees of freedom increases, the shape of the $t$-distribution (black curve) gets closer to the standard normal distribution (red curve). The red curve is identical in the three images below.
$t$-distribution
with $df = 1$
$t$-distribution
with $df = 5$
$t$-distribution
with $df = 15$
(Image credit: Webster West, http://www.stat.tamu.edu/~west/applets/tdemo1.html)
## 3 Hypothesis Tests
### 3.1 Body Temperatures Revisited
We will apply the $t$-distribution to the body temperature data we explored previously.
This hypothesis test is conducted in a manner similar to a test for a single mean where $\sigma$ is known, except that instead of using the population standard deviation, $\sigma$, in the calculations, we estimate this value using the sample standard deviation, $s$. This leads to a $t$-distribution, rather than a normal distribution for the test statistic. We will not need to compute the value of this test statistic by hand. It will be done using Software.
Summarize the relevant background information
We want to conduct a hypothesis test to determine if the mean body temperature is different from 98.6° Fahrenheit. Previously, we assumed that we knew the value of $\sigma$. Actually, this value is not known.
State the null and alternative hypotheses and the level of significance
$$\begin{array}{rl} H_0: \mu = 98.6\\ H_a: \mu \ne 98.6\\ \end{array}$$ $$\alpha = 0.05$$
Describe the data collection procedures
We will use the body temperature data, BodyTemp, collected by Dr. Mackowiak and his colleagues to conduct the test.
Give the relevant summary statistics
$$\begin{array}{l} \bar x = 98.23\\ s = 0.738\\ n = 148 \end{array}$$
Make an appropriate graph (e.g. a histogram) to illustrate the data
Verify the requirements have been met
• We assume that the individuals chosen to participate in the study represent a (simple) random sample from the population.
• $\bar x$ will be normally distributed, because the sample size is large. (Note: We could have also noticed that the body temperature data appears to be normally distributed, so even with a small sample size, $\bar x$ would be normal.)
Give the test statistic and its value
We will need to conduct the analysis using software, so we can report this value.
Instructions for conducting a test for one mean with $sigma$ unknown:
Excel Instructions
Here are the instructions for conducting the one sample t-test in Excel:
The Excel file needed for this analysis is QuantitativeInferentialProcedures.xls. We will use this Excel file to conduct the hypothesis tests for a single mean with $\sigma$ unknown.
• After opening the file, please click on the tab labeled "One-sample t-test".
• Paste the data in the appropriate place in Column A.
• Set the null hypothesis value in cell D10. For this example, the value should be 98.6.
• Click in cell D11 and use the drop down menu to set the alternative hypothesis to: "Not Equal To"
The image below shows the Excel file after these changes have been made.
Consider each of the following alternative hypotheses that could be used in a test for a single mean where the null hypothesis is $H_0: \mu = 98.6$.
• $H_a: \mu \ne 98.6$
Choose "Not Equal To" in the drop-down menu in cell D11.
• $H_a: \mu < 98.6$
Choose "Less Than" in the drop-down menu in cell D11.
• $H_a: \mu > 98.6$
Choose "Greater Than" in the drop-down menu in cell D11.
You will have opportunities to practice using each of these.
The interpretation of the results will follow the pattern established in the previous hypothesis tests. If the $P$-value is less than the $\alpha$ level, we reject the null hypothesis. If the $P$-value is greater than the $\alpha$ level, we fail to reject the null hypothesis. This is true for every hypothesis test.
The test statistic is $t$ and its value is $-6.029$. So, we have $t = -6.029$.
Notice this is the same number we get if we use the formula: \begin{align} t &= \frac{ \bar x - \mu }{ s / \sqrt{n} } \\ &\approx \frac{ 98.234 - 98.6 }{ 0.738 / \sqrt{148} } \\ &\approx -6.03 \end{align} Any differences are due to rounding.
State the degrees of freedom
In Excel, the degrees of freedom ($df$) are given in cell E21.
$$df = 147$$
Find the $P$-value and compare it to the level of significance
The $P$-value is given in the software as "1.2723E-08". Writing this properly and comparing it to $\alpha$, we have: $$P\text{-value} = 1.2723 \times 10^{-8} = 0.000~000~012~723 < 0.05 = \alpha$$
The interpretation of the results will follow the pattern established in the previous hypothesis tests. If the $P$-value is less than the $\alpha$ level, we reject the null hypothesis. If the $P$-value is greater than the $\alpha$ level, we fail to reject the null hypothesis. This is true for every hypothesis test.
Since the $P$-value was less than $\alpha$, we reject the null hypothesis.
Present your conclusion in an English sentence, relating the result to the context of the problem
There is sufficient evidence to suggest that the mean body temperature is not 98.6° Fahrenheit.
### 3.2 Baby Boom
Summarize the relevant background information
The birth weight of a child is an important indicator of their neonatal health. It is important that pediatric health care providers track changes in the birth weights over time. The birth weight of children in Australia has historically had a population mean of 3373 grams. Is this still the mean birth weight of Australian children, or has there been a change? We will use the 0.05 level of significance.
State the null and alternative hypotheses and the level of significance
\begin{align} H_0: &~ \mu=3373 \\ H_a: &~ \mu \ne 3373 \end{align} $$\alpha = 0.05$$
Describe the data collection procedures
The birth weights of all children born on December 18, 1997 at the Mater Mothers' Hospital in Brisbane, Australia were recorded . The time of birth (on a 24 hour clock), gender, and birth weight of each child are given in the file BabyBoom.
Using this data set, test the hypothesis that the mean weight of babies born in Australia is 3373 grams. Use the $\alpha=0.05$ level of significance for this problem. Make an appropriate graph of the data.
1. Give the relevant summary statistics
When the data are entered into the file QuantitativeInferentialProcedures.xls, the following output is generated:
The relevant summary statistics are:
\begin{align} \bar x &= 3275.95 \textrm{g}\\ s &= 528.03 \textrm{g}\\ n &= 44 \end{align}
2. Make an appropriate graph to illustrate the data
Verify the requirements have been met
The data show a left-skewed shape, however the sample size is large. Using the Central Limit Theorem, we can conclude that the sample mean is normally distributed and the requirements are satisfied.
3. Give the test statistic and its value
The test statistic is a $t$. The value of the test statistic is $-1.219$. This was taken from the output above. We have $t = -1.219$.
4. State the degrees of freedom
The degrees of freedom are given in the output above:
$df = 43$
Notice that this is one less than than sample size. For a test for a single mean, the degrees of freedom are always equal to one less than the sample size.
Mark the test statistic and $P$-value on a graph of the sampling distribution
The $P$-value is shaded in green:
Find the $P$-value and compare it to the level of significance
$$P\text{-value}=0.228 > 0.05 = \alpha$$
Since the $P$-value is greater than the level of significance, we fail to reject the null hypothesis.
Present your conclusion in an English sentence, relating the result to the context of the problem
There is insufficient evidence to suggest that the mean weight of babies born in Australia is different from 3373 grams.
## 4 Confidence Intervals
The procedure for finding confidence intervals when $\sigma$ is not known is very similar to the process when $\sigma$ is known. As a reminder, here is the confidence interval (from Lesson 10) when $\sigma$ is known. $$\left( \bar x - z^* \frac{\sigma}{\sqrt{n}}, ~ \bar x + z^* \frac{\sigma}{\sqrt{n}} \right)$$
To construct the confidence interval when $\sigma$ is not known, we replace the population standard deviation, $\sigma$, with its point estimate, $s$. The appropriate distribution is a $t$, rather than a $z$. So, we replace $z^*$ with $t^*$. So the confidence interval when $\sigma$ is not known is $$\left( \bar x - t^* \frac{s}{\sqrt{n}}, ~ \bar x + t^* \frac{s}{\sqrt{n}} \right)$$ where $t^*$ is the $t$-score corresponding to the confidence level and $s$ is the sample standard deviation.
We will use
Excel
to construct confidence intervals when $\sigma$ is unknown. The values of $t^*$ and $s$ will be computed for us. Examples of this are given below. You can see the details by clicking on [Show Hand Calculations] in Step 4.
### 4.1 Automatic Language Translation Programs
Summarize the relevant background information
Computer software is commonly used to translate text from one language to another. As part of his Ph.D. thesis, Philipp Koehn developed a phrase-based translation program called Pharaoh.
The quality of the translation can vary. A good translation system should match a professional human translation. It is important to be able to quantify how good the translations produced by Pharaoh are.
The IBM T. J. Watson Research Center developed methods to measure the quality of a translation from one language to another. One of these is the the BiLingual Evaluation Understudy (BLEU). BLEU is a score ranging from 0 to 1 that indicates how well a computer translation matches a professional human translation of the same text. Higher scores indicate a better match. BLEU helps companies who develop translation software "to monitor the effect of daily changes to their systems in order to weed out bad ideas from good ideas".
Describe the data collection procedures
To test Pharaoh's ability to translate, Koehn took a random sample of 100 blocks of Spanish text, each of which contained 300 sentences, and used Pharaoh to translate each of these to English. The BLEU score was calculated for each of the 100 blocks. The data were extracted from Figure 2 in a paper Koehn published. The 100 BLEU scores are given in BLEU-Scores.
Koehn wants to find an estimate of the true mean BLEU score for text translated by the Pharaoh computer program. He would like to compute a confidence interval, but he does not know the true population standard deviation, $\sigma$.
Give the relevant summary statistics
Excel Instructions
Copying the BLEU-Scores data into the QuantitativeInferentialProcedures.xls file in Excel, we get the following:
The summary statistics are:
\begin{align} \bar x =& 0.2876 \\ s =& 0.0264 \\ n =& 100 \end{align}
Make an appropriate graph to illustrate the data
Here is a histogram of the data showing 10 bins.
Verify the requirements have been met
The requirements for creating a confidence interval for a mean with $\sigma$ unknown are the same as the requirements for this procedure when $\sigma$ is known:
#### 4.1.1 Requirements
There are two requirements that need to be checked when computing a confidence interval for a mean with $\sigma$ unknown:
1. A simple random sample was drawn from the population
2. $\bar x$ is normally distributed
Recall the requirement of normality is satisfied if the data are approximately normally distributed or if the sample size is large.
The data are bell-shaped and fairly symmetric. So, the sample mean, $\bar x$, is approximately normally distributed.
Find the confidence interval
The formula for the confidence interval where $\sigma$ is known was given in the reading titled Inference for One Mean: Sigma Known (Confidence Interval)#11:CIformula-SigmaKnown|Lesson 10: Inference for One Mean: Sigma Known (Confidence Interval)|Inference for One Mean: Sigma Known (Confidence Interval) as $$\left( \bar x - z^* \frac{\sigma}{\sqrt{n}}, ~ \bar x + z^* \frac{\sigma}{\sqrt{n}} \right)$$ It is impossible to know the true standard deviation of the BLEU scores for a new translation program like Pharaoh. Replacing $\sigma$ with $s$ and replacing $z^*$ with $t^*$, we get: $$\left( \bar x - t^* \frac{s}{\sqrt{n}}, ~ \bar x + t^* \frac{s}{\sqrt{n}} \right)$$ The value of $t^*$ depends on the level of confidence and the sample size. It must be computed using software or looked up on a table. If you are asked to compute a confidence interval for a mean where the population standard deviation is unknown, the value of $t^*$ will be given to you. The other numbers ($\bar x$, $s$, and $n$) can all be obtained directly from your data.
If we want to create a 95% confidence interval for $\mu$, with 99 degrees of freedom, then $t^* = 1.9842$.
Using the sample statistics: \begin{align} \bar x =& 0.2876 \\ s =& 0.0264 \\ n =& 100 \end{align}
The 95% confidence interval for $\mu$ is: $$\left( 0.2876 - 1.9842 \frac{0.0264}{\sqrt{100}}, ~ 0.2876 + 1.9842 \frac{0.0264}{\sqrt{100}} \right)$$ which reduces to: $$\left( 0.2824, ~ 0.2928 \right)$$
Excel Instructions
We will use Excel to find the confidence intervals for the mean.
Do the following:
In this case, we are using a 95% confidence level. So, you set the confidence level in the Excel file to 0.95 (i.e., 95%). The confidence interval is given to you in cells G21 and H21.
The 95% confidence interval for the true mean BLEU score is:
$$(0.282, 0.293)$$
Present your observations in an English sentence, relating the result to the context of the problem
We are 95% confident that the true mean BLEU score for all translations by the Pharaoh program is between 0.2824 and 0.2928.
Answer the following questions using the data set BLEU-Scores, which gives the BLEU scores for $n=100$ translations from Spanish to English by the computer program Pharaoh.
5. Use
Excel
to find a 90% confidence interval for the true mean BLEU score for translations by the Pharaoh program. Give your answer accurate to 3 decimal places. Interpret this confidence interval in a complete sentence.
$(0.283, 0.292)$
We are 90% confident that the true mean BLEU score for translations by the Pharaoh program lies between 0.283 and 0.292.
6. Use
Excel
to find a 99% confidence interval for the true mean BLEU score for translations by the Pharaoh program. Give your answer accurate to 3 decimal places. Interpret this confidence interval in a complete sentence.
$(0.281, 0.295)$
We are 99% confident that the true mean BLEU score for translations by the Pharaoh program lies between 0.281 and 0.295.
7. What do you notice about the confidence interval as the confidence level increased from 90% to 99%?
As the confidence level increases, the width of the confidence interval increases. We could also say, as the confidence level increases, the margin of error increases. The center of the confidence interval (the sample mean) is unchanged.
### 4.2 Euro Weights
Summarize the relevant background information
A group of statisticians measured the weights of 2000 Belgian one Euro coins in eight batches. Each batch contains coins that were all minted together. You can learn more about these data at: http://www.amstat.org/publications/jse/datasets/euroweight.txt
Describe the data collection procedures
The coins were "borrowed" from a bank in Belgium, one batch at a time. The weights (in grams) of the coins are given in the file EuroWeight.
8. Give the relevant summary statistics.
The following output was generated using the Excel file QuantitativeInferentialProcedures.xls:
Typically, we round our computations to one more decimal place of precision than was given in the original data. This means we need one more decimal place than is shown in the output above. We can increase the precision by selecting the cell we want to modify and then clicking on the following button in the "Home" menu ribbon in Excel:
When we do this, we get the following summary statistics:
\begin{align} \bar x =& 7.521 \\ s =& 0.034 \\ n =& 2000 \end{align}
9. Make an appropriate graph to illustrate the data
Verify the requirements have been met
We need to check the following two requirements:
1. A simple random sample was drawn from the population
2. $\bar x$ is normally distributed
The coins were taken in batches, but we can think of the batches as a random sample of the possible coins that could be chosen.
The data follow a bell-shaped distribution. There are a few potential outliers, but they are not too frequent or extreme. We can conclude that $\bar x$ is normally distributed.
10. Find the confidence interval: Use
Excel
to create a 99% confidence interval for the true mean.
$(7.519, ~ 7.523)$
11. Present your observations in an English sentence, relating the result to the context of the problem.
We are 99% confident that the true mean weight of all Belgian Euros is between 7.519 grams and 7.523 grams.
## 5 Summary
Remember...
• In practice we rarely know the true standard deviation $\sigma$ and will therefore be unable to calculate a z-score. Student's t-distribution gives us a new test statistic, $t$, that is calculated using the sample standard deviation ($s$) instead.
$$\displaystyle{ t = \frac {\bar x - \mu} {s / \sqrt{n}} }$$
• The $t$-distribution is similar to a normal distribution in that it is bell-shaped and symmetrical, but the exact shape of the $t$-distribution depends on the degrees of freedom ($df$).
$$df=n-1$$
• You will use Excel to carry out hypothesis testing and create confidence intervals involving $t$-distributions. |
# Solving Two-Step Inequalities
## Presentation on theme: "Solving Two-Step Inequalities"— Presentation transcript:
Solving Two-Step Inequalities
Designed by Skip Tyler, Varina High School
Solving an inequality is similar to solving an equation, with one main difference.
If you divide the inequality by a negative number, you switch the direction of the symbol.
1) Solve 5m - 8 > 12 + 8 + 8 5m > 20 5 5 m > 4 5(4) – 8 = 12
5m > 20 m > 4 5(4) – 8 = 12 Draw “the river” Add 8 to both sides Simplify Divide both sides by 5 Check your answer Graph the solution o 4 5 3
2) Solve 12 - 3a > 18 - 12 - 12 -3a > 6 -3 -3 a < -2
-3a > 6 a < -2 12 - 3(-2) = 18 Draw “the river” Subtract 12 from both sides Simplify Divide both sides by -3 Simplify (Switch the inequality!) Check your answer Graph the solution o -2 -1 -3
Which graph shows the solution to 2x - 10 ≥ 4?
3) Solve 5m - 4 < 2m + 11 o -2m -2m 3m - 4 < 11 + 4 + 4
3m < 15 m < 5 5(5) – 4 = 2(5) + 11 Draw “the river” Subtract 2m from both sides Simplify Add 4 to both sides Divide both sides by 3 Check your answer Graph the solution o 5 6 4
4) Solve 2r - 18 ≤ 5r + 3 ● -2r -2r -18 ≤ 3r + 3 - 3 - 3 -21 ≤ 3r 3 3
-21 ≤ 3r -7 ≤ r or r ≥ -7 2(-7) – 18 = 5(-7) + 3 Draw “the river” Subtract 2r from both sides Simplify Subtract 3 from both sides Divide both sides by 3 Check your answer Graph the solution -7 -6 -8
6) Solve -2x + 6 ≥ 3x - 4 x ≥ -2 x ≤ -2 x ≥ 2 x ≤ 2 Answer Now
5) Solve 26p - 20 > 14p + 64 o -14p -14p 12p – 20 > 64 + 20 + 20
12p > 84 p > 7 26(7) – 20 = 14(7) + 64 Draw “the river” Subtract 14p from both sides Simplify Add 20 to both sides Divide both sides by 12 Check your answer Graph the solution o 7 8 6
What are the values of x if 3(x + 4) - 5(x - 1) < 5? |
# NCERT solution class 12 chapter 5 Continuity and Differentiability exercise 5.2 mathematics part 1
## EXERCISE 5.2
#### Question 1:
Differentiate the functions with respect to x.
Let f(x)=sinx2+5, ux=x2+5, and v(t)=sint
Then, vou=vux=vx2+5=tanx2+5=f(x)
Thus, f is a composite of two functions.
Alternate method
#### Question 2:
Differentiate the functions with respect to x.
Thus, is a composite function of two functions.
Put t = u (x) = sin x
By chain rule,
Alternate method
#### Question 3:
Differentiate the functions with respect to x.
Thus, is a composite function of two functions, u and v.
Put t = u (x) = ax + b
Hence, by chain rule, we obtain
Alternate method
#### Question 4:
Differentiate the functions with respect to x.
Thus, is a composite function of three functions, u, v, and w.
Hence, by chain rule, we obtain
Alternate method
#### Question 5:
Differentiate the functions with respect to x.
The given function is, where g (x) = sin (ax + b) and
h (x) = cos (cx d)
∴ is a composite function of two functions, u and v.
Therefore, by chain rule, we obtain
h is a composite function of two functions, p and q.
Put y = p (x) = cx d
Therefore, by chain rule, we obtain
#### Question 6:
Differentiate the functions with respect to x.
The given function is.
#### Question 7:
Differentiate the functions with respect to x.
#### Question 8:
Differentiate the functions with respect to x.
Clearly, is a composite function of two functions, and v, such that
By using chain rule, we obtain
Alternate method
#### Question 9:
Prove that the function given by
is notdifferentiable at x = 1.
The given function is
It is known that a function f is differentiable at a point x = c in its domain if both
are finite and equal.
To check the differentiability of the given function at x = 1,
consider the left hand limit of f at x = 1
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1
#### Question 10:
Prove that the greatest integer function defined byis not
differentiable at x = 1 and x = 2.
The given function f is
It is known that a function f is differentiable at a point x = c in its domain if both
are finite and equal.
To check the differentiability of the given function at x = 1, consider the left hand limit of f at x = 1
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at
x = 1
To check the differentiability of the given function at x = 2, consider the left hand limit
of f at x = 2
Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2
error: Content is protected !! |
# Understanding Linear Algebra
## Section6.5Orthogonal least squares
Suppose we collect some data when performing an experiment and plot it as shown on the left of Figure 6.5.1. Notice that there is no line on which all the points lie; in fact, it would be surprising if there were since we can expect some uncertainty in the measurements recorded. There does, however, appear to be a line, as shown on the right, on which the points almost lie.
In this section, we’ll explore how the techniques developed in this chapter enable us to find the line that best approximates the data. More specifically, we’ll see how the search for a line passing through the data points leads to an inconsistent system $$A\xvec=\bvec\text{.}$$ Since we are unable to find a solution, we instead seek the vector $$\xvec$$ where $$A\xvec$$ is as close as possible to $$\bvec\text{.}$$ Orthogonal projection gives us just the right tool for doing this.
### Preview Activity6.5.1.
1. Is there a solution to the equation $$A\xvec=\bvec$$ where $$A$$ and $$\bvec$$ are such that
\begin{equation*} \begin{bmatrix} 1 \amp 2 \\ 2 \amp 5 \\ -1 \amp 0 \\ \end{bmatrix} \xvec = \threevec5{-3}{-1}\text{.} \end{equation*}
2. We know that $$\threevec12{-1}$$ and $$\threevec250$$ form a basis for $$\col(A)\text{.}$$ Find an orthogonal basis for $$\col(A)\text{.}$$
3. Find the orthogonal projection $$\widehat\bvec$$ of $$\bvec$$ onto $$\col(A)\text{.}$$
4. Explain why the equation $$A\xvec=\widehat\bvec$$ must be consistent and then find its solution.
### Subsection6.5.1A first example
When we’ve encountered inconsistent systems in the past, we’ve simply said there is no solution and moved on. The preview activity, however, shows how we can find approximate solutions to an inconsistent system: if there are no solutions to $$A\xvec = \bvec\text{,}$$ we instead solve the consistent system $$A\xvec = \bhat\text{,}$$ the orthogonal projection of $$\bvec$$ onto $$\col(A)\text{.}$$ As we’ll see, this solution is, in a specific sense, the best possible.
#### Activity6.5.2.
Suppose we have three data points $$(1,1)\text{,}$$ $$(2,1)\text{,}$$ and $$(3,3)$$ and that we would like to find a line passing through them.
1. Plot these three points in Figure 6.5.2. Are you able to draw a line that passes through all three points?
2. Remember that the equation of a line can be written as $$b + mx=y$$ where $$m$$ is the slope and $$b$$ is the $$y$$-intercept. We will try to find $$b$$ and $$m$$ so that the three points lie on the line.
The first data point $$(1,1)$$ gives an equation for $$b$$ and $$m\text{.}$$ In particular, we know that when $$x=1\text{,}$$ then $$y=1$$ so we have $$b + m(1) = 1$$ or $$b + m = 1\text{.}$$ Use the other two data points to create a linear system describing $$m$$ and $$b\text{.}$$
3. We have obtained a linear system having three equations, one from each data point, for the two unknowns $$b$$ and $$m\text{.}$$ Identify a matrix $$A$$ and vector $$\bvec$$ so that the system has the form $$A\xvec=\bvec\text{,}$$ where $$\xvec=\ctwovec bm\text{.}$$
Notice that the unknown vector $$\xvec=\ctwovec bm$$ describes the line that we seek.
4. Is there a solution to this linear system? How does this question relate to your attempt to draw a line through the three points above?
5. Since this system is inconsistent, we know that $$\bvec$$ is not in the column space $$\col(A)\text{.}$$ Find an orthogonal basis for $$\col(A)$$ and use it to find the orthogonal projection $$\widehat\bvec$$ of $$\bvec$$ onto $$\col(A)\text{.}$$
6. Since $$\widehat\bvec$$ is in $$\col(A)\text{,}$$ the equation $$A\xvec = \widehat\bvec$$ is consistent. Find its solution $$\xvec = \ctwovec{b}{m}$$ and sketch the line $$y=b + mx$$ in Figure 6.5.2. We say that this is the line of best fit.
This activity illustrates the idea behind a technique known as orthogonal least squares, which we have been working toward throughout this chapter. If the data points are denoted as $$(x_i, y_i)\text{,}$$ we construct the matrix $$A$$ and vector $$\bvec$$ as
\begin{equation*} A = \begin{bmatrix} 1 \amp x_1 \\ 1 \amp x_2 \\ 1 \amp x_3 \\ \end{bmatrix},\hspace{24pt} \bvec = \threevec{y_1}{y_2}{y_3}\text{.} \end{equation*}
With the vector $$\xvec=\ctwovec bm$$ representing the line $$b+mx = y\text{,}$$ we see that the equation $$A\xvec=\bvec$$ describes a line passing through all the data points. In our activity, it is visually apparent that there is no such line, which agrees with the fact that the equation $$A\xvec=\bvec$$ is inconsistent.
Remember that $$\bhat\text{,}$$ the orthogonal projection of $$\bvec$$ onto $$\col(A)\text{,}$$ is the closest vector in $$\col(A)$$ to $$\bvec\text{.}$$ Therefore, when we solve the equation $$A\xvec=\bhat\text{,}$$ we are finding the vector $$\xvec$$ so that $$A\xvec = \threevec{b+mx_1}{b+mx_2}{b+mx_3}$$ is as close to $$\bvec=\threevec{y_1}{y_2}{y_3}$$ as possible. Let’s think about what this means within the context of this problem.
The difference $$\bvec-A\xvec = \threevec{y_1-(b+mx_1)}{y_2-(b+mx_2)}{y_3-(b+mx_3)}$$ so that the square of the distance between $$A\xvec$$ and $$\bvec$$ is
\begin{align*} \len{\bvec - A\xvec}^2 \amp =\\ \amp \left(y_1-(b+mx_1)\right)^2 + \left(y_2-(b+mx_2)\right)^2 + \left(y_3-(b+mx_3)\right)^2\text{.} \end{align*}
Our approach finds the values for $$b$$ and $$m$$ that make this sum of squares as small as possible, which is why we call this a least-squares problem.
Drawing the line defined by the vector $$\xvec=\ctwovec bm\text{,}$$ the quantity $$y_i - (b + mx_i)$$ reflects the vertical distance between the line and the data point $$(x_i, y_i)\text{,}$$ as shown in Figure 6.5.5. Seen in this way, the square of the distance $$\len{\bvec-A\xvec}^2$$ is a measure of how much the line defined by the vector $$\xvec$$ misses the data points. The solution to the least-squares problem is the line that misses the data points by the smallest amount possible.
### Subsection6.5.2Solving least-squares problems
Now that we’ve seen an example of what we’re trying to accomplish, let’s put this technique into a more general framework.
Given an inconsistent system $$A\xvec = \bvec\text{,}$$ we seek the vector $$\xvec$$ that minimizes the distance from $$A\xvec$$ to $$\bvec\text{.}$$ In other words, $$\xvec$$ satisfies $$A\xvec = \widehat\bvec\text{,}$$ where $$\bhat$$ is the orthogonal projection of $$\bvec$$ onto the column space $$\col(A)\text{.}$$ We know the equation $$A\xvec=\bhat$$ is consistent since $$\bhat$$ is in $$\col(A)\text{,}$$ and we know there is only one solution if we assume that the columns of $$A$$ are linearly independent.
We will usually denote the solution of $$A\xvec = \bhat$$ by $$\xhat$$ and call this vector the least-squares approximate solution of $$A\xvec=\bvec$$ to distinguish it from a (possibly non-existent) solution of $$A\xvec=\bvec\text{.}$$
There is an alternative method for finding $$\xhat$$ that does not involve first finding the orthogonal projection $$\bhat\text{.}$$ Remember that $$\bhat$$ is defined by the fact that $$\widehat\bvec - \bvec$$ is orthogonal to $$\col(A)\text{.}$$ In other words, $$\bhat-\bvec$$ is in the orthogonal complement $$\col(A)^\perp\text{,}$$ which Proposition 6.2.10 tells us is the same as $$\nul(A^T)\text{.}$$ Since $$\bhat-\bvec$$ is in $$\nul(A^T)\text{,}$$ it follows that
\begin{equation*} A^T(\widehat\bvec-\bvec) = \zerovec\text{.} \end{equation*}
Because the least-squares approximate solution is the vector $$\xhat$$ such that $$A\xhat = \bhat\text{,}$$ we can rearrange this equation to see that
\begin{align*} A^T(A\xhat - \bvec) \amp = \zerovec\\ A^TA\xhat - A^T\bvec \amp = \zerovec\\ A^TA\xhat \amp = A^T\bvec\text{.} \end{align*}
This equation is called the normal equation, and we have the following proposition.
#### Example6.5.7.
Consider the equation
\begin{equation*} \begin{bmatrix} 2 \amp 1 \\ 2 \amp 0 \\ -1 \amp 3 \\ \end{bmatrix} \xvec = \threevec{16}{-1}7 \end{equation*}
with matrix $$A$$ and vector $$\bvec\text{.}$$ Since this equation is inconsistent, we will find the least-squares approximate solution $$\xhat$$ by solving the normal equation $$A^TA\xhat = A^T\bvec\text{,}$$ which has the form
\begin{equation*} A^TA\xhat = \begin{bmatrix} 9 \amp -1 \\ -1 \amp 10 \\ \end{bmatrix} = \twovec{23}{37} = A^T\bvec \end{equation*}
and the solution $$\xhat=\twovec34\text{.}$$
#### Activity6.5.3.
The rate at which a cricket chirps is related to the outdoor temperature, as reflected in some experimental data that we’ll study in this activity. The chirp rate $$C$$ is expressed in chirps per second while the temperature $$T$$ is in degrees Fahrenheit. Evaluate the following cell to load the data:
sage.repl.load.load('https://raw.githubusercontent.com/davidaustinm/ula_modules/master/orthogonality.py', globals())
data = [vector(row) for row in df.values]
chirps = vector(df['Chirps'])
temps = vector(df['Temperature'])
print(df)
list_plot(data, color='blue', size=40, xmin=12, xmax=22, ymin=60, ymax=100)
Evaluating this cell also provides:
• the vectors chirps and temps formed from the columns of the dataset.
• the command onesvec(n), which creates an $$n$$-dimensional vector whose entries are all one.
• Remember that you can form a matrix whose columns are the vectors v1 and v2 with matrix([v1, v2]).T.
We would like to represent this relationship by a linear function
\begin{equation*} \beta_0 + \beta_1 C = T\text{.} \end{equation*}
1. Use the first data point $$(C_1,T_1)=(20.0,88.6)$$ to write an equation involving $$\beta_0$$ and $$\beta_1\text{.}$$
2. Suppose that we represent the unknowns using a vector $$\xvec = \twovec{\beta_0}{\beta_1}\text{.}$$ Use the 15 data points to create the matrix $$A$$ and vector $$\bvec$$ so that the linear system $$A\xvec= \bvec$$ describes the unknown vector $$\xvec\text{.}$$
3. Write the normal equations $$A^TA\xhat = A^T\bvec\text{;}$$ that is, find the matrix $$A^TA$$ and the vector $$A^T\bvec\text{.}$$
4. Solve the normal equations to find $$\xhat\text{,}$$ the least-squares approximate solution to the equation $$A\xvec=\bvec\text{.}$$ Call your solution xhat since x has another meaning in Sage.
What are the values of $$\beta_0$$ and $$\beta_1$$ that you found?
5. If the chirp rate is 22 chirps per second, what is your prediction for the temperature?
You can plot the data and your line, assuming you called the solution xhat, using the cell below.
plot_model(xhat, data, domain=(12, 22))
This example demonstrates an approach, called linear regression, in which a collection of data is modeled using a linear function found by solving a least-squares problem. Once we have the linear function that best fits the data, we can make predictions about situations that we haven’t encountered in the data.
If we’re going to use our function to make predictions, it’s natural to ask how much confidence we have in these predictions. This is a statistical question that leads to a rich and well-developed theory 1 , which we won’t explore in much detail here. However, there is one simple measure of how well our linear function fits the data that is known as the coefficient of determination and denoted by $$R^2\text{.}$$
We have seen that the square of the distance $$\len{\bvec-A\xvec}^2$$ measures the amount by which the line fails to pass through the data points. When the line is close to the data points, we expect this number to be small. However, the size of this measure depends on the scale of the data. For instance, the two lines shown in Figure 6.5.8 seem to fit the data equally well, but $$|\bvec-A\xhat|^2$$ is 100 times larger on the right.
The coefficient of determination $$R^2$$ is defined by normalizing $$|\bvec-A\xhat|^2$$ so that it is independent of the scale. Recall that we described how to demean a vector in Section 6.1: given a vector $$\vvec\text{,}$$ we obtain $$\widetilde{\vvec}$$ by subtracting the average of the components from each component.
#### Definition6.5.9.Coefficient of determination.
The coefficient of determination is
\begin{equation*} R^2 = 1 - \frac{|\bvec - A\xhat|^2} {|\widetilde{\bvec}|^2}, \end{equation*}
where $$\widetilde{\bvec}$$ is the vector obtained by demeaning $$\bvec\text{.}$$
A more complete explanation of this definition relies on the concept of variance, which we explore in Exercise 6.5.6.12 and the next chapter. For the time being, it’s enough to know that $$0\leq R^2 \leq 1$$ and that the closer $$R^2$$ is to 1, the better the line fits the data. In our original example, illustrated in Figure 6.5.8, we find that $$R^2 = 0.75\text{,}$$ and in our study of cricket chirp rates, we have $$R^2=0.69\text{.}$$ However, assessing the confidence we have in predictions made by solving a least-squares problem can require considerable thought, and it would be naive to rely only on the value of $$R^2\text{.}$$
### Subsection6.5.3Using $$QR$$ factorizations
As we’ve seen, the least-squares approximate solution $$\xhat$$ to $$A\xvec=\bvec$$ may be found by solving the normal equation $$A^TA\xhat = A^T\bvec\text{,}$$ and this can be a practical strategy for some problems. However, this approach can be problematic as small rounding errors can accumulate and lead to inaccurate final results.
As the next activity demonstrates, there is an alternate method for finding the least-squares approximate solution $$\xhat$$ using a $$QR$$ factorization of the matrix $$A\text{,}$$ and this method is preferable as it is numerically more reliable.
#### Activity6.5.4.
1. Suppose we are interested in finding the least-squares approximate solution to the equation $$A\xvec = \bvec$$ and that we have the $$QR$$ factorization $$A=QR\text{.}$$ Explain why the least-squares approximate solution is given by solving
\begin{align*} A\xhat \amp = QQ^T\bvec \\\\ QR\xhat \amp = QQ^T\bvec \\ \end{align*}
2. Multiply both sides of the second expression by $$Q^T$$ and explain why
\begin{equation*} R\xhat = Q^T\bvec. \end{equation*}
Since $$R$$ is upper triangular, this is a relatively simple equation to solve using back substitution, as we saw in Section 5.1. We will therefore write the least-squares approximate solution as
\begin{equation*} \xhat = R^{-1}Q^T\bvec, \end{equation*}
and put this to use in the following context.
3. Brozak’s formula, which is used to calculate a person’s body fat index $$BFI\text{,}$$ is
\begin{equation*} BFI = 100 \left(\frac{4.57}{\rho} - 4.142\right) \end{equation*}
where $$\rho$$ denotes a person’s body density in grams per cubic centimeter. Obtaining an accurate measure of $$\rho$$ is difficult, however, because it requires submerging the person in water and measuring the volume of water displaced. Instead, we will gather several other body measurements, which are more easily obtained, and use it to predict $$BFI\text{.}$$
For instance, suppose we take 10 patients and measure their weight $$w$$ in pounds, height $$h$$ in inches, abdomen $$a$$ in centimeters, wrist circumference $$r$$ in centimeters, neck circumference $$n$$ in centimeters, and $$BFI\text{.}$$ Evaluating the following cell loads and displays the data.
sage.repl.load.load('https://raw.githubusercontent.com/davidaustinm/ula_modules/master/orthogonality.py', globals())
weight = vector(df['Weight'])
height = vector(df['Height'])
abdomen = vector(df['Abdomen'])
wrist = vector(df['Wrist'])
neck = vector(df['Neck'])
BFI = vector(df['BFI'])
print(df)
In addition, that cell provides:
1. vectors weight, height, abdomen, wrist, neck, and BFI formed from the columns of the dataset.
2. the command onesvec(n), which returns an $$n$$-dimensional vector whose entries are all one.
3. the command QR(A) that returns the $$QR$$ factorization of $$A$$ as Q, R = QR(A).
4. the command demean(v), which returns the demeaned vector $$\widetilde{\vvec}\text{.}$$
We would like to find the linear function
\begin{equation*} \beta_0 + \beta_1w + \beta_2h + \beta_3a + \beta_4r + \beta_5n = BFI \end{equation*}
that best fits the data.
Use the first data point to write an equation for the parameters $$\beta_0,\beta_1,\ldots,\beta_5\text{.}$$
4. Describe the linear system $$A\xvec = \bvec$$ for these parameters. More specifically, describe how the matrix $$A$$ and the vector $$\bvec$$ are formed.
5. Construct the matrix $$A$$ and find its $$QR$$ factorization in the cell below.
6. Find the least-squares approximate solution $$\xhat$$ by solving the equation $$R\xhat = Q^T\bvec\text{.}$$ You may want to use N(xhat) to display a decimal approximation of the vector. What are the parameters $$\beta_0,\beta_1,\ldots,\beta_5$$ that best fit the data?
7. Find the coefficient of determination $$R^2$$ for your parameters. What does this imply about the quality of the fit?
8. Suppose a person’s measurements are: weight 190, height 70, abdomen 90, wrist 18, and neck 35. Estimate this person’s $$BFI\text{.}$$
To summarize, we have seen that
### Subsection6.5.4Polynomial Regression
In the examples we’ve seen so far, we have fit a linear function to a dataset. Sometimes, however, a polynomial, such as a quadratic function, may be more appropriate. It turns out that the techniques we’ve developed in this section are still useful as the next activity demonstrates.
#### Activity6.5.5.
1. Suppose that we have a small dataset containing the points $$(0,2)\text{,}$$ $$(1,1)\text{,}$$ $$(2,3)\text{,}$$ and $$(3,3)\text{,}$$ such as appear when the following cell is evaluated.
sage.repl.load.load('https://raw.githubusercontent.com/davidaustinm/ula_modules/master/orthogonality.py',
globals())
data = [[0, 2], [1, 1], [2, 3], [3, 3]]
list_plot(data, color='blue', size=40)
In addition to loading and plotting the data, evaluating that cell provides the following commands:
• Q, R = QR(A) returns the $$QR$$ factorization of $$A\text{.}$$
• demean(v) returns the demeaned vector $$\widetilde{\vvec}\text{.}$$
Let’s fit a quadratic function of the form
\begin{equation*} \beta_0 + \beta_1 x + \beta_2 x^2 = y \end{equation*}
to this dataset.
Write four equations, one for each data point, that describe the coefficients $$\beta_0\text{,}$$ $$\beta_1\text{,}$$ and $$\beta_2\text{.}$$
2. Express these four equations as a linear system $$A\xvec = \bvec$$ where $$\xvec = \threevec{\beta_0}{\beta_1}{\beta_2}\text{.}$$
Find the $$QR$$ factorization of $$A$$ and use it to find the least-squares approximate solution $$\xhat\text{.}$$
3. Use the parameters $$\beta_0\text{,}$$ $$\beta_1\text{,}$$ and $$\beta_2$$ that you found to write the quadratic function that fits the data. You can plot this function, along with the data, by entering your function in the place indicated below.
list_plot(data, color='blue', size=40) + plot( **your function here**,
0, 3, color='red')
4. What is your predicted $$y$$ value when $$x=1.5\text{?}$$
5. Find the coefficient of determination $$R^2$$ for the quadratic function. What does this say about the quality of the fit?
6. Now fit a cubic polynomial of the form
\begin{equation*} \beta_0 + \beta_1x + \beta_2 x^2 + \beta_3x^3 = y \end{equation*}
to this dataset.
7. Find the coefficient of determination $$R^2$$ for the cubic function. What does this say about the quality of the fit?
8. What do you notice when you plot the cubic function along with the data? How does this reflect the value of $$R^2$$ that you found?
list_plot(data, color='blue', size=40) + plot( **your function here**,
0, 3, color='red')
The matrices $$A$$ that you created in the last activity when fitting a quadratic and cubic function to a dataset have a special form. In particular, if the data points are labeled $$(x_i, y_i)$$ and we seek a degree $$k$$ polynomial, then
\begin{equation*} A = \begin{bmatrix} 1 \amp x_1 \amp x_1^2 \amp \ldots \amp x_1^k \\ 1 \amp x_2 \amp x_2^2 \amp \ldots \amp x_2^k \\ \vdots \amp \vdots \amp \vdots \amp \ddots \amp \vdots \\ 1 \amp x_m \amp x_m^2 \amp \ldots \amp x_m^k \\ \end{bmatrix}. \end{equation*}
This is called a Vandermonde matrix of degree $$k\text{.}$$
#### Activity6.5.6.
This activity explores a dataset describing Arctic sea ice and that comes from Sustainability Math. 2
Evaluating the cell below will plot the extent of Arctic sea ice, in millions of square kilometers, during the twelve months of 2012.
sage.repl.load.load('https://raw.githubusercontent.com/davidaustinm/ula_modules/master/orthogonality.py', globals())
data = [vector([row[0], row[2]]) for row in df.values]
month = vector(df['Month'])
ice = vector(df['2012'])
print(df[['Month', '2012']])
list_plot(data, color='blue', size=40)
In addition, you have access to a few special variables and commands:
• month is the vector of month values and ice is the vector of sea ice values from the table above.
• vandermonde(x, k) constructs the Vandermonde matrix of degree $$k$$ using the points in the vector x.
• Q, R = QR(A) provides the $$QR$$ factorization of $$A\text{.}$$
• demean(v) returns the demeaned vector $$\widetilde{\vvec}\text{.}$$
1. Find the vector $$\xhat\text{,}$$ the least-squares approximate solution to the linear system that results from fitting a degree 5 polynomial to the data.
2. If your result is stored in the variable xhat, you may plot the polynomial and the data together using the following cell.
plot_model(xhat, data)
3. Find the coefficient of determination $$R^2$$ for this polynomial fit.
4. Repeat these steps to fit a degree 8 polynomial to the data, plot the polynomial with the data, and find $$R^2\text{.}$$
5. Repeat one more time by fitting a degree 11 polynomial to the data, creating a plot, and finding $$R^2\text{.}$$
It’s certainly true that higher degree polynomials fit the data better, as seen by the increasing values of $$R^2\text{,}$$ but that’s not always a good thing. For instance, when $$k=11\text{,}$$ you may notice that the graph of the polynomial wiggles a little more than we would expect. In this case, the polynomial is trying too hard to fit the data, which usually contains some uncertainty, especially if it’s obtained from measurements. The error built in to the data is called noise, and its presence means that we shouldn’t expect our polynomial to fit the data perfectly. When we choose a polynomial whose degree is too high, we give the noise too much weight in the model, which leads to some undesirable behavior, like the wiggles in the graph.
Fitting the data with a polynomial whose degree is too high is called overfitting, a phenomenon that can appear in many machine learning applications. Generally speaking, we would like to choose $$k$$ large enough to capture the essential features of the data but not so large that we overfit and build the noise into the model. There are ways to determine the optimal value of $$k\text{,}$$ but we won’t pursue that here.
6. Choosing a reasonable value of $$k\text{,}$$ estimate the extent of Arctic sea ice at month 6.5, roughly at the Summer Solstice.
### Subsection6.5.5Summary
This section introduced some types of least-squares problems and a framework for working with them.
• Given an inconsistent system $$A\xvec=\bvec\text{,}$$ we find $$\xhat\text{,}$$ the least-squares approximate solution, by requiring that $$A\xhat$$ be as possible to $$\bvec$$ as possible. In other words, $$A\xhat = \bhat$$ where $$\bhat$$ is the orthogonal projection of $$\bvec$$ onto $$\col(A)\text{.}$$
• One way to find $$\xhat$$ is by solving the normal equations $$A^TA\xhat = A^T\bvec.$$ This is not our preferred method since numerical problems can arise.
• A second way to find $$\xhat$$ uses a $$QR$$ factorization of $$A\text{.}$$ If $$A=QR\text{,}$$ then $$\xhat = R^{-1}Q^T\bvec$$ and finding $$R^{-1}$$ is computationally feasible since $$R$$ is upper triangular.
• This technique may be applied widely and is useful for modeling data. We saw examples in this section where linear functions of several input variables and polynomials provided effective models for different datasets.
• A simple measure of the quality of the fit is the coefficient of determination $$R^2$$ though some additional thought should be given in real applications.
### Exercises6.5.6Exercises
Evaluating the following cell loads in some commands that will be helpful in the following exercises. In particular, there are commands:
• QR(A) that returns the $$QR$$ factorization of A as Q, R = QR(A),
• onesvec(n) that returns the $$n$$-dimensional vector whose entries are all 1,
• demean(v) that demeans the vector v,
• vandermonde(x, k) that returns the Vandermonde matrix of degree $$k$$ formed from the components of the vector x, and
• plot_model(xhat, data) that plots the data and the model xhat.
sage.repl.load.load('https://raw.githubusercontent.com/davidaustinm/ula_modules/master/orthogonality.py', globals())
#### 1.
Suppose we write the linear system
\begin{equation*} \begin{bmatrix} 1 \amp -1 \\ 2 \amp -1 \\ -1 \amp 3 \end{bmatrix} \xvec = \threevec{-8}5{-10} \end{equation*}
as $$A\xvec=\bvec\text{.}$$
1. Find an orthogonal basis for $$\col(A)\text{.}$$
2. Find $$\bhat\text{,}$$ the orthogonal projection of $$\bvec$$ onto $$\col(A)\text{.}$$
3. Find a solution to the linear system $$A\xvec = \bhat\text{.}$$
#### 2.
Consider the data in Table 6.5.11.
1. Set up the linear system $$A\xvec=\bvec$$ that describes the line $$b + mx = y$$ passing through these points.
2. Write the normal equations that describe the least-squares approximate solution to $$A\xvec=\bvec\text{.}$$
3. Find the least-squares approximate solution $$\xhat$$ and plot the data and the resulting line.
4. What is your predicted $$y$$-value when $$x=3.5\text{?}$$
5. Find the coefficient of determination $$R^2\text{.}$$
#### 3.
Consider the four points in Table 6.5.11.
1. Set up a linear system $$A\xvec = \bvec$$ that describes a quadratic function
\begin{equation*} \beta_0+\beta_1x+\beta_2x^2 = y \end{equation*}
passing through the points.
2. Use a $$QR$$ factorization to find the least-squares approximate solution $$\xhat$$ and plot the data and the graph of the resulting quadratic function.
3. What is your predicted $$y$$-value when $$x=3.5\text{?}$$
4. Find the coefficient of determination $$R^2\text{.}$$
#### 4.
Consider the data in Table 6.5.12.
1. Set up a linear system $$A\xvec = \bvec$$ that describes the relationship
\begin{equation*} \beta_0 + \beta_1 x_1 + \beta_2 x_2 = y. \end{equation*}
2. Find the least-squares approximate solution $$\xhat\text{.}$$
3. What is your predicted $$y$$-value when $$x_1 = 2.4$$ and $$x_2=2.9\text{?}$$
4. Find the coefficient of determination $$R^2\text{.}$$
#### 5.
Determine whether the following statements are true or false and explain your thinking.
1. If $$A\xvec=\bvec$$ is consistent, then $$\xhat$$ is a solution to $$A\xvec=\bvec\text{.}$$
2. If $$R^2=1\text{,}$$ then the least-squares approximate solution $$\xhat$$ is also a solution to the original equation $$A\xvec=\bvec\text{.}$$
3. Given the $$QR$$ factorization $$A=QR\text{,}$$ we have $$A\xhat=Q^TQ\bvec\text{.}$$
4. A $$QR$$ factorization provides a method for finding the least-squares approximate solution to $$A\xvec=\bvec$$ that is more reliable than solving the normal equations.
5. A solution to $$AA^T\xvec = A\bvec$$ is the least-squares approximate solution to $$A\xvec = \bvec\text{.}$$
#### 6.
Explain your response to the following questions.
1. If $$\xhat=\zerovec\text{,}$$ what does this say about the vector $$\bvec\text{?}$$
2. If the columns of $$A$$ are orthonormal, how can you easily find the least-squares approximate solution to $$A\xvec=\bvec\text{?}$$
#### 7.
The following cell loads in some data showing the number of people in Bangladesh living without electricity over 27 years. It also defines vectors year, which records the years in the dataset, and people, which records the number of people.
sage.repl.load.load('https://raw.githubusercontent.com/davidaustinm/ula_modules/master/orthogonality.py', globals())
data = [vector(row) for row in df.values]
year = vector(df['Year'])
people = vector(df['People'])
print(df)
list_plot(data, size=40, color='blue')
1. Suppose we want to write
\begin{equation*} N = \beta_0 + \beta_1 t \end{equation*}
where $$t$$ is the year and $$N$$ is the number of people. Construct the matrix $$A$$ and vector $$\bvec$$ so that the linear system $$A\xvec=\bvec$$ describes the vector $$\xvec=\twovec{\beta_0}{\beta_1}\text{.}$$
2. Using a $$QR$$ factorization of $$A\text{,}$$ find the values of $$\beta_0$$ and $$\beta_1$$ in the least-squares approximate solution $$\xhat\text{.}$$
3. What is the coefficient of determination $$R^2$$ and what does this tell us about the quality of the approximation?
4. What is your prediction for the number of people living without electricity in 1985?
5. Estimate the year in which there will be no people living without electricity.
#### 8.
This problem concerns a dataset describing planets in our Solar system. For each planet, we have the length $$L$$ of the semi-major axis, essentially the distance from the planet to the Sun in AU (astronomical units), and the period $$P\text{,}$$ the length of time in years required to complete one orbit around the Sun.
We would like to model this data using the function $$P = CL^r$$ where $$C$$ and $$r$$ are parameters we need to determine. Since this isn’t a linear function, we will transform this relationship by taking the natural logarithm of both sides to obtain
\begin{equation*} \ln(P) = \ln(C) + r\ln(L). \end{equation*}
Evaluating the following cell loads the dataset and defines two vectors logaxis, whose components are $$\ln(L)\text{,}$$ and logperiod, whose components are $$\ln(P)\text{.}$$
import numpy as np
logaxis = vector(np.log(df['Semi-major axis']))
logperiod = vector(np.log(df['Period']))
print(df)
1. Construct the matrix $$A$$ and vector $$\bvec$$ so that the solution to $$A\xvec=\bvec$$ is the vector $$\xvec=\ctwovec{\ln(C)}r\text{.}$$
2. Find the least-squares approximate solution $$\xhat\text{.}$$ What does this give for the values of $$C$$ and $$r\text{?}$$
3. Find the coefficient of determination $$R^2\text{.}$$ What does this tell us about the quality of the approximation?
4. Suppose that the orbit of an asteroid has a semi-major axis whose length is $$L=4.0$$ AU. Estimate the period $$P$$ of the asteroid’s orbit.
5. Halley’s Comet has a period of $$P=75$$ years. Estimate the length of its semi-major axis.
#### 9.
Evaluating the following cell loads a dataset describing the temperature in the Earth’s atmosphere at various altitudes. There are also two vectors altitude, expressed in kilometers, and temperature, in degrees Celsius.
sage.repl.load.load('https://raw.githubusercontent.com/davidaustinm/ula_modules/master/orthogonality.py', globals())
data = [vector(row) for row in df.values]
altitude = vector(df['Altitude'])
temperature = vector(df['Temperature'])
print(df)
list_plot(data, size=40, color='blue')
1. Describe how to form the matrix $$A$$ and vector $$\bvec$$ so that the linear system $$A\xvec=\bvec$$ describes a degree $$k$$ polynomial fitting the data.
2. After choosing a value of $$k\text{,}$$ construct the matrix $$A$$ and vector $$\bvec\text{,}$$ and find the least-squares approximate solution $$\xhat\text{.}$$
3. Plot the polynomial and data using plot_model(xhat, data).
4. Now examine what happens as you vary the degree of the polynomial $$k\text{.}$$ Choose an appropriate value of $$k$$ that seems to capture the most important features of the data while avoiding overfitting, and explain your choice.
5. Use your value of $$k$$ to estimate the temperature at an altitude of 55 kilometers.
#### 10.
The following cell loads some data describing 1057 houses in a particular real estate market. For each house, we record the living area in square feet, the lot size in acres, the age in years, and the price in dollars. The cell also defines variables area, size, age, and price.
sage.repl.load.load('https://raw.githubusercontent.com/davidaustinm/ula_modules/master/orthogonality.py', globals())
df = df.fillna(df.mean())
area = vector(df['Living.Area'])
size = vector(df['Lot.Size'])
age = vector(df['Age'])
price = vector(df['Price'])
df
We will use linear regression to predict the price of a house given its living area, lot size, and age:
\begin{equation*} \beta_0 + \beta_1~\text{Living Area} + \beta_2~\text{Lot Size} + \beta_3~\text{Age} = \text{Price}. \end{equation*}
1. Use a $$QR$$ factorization to find the least-squares approximate solution $$\xhat\text{.}$$
2. Discuss the significance of the signs of $$\beta_1\text{,}$$ $$\beta_2\text{,}$$ and $$\beta_3\text{.}$$
3. If two houses are identical except for differing in age by one year, how would you predict that their prices compare to each another?
4. Find the coefficient of determination $$R^2\text{.}$$ What does this say about the quality of the fit?
5. Predict the price of a house whose living area is 2000 square feet, lot size is 1.5 acres, and age is 50 years.
#### 11.
We observed that if the columns of $$A$$ are linearly independent, then there is a unique least-squares approximate solution to the equation $$A\xvec=\bvec$$ because the equation $$A\xhat=\bhat$$ has a unique solution. We also said that $$\xhat$$ is the unique solution to the normal equation $$A^TA\xhat = A^T\bvec$$ without explaining why this equation has a unique solution. This exercise offers an explanation.
Assuming that the columns of $$A$$ are linearly independent, we would like to conclude that the equation $$A^TA\xhat=A^T\bvec$$ has a unique solution.
1. Suppose that $$\xvec$$ is a vector for which $$A^TA\xvec = \zerovec\text{.}$$ Explain why the following argument is valid and allows us to conclude that $$A\xvec = \zerovec\text{.}$$
\begin{equation*} \begin{aligned} A^TA\xvec \amp = \zerovec \\ \xvec\cdot A^TA\xvec \amp = \xvec\cdot\zerovec = 0 \\ (A\xvec)\cdot(A\xvec) \amp = 0 \\ \len{A\xvec}^2 \amp = 0. \\ \end{aligned} \end{equation*}
In other words, if $$A^TA\xvec = \zerovec\text{,}$$ we know that $$A\xvec = \zerovec\text{.}$$
2. If the columns of $$A$$ are linearly independent and $$A\xvec = \zerovec\text{,}$$ what do we know about the vector $$\xvec\text{?}$$
3. Explain why $$A^TA\xvec = \zerovec$$ can only happen when $$\xvec = \zerovec\text{.}$$
4. Assuming that the columns of $$A$$ are linearly independent, explain why $$A^TA\xhat=A^T\bvec$$ has a unique solution.
#### 12.
This problem is about the meaning of the coefficient of determination $$R^2$$ and its connection to variance, a topic that appears in the next section. Throughout this problem, we consider the linear system $$A\xvec=\bvec$$ and the approximate least-squares solution $$\xhat\text{,}$$ where $$A\xhat=\bhat\text{.}$$ We suppose that $$A$$ is an $$m\times n$$ matrix, and we will denote the $$m$$-dimensional vector $$\onevec = \fourvec11{\vdots}1\text{.}$$
1. Explain why $$\bbar\text{,}$$ the mean of the components of $$\bvec\text{,}$$ can be found as the dot product
\begin{equation*} \bbar = \frac 1m \bvec\cdot\onevec. \end{equation*}
2. In the examples we have seen in this section, explain why $$\onevec$$ is in $$\col(A)\text{.}$$
3. If we write $$\bvec = \bhat + \bvec^\perp\text{,}$$ explain why
\begin{equation*} \bvec^\perp\cdot\onevec = 0 \end{equation*}
and hence why the mean of the components of $$\bvec^\perp$$ is zero.
4. The variance of an $$m$$-dimensional vector $$\vvec$$ is $$\var(\vvec) = \frac1m \len{\widetilde{\vvec}}^2\text{,}$$ where $$\widetilde{\vvec}$$ is the vector obtained by demeaning $$\vvec\text{.}$$
Explain why
\begin{equation*} \var(\bvec) = \var(\bhat) + \var(\bvec^\perp). \end{equation*}
5. Explain why
\begin{equation*} \frac{\len{\bvec - A\xhat}^2}{\len{\widetilde{\bvec}}^2} = \frac{\var(\bvec^\perp)}{\var(\bvec)} \end{equation*}
and hence
\begin{equation*} R^2 = \frac{\var(\bhat)}{\var(\bvec)} = \frac{\var(A\xhat)}{\var(\bvec)}. \end{equation*}
These expressions indicate why it is sometimes said that $$R^2$$ measures the “fraction of variance explained” by the function we are using to fit the data. As seen in the previous exercise, there may be other features that are not recorded in the dataset that influence the quantity we wish to predict.
6. Explain why $$0\leq R^2 \leq 1\text{.}$$
For example, see Gareth James, Daniela Witten, Trevor Hastie, Robert Tibshirani. An Introduction to Statistical Learning: with Applications in R. Springer, 2013.
sustainabilitymath.org |
# What is 3/334 Simplified?
Are you looking to calculate how to simplify the fraction 3/334? In this really simple guide, we'll teach you exactly how to simplify 3/334 and convert it to the lowest form (this is sometimes calling reducing a fraction to the lowest terms).
To start with, the number above the line (3) in a fraction is called a numerator and the number below the line (334) is called the denominator.
So what we want to do here is to simplify the numerator and denominator in 3/334 to their lowest possible values, while keeping the actual fraction the same.
To do this, we use something called the greatest common factor. It's also known as the greatest common divisor and put simply, it's the highest number that divides exactly into two or more numbers.
In our case with 3/334, the greatest common factor is 1. Once we have this, we can divide both the numerator and the denominator by it, and voila, the fraction is simplified:
3/1 = 3
334/1 = 334
3 / 334
As you can see, 3/334 cannot be simplified any further, so the result is the same as we started with. Not very exciting, I know, but hopefully you have at least learned why it cannot be simplified any further!
So there you have it! You now know exactly how to simplify 3/334 to its lowest terms. Hopefully you understood the process and can use the same techniques to simplify other fractions on your own. The complete answer is below:
3/334
## Convert 3/334 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
3 / 334 = 0.009
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# POOLS, BARBECUES AND GARDENING
Date of the Problem
May 31, 2021
Tate’s neighborhood swimming pool opened this past weekend for the summer. At Tate’s neighborhood pool, it is estimated that the number of people at the pool on opening day is 253% of the daily average number of people who go to the pool throughout the year. If 432 people came to the pool on opening day, what is the average number of people per day throughout the summer? Express your answer to the nearest whole number.
Let x be the average number of people per day at the pool. We can set up the equation 2.53x = 432 to demonstrate the situation on opening day. Dividing both sides by 2.53 gives us that x = 171 people, to the nearest whole number.
Tate’s dad decided this was the perfect weekend to get out the grill for the season. He thought he’d used the current propane tank of gas for approximately 20 hours last summer and thinks each tank lasts approximately 50 hours. If each meal uses an average of 40 minutes of propane, for how many meals will the current tank last?
According to Tate's dad's memory, there is enough propane for 50 – 20 = 30 hours of grilling. This is equivalent to 30(60) = 1800 minutes of grilling, which is enough for 1800/40 = 45 meals.
Tate’s mom figured it was the perfect weekend to do some gardening. She found some herbs that need a four-inch space all around the plant. The plant takes up a circle of radius four inches; the sides of the pot and every other plant in the pot must be at least four inches away from each plant. Her pot has a diameter of 14 inches. What is the greatest number of plants that can be planted in the pot?
If we place our first plant four inches from the side of the pot, no other plants can be planted within four inches of that plant. Notice that the plant is four inches from the side of the pot, and if we were to continue that segment out four inches in the opposite direction, it would create a diameter of 8 inches around that plant and go past the middle of the pot since the pot's radius is only seven inches. Therefore, only 1 plant can be planted in this pot according to the space requirements for the plant.
Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
CCSS (Common Core State Standard)
Difficulty |
# Statement-1 :Variance of first n natural number is n2−112. Statement-2 : S.D. of first n natural number is √n2−112
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## To solve the problem, we need to verify the two statements regarding the variance and standard deviation of the first n natural numbers.Step 1: Understanding VarianceThe variance σ2 of a set of numbers is calculated using the formula:σ2=∑ni=1x2in−(∑ni=1xin)2where xi are the numbers in the set.Step 2: Calculate the Sum of the First n Natural NumbersThe first n natural numbers are 1,2,3,…,n.The sum of the first n natural numbers is given by:n∑i=1xi=n(n+1)2Step 3: Calculate the Sum of the Squares of the First n Natural NumbersThe sum of the squares of the first n natural numbers is given by:n∑i=1x2i=n(n+1)(2n+1)6Step 4: Substitute into the Variance FormulaNow, substituting these sums into the variance formula:σ2=n(n+1)(2n+1)6n−⎛⎜⎝n(n+1)2n⎞⎟⎠2This simplifies to:σ2=(n+1)(2n+1)6−((n+1)2)2Step 5: Simplifying the Variance ExpressionNow we simplify:1. The first term is (n+1)(2n+1)6.2. The second term is (n+1)24.To combine these, we need a common denominator:σ2=(n+1)(2n+1)6−3(n+1)212=2(n+1)(2n+1)−3(n+1)212Factoring out (n+1):=(n+1)(2(2n+1)−3(n+1))12Expanding the terms:=(n+1)(4n+2−3n−3)12=(n+1)(n−1)12Thus, we find:σ2=n2−112Step 6: Conclusion for Statement 1The first statement is true: the variance of the first n natural numbers is indeed n2−112.Step 7: Calculate Standard DeviationThe standard deviation σ is the square root of the variance:σ=√σ2=√n2−112Step 8: Conclusion for Statement 2The second statement is also true: the standard deviation of the first n natural numbers is √n2−112.Final AnswerBoth statements are true.---
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# Factors and Monomials Lessons 4.1, SOL 7.5 We will learn to identify factors of numbers, and decide if an expression is a monomial.
## Presentation on theme: "Factors and Monomials Lessons 4.1, SOL 7.5 We will learn to identify factors of numbers, and decide if an expression is a monomial."— Presentation transcript:
Factors and Monomials Lessons 4.1, SOL 7.5 We will learn to identify factors of numbers, and decide if an expression is a monomial.
REMEMBER: What is a factor? A factor is a number that is multiplied to get a product. Make a factor tree 48 6868 23 2 4 2 2
Why should we know this? There are 72 people coming to your wedding. The small tables could seat 6; while the large tables seat 10 people. Which size table should you use so all the tables are filled evenly? 72 ÷ 6 = 12 72 ÷ 10 = 7.2 Use the small tables!
Why should we know this? There are 48 students in a marching band. For the halftime show we want to put them in equal rows. How many students should be in each row? 48 ÷ 2 = 24 48 ÷ 3 = 16 48 ÷ 4 = 12 48 ÷ 5 = 9.8 48 ÷ 6 = 8
Rules of Divisibility… How do you know if a number is divisible by 5 ? 65? 132? 1,890? A number is divisible by 5 if the ones digit is zero or five
Rules of Divisibility… How do you know if a number is divisible by 10 ? 120? 940? 1,241? A number is divisible by 10 if the ones digit is a zero
Rules of Divisibility… How do you know if a number is divisible by 2 ? 54? 125? 1,238? A number is divisible by 2 if the ones digit is divisible by two
Rules of Divisibility… How do you know if a number is divisible by 3 ? 72? 169? 1,453? A number is divisible by 3 if the sum of its digits is divisible by three 7+2=9 9÷3=3 1+6+9=16 16÷3=5.3333 1+4+5+3=12 12÷3=4
Rules of Divisibility… How do you know if a number is divisible by 6 ? 48? 351? 1,920? A number is divisible by 6 if the number is divisible by two and three 8÷2=4 YES! 4+8=12 12÷3=4 YES! 1÷2=0.5 NO! 3+5+1=9 9÷3=3 YES! 0÷2=0 YES! 1+9+2+0=12 12÷3=4 YES!
What is a monomial? 8 x 8x A monomial is a number, a variable, or a product of a number and/or a variable. 18 18-x 18y 18+y
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Functions and algebra: Work with algebraic expressions using the remainder and the factor theorems
Unit 1: Use the remainder and factor theorem to factorise third degree polynomials
Natashia Bearam-Edmunds
Unit 1 outcomes
By the end of this unit you will be able to:
• Find the factors of a cubic polynomial.
• Find the remainder of cubic polynomial.
• Use division by inspection, synthetic division or long division to factorise and solve cubic polynomials.
What you should know
Before you start this unit, make sure you can:
Introduction
You have already seen examples of polynomial expressions in level 2 subject outcome 2.1, unit 1. Remember that a polynomial is an expression with one or more variables with different coefficients and non-negative integer powers.
We define a polynomial as: $\scriptsize {{a}_{n}}{{x}^{n}}+{{a}_{{n-1}}}{{x}^{{n-1}}}+...+{{a}_{2}}{{x}^{2}}+{{a}_{1}}x+{{a}_{0}}{{x}^{0}}$ where $\scriptsize n\in {{N}_{0}}$. $\scriptsize {{a}_{n}},{{a}_{{n-1}}},{{a}_{2}},{{a}_{1}}$ and $\scriptsize {{a}_{0}}$ are the coefficients of each term and are usually constants.
There is an unlimited variety in the number of terms and the powers of the variable. While the order of the terms in a polynomial is not important for performing operations, we generally arrange the terms in decreasing powers of the variable. This is called the general form.
Notice that the definition of a polynomial states that all exponents of the variables must be elements of the set of whole numbers. If an expression contains terms with exponents that are not whole numbers, then it is not a polynomial.
For example, $\scriptsize \displaystyle \frac{2}{x}+2{{x}^{2}}+1,\text{ }3\sqrt{y}+y$ and $\scriptsize {{k}^{2}}-6k+3{{k}^{{-\tfrac{1}{2}}}}$ are not polynomials. Look at each expression and make sure you understand why these are not polynomials. Can you see that in each case the exponents are not whole numbers?
The degree of a polynomial is the highest power of the variable. If the expression has been written in general form then it is the power of the first variable. The leading term is the term with the highest power of the variable, or the term with the highest degree. The leading coefficient is the coefficient of the leading term.
The polynomial $\scriptsize 2{{x}^{3}}-3{{x}^{2}}+x-1$ has a degree of $\scriptsize 3$, the leading term is $\scriptsize 2{{x}^{3}}$ and the leading coefficient is $\scriptsize 2$.
Exercise 1.1
State whether the following statements are true or false:
1. $\scriptsize 2{{x}^{{-1}}}$ is a monomial because it only has one term.
2. $\scriptsize 45$ is a constant polynomial of degree $\scriptsize 0$.
3. $\scriptsize 2{{x}^{2}}-3x+1$ is a quadratic polynomial of degree $\scriptsize 2$ with a leading coefficient of $\scriptsize 2$.
4. A cubic polynomial always has three terms and all the exponents are natural numbers.
The full solutions are at the end of the unit.
Dividing cubic polynomials
In level 3 subject outcome 2.3 we looked at various methods to factorise and solve quadratic equations. In this unit we will factorise cubic polynomials with one variable. Remember that a cubic polynomial is an algebraic expression with a highest power of $\scriptsize 3$. The standard form of a cubic polynomial is $\scriptsize a{{x}^{3}}+b{{x}^{2}}+cx+d$.
The following activity is useful to remind ourselves of the basics of long division, which can be applied to polynomials too and is needed to factorise cubic polynomials.
Activity 1.1: Investigate simple division
Time required: 10 minutes
What you need:
• a pen and paper
What to do
Consider the following situation and answer the questions that follow.
Six learners are at a product promotion and there are $\scriptsize 15$ free gifts to be given away. Each learner must receive the same number of gifts.
1. How many gifts does each learner get?
2. How many gifts will be left over?
3. Use the following variables to express the above situation as a mathematical equation:
$\scriptsize a=$ total number of gifts
$\scriptsize b=$ total number of learners
$\scriptsize \displaystyle q=$ number of gifts for each learner
$\scriptsize r=$ number of gifts remaining
4. What is each part of the division expression called?
5. What does your equation say in words?
What did you find?
1. Since each learner has to a get the same whole number of gifts, each learner will get $\scriptsize 2$ gifts.
2. In total, the learners will get $\scriptsize 12$ gifts. Therefore, there will be $\scriptsize 3$ left over.
3. Using numerical values we can write this as $\scriptsize \displaystyle \frac{{15}}{6}=2$ remainder $\scriptsize 3$. This is the same as$\scriptsize 15=6\times 2+3$. Using the variables this becomes $\scriptsize a=b\times q+r,\text{ }b\ne 0$.
4. $\scriptsize 15$ is the dividend
$\scriptsize 6$ is the divisor
$\scriptsize 2$ is the quotient
$\scriptsize 3$ is the remainder
5. The dividend is equal to the divisor multiplied by the quotient, plus the remainder.
The activity reminded us that if an integer $\scriptsize a$ is divided by an integer $\scriptsize b$, then the answer is $\scriptsize q$ with a remainder of $\scriptsize r$. Sometimes $\scriptsize r=0$.
For example, $\scriptsize 8$ divided by $\scriptsize 3$ gives a whole number answer of $\scriptsize 2$ with a remainder of $\scriptsize 2$.
We can write this as $\scriptsize \displaystyle \frac{8}{3}=2+\displaystyle \frac{2}{3}$ which is the same as saying that $\scriptsize 8=3\times 2+2$.
Take note!
This rule can be extended to include the division of polynomials; if a polynomial $\scriptsize f(x)$ is divided by a polynomial $\scriptsize g(x)$, then the answer is $\scriptsize Q(x)$ with a remainder $\scriptsize R(x)$.
$\scriptsize f(x)=g(x)\cdot Q(x)+R(x)$ where $\scriptsize g(x)\ne 0$.
We are familiar with the process of long division in ordinary arithmetic. Let’s have a look at the process again to remind us how this is done.
\scriptsize \displaystyle \begin{align*} &3\overset{{59}}{\overline{\left){{178}}\right.}}\\ &\;\underline{{-15}}\downarrow && \text{ Step 1: }5\times 3=15\text{ and }17-15=2\\ &\;\;\;\;\;28 && \text{ Step }2\text{: Bring down the }8\\ &\;\;\underline{{-27}} &&\text{ Step 3: }9\times 3=27 \text{ and }28-27=1\\ &\;\;\;\;\;\;1\text{ }\end{align*}
$\scriptsize \displaystyle \frac{{178}}{3}=59$ remainder $\scriptsize 1$ or $\scriptsize 59\displaystyle \frac{1}{3}\text{ }$.
Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic.
\scriptsize \begin{align*}\text{Dividend } &= \text{ divisor}\cdot \text{quotient + remainder}\\178&=3\cdot 59+1\\&=177+1\\&=178\end{align*}
You can use long division and synthetic division, which is explained in detail shortly, to find the quotient and remainder when a cubic polynomial is divided by another polynomial.
Example 1.1
Use the method of long division to find the quotient and remainder when $\scriptsize f(x)=2{{x}^{3}}+3{{x}^{2}}+x-5$ is divided by $\scriptsize x-1$.
Solution
Write down the known and unknown expressions.
$\scriptsize f(x)=g(x)\cdot Q(x)+R(x)$
$\scriptsize 2{{x}^{3}}+3{{x}^{2}}+x-5=(x-1)\cdot Q(x)+R(x)$
Use long division to find $\scriptsize Q(x)$ and$\scriptsize R(x)$.
Make sure that $\scriptsize f(x)$ and $\scriptsize g(x)$ are written in descending order of the exponents. If a term of a certain degree is missing from$\scriptsize f(x)$, then write the term with a coefficient of $\scriptsize 0$.
\scriptsize \displaystyle \begin{align*}&x-1\overset{{2{{x}^{2}}+5x+4}}{\overline{\left){{2{{x}^{3}}+3{{x}^{2}}+x-5\text{ }}}\right.}} && \text{ Divide }2{{x}^{3}}\text{ by }x\text{ and put the answer in the quotient}\\&\text{ }-\underline{{(2{{x}^{3}}-2{{x}^{2}})}} && \text{ Multiply the quotient by the divisor and subtract from }f(x)\\ & \text{ }5{{x}^{2}}+x && \text{ Bring down the next term from }f(x)\text{ and repeat the process }\\ &\text{ }-\underline{{(5{{x}^{2}}-5x)}}\\ & \text{ 6}x-5\\ &\text{ }-(\underline{{6x-6}})\\ & \text{ }1\\&\text{ }\end{align*}
\scriptsize \begin{align*}Q(x)&=2{{x}^{2}}+5x+6\\R(x)&=1\\ &\therefore 2{{x}^{3}}+3{{x}^{2}}+x-5=(x-1)\cdot (2{{x}^{2}}+5x+6)+1\end{align*}
You can multiply the brackets and simplify to check that the answer is correct.
Exercise 1.2
1. Find the quotient and remainder of the following cubic polynomials by using long division:
1. $\scriptsize 6{{x}^{3}}+{{x}^{2}}-4x+5$ divided by $\scriptsize 2x-1$
2. $\scriptsize {{x}^{3}}+3{{x}^{2}}-x-3$ divided by $\scriptsize x-1$
2. Hence or otherwise, factorise 1b) completely.
The full solutions are at the end of the unit.
As you have seen, long division of polynomials involves many steps and can be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is $\scriptsize 1$.
Example 1.2
Use the method of synthetic division to find the quotient and remainder when:
1. $\scriptsize f(x)=2{{x}^{3}}+3{{x}^{2}}+x-5$ is divided by $\scriptsize x-1$.
2. $\scriptsize g(x)=6{{x}^{3}}+{{x}^{2}}-4x+5$ is divided by $\scriptsize 2x-1$.
Solutions
1. Recall the long division solution from Example 1.1 looked this this:
\scriptsize \displaystyle \begin{align*}x-1\overset{{2{{x}^{2}}+5x+6}}{\overline{\left){{2{{x}^{3}}+3{{x}^{2}}+x-5\text{ }}}\right.}}\text{ }\\\text{ }-\underline{{(2{{x}^{3}}-2{{x}^{2}})}}\text{ }\\\text{ }5{{x}^{2}}+x\text{ }\\\text{ }-\underline{{(5{{x}^{2}}-5x)}}\\\text{ 6}x-5\\\text{ }-(6x-6)\\\text{ }1\\\text{ }\end{align*}
.
There is a lot of repetition in the long division. Let’s see how we can simplify the process.
.
Synthetic division allows us to collapse the ‘table’ of long division by moving each of the rows up to fill any vacant spots. Also, instead of dividing by $\scriptsize -1$, as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the ‘divisor’ to $\scriptsize +1$, multiply and add.
.
This will become clearer by completing the example $\scriptsize f(x)=2{{x}^{3}}+3{{x}^{2}}+x-5$ divided by $\scriptsize x-1$ using synthetic division.
.
The process starts by writing the constant term of the divisor with the opposite sign and the coefficients of the polynomial. So the $\scriptsize -1$ constant term in the divisor becomes $\scriptsize +1$.
.
$\scriptsize 1\left| \!{\underline {\, {2\text{ }3\text{ }1\text{ }-5} \,}} \right.$
.
Bring down the leading coefficient then multiply this leading coefficient by the divisor and write that answer in column $\scriptsize 2$.
.
\scriptsize \begin{align*}1\left| \!{\underline {\, \begin{align*}2\text{ }3\text{ }1\text{ }-5\\\text{ }2\end{align*} \,}} \right. \\\text{ }2\end{align*}
.
Next, add the numbers in the second column and write this answer down. Next, multiply this answer by the divisor and write the result in the third column and add the numbers in the third column.
\scriptsize \begin{align*}1\left| \!{\underline {\, \begin{align*}2\text{ }3\text{ }1\text{ }-5\\\text{ }2\text{ }5\text{ }6\end{align*} \,}} \right. \\\text{ }2\text{ }5\text{ }6\text{ 1}\end{align*}
.
Once again, multiply this result of adding the numbers in the third column by the divisor and write this number in the fourth column. Add the numbers in the fourth column. This final entry in the last column is the remainder.
.
Using synthetic division the quotient is $\scriptsize 2{{x}^{2}}+5x+6$ and the remainder is $\scriptsize 1$ just as it was in Example 1.1.
2. In this case we see that the coefficient of the divisor is not $\scriptsize 1$. To use synthetic division, the coefficient of the divisor must be $\scriptsize 1$. So, we make the leading coefficient of the divisor equal to $\scriptsize 1$ by dividing by a factor of $\scriptsize 2$ to get $\scriptsize 2(x-\displaystyle \frac{1}{2})$.
Now the $\scriptsize -\displaystyle \frac{1}{2}$ constant term in the divisor becomes $\scriptsize +\displaystyle \frac{1}{2}$. The process for synthetic division will follow the same steps as before.
.
$\scriptsize \tfrac{1}{2}\left| \!{\underline {\, {\text{6 1 }-\text{4 }5} \,}} \right.$
.
\scriptsize \displaystyle \begin{align*}\displaystyle \frac{1}{2}\left| \!{\underline {\, \begin{align*}\text{6 1 }-4\text{ }5\\\text{ 3 2 }-1\end{align*} \,}} \right. \\\text{ }6\text{ }4\text{ }-2\text{ }4\end{align*}
.
Using synthetic division the quotient is $\scriptsize \displaystyle 6{{x}^{2}}+4x-2=2(3{{x}^{2}}+2x-1)$ and the remainder is $\scriptsize 4$. But, we are not done yet. Remember, we wrote the divisor as $\scriptsize 2(x-\displaystyle \frac{1}{2})$. Now we need to rewrite the expression to take it back to its original form.
.
\scriptsize \begin{align*}6{{x}^{3}}+{{x}^{2}}-4x+5&=\displaystyle \frac{1}{2}\cdot 2(x-\displaystyle \frac{1}{2})\cdot 2(3{{x}^{2}}+2x-1)+4\\&=2(x-\displaystyle \frac{1}{2})\cdot (3{{x}^{2}}+2x-1)+4\\&=(2x-1)(3{{x}^{2}}+2x-1)+4\end{align*}
Note
For a demonstration of the synthetic division method watch the video “Synthetic Division”.
Synthetic Division (Duration: 05.20)
Synthetic division is an alternative that can only be used when the divisor is a binomial in the form $\scriptsize x-k$ where $\scriptsize k$ is a real number. To use synthetic division, the coefficient of the divisor must be $\scriptsize 1$. Say, for example, the divisor was $\scriptsize 2x-1$. In this case, we make the leading coefficient of the divisor equal to $\scriptsize 1$ by rewriting it as $\scriptsize 2(x-\displaystyle \frac{1}{2})$. In synthetic division, only the coefficients are used in the division process.
Take note!
To divide two polynomials using synthetic division do the following:
1. Write $\scriptsize k$ for the divisor.
2. Write the coefficients of the dividend.
3. Bring the lead coefficient down.
4. Multiply the lead coefficient by $\scriptsize k$. Write the product in the next column.
5. Add the terms of the second column.
6. Multiply the result by $\scriptsize k$. Write the product in the next column.
7. Repeat steps 5 and 6 for the remaining columns.
8. Use the last row of numbers to write the quotient. The number in the last column is the remainder and has degree $\scriptsize 0$. The next number from the right has degree $\scriptsize 1$, the next number from the right has degree $\scriptsize 2$ and so on.
Exercise 1.3
Find the quotient and remainder of the following by using synthetic division:
1. $\scriptsize 2{{x}^{3}}-3{{x}^{2}}+4x+5$ divided by $\scriptsize x+2$
2. $\scriptsize 4{{x}^{3}}+10{{x}^{2}}-6x-20$ divided by $\scriptsize x-2$
3. $\scriptsize 2{{x}^{3}}+5x-4$ divided by $\scriptsize x-1$
The full solutions are at the end of the unit.
Remainder theorem
Now that we know how to divide polynomials, we can use polynomial division to find the value of polynomials using the remainder theorem.
Activity 1.2: Evaluate polynomials using the remainder theorem
Time required: 15 minutes
What you need:
• a pen and paper
What to do:
Given the following functions:
\scriptsize \begin{align*}f(x)&={{x}^{3}}+3{{x}^{2}}+4x+12\\d(x)&=x-1\\g(x)&=4{{x}^{3}}-2{{x}^{2}}+2x-4\\h(x)&=2x+1\end{align*}
Determine $\scriptsize \displaystyle \frac{{f(x)}}{{d(x)}}$ and $\scriptsize \displaystyle \frac{{g(x)}}{{h(x)}}$.
Write your answers in the form $\scriptsize a(x)=b(x)\cdot Q(x)+R(x)$.
Calculate $\scriptsize f(1)$ and $\scriptsize g(-\displaystyle \frac{1}{2})$.
Compare your answers to question 1 and 3. What do you notice?
Write a mathematical equation to describe your conclusions.
Complete the following sentence: a cubic function divided by a linear expression gives a quotient with a degree of ________ and a remainder with a degree of ______, which is called a constant.
What did you find?
1. Using synthetic division or long division we get:
\scriptsize \displaystyle \begin{align*}&\displaystyle \frac{{f(x)}}{{d(x)}}=\displaystyle \frac{{{{x}^{3}}+3{{x}^{2}}+4x+12}}{{x-1}}\\\ &1\left| \!{\underline {\, \begin{align*}1\text{ }3\text{ }4\text{ }12\\\text{ }1\text{ }4\text{ }8\end{align*} \,}} \right. \\& \text{ }1\text{ }4\text{ }8\text{ }20\\&\text{Quotient}={{x}^{2}}+4x+8\\&\text{Remainder}=20\end{align*}
.
\scriptsize \displaystyle \begin{align*}\displaystyle \frac{{g(x)}}{{h(x)}}=\displaystyle \frac{{4{{x}^{3}}-2{{x}^{2}}+2x-4}}{{2x+1}}\\2x+1\overset{{2{{x}^{2}}-2x+2}}{\overline{\left){\begin{align*}4{{x}^{3}}-2{{x}^{2}}+2x-4\\\underline{{-(4{{x}^{3}}+2{{x}^{2}})}}\\\text{ }-4{{x}^{2}}+2x\\\text{ }\underline{{-\text{(}-4{{x}^{2}}-2x)}}\\\text{ 4}x-4\\\text{ }\underline{{\text{ }-\text{(4}x+2)}}\\\text{ }-6\end{align*}}\right.}}\\\text{Quotient}=2{{x}^{2}}-2x+2\\\text{Remainder}=-6\end{align*}
2. $\scriptsize \displaystyle {{x}^{3}}+3{{x}^{2}}+4x+12=(x-1)({{x}^{2}}+4x+8)+20$
$\scriptsize \displaystyle 4{{x}^{3}}-2{{x}^{2}}+2x-4=(2x+1)(2{{x}^{2}}-2x+2)-6$
3. .
\scriptsize \begin{align*}f(1)&={{(1)}^{3}}+3{{(1)}^{2}}+4(1)+12\\&=20\end{align*}
.
\scriptsize \begin{align*}g(-\displaystyle \frac{1}{2})&=4{{(-\displaystyle \frac{1}{2})}^{3}}-2{{(-\displaystyle \frac{1}{2})}^{2}}+2(-\displaystyle \frac{1}{2})-4\\&=-6\end{align*}
4. The remainders using division of the polynomials are the same as finding the function values using the divisor.
5. If a polynomial is divided by $\scriptsize x-a$, the remainder may be found quickly by evaluating the polynomial function at $\scriptsize a$.
So we can say $\scriptsize R=p(a)$.
6. A cubic function divided by a linear expression gives a quotient with a degree of $\scriptsize \underline{2}$ and a remainder with a degree of $\scriptsize \underline{0}$, which is called a constant.
The remainder theorem:
If a polynomial $\scriptsize f(x)$ is divided by $\scriptsize ax-k$, then the remainder is the value $\scriptsize f\left( {\displaystyle \frac{k}{a}} \right)$.
You can also use the remainder theorem to solve for an unknown variable. This is shown in the next example.
Example 1.3
Determine the value of $\scriptsize t$ if $\scriptsize f(x)={{x}^{3}}+t{{x}^{2}}+4x+2$ gives a remainder of $\scriptsize 16$ when divided by $\scriptsize 2x+1$.
Solution
The remainder theorem tells us that the remainder when $\scriptsize f(x)$ is divided by $\scriptsize 2x+1$ can be found using $\scriptsize f(-\displaystyle \frac{1}{2})$. We are also told that $\scriptsize f(-\displaystyle \frac{1}{2})=16$.
$\scriptsize f(x)={{x}^{3}}+t{{x}^{2}}+4x+2$
\scriptsize \begin{align*}f(-\displaystyle \frac{1}{2})&={{(-\displaystyle \frac{1}{2})}^{3}}+t{{(-\displaystyle \frac{1}{2})}^{2}}+4(-\displaystyle \frac{1}{2})+2\\&=-\displaystyle \frac{1}{8}+\displaystyle \frac{1}{4}t-2+2\\&=\displaystyle \frac{1}{4}t-\displaystyle \frac{1}{8}\end{align*}
But $\scriptsize f(-\displaystyle \frac{1}{2})=16$
\scriptsize \displaystyle \begin{align*}&\displaystyle \frac{1}{4}t-\displaystyle \frac{1}{8}=16\\&\therefore 2t-1=128\\&\therefore 2t=129\\&\therefore t=\displaystyle \frac{{129}}{2}\end{align*}
Exercise 1.4
1. Use the remainder theorem to determine the remainder when $\scriptsize f(x)=3{{x}^{3}}+5{{x}^{2}}-x+1$ is divided by:
1. $\scriptsize x+2$
2. $\scriptsize 2x-1$
3. $\scriptsize 3x+1$
2. Calculate the value of $\scriptsize m$ if $\scriptsize 2{{x}^{3}}-7{{x}^{2}}+mx-26$ is divided by $\scriptsize x-2$ and gives a remainder of$\scriptsize -24$.
The full solutions are at the end of the unit.
Using the factor theorem to solve cubic polynomials
You can use the factor theorem to solve for the zeros or roots (x-intercepts) of a cubic function. The factor theorem describes the relationship between the root of a polynomial and a factor of the polynomial.
You have seen in arithmetic division that if an integer $\scriptsize a$ is divided by an integer $\scriptsize b$ and the answer is $\scriptsize q$ with remainder $\scriptsize r=0$ then $\scriptsize b$ is a factor of $\scriptsize a$. For example, $\scriptsize 27$ divided by $\scriptsize 9$ is $\scriptsize 3$ with remainder $\scriptsize 0$. Therefore, $\scriptsize 9$ is a factor of $\scriptsize 27$. This is also true of polynomials.
Recall that if $\scriptsize f(x)\div (x-k)$ then $\scriptsize f(x)=(x-k)\cdot Q(x)+R(x)$.
If $\scriptsize k$ is a root of the function, then the remainder $\scriptsize R$ is zero and $\scriptsize f\left( k \right)=0$. Therefore, $\scriptsize f(x)=(x-k)\cdot Q(x)+0$ or $\scriptsize f(x)=(x-k)\cdot Q(x)$.
Written in this form, where the remainder is zero, we can say that $\scriptsize x-k$ is a factor of $\scriptsize f(x)$. We can conclude if $\scriptsize k$ is a zero of $\scriptsize f(x)$, then $\scriptsize x-k$ is a factor of $\scriptsize f(x)$.
Conversely, if $\scriptsize x-k$ is a factor of $\scriptsize f(x)$ then the remainder is $\scriptsize 0$.
Factor theorem:
If $\scriptsize f(x)$ is divided by $\scriptsize ax-k$ and the remainder, given by $\scriptsize f\left( {\displaystyle \frac{k}{a}} \right)$ is equal to $\scriptsize 0$, then $\scriptsize ax-k$ is a factor of $\scriptsize f(x)$.
Converse: If $\scriptsize ax-k$ is a factor of $\scriptsize f(x)$ then $\scriptsize f\left( {\displaystyle \frac{k}{a}} \right)=0$.
Cubic polynomials are factorised by using long division, division by inspection or synthetic division. You may use whichever method you are most comfortable with to factorise cubic polynomials. Remember that the degree of the polynomial tells you how many roots the function will have at most. So, a cubic polynomial will have at most three roots, a quadratic function at most two roots and so on.
Example 1.4
Show that $\scriptsize x+2$ is a factor of $\scriptsize g(x)={{x}^{3}}-6{{x}^{2}}-x+30$. Find the remaining factors of $\scriptsize g(x)$. Use the factors to determine the zeros of the cubic polynomial.
Solution
Use the remainder theorem to show that $\scriptsize g(-2)=0$. If$\scriptsize g(-2)=0$ then $\scriptsize x+2$ is a factor of $\scriptsize g(x)$.
\scriptsize \begin{align*}g(-2)&={{(-2)}^{3}}-6{{(-2)}^{2}}-(-2)+30\\&=-8-24+2+30\\&=0\end{align*}
Since $\scriptsize g(-2)=0$ then $\scriptsize x+2$ is a factor of $\scriptsize {{x}^{3}}-6{{x}^{2}}-x+30$.
Now, that we have a linear factor we can divide $\scriptsize g(x)$ by $\scriptsize x+2$ to find the quadratic factor.
You can use long division or synthetic division to factorise further but one of the quickest methods to factorise a cubic polynomial is division by inspection.
We know that $\scriptsize {{x}^{3}}-6{{x}^{2}}-x+30=(x+2)(...)$.
The first term in the second bracket must be $\scriptsize {{x}^{2}}$ to give $\scriptsize {{x}^{3}}$ when we multiply the brackets to make the polynomial a cubic.
The last term in the second bracket must be $\scriptsize 15$ because $\scriptsize 15\times 2=30$. The middle term must be some $\scriptsize ax$ term.
So far we have $\scriptsize {{x}^{3}}-6{{x}^{2}}-x+30=(x+2)({{x}^{2}}+ax+15)$.
Now, we must find the coefficient of the middle term in the second bracket. When you multiply the brackets out, the terms that you will multiply together and collect to get the $\scriptsize -6{{x}^{2}}$ term are: $\scriptsize 2\times {{x}^{2}}$ and $\scriptsize x\times a{{x}^{2}}$.
$\scriptsize (x+2)({{x}^{2}}+ax+15)$
$\scriptsize 2\times {{x}^{2}}+a\times {{x}^{2}}=-6{{x}^{2}}$
Solve this simple equation (which can be done in your head without showing working) to find $\scriptsize a$.
\scriptsize \displaystyle \begin{align*}2{{x}^{2}}+a{{x}^{2}} & =-6{{x}^{2}}\\a{{x}^{2}} & =-8{{x}^{2}}\\\therefore a & =-8\end{align*}
So the coefficient of the x-term is $\scriptsize -8$.
$\scriptsize \therefore {{x}^{3}}-6{{x}^{2}}-x+30=(x+2)({{x}^{2}}-8x+15)$
We must factorise the quadratic factor further to fully factorise the cubic polynomial.
\scriptsize \begin{align*}{{x}^{3}}-6{{x}^{2}}-x+30&=(x+2)({{x}^{2}}-8x+15)\\&=(x+2)(x-3)(x-5)\end{align*}
The zeros of the function are found by solving the previous equation.
\scriptsize \begin{align*}&(x+2)(x-3)(x-5)=0\\&\therefore x=-2,3\text{ or }5\end{align*}
Take note!
To factorise a cubic polynomial we do the following:
1. Find one linear factor by trial and error. Note: generally only try factors that would divide into the constant term. For example, if the constant term is $\scriptsize 3$ then the only factors you need to try are numbers that divide into $\scriptsize 3:-3,-1,1\text{ and }3$.
2. Use the factor theorem to confirm that your answer in question 1 is a factor by showing that $\scriptsize f(k)=0$.
3. Divide the given cubic polynomial by your linear factor to get the quadratic factor.
4. Factorise the quadratic trinomial to find the other two factors of the cubic polynomial.
Exercise 1.5
1. Factorise:
1. $\scriptsize f(x)={{x}^{3}}+{{x}^{2}}-9x-9$
2. $\scriptsize g(x)={{x}^{3}}-3{{x}^{2}}+4$
2. $\scriptsize f(x)=2{{x}^{3}}+{{x}^{2}}-5x+2$
1. Find $\scriptsize f(1)$.
2. Factorise $\scriptsize f(x)$ and list the zeros of the function.
The full solutions are at the end of the unit.
Sometimes it is not possible to factorise a quadratic expression using inspection, in which case we use the quadratic formula to fully factorise and solve the cubic equation.
Summary
In this unit you have learnt the following:
• How to divide polynomials using long division to find the quotient and remainder.
• How to divide polynomials using synthetic division to find the quotient and remainder.
• How to use the remainder theorem to find the remainder of a polynomial.
• How to use the remainder theorem to find an unknown value.
• How to find a linear factor of a cubic polynomial using the factor theorem.
• How to fully factorise a cubic polynomial using division by inspection.
• How to solve equations with cubic polynomials.
Unit 1: Assessment
Suggested time to complete: 45 minutes
1. The volume of a rectangular solid is given by the polynomial $\scriptsize 3{{x}^{4}}-3{{x}^{3}}-33{{x}^{2}}+54x$. The length is $\scriptsize 3x$ and the width is $\scriptsize x-2$. Find the height, h.
2. When $\scriptsize f(x)=3{{x}^{5}}+p{{x}^{4}}+10{{x}^{2}}-21x+12$ is divided by $\scriptsize x-2$ it leaves a remainder of $\scriptsize 10$. Find the value of $\scriptsize p$.
3. Solve $\scriptsize {{x}^{3}}+{{x}^{2}}-16x=16$.
4. Let $\scriptsize g(x)$ be a polynomial in $\scriptsize x$.
1. If $\scriptsize g(x)$ is divided by $\scriptsize x-k$, what does $\scriptsize g(k)$ represent?
2. If $\scriptsize g(k)=0$ what can be said about $\scriptsize x-k$?
5. The polynomial $\scriptsize f(x)={{x}^{3}}+p{{x}^{2}}-qx-6$ is exactly divisible by $\scriptsize {{x}^{2}}+x-2$.
1. Factorise $\scriptsize {{x}^{2}}+x-2$.
2. Calculate the values of $\scriptsize p$ and $\scriptsize q$.
The full solutions are at the end of the unit.
Unit 1: Solutions
Exercise 1.1
1. False. It is not a polynomial since the exponent is not a whole number.
2. True.
3. True.
4. False. A cubic polynomial is an expression with a degree of $\scriptsize 3$ and does not always have three terms.
Back to Exercise 1.1
Exercise 1.2
1. .
1. .
\scriptsize \displaystyle \begin{align*}2x-1\overset{{3{{x}^{2}}+2x-1}}{\overline{\left){\begin{align*}6{{x}^{3}}+{{x}^{2}}-4x+5\\-\underline{{(6{{x}^{3}}-3{{x}^{2}})}}\end{align*}}\right.}}\\\text{ }4{{x}^{2}}-4x\\\text{ }-\underline{{(\text{ }4{{x}^{2}}-2x)}}\text{ }\\\text{ }-2x+5\\\text{ }-\underline{{(2x+1)}}\\\text{ }4\end{align*}
\scriptsize \begin{align*}Q(x)&=3{{x}^{2}}+2x-1\\R(x)&=4\\\therefore 6{{x}^{3}}&+{{x}^{2}}-4x+5=(2x-1)\cdot (3{{x}^{2}}+2x-1)+4\end{align*}
2. .
\scriptsize \begin{align*}x-1\overset{{{{x}^{2}}+4x+3}}{\overline{\left){{{{x}^{3}}+3{{x}^{2}}-x-3}}\right.}}\\\text{ }-\underline{{({{x}^{3}}-{{x}^{2}})}}\\\text{ }4{{x}^{2}}-x\\\text{ }-\underline{{(4{{x}^{2}}-4x)\text{ }}}\text{ }\\\text{ }3x-3\\\text{ }-\underline{{(3x-3)}}\\\text{ 0}\end{align*}
\scriptsize \begin{align*}&Q(x)={{x}^{2}}+4x+3\\&R(x)=0\end{align*}
\scriptsize \begin{align*}\therefore {{x}^{3}}+3{{x}^{2}}-x-3&=(x-1)\cdot ({{x}^{2}}+4x+3)+0\\ &=(x-1)\cdot ({{x}^{2}}+4x+3)\end{align*}
2. Since there is no remainder when $\scriptsize \displaystyle {{x}^{3}}+3{{x}^{2}}-x-3$ is divided by $\scriptsize \displaystyle (x-1)$, we can factorise the divisor multiplied by the quotient completely and we will get the dividend.
\scriptsize \displaystyle \begin{align*}{{x}^{3}}+3{{x}^{2}}-x-3=(x-1)\cdot ({{x}^{2}}+4x+3)\\\text{ }=(x-1)(x+1)(x+3)\end{align*}
Back to Exercise 1.2
Exercise 1.3
1. .
\scriptsize \begin{align*}-2\left| \!{\underline {\, \begin{align*}2\text{ }-3\text{ }4\text{ }5\\\text{ }-4\text{ 14 }-36\end{align*} \,}} \right. \\\text{ }2\text{ }-7\text{ 18 }-31\end{align*}
\scriptsize \begin{align*}&\text{Quotient}=2{{x}^{2}}-7x+18\\&\text{Remainder}=-31\end{align*}
2. .
\scriptsize \begin{align*}2\left| \!{\underline {\, \begin{align*}4\text{ }10\text{ }-6\text{ }-20\\\text{ }8\text{ }36\text{ }60\end{align*} \,}} \right. \\\text{ 4 }18\text{ }30\text{ }40\end{align*}
\scriptsize \begin{align*}\text{Quotient}=4{{x}^{2}}+18x+30\\\text{Remainder}=40\end{align*}
3. .
Remember if a term is missing in the dividend we write the coefficient as $\scriptsize 0$.
\scriptsize \begin{align*}1\left| \!{\underline {\, \begin{align*}2\text{ }0\text{ }5\text{ }-4\\\text{ }2\text{ 2 7}\end{align*} \,}} \right. \\\text{ }2\text{ }2\text{ }7\text{ 3}\end{align*}
\scriptsize \begin{align*}&\text{Quotient}=2{{x}^{2}}+2x+7\\&\text{Remainder}=3\end{align*}
Back to Exercise 1.3
Exercise 1.4
1. .
1. .
\scriptsize \begin{align*}f(x)&=3{{x}^{3}}+5{{x}^{2}}-x+1\\f(-2)&=3{{(-2)}^{3}}+5{{(-2)}^{2}}-(-2)+1\\&=-24+20+2+1\\&=-1\end{align*}
2. .
\scriptsize \begin{align*}f(x)&=3{{x}^{3}}+5{{x}^{2}}-x+1\\f(\displaystyle \frac{1}{2})&=3{{\left( {\displaystyle \frac{1}{2}} \right)}^{3}}+5{{\left( {\displaystyle \frac{1}{2}} \right)}^{2}}-\left( {\displaystyle \frac{1}{2}} \right)+1\\&=\displaystyle \frac{3}{8}+\displaystyle \frac{5}{4}-\displaystyle \frac{1}{2}+1\\&=\displaystyle \frac{{17}}{8}\end{align*}
3. .
\scriptsize \begin{align*}f(x)&=3{{x}^{3}}+5{{x}^{2}}-x+1\\f(-\displaystyle \frac{1}{3})&=3{{(-\displaystyle \frac{1}{3})}^{3}}+5{{(-\displaystyle \frac{1}{3})}^{2}}-(-\displaystyle \frac{1}{3})+1\\&=-\displaystyle \frac{1}{9}+\displaystyle \frac{5}{9}+\displaystyle \frac{1}{3}+1\\&=\displaystyle \frac{{16}}{9}\end{align*}
2. .
$\scriptsize \text{Let }g(x)=2{{x}^{3}}-7{{x}^{2}}+mx-26$
Then $\scriptsize g(2)=-24$
\scriptsize \begin{align*}&g(2)=2{{(2)}^{3}}-7{{(2)}^{2}}+m(2)-2\\&\therefore -24=16-28+2m-2\\&\therefore -2m=10\\&\therefore m=-5\end{align*}
Back to Exercise 1.4
Exercise 1.5
1. .
1. Find a factor by trial and error.
\scriptsize \begin{align*}f(3)&={{(3)}^{3}}+{{(3)}^{2}}-9(3)-9\\&=0\end{align*}
\scriptsize \begin{align*}&f(3)=0\\&\therefore x-3\text{ is a factor of }f(x)\end{align*}
Divide to find the other factors. Use division by inspection, long division or synthetic division.
$\scriptsize {{x}^{3}}+{{x}^{2}}-9x-9=(x-3)({{x}^{2}}+ax+3)$
\scriptsize \begin{align*}{{x}^{2}}&=-3{{x}^{2}}+a{{x}^{2}}\\4{{x}^{2}}&=a{{x}^{2}}\\4&=a\end{align*}
.
\scriptsize \begin{align*}\therefore {{x}^{3}}+{{x}^{2}}-9x-9&=(x-3)({{x}^{2}}+4x+3)\\\text{ }&=(x-3)(x+1)(x+3)\end{align*}
To solve, let $\scriptsize (x-3)({{x}^{2}}+1)(x+3)=0$
$\scriptsize \therefore x=3,-1$ or $\scriptsize -3$
2. .
\scriptsize \begin{align*}g(x)&={{x}^{3}}-3{{x}^{2}}+4\\g(-1)&=-1-3+4\\&=0\end{align*}
$\scriptsize \therefore (x+1)\text{ is a factor of }g(x)$
Remember to include the $\scriptsize 0x$ term especially if using long division or synthetic division.
\scriptsize \begin{align*}g(x)&={{x}^{3}}-3{{x}^{2}}+0x+4\\&=(x+1)({{x}^{2}}+ax+4)\end{align*}
To find coefficient of the middle term:
\scriptsize \begin{align*}{{x}^{2}}+a{{x}^{2}}&=-3{{x}^{2}}\\a{{x}^{2}}&=-4{{x}^{2}}\\a&=-4\end{align*}
\scriptsize \begin{align*}{{x}^{3}}-3{{x}^{2}}+4&=(x+1)({{x}^{2}}-4x+4)\\&=(x+1)(x-2)(x-2)\end{align*}
Solve:
\scriptsize \begin{align*}&(x+1)(x-2)(x-2)=0\\&\therefore x=-1\text{ or }2\end{align*}
2. .
1. .
\scriptsize \begin{align*}f(1)&=2{{(1)}^{3}}+{{(1)}^{2}}-5(1)+2\\&=2+1-5+2\\&=0\end{align*}
2. .
$\scriptsize x-1$ is a factor of $\scriptsize f(x)$
\scriptsize \displaystyle \begin{align*}2{{x}^{3}}+{{x}^{2}}-5x+2 & =(x-1)(2{{x}^{2}}+ax-2)\\-2{{x}^{2}}+a{{x}^{2}} & ={{x}^{2}}\\a{{x}^{2}} & =3{{x}^{2}}\\\therefore a & =3\\2{{x}^{3}}+{{x}^{2}}-5x+2 & =(x-1)(2{{x}^{2}}+3x-2)\\ & =(x-1)(2x-1)(x+2)\end{align*}
The zeros of the function are $\scriptsize \displaystyle 1,\displaystyle \frac{1}{2}\text{ or}-2$
Back to Exercise 1.5
Unit 1: Assessment
1. .
\scriptsize \displaystyle \begin{align*}v&=l\cdot w\cdot h\\&\therefore 3{{x}^{4}}-3{{x}^{3}}-33{{x}^{2}}+54x=3x(x-2)h\\&\therefore (x-2)h=\displaystyle \frac{{3{{x}^{4}}-3{{x}^{3}}-33{{x}^{2}}+54x}}{{3x}}=\displaystyle \frac{{3x({{x}^{3}}-{{x}^{2}}-11x+18)}}{{3x}}={{x}^{3}}-{{x}^{2}}-11x+18\\&\therefore h=\displaystyle \frac{{{{x}^{3}}-{{x}^{2}}-11x+18}}{{x-2}}\end{align*}
$\scriptsize (x-2)$ is a factor of $\scriptsize \displaystyle {{x}^{3}}-{{x}^{2}}-11x+18$
.
\scriptsize \begin{align*}&{{x}^{3}}-{{x}^{2}}-11x+18=(x-2)({{x}^{2}}+ax-9)\\&\therefore -2{{x}^{2}}+a{{x}^{2}}=-{{x}^{2}}\\&\therefore a{{x}^{2}}={{x}^{2}}\\&\therefore a=1\end{align*}
$\scriptsize \displaystyle \therefore {{x}^{3}}-{{x}^{2}}-11x+18=(x-2)({{x}^{2}}+x-9)$
\scriptsize \displaystyle \begin{align*}&h=\displaystyle \frac{{(x-2)({{x}^{2}}+x-9)}}{{(x-2)}}\\&\therefore h={{x}^{2}}+x-9\end{align*}
2. .
$\scriptsize f(x)=3{{x}^{5}}+p{{x}^{4}}+10{{x}^{2}}-21x+12$
But $\scriptsize f(2)=10$
.
\scriptsize \begin{align*}&\therefore 3{{(2)}^{5}}+p{{(2)}^{4}}+10{{(2)}^{2}}-21(2)+12=10\\&\therefore 96+16p+40-42+12=10\\&\therefore 16p=10-106=-96\\&\therefore p=-6\end{align*}
3. .
$\scriptsize {{x}^{3}}+{{x}^{2}}-16x-16=0$
Find a factor.
\scriptsize \begin{align*}&{{(4)}^{3}}+{{(4)}^{2}}-16(4)-16=0\\&\therefore x-4\text{ is factor}\end{align*}
.
\scriptsize \begin{align*}{{x}^{3}}+{{x}^{2}}-16x-16&=(x-4)({{x}^{2}}+ax+4)\\&=(x-4)({{x}^{2}}+5x+4)\\&=(x-4)(x+1)(x+4)\end{align*}
\scriptsize \begin{align*}&(x-4)(x+1)(x+4)=0\\&\therefore x=4;-1\text{ and }-4\end{align*}
4. Let $\scriptsize g(x)$ be a polynomial in $\scriptsize x$.
1. $\scriptsize g(k)$ represents the remainder.
2. If then $\scriptsize x-k$ is a factor of $\scriptsize g(x)$.
5. The polynomial $\scriptsize f(x)={{x}^{3}}+p{{x}^{2}}-qx-6$ is exactly divisible by $\scriptsize {{x}^{2}}+x-2$.
1. .
$\scriptsize {{x}^{2}}+x-2=(x+2)(x-1)$
2. .
$\scriptsize f(-2)=0$
\scriptsize \displaystyle \begin{align*}&\therefore {{(-2)}^{3}}+p{{(-2)}^{2}}-q(-2)-6=0\\&\therefore -8+4p+2q-6=0\\&\therefore 4p+2q-14=0\\&\therefore 2p+q-7=0\end{align*}
.
$\scriptsize \displaystyle f(1)=0$
\scriptsize \displaystyle \begin{align*}&\therefore {{(1)}^{3}}+p{{(1)}^{2}}-q(1)-6=0\\&\therefore 1+p-q-6=0\\&\therefore p-q-5=0\end{align*}
Solve simultaneously
$\scriptsize \displaystyle 2p+q-7=0\text{ (}1)$
$\scriptsize \displaystyle p-q-5=0\text{ (2)}$
.
From (2):$\scriptsize p=q+5(3)$
Substitute into (1):
\scriptsize \displaystyle \begin{align*}&\therefore 2(q+5)+q-7=0\\&\therefore 2q+10+q-7=0\\&\therefore 3q+3=0\\&\therefore q=-1\end{align*}
Substitute into (3):
$\scriptsize p=-1+5=4\text{ }$
Back to Unit 1: Assessment |
## EXAMPLE PROBLEMS ON BASIC PROPORTIONALITY THEOREM
Basic Proportionality Theorem :
If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
## Example Problems on Basic Proportionality Theorem
Example 1 :
In a triangle ABC,D and E are points on the sides AB and AC respectively such that DE is parallel to BC.
(i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, then find AC.
Solution :
In the given triangle ABC the sides DE is parallel to the side BC. By using “Thales theorem”, we get
(6/9) = (8/EC)
EC = (9 8)/6
EC = (3 4)
EC = 12 cm
Example 2 :
In triangle ABC, if AD = 8 cm, AB = 12 cm and AE = 12 cm, then find CE.
Solution :
In the given triangle ABC the sides DE is parallel to the side BC. By using “Thales theorem”, we get
12 = 8 + DB
DB = 4 cm
(8/4) = (12/EC)
EC = (12 4)/8
EC = 48/8
EC = 6 cm
Example 3 :
In triangle ABC, if AD = 4 x – 3, BD = 3 x – 1, AE = 8 x – 7 and EC = 5 x – 3, then find the value of x.
Solution :
In the given triangle ABC the sides DE is parallel to the side BC. By using “Thales theorem”, we get
(4 x – 3)/(3 x – 1) = (8 x – 7)/(5 x – 3)
(4 x – 3) (5 x – 3) = (8 x – 7) (3 x – 1)
20 x ² – 12 x – 15 x + 9 = 24 x ² – 8 x – 21 x + 7
20 x ² – 24 x + 9 = 8 x ² – 29 x + 7
20 x ² - 24 x ² – 27 x + 29 x + 9 – 7 = 0
- 4 x ² + 2 x + 2 = 0
Multiply by - 2
2 x ² – x - 1 = 0
(x – 1) (2 x + 1) = 0
x - 1 = 0x = 1 2x + 1 = 02x = -1x = -1/2
So the value of x is 1.
After having gone through the stuff given above, we hope that the students would have understood, examples on basic proportionality theorem.
Apart from the stuff given in this section , if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
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# Question 5998a
Aug 3, 2017
$\text{molarity} = 14.6$ $M$
$\text{volume} = 13.7$ $\text{mL}$
#### Explanation:
We're asked to find (a) the molarity of the ${\text{H"_2"SO}}_{4}$ solution, and (b) the volume of that acid solution required to make $1$ $\text{L}$ of a $0.2$ $M$ solution.
(a)
To find the molarity of the solution, we can first recognize that in an 80% by mass solution, and we assume a $100$-$\text{g}$ sample, there would be
• $80.0$ ${\text{g H"_2"SO}}_{4}$
• $20.0$ $\text{g H"_2"O}$
Also in a $100$-$\text{g}$ sample, using the given density, the volume (in liters) of the solution is
V = 100cancel("g soln")((1cancel("mL soln"))/(1.787cancel("g soln")))((1color(white)(l)"L soln")/(10^3cancel("mL soln"))) = color(red)(ul(0.0560color(white)(l)"L soln"
Remember that the molarity equation is
ul("molarity" = "mol solute"/"L soln"
We just found the liters of solution, so let's use the $80$ ${\text{g H"_2"SO}}_{4}$ and the molar mass of ${\text{H"_2"SO}}_{4}$ to find the moles:
80.0cancel("g H"_2"SO"_4)((1color(white)(l)"mol H"_2"SO"_4)/(98.078cancel("g H"_2"SO"_4))) = color(green)(ul(0.816color(white)(l)"mol H"_2"SO"_4
The molarity of the solution is thus
"molarity" = color(green)(0.816color(white)(l)"mol H"_2"SO"_4)/(color(red)(0.0560color(white)(l)"L soln")) = color(blue)(ulbar(|stackrel(" ")(" "14.6color(white)(l)M" ")|)
(b)
Now, we can use the dilution equation to find the volume of this solution required to make $1$ $\text{L}$ of a $0.2$ $M$ solution.
The dilution equation is
ul(M_1V_1 = M_2V_2
where
• ${M}_{1}$ and ${M}_{2}$ are the molarities of the two solutions
• ${V}_{1}$ and ${V}_{2}$ are the respective volumes of the two solutions
We know:
• M_1 = color(blue)(14.6color(white)(l)M
• V_1 = ?
• ${M}_{2} = 0.2 \textcolor{w h i t e}{l} M$
• ${V}_{2} = 1 \textcolor{w h i t e}{l} \text{L}$
Let's rearrange the equation to solve for the unknown volume, ${V}_{1}$:
${V}_{1} = \frac{{M}_{2} {V}_{2}}{{M}_{1}}$
Plugging in known values:
V_1 = ((0.2cancel(M))(1color(white)(l)"L"))/(color(blue)(14.6)cancel(color(blue)(M))) = color(blue)(0.0137color(white)(l)"L" = color(blue)(ulbar(|stackrel(" ")(" "13.7color(white)(l)"mL"" ")|)# |
# 6.2 Graphs of the other trigonometric functions (Page 8/9)
Page 8 / 9
## Using the graphs of trigonometric functions to solve real-world problems
Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function .
## Using trigonometric functions to solve real-world scenarios
Suppose the function $\text{\hspace{0.17em}}y=5\mathrm{tan}\left(\frac{\pi }{4}t\right)\text{\hspace{0.17em}}$ marks the distance in the movement of a light beam from the top of a police car across a wall where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the time in seconds and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is the distance in feet from a point on the wall directly across from the police car.
1. Find and interpret the stretching factor and period.
2. Graph on the interval $\text{\hspace{0.17em}}\left[0,5\right].$
3. Evaluate $\text{\hspace{0.17em}}f\left(1\right)\text{\hspace{0.17em}}$ and discuss the function’s value at that input.
1. We know from the general form of $\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bt\right)\text{\hspace{0.17em}}$ that $\text{\hspace{0.17em}}|A|\text{\hspace{0.17em}}$ is the stretching factor and $\text{\hspace{0.17em}}\frac{\pi }{B}\text{\hspace{0.17em}}$ is the period.
We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period.
The period is $\text{\hspace{0.17em}}\frac{\pi }{\frac{\pi }{4}}=\frac{\pi }{1}\cdot \frac{4}{\pi }=4.\text{\hspace{0.17em}}$ This means that every 4 seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches.
2. To graph the function, we draw an asymptote at $\text{\hspace{0.17em}}t=2\text{\hspace{0.17em}}$ and use the stretching factor and period. See [link]
3. period: $\text{\hspace{0.17em}}f\left(1\right)=5\mathrm{tan}\left(\frac{\pi }{4}\left(1\right)\right)=5\left(1\right)=5;\text{\hspace{0.17em}}$ after 1 second, the beam of has moved 5 ft from the spot across from the police car.
Access these online resources for additional instruction and practice with graphs of other trigonometric functions.
## Key equations
Shifted, compressed, and/or stretched tangent function $y=A\text{\hspace{0.17em}}\mathrm{tan}\left(Bx-C\right)+D$ Shifted, compressed, and/or stretched secant function $y=A\text{\hspace{0.17em}}\mathrm{sec}\left(Bx-C\right)+D$ Shifted, compressed, and/or stretched cosecant function $y=A\text{\hspace{0.17em}}\mathrm{csc}\left(Bx-C\right)+D$ Shifted, compressed, and/or stretched cotangent function $y=A\text{\hspace{0.17em}}\mathrm{cot}\left(Bx-C\right)+D$
## Key concepts
• The tangent function has period $\text{\hspace{0.17em}}\pi .$
• $f\left(x\right)=A\mathrm{tan}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ is a tangent with vertical and/or horizontal stretch/compression and shift. See [link] , [link] , and [link] .
• The secant and cosecant are both periodic functions with a period of $\text{\hspace{0.17em}}2\pi .\text{\hspace{0.17em}}$ $f\left(x\right)=A\mathrm{sec}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ gives a shifted, compressed, and/or stretched secant function graph. See [link] and [link] .
• $f\left(x\right)=A\mathrm{csc}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ gives a shifted, compressed, and/or stretched cosecant function graph. See [link] and [link] .
• The cotangent function has period $\text{\hspace{0.17em}}\pi \text{\hspace{0.17em}}$ and vertical asymptotes at $\text{\hspace{0.17em}}0,±\pi ,±2\pi ,....$
• The range of cotangent is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),\text{\hspace{0.17em}}$ and the function is decreasing at each point in its range.
• The cotangent is zero at $\text{\hspace{0.17em}}±\frac{\pi }{2},±\frac{3\pi }{2},....$
• $f\left(x\right)=A\mathrm{cot}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ is a cotangent with vertical and/or horizontal stretch/compression and shift. See [link] and [link] .
• Real-world scenarios can be solved using graphs of trigonometric functions. See [link] .
## Verbal
Explain how the graph of the sine function can be used to graph $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x.$
Since $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is the reciprocal function of $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ you can plot the reciprocal of the coordinates on the graph of $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to obtain the y -coordinates of $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ The x -intercepts of the graph $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ are the vertical asymptotes for the graph of $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x.$
can you not take the square root of a negative number
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
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
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
Where do the rays point?
Spiro
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas |
In this article you’ll learn about Permutation and Combination problems: Definition, formulas, solved examples and a quiz with practice questions.
## Permutations
#### Definition
Permutations are the different ways in which a collection of items can be arranged.
For example:
The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are ABC, ACB, BCA, CBA, CAB, BAC.
Note that ABC and CBA are not same as the order of arrangement is different. The same rule applies while solving any problem in Permutations.
The number of ways in which n things can be arranged, taken all at a time, nPn = n!, called ‘n factorial.’
#### Factorial Formula
Factorial of a number n is defined as the product of all the numbers from n to 1.
For example, the factorial of 5, 5! = 5*4*3*2*1 = 120.
Therefore, the number of ways in which the 3 letters can be arranged, taken all a time, is 3! = 3*2*1 = 6 ways.
Number of permutations of n things, taken r at a time, denoted by:
nPr = n! / (n-r)!
For example:
The different ways in which the 3 letters, taken 2 at a time, can be arranged is 3!/(3-2)! = 3!/1! = 6 ways.
### Important Permutation Formulas
1! = 1
0! = 1
Let us take a look at some examples:
Problem 1: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘CHAIR’.
Solution:
‘CHAIR’ contains 5 letters.
Therefore, the number of words that can be formed with these 5 letters = 5! = 5*4*3*2*1 = 120.
Problem 2: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘INDIA’.
Solution:
The word ‘INDIA’ contains 5 letters and ‘I’ comes twice.
When a letter occurs more than once in a word, we divide the factorial of the number of all letters in the word by the number of occurrences of each letter.
Therefore, the number of words formed by ‘INDIA’ = 5!/2! = 60.
Problem 3: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘SWIMMING?
Solution:
The word ‘SWIMMING contains 8 letters. Of which, I occurs twice and M occurs twice.
Therefore, the number of words formed by this word = 8! / (2!*2!) = 10080.
Problem 4: How many different words can be formed with the letters of the word ‘SUPER’ such that the vowels always come together?
Solution:
The word ‘SUPER’ contains 5 letters.
In order to find the number of permutations that can be formed where the two vowels U and E come together.
In these cases, we group the letters that should come together and consider that group as one letter.
So, the letters are S,P,R, (UE). Now the number of words are 4.
Therefore, the number of ways in which 4 letters can be arranged is 4!
In U and E, the number of ways in which U and E can be arranged is 2!
Hence, the total number of ways in which the letters of the ‘SUPER’ can be arranged such that vowels are always together are 4! * 2! = 48 ways.
Problem 5: Find the number of different words that can be formed with the letters of the word ‘BUTTER’ so that the vowels are always together.
Solution:
The word ‘BUTTER’ contains 6 letters.
The letters U and E should always come together. So the letters are B, T, T, R, (UE).
Number of ways in which the letters above can be arranged = 5!/2! = 60 (since the letter ‘T’ is repeated twice).
Number of ways in which U and E can be arranged = 2! = 2 ways
Therefore, total number of permutations possible = 60*2 = 120 ways.
Problem 6: Find the number of permutations of the letters of the word ‘REMAINS’ such that the vowels always occur in odd places.
Solution:
The word ‘REMAINS’ has 7 letters.
There are 4 consonants and 3 vowels in it.
Writing in the following way makes it easier to solve these type of questions.
(1) (2) (3) (4) (5) (6) (7)
No. of ways 3 vowels can occur in 4 different places = 4P3 = 24 ways.
After 3 vowels take 3 places, no. of ways 4 consonants can take 4 places = 4P4 = 4! = 24 ways.
Therefore, total number of permutations possible = 24*24 = 576 ways.
## Combinations
#### Definition
The different selections possible from a collection of items are called combinations.
For example:
The different selections possible from the alphabets A, B, C, taken 2 at a time, are AB, BC and CA.
It does not matter whether we select A after B or B after A. The order of selection is not important in combinations.
To find the number of combinations possible from a given group of items n, taken r at a time, the formula, denoted by nCr is
nCr = n! / [r! * (n-r)!]
For example, verifying the above example, the different selections possible from the alphabets A, B, C, taken two at a time are
3C2 = 3! / (2! * (3-2)!) = 3 possible selections (i.e., AB, BC, CA)
### Important Combination formulas
nCn = 1
nC0 = 1
nC1 = n
nCr = nC(n-r)
The number of selections possible with A, B, C, taken all at a time is 3C3 = 1 (i.e. ABC)
### Solved examples of Combination
Let us take a look at some examples to understand how Combinations work:
Problem 1: In how many ways can a committee of 1 man and 3 women can be formed from a group of 3 men and 4 women?
Solution:
No. of ways 1 man can be selected from a group of 3 men = 3C1 = 3! / 1!*(3-1)! = 3 ways.
No. of ways 3 women can be selected from a group of 4 women = 4C3 = 4! / (3!*1!) = 4 ways.
Problem 2: Among a set of 5 black balls and 3 red balls, how many selections of 5 balls can be made such that at least 3 of them are black balls.
Solution:
Selecting at least 3 black balls from a set of 5 black balls in a total selection of 5 balls can be
3 B and 2 R
4 B and 1 R and
5 B and 0 R balls.
Therefore, our solution expression looks like this.
5C3 * 3C2 + 5C4 * 3C1 + 5C5 * 3C0 = 46 ways .
Problem 3: How many 4 digit numbers that are divisible by 10 can be formed from the numbers 3, 5, 7, 8, 9, 0 such that no number repeats?
Solution:
If a number is divisible by 10, its units place should contain a 0.
_ _ _ 0
After 0 is placed in the units place, the tens place can be filled with any of the other 5 digits.
Selecting one digit out of 5 digits can be done in 5C1 = 5 ways.
After filling the tens place, we are left with 4 digits. Selecting 1 digit out of 4 digits can be done in 4C1 = 4 ways.
After filling the hundreds place, the thousands place can be filled in 3C1 = 3 ways.
Therefore, the total combinations possible = 5*4*3 = 60.
## Permutations and Combinations Quiz
Try these practice problems.
Problem 1
Solve the following.
i) 30P2
ii) 30C2
A. 870, 435
B. 435, 870
C. 870, 470
D. 435, 835
A
Explanation:
30P2 = 30! / 28! = 30*29*28! / 28! = 30*29 = 870.
30C2 = 30! / (2!*28!) = 435.
Problem 2
How many different possible permutations can be made from the word ‘BULLET’ such that the vowels are never together?
A. 360
B. 120
C. 480
D. 240
D.
Explanation:
The word ‘BULLET’ contains 6 letters of which 1 letter occurs twice = 6! / 2! = 360
No. of permutations possible with vowels always together = 5! * 2! / 2! = 120
No. of permutations possible with vowels never together = 360-120 = 240.
Problem 3
In how many ways can a selection of 3 men and 2 women can be made from a group of 5 men and 5 women ?
A. 10
B. 20
C. 30
D. 100
D.
Explanation:
5C3 * 5C2 = 100
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# Ratio And Proportion - MCQ 3
## 20 Questions MCQ Test Quantitative Techniques for CLAT | Ratio And Proportion - MCQ 3
Description
This mock test of Ratio And Proportion - MCQ 3 for Quant helps you for every Quant entrance exam. This contains 20 Multiple Choice Questions for Quant Ratio And Proportion - MCQ 3 (mcq) to study with solutions a complete question bank. The solved questions answers in this Ratio And Proportion - MCQ 3 quiz give you a good mix of easy questions and tough questions. Quant students definitely take this Ratio And Proportion - MCQ 3 exercise for a better result in the exam. You can find other Ratio And Proportion - MCQ 3 extra questions, long questions & short questions for Quant on EduRev as well by searching above.
QUESTION: 1
### Divide Rs.2340 into three parts, such that first part be double that of second part and second part be 1/3 of the third part.Find the Third part amount?
Solution:
Answer – B.Rs.1170 Explanation : First: Second: Third = 2:1:3 Third part = 3*2340/6 = 1170
QUESTION: 2
### The ratio of income of A and B is 2:3. The sum of their expenditure is Rs.8000 and the amount of savings of A is equal to the amount of expenditure of B.What is the their ratio of sum of income to their sum of savings?
Solution:
Answer -A.5:3 Explanation : 2I-E + E = 8000
I = 4000
Sum of their Income = 5*I = 5*4000 = 20,000 Sum of their Savings = 20000-8000 = 12,000 20000:12000 = 5:3
QUESTION: 3
### There are 2 containers of equal capacity. The ratio of milk to water in the first container is 4:5 and in the second container is 3:7.If they are mixed up then the ratio of milk to water in the mixture will be
Solution:
Answer – D.67:113 Explanation : 4+5 = 9=> 40:50
3+7 = 10=> 27:63
40+27 : 50:63 = 67:113
QUESTION: 4
There are two numbers. When 25% of the first number is added to the second number, the resultant number is 1.5times th first number.What is the ratio of 1st number to the 2nd number ?
Solution:
Answer – C.4:5 Explanation : A+25/100 + B = 1.5A
A/4 + B = 15A/10
10A+40B/40 =60A/40
10A+40B = 60A
50A = 40B
A/B = 4/5
QUESTION: 5
A bag contains 10p,25p and Rs50p coins in the ratio of 5:2:1 respectively. If the total money in the bag is Rs.120.Find the number of 25p coins in that bag?
Solution:
Answer – A.160 Explanation : 10*5 : 25*2 : 50*1 = 50:50:50 = 1:1:1
120/3 = Rs.40 Rs. 1 = 4 Rs.40 = 4*40 = 160 coins
QUESTION: 6
The ratio of Ganesh’s age and his mother’s age is 5:12.The difference of their ages is 21.The ratio of their ages after 4 years will be
Solution:
Answer – D.19:40 Explanation : 12x – 5x = 21 7x = 21 X = 3
5:12 = 15:36
After 4 years = 19:40
QUESTION: 7
The ratio of students of three classes is 2:3:4. If 15 students are increased in each classes then their ratio turns into 13:18:23. What was the total number of students in all the three classes originally ?
Solution:
Answer – C.225 Explanation : 50:75:100
15 students increased 65:90:115 => 13:18 :23
Total no of students = 50+75+100 = 225
QUESTION: 8
Ravi and Govind have money in the ratio 5 : 12 and Govind and Kiran also have money in the same ratio 5 : 12. If Ravi has Rs. 500, Kiran has
Solution:
Answer – B.Rs.2880 Explanation : Ravi : Kiran = 5/12* 5/12 = 25/144 Kiran = 144*500/25 = 2880
QUESTION: 9
A town with a population of 1000 has provision for 30days, after 10 days 600 more men added, how long will the food last at the same rate ?
Solution:
Answer – C.12 ½ days Explanation : 1000*20/1600 = 12 1/2 days
QUESTION: 10
A man spends Rs.2480 to buy lunch box Rs.120 each and bottles at Rs.80 each,What will be the ratio of maximum number of bottles to lunch box are bought ?
Solution:
Answer – A.13:12 Explanation : Check the ans using option 13*80+ 12*120 = 1040+1440 = 2480
QUESTION: 11
Three cars travel same distance with speeds in the ratio 2 : 4 : 7. What is the ratio of the times taken by them to cover the distance?
Solution:
B) 14 : 7 : 4
Explanation: s = d/t Since distance is same, so ratio of times: 1/2 : 1/4 : 1/7 = 14 : 7 : 4
QUESTION: 12
Section A and section B of 7th class in a school contains total 285 students.Which of the following can be a ratio of the ratio of the number of boys and number of girls in the class?
Solution:
B) 10 : 9
Explanation: The number of boys and girls cannot be in decimal values, so the denominator should completely divide number of students (285).
Check each option: 6+5 = 11, and 11 does not divide 285 completely. 10+9 = 19, and only 19 divides 285 completely among all.
QUESTION: 13
180 sweets are divided among friends A, B, C and D in which B and C are brothers also such that sweets divided between A and B are in the ratio 2 : 3, between B and C in the ratio 2 : 5 and between C and D in ratio 3 : 4. What is the number of sweets received by the brothers together?
Solution:
B) 84
Explanation: A/B = N1/D1 B/C = N2/D2 C/D = N3/D3
A : B : C : D = N1*N2*N3 : D1*N2*N3 : D1*D2*N3 : D1*D2*D3
A/B = 2/3 B/C = 2/5 C/D = 3/4
A : B : C : D
2*2*3 : 3*2*3 : 3*5*3 : 3*5*4
4 : 6 : 15 : 20
B and C together = [(6+15)/(4+6+15+20)] * 180
QUESTION: 14
Number of students in 4th and 5th class is in the ratio 6 : 11. 40% in class 4 are girls and 48% in class 5 are girls. What percentage of students in both the classes are boys?
Solution:
B) 54.8%
Explanation: Total students in both = 6x+11x = 17x Boys in class 4 = (60/100)*6x = 360x/100 Boys in class 5 = (52/100)*11x = 572x/100 So total boys = 360x/100 + 572x/100 = 932x/100 = 9.32x % of boys = [9.32x/17x] * 100
QUESTION: 15
Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture?
Solution:
E) 5 : 7
Explanation: Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg Total brass = 30+40 = 70 kg So copper in mixture is (50+70) – 70 = 50 kg So copper to brass = 50 : 70
QUESTION: 16
Ratio of A and B is in the ratio 5 : 8. After 6 years, the ratio of ages of A and B will be in the ratio 17 : 26. Find the present age of B.
Solution:
A) 72
Explanation: A/B = 5/8 , A+6/B+6 = 17/26
Solve both, B = 72
QUESTION: 17
A bag contains 25p, 50p and 1Re coins in the ratio of 2 : 4 : 5 respectively. If the total money in the bag is Rs 75, find the number of 50p coins in the bag.
Solution:
D) 40
Explanation: 2x, 4x, 5x (25/100)*2x + (50/100)*4x + 1*5x = 75 x = 10, so 50 p coins = 4x = 40
QUESTION: 18
A is directly proportional to B and also directly proportional to C. When B = 6 and C = 2, A = 24. Find the value of A when B = 8 and C = 3.
Solution:
D) 48
Explanation: A directly proportional B, A directly proportional to C: A = kB, A = kC Or A = kBC When B = 6 and C = 2, A = 24: 24 = k*6*2 k = 2 Now when B = 8 and C = 3: A = 2*8*3
QUESTION: 19
A is directly proportional to B and also inversely proportional to the square of C.When B = 16 and C = 2, A = 36. Find the value of A when B = 32 and C = 4.
Solution:
C) 18
Explanation: A = kB, A = k/C2 Or A = kB/ C2
When B = 16 and C = 2, A = 36: 36 = k*16/ 22 k = 9 Now when B = 32 and C = 4: A = 9*32/ 42
QUESTION: 20
A is directly proportional to the inverse of B and also inversely proportional to C. When B = 36 and C = 9, A = 42. Find the value of A when B = 64 and C = 21.
Solution:
A) 24
Explanation: A = k√B, A = k/C Or A = k√B/C When B = 36 and C = 9, A = 42: 42 = k√36/9 k = 63 Now when B = 64 and C = 21: A = 63*√64/21 |
1 / 26
# PHY111: Summer 201253 - PowerPoint PPT Presentation
PHY111: Summer 201253. Lesson 02 : Energy Part I Work-Energy Theorem Energy Transfer & Transformation. 1/26. Applied Force. INTRODUCTION to Work & Energy (“The Work/Energy Theorem”). 2/26. Applied Force. Displacement. INTRODUCTION to Work & Energy.
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### PHY111: Summer 201253
Lesson 02: Energy Part I
Work-Energy Theorem
Energy Transfer & Transformation
1/26
INTRODUCTION to
Work & Energy
(“The Work/Energy Theorem”)
2/26
Displacement
INTRODUCTION to
Work & Energy
WORK is defined as the Applied Force times the Displacement
W = F . d
Work = 2 N * 3 m
Work = 6 N.m
Work = 6 Joules
F = 2 N
d = 3 m
3/26
If a Force is Applied, but no displacement or movement results, then NO Work is done on the Object.
4/26
The person might be expending ENERGY, but NO WORK is being done ON the object.
5/26
Discuss: What could have kept the box from moving?
Think:
If the box had never moved, how much work would have been done?
W=Fxd=(4N)(0m)=0N.m=0J
“Does this mean that no Energy was expended??”
Answer: No, it simply means that even though Energy was expended or used, no work was done on the object. In order for work to be done on an object, there has to be a force applied and there has to be a displacement parallel to the direction of the force.
6/26
The person might be exerting a Force, but NO WORK is being done ON the object, since it is not moving
7/26
The person might be exerting a Force, but NO WORK is being done ON the object, since it is not moving
We can just as easily replace the person with a metal pole, which exerts a force but expends no energy and does no work!
8/26
If the person is carrying the box horizontally, exerting a Force upward…
9/26
Further thoughts on WORK
No work is done on the box since the force is not in the direction of the motion…
10/26
Further thoughts on WORK
A force perpendicular to the displacement direction does no work on the object!
11/26
Even though the person is expending energy.
12/26
Work and Energy are Conserved
If we use a Lever to move a Box upward… we exert a Force downward…
The Load we are lifting is 40 N…
40 N
13/26
F = 10 N
So work Input is 10 N x 2.0 m = 20 J
2.0 m
40 N
0.5 m
Work and Energy are Conserved
The Box rises 0.5 m…
The Force was applied through a distance of 2.0 m…
14/26
Work Output = 20 J
Work Input = Work Output
Conservation of Energy (or Work)
It is plain that the Energy to do Work is conserved: “You never get more Work out of something than the amount of energy you put in.”
Think: What was the efficiency of this simple machine?
… Is this possible? Why/Why not?
… What was most likely the case, then?
15/26
• Below are some online simulations for you to explore that will help you gain a fuller understanding of Work and Energy concepts: [For the first two, just hit the Green “Run Now!” button, for the others hit “Play”].
• http://physics.bu.edu/~duffy/classroom.html [“U” is potential energy]
• http://physics.bu.edu/~duffy/classroom.html [again, “U” is potential energy, but in this case it is Gravitational PE instead of Elastic PE]
16/26
• I have put together an online lab-quest for you. Originally I made this with small lab groups in mind, but it is very easily done by individuals, as well. (The only difference is the level of discussion, for the most part).
• On the following slides I will lead you through the assignment. This assignment should not take you more than the amount of time it would take to finish a class. However, since it is being given as an online assignment you may use as much time as you need (until Wednesday’s class, at which point you should turn this in for a grade).
• Note that even though I gave you a link, these same files were uploaded to Blackboard yesterday (Sunday) for you.
17/26
• Print out (or organize a blank sheet for the responses) the following document:
• http://mrlafazia.com/labs/quests/EnergyTransformations/EnergyTransformations.doc
• For the text-files, I have written out mini-lessons (as if I were giving you a small “lecture”). You will then answer the “subquests” for each text-file.
• For the video files, I have written questions on the “Energy Transformations” document, itself (as opposed to just leaving blank spaces). You may have to download the files first to view them or type out the URLs or download different codices or video players.
• FIRST: Read Sections 5.7 – 5.12 in the text. (You have probably already read this. You may also want to supplement with online resources).
18/26
http://mrlafazia.com/labs/quests/EnergyTransformations/01.txt
Read the mini-lesson from the above text-file and answer the Sub-Quest questions (#’s 1-5)
19/26
http://mrlafazia.com/labs/quests/EnergyTransformations/02.asx
Watch the video-clip at the above link and answer the questions which are on the word document.
20/26
http://mrlafazia.com/labs/quests/EnergyTransformations/03.txt
Read the mini-lesson from the above text-file and answer the Sub-Quest questions (#’s 1-4)
21/26
http://mrlafazia.com/labs/quests/EnergyTransformations/04.asx
Watch the video-clip at the above link and answer the questions which are on the word document.
22/26
http://mrlafazia.com/labs/quests/EnergyTransformations/05.asx
Watch the video-clip at the above link and answer the questions which are on the word document.
23/26
http://mrlafazia.com/labs/quests/EnergyTransformations/06.txt
Read the mini-lesson from the above text-file and answer the Sub-Quest questions (#’s 1-3)
24/26 |
SSAT Middle Level Math : How to find a ratio
Example Questions
Example Question #1 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c
Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?
Explanation:
We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:
Reduce.
We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:
Now, we can create a proportion using our two ratios.
Cross multiply and solve for .
Simplify.
Divide both sides of the equation by .
Solve.
There are red cars in the parking lot.
Example Question #1 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c
Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?
Explanation:
We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:
Reduce.
We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:
Now, we can create a proportion using our two ratios.
Cross multiply and solve for .
Simplify.
Divide both sides of the equation by .
Solve.
There are red cars in the parking lot.
Example Question #2 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c
Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?
Explanation:
We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:
Reduce.
We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:
Now, we can create a proportion using our two ratios.
Cross multiply and solve for .
Simplify.
Divide both sides of the equation by .
Solve.
There are red cars in the parking lot.
Example Question #1 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c
Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?
Explanation:
We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:
Reduce.
We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:
Now, we can create a proportion using our two ratios.
Cross multiply and solve for .
Simplify.
Divide both sides of the equation by .
Solve.
There are red cars in the parking lot.
Example Question #1 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c
Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?
Explanation:
We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:
Reduce.
We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:
Now, we can create a proportion using our two ratios.
Cross multiply and solve for .
Simplify.
Divide both sides of the equation by .
Solve.
There are red cars in the parking lot.
Example Question #2 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c
Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?
Explanation:
We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:
Reduce.
We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:
Now, we can create a proportion using our two ratios.
Cross multiply and solve for .
Simplify.
Divide both sides of the equation by .
Solve.
There are red cars in the parking lot.
Example Question #1 : Find A Percent Of A Quantity As A Rate Per 100: Ccss.Math.Content.6.Rp.A.3c
Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?
Explanation:
We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:
Reduce.
We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:
Now, we can create a proportion using our two ratios.
Cross multiply and solve for .
Simplify.
Divide both sides of the equation by .
Solve.
There are red cars in the parking lot.
Example Question #71 : Ratios & Proportional Relationships
Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?
Explanation:
We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:
Reduce.
We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:
Now, we can create a proportion using our two ratios.
Cross multiply and solve for .
Simplify.
Divide both sides of the equation by .
Solve.
There are red cars in the parking lot.
Example Question #1 : Solving Word Problems With One Unit Conversions
A carpenter is making a model house and he buys of crown moulding to use as accent pieces. He needs of the moulding for the house. How many feet of the material does he need to finish the model?
Explanation:
We can solve this problem using ratios. There are in . We can write this relationship as the following ratio:
We know that the carpenter needs of material to finish the house. We can write this as a ratio using the variable to substitute the amount of feet.
Now, we can solve for by creating a proportion using our two ratios.
Cross multiply and solve for .
Simplify.
Divide both sides by .
Solve.
Reduce.
The carpenter needs of material.
Example Question #81 : Ratios & Proportional Relationships
A carpenter is making a model house and he buys of crown moulding to use as accent pieces. He needs of the moulding for the house. How many feet of the material does he need to finish the model?
Explanation:
We can solve this problem using ratios. There are in . We can write this relationship as the following ratio:
We know that the carpenter needs of material to finish the house. We can write this as a ratio using the variable to substitute the amount of feet.
Now, we can solve for by creating a proportion using our two ratios.
Cross multiply and solve for .
Simplify.
Divide both sides by .
Solve.
The carpenter needs of material. |
# Thread: Number of ways in which 7 distinct objects can be put..?
1. ## Number of ways in which 7 distinct objects can be put..?
Find the number of ways in which 7 distinct objects can be put in three identical boxes so that no box remains empty.
2. This is similar to a problem i had in an old course where you had to find integers x,y,z s.t. x + y + z = some +ve integer.
So i think this one can be reworded as the 3 boxes being x,y and z. And n being 7 distinct objects.
Since you want +ve integers what to do is set x' = x-1 etc and then you get x' + y' + z' = 4.
This is done by using the binomial expression $\left(\begin{array}{cc}n + k-1\\k-1\end{array}\right)$ where k is the number of 'boxes'
So we get $\left(\begin{array}{cc}4+3-1\\3-1\end{array}\right) = \left(\begin{array}{cc}6\\2\end{array}\right)$ = 15.
This is similar to a problem i had in an old course where you had to find integers w,x,y,z s.t. w + x + y + z = some +ve integer.
So i think this one can be reworded as the 4 boxes being w,x,y and z. And n being 7 distinct objects.
Since you want +ve integers what to do is set w' = w-1 etc and then you get w + x + y + z = 5.
This is done by using the binomial expression $\left(\begin{array}{cc}n + k-1\\k-1\end{array}\right)$ where k is the number of 'boxes'
So we get $\left(\begin{array}{cc}5+4-1\\4-1\end{array}\right) = \left(\begin{array}{cc}8\\3\end{array}\right)$ = 56.
its $3$ boxes.
4. Ah grim. Fixed!
5. Actually my first post was totally full of fail thanks for noticing that boxes thing...
6. The answer is 322. Are we missing something?
7. Yeah i think my way was if there was 7 identical objects... 15 is waaaay to low.
But how did you get 322? I just did a 'brute force' method and came up about 99 but didnt really check it.
Since each box must contain 1 object it can be simplified down to how many ways can 4 objects be distributed between 3 boxes... This questions bugging me now...
EDIT: Is the answer just $3^4 = 81$?
8. Well,this is how it goes.
Since the boxes are identical the problem reduces to the fact $7$ distinct objects have to be divided(not distributed) into $3$ groups as arrangement among the boxes are immaterial since the boxes are identical
The groupings can be done as follows $1,3,3),(1,2,4),(2,2,3),(1,1,5)" alt="1,3,3),(1,2,4),(2,2,3),(1,1,5)" />
This can be done in following ways:
= $\frac{7!}{1!3!3!}\frac{1}{2!}+\frac{7!}{1!2!4!}+\f rac{7!}{2!2!3!}\frac{1}{2!}+\frac{7!}{1!1!5!}\frac {1}{2!}$
$=70+105+105+21$
$
=301
$
Either the answer is 301 or I have missed a grouping which is not likely
9. Originally Posted by pankaj
Well,this is how it goes.
Since the boxes are identical the problem reduces to the fact $7$ distinct objects have to be divided(not distributed) into $3$ groups as arrangement among the boxes are immaterial since the boxes are identical
The groupings can be done as follows $1,3,3),(1,2,4),(2,2,3),(1,1,5)" alt="1,3,3),(1,2,4),(2,2,3),(1,1,5)" />
This can be done in following ways:
= $\frac{7!}{1!3!3!}\frac{1}{2!}+\frac{7!}{1!2!4!}+\f rac{7!}{2!2!3!}\frac{1}{2!}+\frac{7!}{1!1!5!}\frac {1}{2!}$
$=70+105+105+21$
$
=301
$
Either the answer is 301 or I have missed a grouping which is not likely
What about $(1,4,2)$? Or $(1,5,1)$? Also why did you divide by $2!$?
10. 301 is correct.
The number is also $S(7,3)$, where S is the Stirling Number of the 2nd Kind.
11. Originally Posted by Sampras
What about $(1,4,2)$? Or $(1,5,1)$? Also why did you divide by $2!$?
(1,2,4) is same as (1,4,2) since boxes are identical.There is no such thing as the first box or the second or the third .
If 2 groups contain same number of objects then it is required that we divide by 2!.
For example if {a,b,c,d,e,f,g} are the 7 obects.One grouping containing (2,2,3) objects can be described as follows.
{a,b},{c,d}{e,f,g} and another can be {c,d},{a,b}{e,f,g}.
Now arent these groupings same. |
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# Compound Inequality
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In mathematics, it can be useful limit the solution or even have multiple solutions for an inequality. For this we use a compound inequality, inequalities with multiple inequality signs. When solving compound inequalities, we use some of the same methods used in solving multi-step inequalities. The solutions to compound inequalities can be graphed on a number line, and can be expressed as intervals.
Solving a three part inequality, basically it's the same as solving a two part inequality but instead of now dealing with just one side we're dealing with another side as well. So first step is always just to get our x's by itself, subtract one over we're trying to solve for this x so just subtract 1 and instead of doing it from both sides we now have to do it from all three. So 11 minus 1, 10 less than 3x and then lastly it equal to 6 okay. Solving for x, we need to divide by negative 3, remember when we are in an inequality form if we divide by a negative we have to actually flip that sign. So 10 divided by a negative 3 we can't simplify that so we're just left with negative 10 thirds our sign switches x this sign has to switch as well 6 divided by negative 3 is negative 2 okay.
So what we actually have is x has to be between negative 10 thirds and -2, so this answer right now is in inequality form okay. We want to throw it into another form we could do our brackets and remember this is called "interval noation" we are not including negative ten thirds so we have a soft bracket here we are including -2 so we have this hard bracket here okay. We could also plot this on a number line we are going from negative ten thirds to negative 2, including negative 2 so that gets filled in not including ten thirds so that doesn't, filling an outline okay. Set bill notation again just taking the same exact thing here with a x such that bracket, x such that negative ten thirds is less than x is less than equal to -2. So solved it out and the four different ways of representing the same exact answer.
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## Intermediate Algebra (12th Edition)
$5 \text{ units}$
$\bf{\text{Solution Outline:}}$ Use the Distance Formula to find the distance between the given points $\left( 4.7,2.3 \right)$ and $\left( 1.7,-1.7 \right)$. $\bf{\text{Solution Details:}}$ With the given points, then $x_1= 4.7 ,$ $x_2= 1.7 ,$ $y_1= 2.3 ,$ and $y_2= -1.7 .$ Using the Distance Formula which is given by $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} ,$ then \begin{array}{l}\require{cancel} d=\sqrt{(4.7-1.7)^2+(2.3-(-1.7))^2} \\\\ d=\sqrt{(4.7-1.7)^2+(2.3+1.7)^2} \\\\ d=\sqrt{(3)^2+(4)^2} \\\\ d=\sqrt{9+16} \\\\ d=\sqrt{25} \\\\ d=\sqrt{(5)^2} \\\\ d=5 .\end{array} Hence, the distance is $5 \text{ units} .$ |
# What is 125/110 as a decimal?
## Solution and how to convert 125 / 110 into a decimal
125 / 110 = 1.136
125/110 converted into 1.136 begins with understanding long division and which variation brings more clarity to a situation. Both represent numbers between integers, in some cases defining portions of whole numbers In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. If we need to convert a fraction quickly, let's find out how and when we should.
## 125/110 is 125 divided by 110
Converting fractions to decimals is as simple as long division. 125 is being divided by 110. For some, this could be mental math. For others, we should set the equation. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. We use this as our equation: numerator(125) / denominator (110) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. This is our equation:
### Numerator: 125
• Numerators are the parts to the equation, represented above the fraction bar or vinculum. 125 is one of the largest two-digit numbers you'll have to convert. 125 is an odd number so it might be harder to convert without a calculator. Values closer to one-hundred make converting to fractions more complex. Now let's explore X, the denominator.
### Denominator: 110
• Denominators differ from numerators because they represent the total number of parts which can be found below the vinculum. 110 is one of the largest two-digit numbers to deal with. And it is nice having an even denominator like 110. It simplifies some equations for us. Overall, two-digit denominators are no problem with long division. Next, let's go over how to convert a 125/110 to 1.136.
## Converting 125/110 to 1.136
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 110 \enclose{longdiv}{ 125 }$$
We will be using the left-to-right method of calculation. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Solve for how many whole groups you can divide 110 into 125
$$\require{enclose} 00.1 \\ 110 \enclose{longdiv}{ 125.0 }$$
Since we've extended our equation we can now divide our numbers, 110 into 1250 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiply by the left of our equation (110) to get the first number in our solution.
### Step 3: Subtract the remainder
$$\require{enclose} 00.1 \\ 110 \enclose{longdiv}{ 125.0 } \\ \underline{ 110 \phantom{00} } \\ 1140 \phantom{0}$$
If you don't have a remainder, congrats! You've solved the problem and converted 125/110 into 1.136 If you still have a remainder, continue to the next step.
### Step 4: Repeat step 3 until you have no remainder
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 125/110 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. And the same is true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But they all represent how numbers show us value in the real world. Here are just a few ways we use 125/110, 1.136 or 113% in our daily world:
### When you should convert 125/110 into a decimal
Investments - Comparing currency, especially on the stock market are great examples of using decimals over fractions.
### When to convert 1.136 to 125/110 as a fraction
Carpentry - To build a table, you must have the right measurements. When you stretch the tape measure across the piece of wood, you won't see 10.6 inches. You'll see a tick mark at 10 and 3/5 inches.
### Practice Decimal Conversion with your Classroom
• If 125/110 = 1.136 what would it be as a percentage?
• What is 1 + 125/110 in decimal form?
• What is 1 - 125/110 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 1.136 + 1/2? |
# Construct a Parallelogram – Explanation and Examples
Constructing a parallelogram using a compass and straightedge means we construct a space enclosed by two pairs of parallel lines.
A parallelogram will have opposite sides of equal length that are parallel to each other. Rhombi and squares are specific types of parallelograms.
To do this, we must be able to construct parallel lines. Make sure to review this topic and general rules for construction before moving on.
In this section, we will cover:
• How to Construct a Parallelogram
• How to Construct a Parallelogram Given Two Sides
## How to Construct a Parallelogram
Recall that we can construct a line parallel to another line. To do this, we use the fact that two lines are parallel if and only if a transversal makes the alternate angles equal. In this case, a transversal is a line that cuts through both of the other lines. Alternate angles, then, are the interior angles on opposite sides.
Since these alternate angles are equal, we can construct a line that crosses a given line. This will be our transversal. Then, we copy the angle formed by the transversal and our original line. We can paste this angle so that it is upside down and on the opposite side of the transversal line. If we extend the leg of the angle that is not the transversal, we have a line parallel to our original line.
In the figure above, the line segment TC is parallel to the line AB. The transversal is the line EJ, and the alternate angles are EDF and TCJ.
This figure also shows that to copy an angle, we construct a triangle on the angle. Then, since only one triangle can be created using any three side lengths, we copy this triangle. After that, we know that the new triangle’s corresponding angle will be equal to the angle we wanted to copy.
We can use this to construct a parallelogram.
### What is a Parallelogram
A parallelogram is a type of quadrilateral or four-sided closed figure created with straight lines. A parallelogram has two sets of parallel sides by definition. A rhombus is a parallelogram where all four sides have the same length. Similarly, a rectangle is a parallelogram where all sides have right angles. A square, then, is a parallelogram with right angles and four equal sides.
Parallelograms have some other interesting features. Their opposite angles are equal in measure, and their opposite sides are equal in length.
### How to Construct a Parallelogram Given Two Sides
If we are given two adjacent sides of a parallelogram, say AB and BC, we have to construct a line parallel to AB that intersects C and a line parallel to BC that intersects A.
We can then call the intersection of these two new lines D. The quadrilateral ABCD will be a parallelogram because its opposite sides are parallel.
If we are instead given two parallel lines, we can easily construct a parallelogram. Note, however, that ABCD is a unique parallelogram that uses AB and BC in the situation outlined above. There are infinitely many parallelograms that can be constructed given two parallel lines.
All we have to do is pick a random point on the first parallel line. Let’s call it E. Then, connect A to a random point on the second parallel line, F.
Next, construct any line parallel to EF. We’ll call this line GH. EFGH will also be a parallelogram.
#### Using Congruent Triangles to Construct a Parallelogram
We can also construct a parallelogram using congruent triangles. If we connect two opposite angles of a parallelogram with a line segment, we actually divide the parallelogram into two congruent triangles.
Thus, if we have a triangle, we can copy the triangle so that it is essentially reflected over one of its sides. The new triangle will share a side with the old one and be the mirror image, as shown.
## Examples
In this section we will go over common examples of problems involving parallelograms and their step-by-step solutions.
### Example 1
Construct a parallelogram on the line segment AB.
### Example 1 Solution
Since we are only given one line segment to work with, we have a lot of freedom in constructing a parallelogram. In fact, there are infinitely many parallelograms that we can construct that have a base AB.
We’ll begin by extending the line AB in both directions.
Next, we need to select another vertex for our parallelogram, C, somewhere in the plane that is not on the line AB.
Now, we need to construct a line segment, BC. To keep the construction clear, we will also label a point to the right of A on the line AB as A’ and a point to B’s right on the line AB B’.
Now, since we know that opposite sides of parallelograms are parallel and equal in length, our goal is to construct a segment DC that is parallel to AB and equal in length to construct a parallelogram ABCD. Note, however, that we could have chosen instead to connect the segment AC. In this case, we could have constructed a segment CD’ parallel to AB and equal in length to construct a parallelogram ABD’C. This parallelogram would have had the angle CAB smaller than the corresponding angle DAB in the parallelogram ABCD.
Our first step, then, is to construct a line parallel to AB that goes through C. To do this, we must copy the angle CBB’ so that it becomes BCE.
#### Copy the Angle
Recall that we do this by constructing a triangle on the angle CBB’. In this case, let’s connect C and B’.
Now, we copy the triangle onto the segment CB so that it CBB’ → BCE.
We can now extend CB to an infinite line. We cut off a segment equal to BB’ with endpoint C pointing away from B to a point F. Then, we cut off a segment equal in length to CB’ with endpoint B pointing away from C to a point J.
Then, we can create two circles. One will have center C and radius CF, while the other will have a center B and radius JB. Label the intersection of these two circles as I, then draw the line CI.
The lines AB and CI are parallel. Now, we need to cut off a segment of CI with length AB and endpoint C. Call this segment DC.
Now, we can connect AD. The figure ABCD will be a parallelogram.
### Example 2
Given the two line segments AB and BC, construct a parallelogram.
### Example 2 Solution
As before, we extend the lines AB and BC. Label the point D on the line BC on the opposite side of C.
Now, copy the angle DBA so that A lines up with B, B lines up with A, and D lines up with a new point E. Then, extend the line AE.
AE is parallel to BC, so we need to cut off a segment equal in length to BC with A as an endpoint. Call this new line segment AD.
Now, we can connect DC. This creates the parallelogram ABCD as required. |
# Grade 12 Math Practice Test
Grade 12 math practice test questions are presented along with their solutions on videos.
1. Solve the inequality and present the solution set using intervals, number line and inequality symbols.
Solution on video at rational inequality, question 1
2. Solve the equation
Solution on video at trigonometric equation, question 2
3. Solve the equation
Solution on video at equation with logarithms, question 3
4. Solve the equation
Solution on video at equation with exponentials, question 4
5. Verify the identity :
Solution on video at verify trigonometric identities, question 5
6. Find the exact value of : $\displaystyle \quad \tan \left(\frac{13\pi}{12}\right)$
Solution on video at find exact value of trigonometric function, question 6
7. When polynomial $P(x)$ is divided by $x + 1$, the remainder is equal to $4$ and when $P(x)$ is divided by $x - 2$ it gives a remainder equal to $4$. Polynomial $p(x)$ has degree $3$ and has $x - 1$ as a factor. The leading coefficient of $P(x)$ is equal to $1$. Find $P(x)$
Solution on video at find polynomial given remainders and a factor, question 7
8. Function $f$ is defined by $f(x) = - x^4 - 5x^3 - 3x^2+9x$
a) Factor $f(x)$ completely.
b) Use the zeros to sketch the graph of $f$.
Solution on video at Factor completeley and sketch a polynomial, question 8
9. Find the equation of the polynomial function $g$ whose degree is equal to $4$ and whose graph is shown below and touches (does not cut) the x-axis at $x = -1$.
Solution on video at Find the equation of a polynomial given its graph, question 9
10. For the function $y = - 0.5 \sin \left( 4(x+\frac{\pi}{16}) \right) + 2.5$, make a table of values over 1 period and sketch the graph over 2 periods.
Solution on video at make a table of values and sketch , question 10
11. The velocity $V$ in meters ( $m$ ) of an object is given by the graph below. Write $V$ as a function of time $t$ in seconds ( $s$ ) as a cosine function.
Solution on video at find an equation to a trigonometric equation given by its graph , question 11
12. Given the function $y = \frac{2 x - 4}{x+2}$
a) Find the domain of the function
b) Find the x and y intercepts of the graph of the function
c) Find the equations of all the asymptotes of the function and any intercepts with the graph of the function
d) Make a table of signs and sketch the graph of the function
Solution on video at sketch the graph of the rational function y = (2x - 4) / (x + 2) , question 12
13. Given the function $y = \frac{x^2-9}{x+2}$
a) Find the domain of the function
b) Find the x and y intercepts of the graph of the function
c) Find the equations of all the asymptotes of the function
d) Make a table of signs and sketch the graph of the function
14. Find the equation of the rational function $h(x)$ whose graph is shown below and whose denominator has a polynomial of degree 2.
AN: (2x-4)/[(x-1)(x-2)] with hole
15. Given the function $f(x) = -0.5 \log_2(x^2 - 1)-1$
a) Find the domain of the function
b) Find the x and y intercepts, if any, of the graph of the function
c) Find the equations of all the asymptotes, if any, of the function
d) Make a table of values and sketch the graph of the function
16. Given the function $h(x) = 2 + e^{(x-2)}$
a) Find the domain of the function
b) Find the x and y intercepts, if any, of the graph of the function
c) Find the equations of all the asymptotes, if any, of the function
d) Make a table of values and sketch the graph of the function
17. Given the function $h(x) = \ln (2x - 1) + 2$
a) Find the domain and range of function $h$.
b) Find the inverse of function $h$ and specify its domain and range. |
# Do you Multiply the diameter by pi to find the circumference?
## Do you Multiply the diameter by pi to find the circumference?
Use the formula C = πd to find the circumference if you know the diameter. In this equation, “C” represents the circumference of the circle, and “d” represents its diameter. That is to say, you can find the circumference of a circle just by multiplying the diameter by pi.
## Does the circumference change when the diameter is multiplied?
If the diameter is doubled (say the original length was 2, now the length is 4), the circumference will also double (the original circumference is 2π , now the circumference is 4π ).
## Why do we multiply the diameter by pi to find the circumference?
We’ve learned that the circumference is the distance around the circle and the diameter is the distance across the circle going through the center. Both are parts of a circle. The circumference formula, C = pi*d, tells you how the circumference and diameter are related.
## How does the circumference change if the circle is multiplied by 10?
How does the circumference change? A It is divided by 10. It does not change.
## What is the diameter of a circle multiplied by to get the circumference?
π
Multiply the diameter by π, or 3.14. The result is the circle’s circumference.
## Why is pi used to find circumference?
Succinctly, pi—which is written as the Greek letter for p, or π—is the ratio of the circumference of any circle to the diameter of that circle. Regardless of the circle’s size, this ratio will always equal pi. Hence, it is useful to have shorthand for this ratio of circumference to diameter.
## What is π circumference of a circle?
Circles are all similar, and “the circumference divided by the diameter” produces the same value regardless of their radius. This value is the ratio of the circumference of a circle to its diameter and is called π (Pi).
## How do you find the circumference of a circle using PI?
Given the radius or diameter and pi you can calculate the circumference. The diameter is the distance from one side of the circle to the other at its widest points. The diameter will always pass through the center of the circle. The radius is half of this distance.
## How to calculate circumference to diameter?
Circumference to Diameter Calculator. Circumference of a circle is defined as the distance around it. It is calculated just by multiplying the diameter of the circle with π value. The diameter of a circle is known as the straight line segment which passes through the center of the circle. This is also known as the longest chord of the circle.
## How do you find the constant pi?
The constant pi, designated by the Greek letter π, is the ratio of the circumference to the diameter of a circle. For any circle, if you divide the circumference by the diameter you get pi, an irregular number usually rounded to 3.14. where C = circumference, π = 3.14 and d = diameter.
## How do you find the radius of a circle in math?
How do I find the radius from the circumference? To find the radius from the circumference of a circle, you have to do the following: Divide the circumference by π, or 3.14 for an estimation. The result is the circle’s diameter. Divide the diameter by 2. There you go, you found the circle’s radius. |
# Time and Work: Techniques and examples with solutions Today I'm going to discuss a very important topic i.e. Time and Work of quantitative aptitude. In almost every exam at least 2-3 question are asked every time. In this chapter, I will tell you about a definite relationship between Time and work and easy method to solve the problems.
Question
Time and Work: Techniques and examples with solutions Today I'm going to discuss a very important topic i.e. Time and Work of quantitative aptitude. In almost every exam at least 2-3 question are asked every time. In this chapter, I will tell you about a definite relationship between Time and work and easy method to solve the problems.
2021-04-21
## Three main factors of Time and Work
There is a definite relationship between Time and Work. In this concept, there are only three factors:
• Time taken to complete a certain job
• Unit of work done
• Number of persons doing the job
There is a fundamental relationship between these three, discussed as follows:
Work done(W) = Number of days(Time taken)(T or D) times Number of men(M)
W = D times M
### Some basic points
More number of men can do more work i.e. both are directly proportional
More number of men take less time to complete certain job i.e. both are inversely proportional
By summarizing, we get
(W_1)(W_2) = ((M_1)(M_2)) times ((D_1)(D_2))
Lets start solving some examples:
Example1: 10 men can cut 8trees in 16 days. In how many days can 6men cut 10trees?
Solution: This is a very simple example. You are given:
W_1 = 8
W_2 = 10
M_1 = 10
M_2 = 6
D_1 = 16
D_2 = ?
Using formula,
(W_1)(W_2) = ((M_1)(M_2)) times ((D_1)(D_2))
8(10) = (10)/6 times (16)(D_2)
⇒D_2 = 33.3
## Concept of efficiency
This means, "How much work one person can do in one day (expressed in percentage)"
For example: A person can do a job in 2days ⇒ He can do 50% work in one day Therefore, his efficiency will be 50%
### Just a 2-step concept
This concept involves two steps to calculate efficiency:
• Convert into fraction i.e. per day work
• Multiply with 100 i.e. convert into percentage
Try the following example first, then re-read above points
Example2: If a person can complete his work in 5 days. What will be his efficiency?
Solution: Number of days a person take to complete his work = 5
⇒ He is doing 1/5 th work per day (converted into fraction)
Convert it into percentage:
(100)/5 = 20%
Therefore, his efficiency is 20%.
Summarizing, If a person can do his job in n days, efficiency will be
Efficiency = (100)/n %
### Note: Negative efficiency cancels the positive efficiency
For Example: Positive efficiency = 5% Negative efficiency = 1.5% Net efficiency = 5- 1.5 = 3.5%
As we all know, in competitive exams time management is very important. I suggest you to learn the fractions till 15.
Number of days required to complete work Work/Day Efficiency (%) N 1/n 100/n 1 1 100 2 1/2 50 3 1/3 33.33 4 1/4 25 5 1/5 20 6 1/6 16.66 7 1/7 14.28 8 1/8 12.5 9 1/9 11.11 10 1/10 10 11 1/11 9.09 12 1/12 8.33 13 1/13 7.69 14 1/14 7.14 15 1/15 6.66
Example3: A can do a job in 10 days. B can do a job in 5 days. In how many days they can complete the job if they work together?
Solution: Consider the above table
A's efficiency = 10% ((100)(10))
B's efficiency = 20%
A+ B efficiency = 10 + 20 = 30%
This means, In one day A and B together can do 30% of work.
Therefore,Number of days A and B together take to do 100% of work = (100)(30)
⇒3.33 days
Example4: A and B together can do a job in 4 days. If A can do job in 12 days if he works alone, then how many days B alone take to complete the job?
Solution: A+B take = 4 days
⇒ A+B's efficiency = 25% i.e. they together do 25% of work in one day
A takes = 12 days
⇒ A's efficiency = 8.33%
B's efficiency = (A+B) - (A)
⇒ 25% - 8.33% = 16.66%
This means, B can do 16.66% work in one day
Therefore, to complete the job he will take = (100)(16.66) days
⇒ 6days
Example5: A and B can do job in 8 days. B and C can do same job in 12 days. A, B and C together can do same job in 6 days. In how many days A and C together can complete the job?
Solution: You are given that:
A+B's efficiency = 12.5%
B+C's efficiency = 8.33%
A+B+C's efficiency = 16.66%
we need to find A+C
Consider, 2(A+B+C) = (A+B) + (B+C) + (C+A)
⇒2(16.66) = 12.5+ 8.33 + (C+A)
⇒ C+A = 12.49 = 12.5%
Therefore, A and C takes= (100)(12.5) = 8 days
Hope you all understand this topic. I will soon update questions for your practice.
### Relevant Questions
Basic facts and techniques of Boats and Streams of Quantitative Aptitude Boats and Streams is a part of the Quantitative aptitude section. This is just a logical extension of motion in a straight line. One or two questions are asked from this chapter in almost every exam. Today I will tell you some important facts and terminologies which will help you to make better understanding about this topic.
Solved examples of number series in Quantitative aptitude As we know, questions related to number series are very important in Quantitative aptitude section, So, today I’m going to discuss some problems of number series. These are just for your practice. I have already discussed this chapter in previous session i.e. Sequence and Series. Read this article first, then go through these examples.
Tricks to solve problems related to Series in Quantitative Aptitude Today, I'm going to discuss a very important topic of Quantitative aptitude i.e. Sequence and Series. Sequence and Series is a mathematical concept and basically it is a logical concept.
Tricks to solve Ratio and Proportion Problems Today I'm going to discuss a very helpful trick of ratio and proportion of Quantitative Aptitude section. I'm sure this would be very helpful for you and this trick will save your more than half time, you generally take to solve the question. |
# How do you simplify \frac { 5^ { - 5} } { 5^ { 8} }?
##### 1 Answer
Jan 30, 2017
See the entire simplification process below:
#### Explanation:
The first rule of exponents we will use is:
${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$
${5}^{\textcolor{red}{- 5}} / {5}^{8} = \frac{1}{{5}^{\textcolor{red}{- - 5}} \cdot {5}^{8}} = \frac{1}{{5}^{5} \cdot {5}^{8}}$
The next rule of exponents to use is:
${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$
$\frac{1}{{5}^{\textcolor{red}{5}} \cdot {5}^{\textcolor{b l u e}{8}}} = \frac{1}{{5}^{\textcolor{red}{5} + \textcolor{b l u e}{8}}} = \frac{1}{5} ^ 13 = \frac{1}{1 , 220 , 703 , 125}$ |
# Search by Topic
#### Resources tagged with Odd and even numbers similar to Crossings:
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Broad Topics > Numbers and the Number System > Odd and even numbers
### Crossings
##### Age 7 to 11 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Multiplication Series: Number Arrays
##### Age 5 to 11
This article for teachers describes how number arrays can be a useful reprentation for many number concepts.
### Number Tracks
##### Age 7 to 11 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### Always, Sometimes or Never?
##### Age 5 to 11 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### What Do You Need?
##### Age 7 to 11 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### A Mixed-up Clock
##### Age 7 to 11 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### The Thousands Game
##### Age 7 to 11 Challenge Level:
Each child in Class 3 took four numbers out of the bag. Who had made the highest even number?
### Number Differences
##### Age 7 to 11 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Curious Number
##### Age 7 to 11 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### Becky's Number Plumber
##### Age 7 to 11 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Square Subtraction
##### Age 7 to 11 Challenge Level:
Look at what happens when you take a number, square it and subtract your answer. What kind of number do you get? Can you prove it?
##### Age 11 to 14 Challenge Level:
Great Granddad is very proud of his telegram from the Queen congratulating him on his hundredth birthday and he has friends who are even older than he is... When was he born?
### Break it Up!
##### Age 5 to 11 Challenge Level:
In how many different ways can you break up a stick of 7 interlocking cubes? Now try with a stick of 8 cubes and a stick of 6 cubes.
### More Carroll Diagrams
##### Age 7 to 11 Challenge Level:
How have the numbers been placed in this Carroll diagram? Which labels would you put on each row and column?
### Three Spinners
##### Age 7 to 11 Challenge Level:
These red, yellow and blue spinners were each spun 45 times in total. Can you work out which numbers are on each spinner?
### Always, Sometimes or Never? Number
##### Age 7 to 11 Challenge Level:
Are these statements always true, sometimes true or never true?
### Play to 37
##### Age 7 to 11 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
### Sets of Four Numbers
##### Age 7 to 11 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
### Diagonal Trace
##### Age 7 to 11 Challenge Level:
You can trace over all of the diagonals of a pentagon without lifting your pencil and without going over any more than once. Can the same thing be done with a hexagon or with a heptagon?
### More Carroll Diagrams
##### Age 7 to 11 Challenge Level:
How have the numbers been placed in this Carroll diagram? Which labels would you put on each row and column?
### Odds and Threes
##### Age 7 to 11 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Sets of Numbers
##### Age 7 to 11 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### Picturing Square Numbers
##### Age 11 to 14 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### Down to Nothing
##### Age 7 to 11 Challenge Level:
A game for 2 or more people. Starting with 100, subratct a number from 1 to 9 from the total. You score for making an odd number, a number ending in 0 or a multiple of 6.
### Seven Flipped
##### Age 7 to 11 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Magic Crosses
##### Age 7 to 14 Challenge Level:
Can you find examples of magic crosses? Can you find all the possibilities?
### Magic Letters
##### Age 11 to 14 Challenge Level:
Charlie has made a Magic V. Can you use his example to make some more? And how about Magic Ls, Ns and Ws?
### Take Three Numbers
##### Age 7 to 11 Challenge Level:
What happens when you add three numbers together? Will your answer be odd or even? How do you know?
### Magic Vs
##### Age 7 to 11 Challenge Level:
Can you put the numbers 1-5 in the V shape so that both 'arms' have the same total?
### Carroll Diagrams
##### Age 5 to 11 Challenge Level:
Use the interactivities to fill in these Carroll diagrams. How do you know where to place the numbers?
### Make 37
##### Age 7 to 14 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### Venn Diagrams
##### Age 5 to 11 Challenge Level:
Use the interactivities to complete these Venn diagrams.
### Red Even
##### Age 7 to 11 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Odd Squares
##### Age 7 to 11 Challenge Level:
Think of a number, square it and subtract your starting number. Is the number you’re left with odd or even? How do the images help to explain this?
### Arrangements
##### Age 7 to 11 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### Number Detective
##### Age 5 to 11 Challenge Level:
Follow the clues to find the mystery number.
### Score
##### Age 11 to 14 Challenge Level:
There are exactly 3 ways to add 4 odd numbers to get 10. Find all the ways of adding 8 odd numbers to get 20. To be sure of getting all the solutions you will need to be systematic. What about. . . .
### Take One Example
##### Age 5 to 11
This article introduces the idea of generic proof for younger children and illustrates how one example can offer a proof of a general result through unpacking its underlying structure.
### Odds, Evens and More Evens
##### Age 11 to 14 Challenge Level:
Alison, Bernard and Charlie have been exploring sequences of odd and even numbers, which raise some intriguing questions...
### Part the Piles
##### Age 7 to 11 Challenge Level:
Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy?
### Various Venns
##### Age 7 to 11 Challenge Level:
Use the interactivities to complete these Venn diagrams.
### Prime Magic
##### Age 7 to 16 Challenge Level:
Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find?
### Impossible Sandwiches
##### Age 11 to 18
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
### Odds and Evens
##### Age 11 to 14 Challenge Level:
Is this a fair game? How many ways are there of creating a fair game by adding odd and even numbers? |
Paul's Online Notes
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Section 7.4 : More on the Augmented Matrix
3. For the following system of equations convert the system into an augmented matrix and use the augmented matrix techniques to determine the solution to the system or to determine if the system is inconsistent or dependent.
\begin{align*}3x + 9y & = - 6\\ - 4x - 12y & = 8\end{align*}
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Start Solution
The first step is to write down the augmented matrix for the system of equations.
$\left[ {\begin{array}{rr|r}3&9&{ - 6}\\{ - 4}&{ - 12}&8\end{array}} \right]$ Show Step 2
We need to make the number in the upper left corner a one. In this case we can quickly do that by dividing the top row by 3.
$\left[ {\begin{array}{rr|r}3&9&{ - 6}\\{ - 4}&{ - 12}&8\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{3}{R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&3&{ - 2}\\{ - 4}&{ - 12}&8\end{array}} \right]$ Show Step 3
Next, we need to convert the -4 below the 1 into a zero and we can do that with the following elementary row operation.
$\left[ {\begin{array}{rr|r}1&3&{ - 2}\\{ - 4}&{ - 12}&8\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} + 4{R_{\,1}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&3&{ - 2}\\0&0&0\end{array}} \right]$ Show Step 4
The minute we see the bottom row of all zeroes we know that the system if dependent. We can convert the top row into an equation and solve for $$x$$ as follows,
$x + 3y = - 2\hspace{0.25in} \to \hspace{0.25in}x = - 3y - 2$
From this we can write the solution as,
$\begin{array}{l}{x = - 3t - 2}\\{y = t}\end{array}\hspace{0.25in}t{\mbox{ is any number}}$ |
## Practice Problem Set 5 – bivariate normal distribution
This post provides practice problems to reinforce the concept of bivariate normal distribution discussed in two posts – one is a detailed introduction to bivariate normal distribution and the other is a further discussion that brings out more mathematical properties of the bivariate normal distribution. The properties discussed in these two posts form the basis for the calculation behind the practice problems presented here.
The practice problems presented here are mostly on calculating probabilities. The normal probabilities can be obtained using a normal table or a calculator that has a function for normal distribution (such as TI84+). The answers for normal probabilities given at the end of the post have two versions – one using a normal table (found here) and the other one using TI84+.
.
Practice Problem 5-A Suppose that $X$ and $Y$ follow a bivariate normal distribution with parameters $\mu_X=15$, $\sigma_X=4$, $\mu_Y=20$, $\sigma_Y=5$ and $\rho=-0.7$. Determine the following. Compute the probability $P[12 For $X=20$, determine the mean and standard deviation of the conditional distribution of $Y$ given $X=20$. Determine $P[12, the probability that $12 given $X=20$.
.
Practice Problem 5-B Suppose that $X$ and $Y$ follow a bivariate normal distribution with parameters $\mu_X=6$, $\sigma_X=1.6$, $\mu_Y=4$, $\sigma_Y=1.2$ and $\rho=0.8$. Determine the following. Compute the probability $P[3 Determine $E[Y \lvert X=x]$, the mean of the conditional distribution of $Y$ given $X=x$. Determine $\sigma_{Y \lvert x}^2=Var[Y \lvert X=x]$ and $\sigma_{Y \lvert x}$, the variance and the standard deviation of the conditional distribution of $Y$ given $X=x$. For each of the $x$ values 6, 8, 10 and 12, determine the 99.7% interval $(a,b)$ for the conditional distribution of $Y$ given $x$, i.e. $a$ is three standard deviations below the mean and $b$ is 3 standard deviations above the mean. For each of the $x$ values 6, 8, 10 and 12, determine $P[3. Explain the magnitude of each of these probabilities based on the intervals in 6.
$\text{ }$
Practice Problem 5-C Let $X$ and $Y$ have a bivariate normal distribution with parameters $\mu_X=50$, $\sigma_X=10$, $\mu_Y=60$, $\sigma_Y=5$ and $\rho=0.6$. Determine the following. Calculate $P[100 Determine the 5 parameters of the bivariate normal random variables $L=X+Y$ and $M=X-Y$. Calculate $P[100
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Practice Problem 5-D Suppose $X$ is the height (in inches) and $Y$ is the weight (in pounds) of a male student in a large university. Furthermore suppose that $X$ and $Y$ follow a bivariate normal distribution with parameters $\mu_X=69$, $\mu_Y=155$, $\sigma_X=2.5$, $\sigma_Y=20$ and $\rho=0.55$. What is the distribution of the weights of all male students what are 5 feet 11 inches tall (71 inches)? For a randomly chosen male student who is 71 inches tall, what is the probability that his weight is between 170 and 200 pounds? For male students who are 71 inches tall, what is the 90th percentile of weight?
$\text{ }$
Practice Problem 5-E Suppose that $X$ and $Y$ have a bivariate normal distribution with parameters $\mu_X=70$, $\mu_Y=70$, $\sigma_X=5$, $\sigma_Y=10$ and $\rho>0$. Further suppose that $P[58.24. Determine $\rho$.
$\text{ }$
Practice Problem 5-F Suppose that $X$ and $Y$ have a bivariate normal distribution with parameters $\mu_X=70$, $\mu_Y=60$, $\sigma_X=10$, $\sigma_Y=12$ and $\rho=0.8$. Compute $P[45 When $X=60$, 4 values of $Y$ are observed. Compute $P[45<\overline{Y}<55 \lvert X=60]$ where $\overline{Y}$ is the mean of the sample of size 4.
$\text{ }$
Practice Problem 5-G Let $X$ and $Y$ have a bivariate normal distribution with parameters $\mu_X=70$, $\mu_Y=50$, $\sigma_X=10$, $\sigma_Y=12$ and $\rho=-0.65$. Determine the following. $P[X-Y<50]$ $\displaystyle P[55<\frac{X+Y}{2}<65]$
$\text{ }$
Practice Problem 5-H Let $X$ and $Y$ have a bivariate normal distribution with parameters $\mu_X=70$, $\sigma_X=5$, $\mu_Y=50$, $\sigma_Y=10$ and $\rho=0.75$. Determine the following probabilities. $P \biggl[ \frac{X+Y}{2}<68 \biggr]$ $P \biggl[ \frac{X+Y}{2}<68 \ \biggl \lvert Y=60 \biggr]$
$\text{ }$
Practice Problem 5-I For a couple from a large population of married couples, let $X$ be the height (in inches) of the husband and let $Y$ be the height (in inches) of the wife. Suppose that $X$ and $Y$ have a bivriate normal distribution with parameters $\mu_X=68$, $\mu_Y=65$, $\sigma_X=2.2$, $\sigma_Y=2.5$ and $\rho=0.5$. For a randomly selected wife from this population, determine the probability that her height is between 68 inches and 72 inches. For a randomly selected wife from this population whose husband is 72 inches tall, determine the probability that her height is between 68 inches and 72 inches. For a randomly selected couple from this population, determine the probability that the wife is taller than the husband.
$\text{ }$
Practice Problem 5-J The annual revenues of Company X and Company Y are positively correlated since the correlation coefficient between the two revenues is 0.65. The annual revenue of Company X is, on average, 4,500 with standard deviation 1,500. The annual revenue of Company Y is, on average, 5,500 with standard deviation 2,000. Calculate the probability that annual revenue of Company X is less than 6,800 given that the annual revenue of Company Y is 6,800. Calculate the probability that the annual revenue of Company X is greater than that of Company Y given that their total revenue is 12,000.
.
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
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$\text{ }$
5-A
1. $P[12 (table), 0.8904014212 (TI84+)
2. $E[Y \lvert X=20]=15.625$, $Var[Y \lvert X=20]=12.75$
3. $P[12 (table), 0.8447309876 (TI84+)
5-B
• $P[3 (table),0.5953433508 (TI84+)
• $\displaystyle E[Y \lvert x]=0.4+0.6 \ x$
• $\displaystyle Var[Y \lvert x]=0.5184$, standard deviation = 0.72
• For x = 6, (1.84, 6.16)
• For x = 8, (3.04, 7.36)
• For x = 10, (4.24, 8.56)
• For x = 12, (5.44, 9.76)
• $P[3 (table), 0.8351333522 (TI84+)
• $P[3 (table), 0.3894682472 (TI84+)
• $P[3 (table), 0.025919702 (TI84+)
• $P[3 (table), 0.0001524802646 (TI84+)
5-C
• $P[100 (table), 0.7551912515 (TI84+)
• $\displaystyle \mu_L=110 \ \ \ \sigma_L=\sqrt{185} \ \ \ \mu_M=-10 \ \ \ \sigma_M=\sqrt{65} \ \ \ \rho_{L,M}=\frac{75}{\sqrt{185} \sqrt{65}}$
• $P[100 (table), 0.8966089617 (TI84+)
5-D
• Normal with mean 163.8 and standard deviation $\sqrt{279}$.
• $P[170 (table), 0.3401418637 (TI84+)
• 90th percentile = 185.18 (table), 185.2061314 (TI84+)
5-E
• 0.8
5-F
• $P[45 (table), 0.5119251771 (TI84+)
• $P[45<\overline{Y}<55 \lvert X=60]=0.8329$ (table), 0.8325288097 (TI84+)
5-G
• $P[X-Y<50]=0.9332$ (table), 0.9331927713 (TI84+)
• $\displaystyle P[55<\frac{X+Y}{2}<65]=0.7154$ (table), 0.7135779177 (TI84+)
5-H
• $P \biggl[ \frac{X+Y}{2}<68 \biggr]=0.8708$ (table), 0.8710504336 (TI84+)
• $P \biggl[ \frac{X+Y}{2}<68 \ \biggl \lvert Y=60 \biggr]=0.7517$ (table), 0.7518542213 (TI84+)
5-I
• 0.1125 (table), 0.1125145409 (TI84+)
• 0.3523 (table), 0.3539664536 (TI84+)
• 0.1020 (table), 0.1022447094 (TI84+)
5-J
• 0.9279 (table), 0.9280950079 (TI84+)
• 0.1736 (table), 0.1736950626 (TI84+)
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## Calculating bivariate normal probabilities
This post extends the discussion of the bivariate normal distribution started in this post from a companion blog. Practice problems are given in the next post.
Suppose that the continuous random variables $X$ and $Y$ follow a bivariate normal distribution with parameters $\mu_X$, $\sigma_X$, $\mu_Y$, $\sigma_Y$ and $\rho$. What to make of these five parameters? According to the previous post, we know that
• $\mu_X$ and $\sigma_X$ are the mean and standard deviation of the marginal distribution of $X$,
• $\mu_Y$ and $\sigma_Y$ are the mean and standard deviation of the marginal distribution of $Y$,
• and finally $\rho$ is the correlation coefficient of $X$ and $Y$.
So the five parameters of a bivariate normal distribution are the means and standard deviations of the two marginal distributions and the fifth parameter is the correlation coefficient that serves to connect $X$ and $Y$. If $\rho=0$, then $X$ and $Y$ are simply two independent normal distributions.
When calculating probabilities involving a bivariate normal distribution, keep in mind that both marginal distributions are normal. Furthermore, the conditional distribution of one variable given a value of the other is also normal. Much more can be said about the conditional distributions.
The conditional distribution of $Y$ given $X=x$ is usually denoted by $Y \lvert X=x$ or $Y \lvert x$. In additional to being a normal distribution, it has a mean that is a linear function of $x$ and has a variance that is constant (it does not matter what $x$ is, the variance is always the same). The linear conditional mean and constant variance are given by the following:
$\displaystyle E[Y \lvert X=x]=\mu_Y+\rho \ \frac{\sigma_Y}{\sigma_X} \ (x-\mu_X)$
$\displaystyle Var[Y \lvert X=x]=\sigma_Y^2 \ (1-\rho^2)$
Similarly, the conditional distribution of $X$ given $Y=y$ is usually denoted by $X \lvert Y=y$ or $X \lvert y$. In additional to being a normal distribution, it has a mean that is a linear function of $x$ and has a variance that is constant. The linear conditional mean and constant variance are given by the following:
$\displaystyle E[X \lvert Y=y]=\mu_X+\rho \ \frac{\sigma_X}{\sigma_Y} \ (y-\mu_Y)$
$\displaystyle Var[X \lvert Y=y]=\sigma_X^2 \ (1-\rho^2)$
The information about the conditional distribution of $Y$ on $X=x$ is identical to the information about the conditional distribution of $X$ on $Y=y$, except for the switching of $X$ and $Y$. An example is helpful.
Example 1
Suppose that the continuous random variables $X$ and $Y$ follow a bivariate normal distribution with parameters $\mu_X=10$, $\sigma_X=10$, $\mu_Y=20$, $\sigma_Y=5$ and $\rho=0.6$. The first two parameters are the mean and standard deviation of the marginal distribution of $X$. The next two parameters are the mean and standard deviation of the marginal distribution of $Y$. The parameter $\rho$ is the correlation coefficient of $X$ and $Y$. Both marginal distributions are normal.
Let’s focus on the conditional distribution of $Y$ given $X=x$. It is normally distributed. Its mean and variance are:
\displaystyle \begin{aligned} E[Y \lvert X=x]&=\mu_Y+\rho \ \frac{\sigma_Y}{\sigma_X} \ (x-\mu_X) \\&=20+0.6 \ \frac{5}{10} \ (x-10) \\&=20+0.3 \ (x-10) \\&=17+0.3 \ x \end{aligned}
$\displaystyle \sigma_{Y \lvert x}^2=Var[Y \lvert X=x]=\sigma_Y^2 (1-\rho^2)=25 \ (1-0.6^2)=16$
$\displaystyle \sigma_{Y \lvert x}=4$
The line $y=17+0.3 \ x$ is also called the least squares regression line. It gives the mean of the conditional distribution of $Y$ given $x$. Because $X$ and $Y$ are positively correlated, the least squares line has positive slope. In this case, the larger the $x$, the larger is the mean of $Y$. The standard deviation of $Y$ given $x$ is constant across all possible $x$ values.
With mean and standard deviation known, we can now compute normal probabilities. Suppose the realized value of $X$ is 25. Then the mean of $Y \lvert 25$ is $E[Y \lvert 25]=24.5$. The standard deviation, as indicated above, is 4. In fact, for any other $x$, the standard deviation of $Y \lvert x$ is also 4. Now calculate the probability $P[20. We first calculate it using a normal table found here.
\displaystyle \begin{aligned} P[20
Using a TI84+ calculator, $P[20. In contrast, the probability $P[20 is (using the table found here):
\displaystyle \begin{aligned} P[20
Using a TI84+ calculator, $P[20. Note that $P[20 is for the marginal distribution of $Y$. It is not conditioned on any realized value of $X$.
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## Saturday, 19 November 2011
### NSS Mathematics: Co-geom techniques
This is a question from the latest test. A, B H are points (-6,0), (4,0), (0,10) respectively. C1 and C2 are circles with AO and BO as diameter respectively, AH and BH meet C1 and C2 at F and G respectively.
ai) Show OGHF concyclic.
ii) Show AFGB concyclic.
bi) Find coordinate of G.
ii) Find circle containing AFGB.
We will focus on part b as it includes all method that we can solve a problem related to circle in the co-geom plane.
bi) Find coordinate of G.
Method 1: Circle-line intersection
Recall the equation of C2: $(x-2)^2+y^2=4$
Equation of line BH $y=mx+c$, putting $(0,10), (4,0)$ we get $y=-\frac{5}{2}x+10$.
Putting equation of line BH into equation C2: $(x-2)^2+(-\frac{5}{2}x+10)^2=4$.
By simplification we have $\frac{29}{4}x^2-54x+100=\frac{1}{4}(x-4)(29x-100)=0$. (4,0) is B so x=4 is rejected. Then $x=\frac{100}{29}$, and we can easily get $y=(\frac{100}{29})(\frac{-5}{2})+10=\frac{40}{29}$.
Method 2: (the easiest one) perpendicular line interesection
Observe that OG is perpendicular to BH due to angles in semi-circle,
$m_{BH}=\frac{-5}{2}$, $m_{OG}=-(m_{BH})^{-1}=\frac{2}{5}$. Since OG passes through the origin, $OG:y=\frac{2}{5}x$.
The intersection between line BH and line OG:
$y=\frac{-5}{2}x+10=\frac{2}{5}x$, we can easily obtain $(x,y)=(\frac{100}{29},\frac{40}{29})$.
Method 3: perpendicular line-circle intersection
The intersection between OG and C2 might me a bit easier than intersection between BH and C2:
$(x-2)^2+y^2=(x-2)^2+(\frac{2}{5}x)^2=4$, x = 0 (rej. since it's O) and the same result as before.
Method 4: trigonmetry method
Observe the triangles HOB, HGO and OBG are similar. Let angle GOB = angle HOB be $\theta$ and the coordinate of G is $(OG_x,OG_y)$.
$OG_x=|OG|\cos \theta = |OB|\cos ^2 \theta$, similarly $OG_y=|OG|\sin \theta =|OB|\sin \theta \cos \theta$.
By definition we have $\cos \theta = \frac{10}{\sqrt{116}}$, $\sin \theta = \frac{4}{\sqrt{116}}$, by putting |OB| = 4 we have the same result.
(Note: |XY| is the length of line segment XY.)
bii) Find the equation of circle:
Method 1: general equation of circle
Assume the equation is $(x-x_0)^2+(y-y_0)^2=r^2$.
Putting point A and B we have $x_0=-1$, therefore the equation becomes $(x+1)^2+(y-y_0)^2=r^2$ and $25+y_0^2=r^2$.
Putting $(x,y)=(\frac{100}{29},\frac{40}{29})$, we have $(\frac{100}{29}+1)^2+(\frac{40}{29}-y_0)^2=r^2=25+y_0^2$
After a bunch of complex calculation (ugly number), we have $(x_0,y_0,r^2)=(-1,-1.2,26.44)$, therefore $(x+1)^2+(y+1.2)^2=26.44$ is the desired equation.
Method 2: Perpendicular bisector method
Recall the way you determine the circumcenter: it's the intersection point among three perpendicular bisectors. Therefore if four points are concyclic, choose two perpendicular bisector of them, then insection point is the center of circle. The radius can be easily determined by distance formula.
The perpendicular bisector of AB is trivially $x=-1$.
The perpendicular bisector of BG is given by the locus of P that $PG=PB$
$(x-\frac{100}{29})^2+(y-\frac{40}{29})^2=(x-4)^2+y^2$
$y=\frac{2}{5}x-\frac{4}{5}$
By putting x = -1, we have y = -6/5, and $OA^2=(-6+1)^2+(-6/5)^2=26.44$, and the same result is given.
Conclusion:
1) Change circle problems into linear problems whenever possible.
2) Trigonometry is powerful when there's perpendicular pair of lines.
3) Finding intersection between circles is stupid.
4) Find the equation of circle in terms of $x^2+y^2+ax+by+c=0$ is stupid.
5) Perpendicular bisector is our new method to find equation of circle.
fin. |
#### 4.1: Students understand the place value of whole numbers and decimals to two decimal places and how whole numbers and decimals relate to simple fractions.
4.1.3: Round whole numbers up to 10,000 to the nearest ten, hundred, and thousand.
4.1.4: Order and compare whole numbers using symbols for ?less than? (<), ?equal to? (=), and ?greater than? (>).
4.1.8: Write tenths and hundredths in decimal and fraction notations. Know the fraction and decimal equivalents for halves and fourths (e.g., 1/2 = 0.5 = 0.50, 7/4 = 1 3/4 = 1.75).
#### 4.2: Students solve problems involving addition, subtraction, multiplication, and division of whole numbers and understand the relationships among these operations. They extend their use and understanding of whole numbers to the addition and subtraction of simple fractions and decimals.
4.2.1: Understand and use standard algorithms for addition and subtraction.
4.2.2: Represent as multiplication any situation involving repeated addition.
4.2.3: Represent as division any situation involving the sharing of objects or the number of groups of shared objects.
4.2.5: Use a standard algorithm to multiply numbers up to 100 by numbers up to 10, using relevant properties of the number system.
4.2.6: Use a standard algorithm to divide numbers up to 100 by numbers up to 10 without remainders, using relevant properties of the number system.
4.2.9: Add and subtract decimals (to hundredths), using objects or pictures.
4.2.10: Use a standard algorithm to add and subtract decimals (to hundredths).
4.2.11: Know and use strategies for estimating results of any whole-number computations.
#### 4.3: Students use and interpret variables, mathematical symbols, and properties to write and simplify numerical expressions and sentences. They understand relationships among the operations of addition, subtraction, multiplication, and division
4.3.4: Understand that an equation such as y = 3x + 5 is a rule for finding a second number when a first number is given.
4.3.5: Continue number patterns using multiplication and division.
4.3.6: Recognize and apply the relationships between addition and multiplication, between subtraction and division, and the inverse relationship between multiplication and division to solve problems.
4.3.7: Relate problem situations to number sentences involving multiplication and division.
#### 4.4: Students show an understanding of plane and solid geometric objects and use this knowledge to show relationships and solve problems.
4.4.3: Identify, describe, and draw parallelograms, rhombuses, and trapezoids, using appropriate mathematical tools and technology.
4.4.4: Identify congruent quadrilaterals and give reasons for congruence using sides, angles, parallels and perpendiculars.
#### 4.5: Students understand perimeter and area, as well as measuring volume, capacity, time, and money.
4.5.1: Measure length to the nearest quarter-inch, eighth-inch, and millimeter.
4.5.4: Know and use formulas for finding the areas of rectangles and squares.
4.5.6: Understand that rectangles with the same area can have different perimeters and that rectangles with the same perimeter can have different areas.
4.5.7: Find areas of shapes by dividing them into basic shapes such as rectangles.
#### 4.6: Students organize, represent, and interpret numerical and categorical data and clearly communicate their findings. They show outcomes for simple probability situations.
4.6.1: Represent data on a number line and in tables, including frequency tables.
4.6.3: Summarize and display the results of probability experiments in a clear and organized way.
#### 4.7: Problem Solving
4.7.1: Analyze problems by identifying relationships, telling relevant from irrelevant information, sequencing and prioritizing information, and observing patterns.
4.7.4: Use a variety of methods, such as words, numbers, symbols, charts, graphs, tables, diagrams, tools, and models to solve problems, justify arguments, and make conjectures.
4.7.7: Know and use appropriate methods for estimating results of whole-number computations.
Correlation last revised: 1/20/2017
This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information. |
Tuesday, January 19, 2010
Polynomial Fuctions
Hey guys,
this post will cover just some of the basics of polynomial functions. This post will focus on the divisioin of polynomials through long division and Factor Theorum.
Using Long Divison:
As many of you should alrady know how to divide, dividing with exponents isn't that much harder. So lets start with this example first. Use long division to divide 36x^3 - 6x^2 + 3x - 33 by 3x + 1 (Note: BY 3x + 1 is what they want you to divide with). So how do we do it?
1. Write down the question first
2. Like you would do for normal divison, ask yourself how many times can 3x go into 36x^3? (Answer: 12x^2)
3. After finding that out, treat the answer you found in #2 like 12x^2(3x + 1) in order to progress.
5. Bring down the 3x from the question.
6. Repeat the same processas you did for 2. (how many times does 3x go into -18x^2? Answer: -6x)
7. Repeat steps 3, 4, 5, 6.
8. Once you have reached the remainder (-30R), you can now write down the division statement (Dividend) = (Divisor)(Quotient) + Remainder
Factor Theorum:
The next thing we'll look at is how to use factor theorum to solve for a polynomial (FYI: MUCH fast and WAY easier).
If you are given a question such as: Factor x^4 + 4x^3 - x^2 - 16x - 12
So how do you do it?
1. To find the first zero, find a number for x which can cause the polynomial to equal to zero (in this case -2)
2. Your new polynomial should look like (x+2)(x^3 + 2x^2 - 5x - 6) But your not done yet.
3. Repeat step one, your zero's should now become more clear. (x + 2) (x + 3) (x^2 - x - 2)
4. Finally, factor out x^2 - x - 2 (product/sum)
Well, that's about all you need to know about long division of polynomials and factor theorum. Hope this helped.
Review: Trigonometic Functions
Trigonometic Funtions Part 1
• to convert degrees to radians, multiply degree by (/180)
Special Angles
• you can use the unit circle to solve
• use special angles with unit circle
Equivalent Trigonometric Expressions
• use the cofunction identity sinx = cos (/2 - x) to determine equivalent trigonometric expressions
Compond Angle Formula
• apply equivalent trigonometric expressions to compund angle formula
• determine exact trigonometric ratios for angles expressed as sums or differences or special angles
Identities
• treat each side independantly and transform expression on one side into exact form of the other
Trigonometric Functions Part 2
Sine, Cosine, and Tan Graphs
• y=sinx and y=cosx graphs have amplitude of 1 and a period of 2 and can be transformedy=tanx has no amplitude and no maximum or minimum
Reciprocals
• reciprocals are different from it's inverse
• csc means 1/sinx
• sec means 1/cosx
• cot means 1/tanx
Mapping Rule
• g(x) = af [k(x-d)] + c
• a represents amplitude
• d represents phase shift
• c respresents vertical translation
Period
• 2/k
Amplitude
• max - min / 2
Rational Functions: Exam Review
So Exams are coming up and basically this post is going to cover some things you need to know for Rational Functions. Things such as key features and skills you need to have when dealing with rational functions.
Rational Functions
f(x) = p(x)/q(x), q(x) cannot equal 0
f(x) = 1/x this is a simple rational, it is also the reciprocal of a linear functions.
therefore...
f(x) = 1/x^2 is reciprocal of a quadratic f(x) = 1/x^3 is a reciprocal of a cubic and etc.
Special Skills Required
Things you would need to know when dealing with rationals are...
- Finding a Vertical and/or Horizontal Asymptote
- Find any y-intercepts, zero's, holes (if there are any)
- State the Domain and Range
- Graph a rational function
- Factor into factored form
Key Features
- As stated above, the denominator cannot equal 0, anything being divided by 0 causes problems.
Horizontal Asymptote (H.A.)
- if n < m, H.A. is y =0
- if n = m, H.A. is y = coefficient of x^n/ coefficient of x^m
- if n > m, there is no H.A.
Vertical Asymptote (V.A.)
- find the zero's of the denominator
Y-intercept
- Lets x=0
X-intercept
-Let y=0
Example
f(x) = x^2 - 3x - 10 / x^2 + 9x + 14
Factoring
f(x) = x^2 - 3x - 10/x^2 + 9x + 14
Quadratic formula on top and bottom
f(x) = (x-5) (x+2) /(x+7)(x+2)
the (x+2) cancels each other out
f(x) = (x-5)/(x+7) , x cannot equal 7
Horizontal Asymptotes
- n = m so...
y= coefficient of x^n/coefficient of x^m
y= 1/1
y= 1
Vertical Asymptotes
- solve for the zero's of the denominator
x= -7
Y-intercept
-let x = 0
y= (0-5)/(0+7)
y= -5/7
X-intercept
-let y = 0
0= (x-5)/(x+7)
0= x-5
x=5
Holes
x= -2
Graph
Domain and Range
D= {x| x cannot equal -7, xE R}
R= {y| y cannot equal 1,> yER}
Saturday, December 26, 2009
7.2 Solving Exponential and Log Equations
Hey Guys,
This post is going to cover chapter 7.2, which we covered in class on Wednesday Decmber 16th 2009.
Most of the lesson was based on Radioactive decay and powers of different bases and applying quadratic formula.
Radioactive Decay is the process by which element transforms into a different element. This can be modelled by the following equation:
A(t) = A0(1/2)^t/h ,
where A(t) represents the mass of a substance, Ao represents the initial amount, t represents time, and h represents half life.
In order to solve this equation, 3 variables are needed, so you may isolate the 4th and missing variable.
For more help and for an example of this problem, check out pg. 370 of the textbook.
Powers with Different Bases
When solving powers with different bases you use an algebraic reasoning, where you take the logaritm of both sides and apply the lower law of logs to remove the varibles from the exponent.
Ex: 4^ x+1 = 64^2x
log (4^x+1) = log (64^2x)
(x+1) log 4 = 2x log 64
x+1 = 2x log 64
log 4
x+1 = 2x (3)
x+1 = 6x
1= 5x
1/5 = x
The third method we used to solve is using the quadratic formula
example:
2^x – 2^-x = 4
To use the quadratic formula to solve this equatation, you must multiply both sides by 2^x so that a quadratic equation is obtained in terms of 2^x
2^x(2^x – 2^-x) = 2^x(4)
2^x(2^x) - 2^x(2^-x) = 2^x(4)
2^2x – 2^0 = 2^x(4)
2^2x – 1 = 2^x(4)
Then apply the power of a power law to the term 2^x:
(2^x)^2 – 1 = 2^x(4)
And then write in standard form (az^2+bz+c = 0)
(2^x)^2 – 4(2^x) – 1 = 0
Then you apply the quadratic formula to determine the roots.
This will result in 2 + sqrt(5) and 2 - sqrt(5).
Alternatively, you could have set b = 2^x and then solved for b which would have yielded
2^x = 2 + sqrt(5) &
2^x = 2 - sqrt(5)
At time point you must use the logarithm of both sides by using the power law of logs, and then dividing both sides by log2. This will yeild 2.08 which would be one root of this equation.
The second equation however yeilds a number smaller than 0 which results in no answer since a power must always be a positive number.
Hope that helped ! :)
Monday, December 21, 2009
6.1 The Exponential Function and its Inverse
This is our second last unit of our Advanced Functions course. YAYYYY ^^
In the lesson, it will be separated into two parts. The first part will talk about the exponential functions and the second part will talk about its inverse.
Since this is the first lesson of the new logarithmic unit, most of the information presented should be learned by students.
PART 1- EXPONENTIAL FUNCTIONS
In part 1, we will learn how to solve for the exponent with different bases. The "log" function on your calculators will be used greatly in this unit.
Side note: the default base of the "log" key on your calculators is ALWAYS 10.
PART 2- Inverse
In part 2, we will be pretty much doing the same thing as grade 11. Switch x and y in the same equation and isolate for X again.
Exponential functions can be represented in 3 DIFFERENT ways.
1: General Form- the simplest form of representation of an exponential equation
y=5x or y=1/5x
2. Table of Values- a table which a list of the "x" values with the corresponding "y" values
REMEMBER: the first ratio of differences of the x and y values are always constant.
How do we find the first ratio of differences?
The values of : Y2-Y1 = answer.
3: Exponential functions- this is a graph representation of the exponential functions. However, just by looking at a graph cannot help you determine whether an exponential function is or not.
Rate of Change:
To determine the rage of change of an exponential function is really much similar to finding the rate of change of any other functions.
1) Pick the x values that are really close to the value you are given to find the IROC.
2) For example: if the value that is given is 2. You may go to the left of the value or right to the value which results in x = 2.0001 or x = 1.9999.
3) Sub these values back into the original equation and find the “y” values
4) To choose the two closet points that lead to the given value will give you the best and most accurate slope of secant that is required for the answer.
5) After having the “x” values and the “y” values. Put it into the general equation used to find slopes which is:
AROC [x value – x value] = y2-y1 / x2-x1
6) To conclude your answer, you must write, “Therefore, the IROC of the given value is approximately at “answer of AROC of the 2 x values”.
Inverse functions:
The graph of the inverse function of a logarithmic function is a reflection on the y= x function.
To find an inverse function of a logarithmic function, switch x and y from the equation and solve for x again.
By switching (x,y) -> (y,x) , these can be your new inverse values that can be used to apply transformations.
Differences and similarities between an inverse function and an exponential function
1) A exponential function will have an increasing slope while the inverse function will have an decreasing slope
2) A horizontal asymptote (y = a value) is present in an exponential function. A vertical asymptote (x = a value) is present in the inverse function
3) Depending on the transformations, the horizontal asymptote or the vertical asymptote is affected by the vertical or horizontal shifts. |
How do you simplify sqrt3^2?
Feb 3, 2016
Ambiguous
Explanation:
I cannot understand if the square is inside or outside the radix. Anyhow, if it's inside (the most plausible one...)
$\sqrt{{3}^{2}}$ it's one of the two numbers $\pm a$ such that ${\left(\pm a\right)}^{2} = {3}^{2} = 9$, so they are $\sqrt{{3}^{2}} = \pm 3$
On the other hand, if we consider ${\left(\sqrt{3}\right)}^{2}$, you gotta take both $\pm \sqrt{3}$ and take the square. But, by definition, $\pm \sqrt{3}$ are the only two numbers such that ${\left(\pm \sqrt{3}\right)}^{2} = 3$, so the answer is $3$
Feb 3, 2016
Ambiguous
Explanation:
I cannot understand if the square is inside or outside the radix. Anyhow, if it's inside (the most plausible one...)
$\sqrt{{3}^{2}}$ it's one of the two numbers $\pm a$ such that ${\left(\pm a\right)}^{2} = {3}^{2} = 9$, so they are $\sqrt{{3}^{2}} = \pm 3$
On the other hand, if we consider ${\left(\sqrt{3}\right)}^{2}$, you gotta take both $\pm \sqrt{3}$ and take the square. But, by definition, $\pm \sqrt{3}$ are the only two numbers such that ${\left(\pm \sqrt{3}\right)}^{2} = 3$, so the answer is $3$ |
Chapter 6 Class 9 Lines and Angles
Class 9
Important Questions for Exam - Class 9
### Transcript
Question 2 In figure, the sides AB and AC of ∆ABC are produced to points E and D respectively. If bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O, then prove that ∠BOC = 90° − 1/2∠ BAC. BO is the bisector of ∠ CBE So, ∠ CBO = ∠ EBO = 1/2 ∠ CBE Similarly, CO is the bisector of ∠ BCD So, ∠ BCO = ∠ DCO = 1/2 ∠ BCD ∠ CBE is the exterior angle of Δ ABC Hence, ∠ CBE = x + z ∠ CBE = x + z 1/2(∠ CBE) = 1/2 (x + z) ∠ CBO = 1/2 (x + z) Similarly, ∠ BCD is the exterior angle of Δ ABC Hence, ∠ BCD = x + y 1/2(∠ BCD) = 1/2 (x + y) ∠ BCO = 1/2 (x + y) In Δ OBC ∠ BOC + ∠ BCO + ∠ CBO = 180° ∠ BOC + 1/2 (x + y) + 1/2 (x + z) = 180° ∠ BOC + 1/2 (x + y) + 1/2 (x + z) = 180° ∠ BOC + 1/2 (x + y + x + z) = 180° In Δ ABC x + y + z = 180° Putting (2) in (1) ∠ BOC + 1/2 (x + y + x + z) = 180° ∠ BOC + 1/2 (x + 180° ) = 180° ∠ BOC + 𝑥/2 + 1/2 × 180° = 180° ∠ BOC + 𝑥/2 + 90° = 180° ∠ BOC = 180° – 90° – 𝑥/2 ∠ BOC = 90° – 𝑥/2 ∠ BOC = 90° – 𝑥/2 ∠ BOC = 90° – 1/2 × ∠ BAC Hence proved |
Site Map Math Index Glossary Timeline Questions & Answers Lesson Plans #11a Trig proficiency drill
(M-12) The Tangent
The tangent is a tool of trigonometry, related to the sine and cosine. On this web site it is used in connection with the cross staff.
You already know that right angled triangles are basic to trigonometry. Let ABC be such a triangle (drawing) with C = 90° the right angle and (A, B) the sharp ("acute") angles. Also, let a, b and c be the lengths of its three sides--a of the side across from A, b of the one across from B, and c the longest one, across from C.
Two useful ratios associated with the angle A ("trigonometric functions of A") are
The sine of A sin A= a/c (involving the side a across from A)
The cosine of A cosA = b/c (involving the side b besides A)
Both these ratios involve the long side c ("hypotenuse" in mathspeak), and since both a and b must be smaller than that side, these ratios are always numbers less than 1. We now add two more ratios to our collection--the tangent and the cotangent:
The tangent of A
tan A = a/b (some write "tg A")
.
And the cotangent of A
cotan A = b/a = 1/tan A
A simple relation exists between this pair and the earlier ones. We have
sinA/ cosA= (a/c)/ (b/c)
Multiply top and bottom by c (it is the same as multiplying the fraction by (c/c)=1) and get
sinA / cosA = (a/b) = tan A
Inverting
cosA/ sinA = 1/tanA = cotanA
The calculators which give sines and cosines, and the books which tabulate their values, are also able to give tangents and cotangents.
A Simple Application
At noon a vertical flagpole of height 50' (50 feet) has a shadow of length 18 feet. What is the angle A of the Sun above the horizon? (As explained in the section "Navigation, " that angle allows one to calculate the latitude at the location.) From the drawing:
tanA = 50/18 = 2.7778
If you have a table of tangents, you can now look for the angles whose tangents just above and just below that value, and estimate where between them A should be ("interpolate"). Calculators usually have a button "tan" which, if you punch in the angle, gives you the tangent. But many also have a button "tan -1 " which does the reverse--given the tangent, it brings up the angle. (It may be the same button, deriving tan-1 if first you hit a colored "special mode" button; tan-1 is also called "inverse tangent" or "arc-tangent"). In this example,
tan–1 2.7778 = 70.2°
P.S.: A tangent to a circle is a line grazing it at just one point. If you ever wondered how the name "tangent" entered trigonometry, click here.
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated 25 November 2001 |
Learning Library
### EL Support Lesson
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This lesson can be used as a pre-lesson for the Fractions of a WholeLesson plan.
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This lesson can be used as a pre-lesson for the Fractions of a WholeLesson plan.
Students will be able to use bar models and multiplication to find a fraction of a whole number.
##### Language
Students will be able to explain the strategy used to solve a problem using peer interaction and sentence stems.
(2 minutes)
• Read aloud the content and language objective. Have students repeat or rephrase the student-facing version of the objectives to a table partner.
• Tell students that they will practise describing and explaining how they solved a problem involving multiplying a fraction by a whole number.
(8 minutes)
• Write a problem on the board and read it aloud: "Mary made 24 cupcakes. 1/3 are chocolate cupcakes and the rest are vanilla. How many chocolate cupcakes are there?"
• Tell students that we can solve this problem a variety of ways. One way is to use a bar model. Draw a bar model, and explain that we need to divide it into thirds so that we can figure out how many cupcakes is equal to a third of the 24. Show how you divide the bar into 3 equal sections and write 24 on top of the bar to show that the whole bar represents 24 cupcakes. Tell students that each section represents one third of the cupcakes. So we have to ask ourselves, what is a third of 24? Or in other words, what is 24 divided by 3? Tell students that the answer is 8. There are 8 chocolate cupcakes.
• Demonstrate that you can also solve this problem mentally with repeated addition by thinking of a number that you can add three times to get to 24. If students are comfortable with repeated addition and can decompose numbers, they can figure out that 8 + 8 + 8 = 24.
• Another way to solve this problem is by multiplying the numerator by the whole number to get 24/3. Then divide the new numerator by the denominator to get 8 as the answer.
• Repeat this process by modeling multiple strategies to solve another problem: "There are 36 pounds of rice in the food bank. Three-fourths of this was used to feed people in November. How many pounds of rice was used in November?"
• Be sure to use ample maths vocabulary to clearly describe the process of solving this problem in multiple ways.
• Display the paragraph frame of transition words such as "First, I... Then, I ... Finally, I..." for students to use throughout the lesson.
(10 minutes)
• Distribute the Discussing Fractions of a Whole worksheet to students and display a teacher copy.
• Review the steps involved in the two strategies demonstrated at the top of the worksheet. Tell students that these are not the only ways to solve this type of problem. There are also other strategies such as repeated additions. Students are welcome to use any strategy that works well for them as long as they are able to accurately describe and explain their maths thinking.
• Model how to solve the first problem in the worksheet with the strategy of your choice. Write a description of the procedure in complete sentences and read it aloud to students.
• Instruct students to solve the second problem on the worksheet on their own. Place students into partnerships and have them verbally explain the steps they used to solve the problem. Each pair should listen to each other's strategies before writing down the steps they took in complete sentences.
• Provide the following questions for students to use as they share their description with their classmate.
• What strategy did you use to solve the problem?
• Why did you choose this strategy?
• What did you do when you got to this point?
• Why did you do that?
• Invite one or two pairs of students to model their sharing of the solution to the second problem in front of the class.
(12 minutes)
• Instruct students to independently complete the remaining problems (3–5) on the worksheet and write a description. Tell students to try different strategies with different problems so they are comfortable with a variety of ways to solve for a fraction of a whole.
• Once everyone has completed the worksheet, have students walk around the room with their worksheet and share one problem solution and description with a partner. They may choose any problem to share but each pair must share their work on the same problem.
• Use the Formative Assessment: Peer Explanations Checklist to listen in on students' discussions and note their progress.
• Tell students to use the clarifying questions provided in the previous section in their conversations with their peers. Inform students that they may need to return to their desk to revise or edit their descriptions of the solutions as they talk about their strategy with their partner. Tell students to modify their descriptions as needed. Continue this process until everyone has shared all the problems and made edits/revisions to their descriptions.
• Have a few students share their descriptions and mention the changes made throughout the process.
Beginning
• Allow students to explain the process of solving the problems in their home language before rephrasing, using sentence stems/frames, in English.
• Have students work in a smaller, teacher-led group during group work.
• Create and display a word/phrase bank with helpful terms from the lesson for students to refer to, with images if applicable.
• Provide students with a word bank to refer to when they complete the sentence stems/frames.
• Encourage students to explain the directions in their own words and/or write them down in their maths journals prior to group work.
• Have students share their answers aloud without referring to the sentence stems/frames for support.
• Encourage students to rephrase the directions and key learning points from the lesson.
(4 minutes)
• Show students the following problem and solution, and tell students to write a description of the steps taken to solve the problem using complete sentences:
• Problem: "Carla and her friends made 75 baby hats. They will donate 2/3 of the bracelets to a hospital. How many hats were donated?"
• Solution: a bar model with a total of 75, divided into 3 equal sections and the written answer of 50 hats.
• Have students orally describe the solution to a partner and invite a few students to share their descriptions with the whole class. (For example, "First, they drew a bar model with a total of 75. Then, they divided the bar model into thirds and divided 75 by 3 to get 25. Finally, they multiplied 25 by 2 since they need to figure out what two-thirds of 75 is and they got the answer 50.")
(4 minutes)
• Ask students the following question: How helpful was it to share the explanation of your solution to your peers and receive feedback? Tell them to show their response on their fingers, with one finger indicating not very helpful and five fingers indicating extremely helpful.
• Invite a few students to share their reasoning using the following sentence stems:
• "I found it useful to share with my peers because..."
• "I found it not very helpful to share with my peers because..."
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Math Integration rules Integration by linear substitution
# Integration by linear substitution
When integrating composite functions of the form $f(g(x))$ with a linear inner function, one uses the integration by linear substitution:
$\int f(mx+n) \, \mathrm{d}x$ $=\frac1m F(mx+n)+C$
!
### Remember
Integration by linear substitution may only be used if the inner function $g(x)$ is a linear function (i.e.: $g(x)=mx+n$).
$f(g(x))$ $=f(mx+n)$
i
### Hint
In addition to the integration by linear substitution, there is the integration by nonlinear substitution for arbitrarily composite functions.
Linear substitution is actually only a special case of general substitution, but it is sufficient for most tasks.
### Examples
$\int (2x+4)^2 \, \mathrm{d}x$
1. #### Split function into subfunctions
$f(x)=x^2$ and $g(x)=\color{red}{2}x+4$
$\color{red}{m=2}$
2. #### Integrate $f(x)$
Apply the power rule
$F(x)=\color{blue}{\frac13}x\color{blue}{^3}$
3. #### Insert
$\int f(mx+n) \, \mathrm{d}x$ $=\frac{1}{\color{red}{m}} \cdot\color{blue}{F}(mx+n)+C$
$\int (2x+4)^2 \, \mathrm{d}x$ $=\frac{1}{\color{red}{2}} \cdot \color{blue}{\frac13}(2x+4)^\color{blue}{3}+C$ $=\frac16(2x+4)^3+C$
$\int e^{2x} \, \mathrm{d}x$
1. #### Split function into subfunctions
$f(x)=e^x$ and $g(x)=\color{red}{2}x$
$\color{red}{m=2}$
2. #### Integrate $f(x)$
Apply the power rule
$F(x)=\color{blue}{e}^x$
3. #### Insert
$\int f(mx+n) \, \mathrm{d}x$ $=\frac{1}{\color{red}{m}} \cdot\color{blue}{F}(mx+n)+C$
$\int e^{2x} \, \mathrm{d}x$ $=\frac{1}{\color{red}{2}} \cdot \color{blue}{e}^{2x}+C$
i
### Hint
For exponential functions (second example) there is a "trick" to solve the integral even faster:
The antiderivative of an exponential function with linear inner function is obtained by dividing by the derivative of the inner function.
$\int e^{g(x)} \, \mathrm{d}x=\frac{e^{g(x)}}{g'(x)}+C$
Example: $\int e^{2x} \, \mathrm{d}x=\frac{e^{2x}}{2}+C$ |
GCF the 6 and 12 is the largest possible number that divides 6 and also 12 exactly without any type of remainder. The components of 6 and also 12 room 1, 2, 3, 6 and 1, 2, 3, 4, 6, 12 respectively. There are 3 frequently used techniques to uncover the GCF the 6 and 12 - lengthy division, Euclidean algorithm, and also prime factorization.
You are watching: What is the greatest common factor of 6 and 12
1 GCF of 6 and 12 2 List of Methods 3 Solved Examples 4 FAQs
Answer: GCF the 6 and also 12 is 6.
Explanation:
The GCF of two non-zero integers, x(6) and also y(12), is the biggest positive integer m(6) the divides both x(6) and also y(12) without any kind of remainder.
The methods to discover the GCF that 6 and also 12 are described below.
Long department MethodPrime administrate MethodListing typical Factors
### GCF that 6 and also 12 by lengthy Division
GCF of 6 and 12 is the divisor that we get when the remainder i do not care 0 ~ doing long division repeatedly.
Step 2: since the remainder = 0, the divisor (6) is the GCF of 6 and also 12.
The equivalent divisor (6) is the GCF the 6 and 12.
### GCF of 6 and also 12 by prime Factorization
Prime administer of 6 and 12 is (2 × 3) and also (2 × 2 × 3) respectively. As visible, 6 and also 12 have usual prime factors. Hence, the GCF that 6 and 12 is 2 × 3 = 6.
### GCF of 6 and also 12 by Listing typical Factors
Factors that 6: 1, 2, 3, 6Factors of 12: 1, 2, 3, 4, 6, 12
There room 4 typical factors of 6 and also 12, that space 1, 2, 3, and 6. Therefore, the greatest usual factor that 6 and also 12 is 6.
☛ also Check:
## GCF that 6 and also 12 Examples
Example 1: find the GCF of 6 and 12, if your LCM is 12.
Solution:
∵ LCM × GCF = 6 × 12⇒ GCF(6, 12) = (6 × 12)/12 = 6Therefore, the greatest usual factor that 6 and also 12 is 6.
Example 2: For 2 numbers, GCF = 6 and also LCM = 12. If one number is 6, uncover the various other number.
Solution:
Given: GCF (y, 6) = 6 and also LCM (y, 6) = 12∵ GCF × LCM = 6 × (y)⇒ y = (GCF × LCM)/6⇒ y = (6 × 12)/6⇒ y = 12Therefore, the other number is 12.
Example 3: The product of 2 numbers is 72. If their GCF is 6, what is their LCM?
Solution:
Given: GCF = 6 and also product of number = 72∵ LCM × GCF = product the numbers⇒ LCM = Product/GCF = 72/6Therefore, the LCM is 12.
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## FAQs on GCF that 6 and also 12
### What is the GCF that 6 and 12?
The GCF of 6 and also 12 is 6. To calculate the GCF (Greatest usual Factor) of 6 and also 12, we require to variable each number (factors the 6 = 1, 2, 3, 6; factors of 12 = 1, 2, 3, 4, 6, 12) and choose the greatest element that exactly divides both 6 and 12, i.e., 6.
### What is the Relation between LCM and also GCF that 6, 12?
The complying with equation can be offered to to express the relation between LCM and also GCF the 6 and 12, i.e. GCF × LCM = 6 × 12.
### If the GCF that 12 and also 6 is 6, find its LCM.
GCF(12, 6) × LCM(12, 6) = 12 × 6Since the GCF of 12 and also 6 = 6⇒ 6 × LCM(12, 6) = 72Therefore, LCM = 12☛ GCF Calculator
### What room the approaches to uncover GCF the 6 and also 12?
There space three commonly used approaches to find the GCF the 6 and 12.
See more: How Many Sides Does A Circle Have A Face S Does A Circle Have
By Listing typical FactorsBy element FactorizationBy lengthy Division
### How to find the GCF the 6 and 12 by Long department Method?
To discover the GCF the 6, 12 using long division method, 12 is separated by 6. The matching divisor (6) as soon as remainder amounts to 0 is taken together GCF.
### How to uncover the GCF that 6 and 12 by prime Factorization?
To uncover the GCF of 6 and also 12, us will discover the element factorization that the provided numbers, i.e. 6 = 2 × 3; 12 = 2 × 2 × 3.⇒ due to the fact that 2, 3 are typical terms in the prime factorization that 6 and also 12. Hence, GCF(6, 12) = 2 × 3 = 6☛ What is a prime Number? |
# Factorising
GCSEKS3Level 4-5AQACambridge iGCSEEdexcelEdexcel iGCSEOCRWJEC
## Factorising into Single Brackets
Factorising is putting expressions into brackets, this is the reverse of expanding brackets.
Make sure you are happy with the following topics before continuing.
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
## Factorising into single brackets – 3 Key steps
Example:
Fully factorise the following:
$\textcolor{red}{12}\textcolor{limegreen}{x^2} +\textcolor{red}{8}\textcolor{limegreen}{x}$
Step 1 – Take out the largest common factor of both the numbers, and place it in front of the brackets.
Factors of $\textcolor{red}{12}$ are $1, 2, 3, \textcolor{blue}{4}, 6, 12$
Factors of $\textcolor{red}{8}$ are $1, 2, \textcolor{blue}{4}, 8$
The largest common factor is $\bf{\textcolor{blue}{4}}$
$\textcolor{red}{12} = \textcolor{blue}{4} \times \textcolor{purple}{3}$
$\textcolor{red}{8} = \textcolor{blue}{4} \times \textcolor{purple}{2}$
Step 2 – Take out the highest power of the “Letter” which is a part of every term.
\begin{aligned}\textcolor{red}{12}\textcolor{limegreen}{x^2} = & \textcolor{red}{12} \times \textcolor{limegreen}{x} \times \xcancel{\textcolor{limegreen}{x}}\\ \textcolor{red}{8}\textcolor{limegreen}{x} = & \textcolor{red}{8} \times \xcancel{\textcolor{limegreen}{x}}\end{aligned}
We can take out one $\textcolor{limegreen}{x}$ from each term, placing in front of the brackets.
Step 3 – Place the highest factor and highest power of the letter in front of the bracket, then add the remaining terms inside the bracket
$\textcolor{blue}{4}\textcolor{limegreen}{x}(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)$
$\textcolor{blue}{4}\textcolor{limegreen}{x}(\textcolor{purple}{3}\textcolor{limegreen}{x} +\textcolor{purple}{2})$
To check, you can multiply the bracket back out to see if you have the right answer.
$4x(3x+2) = 12x^2 +8x$
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
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## Example 1: Factorising Two Terms
Fully Factorise the following, $\textcolor{red}{3}\textcolor{limegreen}{x}\textcolor{Orange}{y} + \textcolor{red}{6}\textcolor{limegreen}{x^2}$.
[2 marks]
Step 1 – Take out the largest common factor of both the numbers, and place it in front of the brackets.
Factors of $\textcolor{red}{3}$ are $1, \textcolor{blue}{3}$
Factors of $\textcolor{red}{6}$ are $1, 2, \textcolor{blue}{3}, 6$
The largest common factor is $\bf{\textcolor{blue}{3}}$
$\textcolor{red}{3} = \textcolor{blue}{3} \times \textcolor{purple}{1}$
$\textcolor{red}{6} = \textcolor{blue}{3} \times \textcolor{purple}{2}$
Step 2 – Take out the highest power of the “Letter” which is a part of every term.
\begin{aligned}\textcolor{red}{3}\textcolor{limegreen}{x}\textcolor{Orange}{y} = & \textcolor{red}{3} \times \xcancel{\textcolor{limegreen}{x}} \times \textcolor{orange}{y}\\ \textcolor{red}{6}\textcolor{limegreen}{x^2} = & \textcolor{red}{6} \times \textcolor{limegreen}{x} \times \xcancel{\textcolor{limegreen}{x}}\end{aligned}
We can take out one $\textcolor{limegreen}{x}$ from each term, placing it in front of the brackets.
Step 3 – Place the highest factor and highest letter in front of the bracket, then add the remaining terms inside the bracket
$\textcolor{blue}{3}\textcolor{limegreen}{x}(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)$
$\textcolor{blue}{3}\textcolor{limegreen}{x}(\textcolor{orange}{y} +\textcolor{purple}{2}\textcolor{limegreen}{x})$
To check you can multiply the bracket back out to see if you have the right answer.
$3x(y+2x) = 3xy+6x^2$
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
## Example 2: Factorising with three terms
Fully factorise the following, $\textcolor{red}{8}\textcolor{limegreen}{x}\textcolor{Orange}{y} + \textcolor{red}{12}\textcolor{limegreen}{x^2}\textcolor{Orange}{y} - \textcolor{red}{4}\textcolor{limegreen}{x^2}\textcolor{Orange}{y^2}$.
[3 marks]
Step 1 – Take out the largest common factor of all of the numbers, and place it in front of the brackets.
Factors of $\textcolor{red}{8}$ are $1, 2, \textcolor{blue}{4}, 8$
Factors of $\textcolor{red}{12}$ are $1, 2, 3, \textcolor{blue}{4}, 6, 12$
Factors of $\textcolor{red}{4}$ are $1, 2, \textcolor{blue}{4}$
The highest common factor of all three is $\textcolor{blue}{4}$
$\textcolor{red}{8} = \textcolor{blue}{4} \times \textcolor{purple}{2}$
$\textcolor{red}{12} = \textcolor{blue}{4} \times \textcolor{purple}{3}$
$\textcolor{red}{4}= \textcolor{blue}{4} \times \textcolor{purple}{1}$
Step 2 – Take out the highest power of the “Letter” which is a part of every term.
\begin{aligned}\textcolor{red}{8}\textcolor{limegreen}{x}\textcolor{Orange}{y} = & \textcolor{red}{8} \times \xcancel{\textcolor{limegreen}{x}} \times \xcancel{\textcolor{Orange}{y}}\\ \textcolor{red}{12}\textcolor{limegreen}{x^2}\textcolor{Orange}{y} = & \textcolor{red}{12} \times \xcancel{\textcolor{limegreen}{x}} \times\textcolor{limegreen}{x} \times \xcancel{\textcolor{Orange}{y}} \\ \textcolor{red}{4}\textcolor{limegreen}{x^2}\textcolor{Orange}{y^2} = & \textcolor{red}{4} \times \xcancel{\textcolor{limegreen}{x}} \times \textcolor{limegreen}{x} \times \xcancel{\textcolor{Orange}{y}} \times \textcolor{Orange}{y} \end{aligned}
We can take out one $\textcolor{limegreen}{x}$ and one $\textcolor{Orange}{y}$ from each term.
Step 3 – Place the highest factor and highest letter in front of the bracket, then add the remaining terms inside the bracket.
$\textcolor{blue}{4}\textcolor{limegreen}{x}\textcolor{Orange}{y}(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)$
$\textcolor{blue}{4}\textcolor{limegreen}{x}\textcolor{Orange}{y}(\textcolor{purple}{2} + \textcolor{purple}{3}\textcolor{limegreen}{x} - \textcolor{limegreen}{x}\textcolor{Orange}{y})$
To check you can multiply the bracket back out to see if you have the right answer.
$4xy(2 + 3x-xy) = 8xy + 12x^2y-4x^2y^2$
Level 4-5GCSEAQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
## Factorising Example Questions
Take out a factor of 5 from both terms to get
$5(2pq + 3pqr)$
There is both a $p$ and a $q$ in the two terms inside the bracket. Taking out both $p$ and $q$, we get
$5pq(2 + 3r)$
The two numbers in the bracket have nothing more in common so we are done.
Gold Standard Education
Take out a factor of $u$ from both terms to get,
$u(u^2+3v^3+2)$
The terms inside the bracket have no more common factors, so we are done.
Gold Standard Education
The first and last term have a factor of 4 in common, but the middle term doesn’t, so we can’t take any numbers out as factors.
All 3 terms have a factor of $y$ in them. Specifically, the highest power of $y$ that all 3 terms have in common is $y^5$. Taking $y^5$ out as a factor, we get,
$y^5(4x + 1 + 12y^2)$
The terms in the bracket have no more common factors, so we are done.
Gold Standard Education
Take out a factor of 5 from every term to get
$5(xy^2-x^2y-x^2y^2)$
Now, clearly each term has a factor of $x$ and $y$, so we just need to determine what the highest power of each factor we can take out is,
$5xy(y-x-xy)$
The terms inside the bracket have no more common factors, so we are done.
Gold Standard Education
Take out a factor of 7 from every term to get
$7(abc + 2a^{2}bc + 3ab^{2}c + 7abc^3)$
Now, clearly each term has a factor of $a, b$, and $c$, so we just need to determine what the highest power of each factor we can take out is.
The first term only has the three factors $a, b$, and $c$ to the power of 1 (note that we don’t write the power of 1 since $x^1 = x$), which means that this is the highest power of each factor we can take out is 1. Taking out a factor of $a, b$, and $c$, we get
$7abc(1 + 2a + 3b + 7c^2)$
The terms inside the bracket have no more common factors, so we are done.
Gold Standard Education
## Factorising Worksheet and Example Questions
### (NEW) Factorising (Foundation) Exam Style Questions - MME
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Math 235
Suggested Homework Problems from Bretcher's fifth edition, Fall 2014
Note: The homework problems listed here will give the student experience with the ideas and methods of linear algebra. They are not the only such problems; Exams and the homework assigned by your section's instructor may contain questions rather different from these.
Carfully JUSTIFY all your answers to all homework problems. Answers without sufficient justification will not get credit.
Starred problems are challenge problems
Numbered problems are from the text: Linear Algebra with applications, Fifth Edition, by Otto Bretscher, Pearson Education 2012.
Starred problems are challenge problems
• Section 1.1 page 5: 1, 3, 7, 12, 14, 18, 19, 24, 30, 31, 33.
• Section 1.2 page 18: 5, 10, 18, 24, 26, 39 (In 39 set up the system of linear equations, find the corresponding augmented matrix. You are not asked to solve it by hand).
• Section 1.3 page 34: 1, 4, 6, 8, 10, 18, 24, 34, 36, 47, 55, 58
• Section 2.1 page 53: 4, 6, 10, 12, *13, 16, 19, 22, 34, 36, 37, 42, 47
• Section 2.2 page 71: 1, 2, 4, 6, 7, 10, 11, 13, 14, 15 (see Example 3), 16, 25, 32, 37, 40, 43.
• Section 2.3 page 85: 2, 4, 8, 14, 17, 18, 20, 29, 30, 40, 41, 50, 66
• Section 2.4 page 97: 1, 2, 4, 16, 20, 21, 22, 23, 25, 26, 34, 36 (use upper triangular form), 41, 80, 81
Hint for problem 80 in section 2.4: The line from P_1 to P_3 should be dotted in the figure, being in the back. Note that the plane through P_0, P_1, P_3 is orthogonal to the line spanned by P_2, since P_1-P_0 is orthogonal to P_2 and P_3-P_0 is orthogonal to P_2. Furthermore, the line spanned by P_2 intersects this plane at the center (P_0+P_1+P_3)/3=-(1/3)P_2 of the triangle with vertices P_0, P_1, P_3. Finally, the latter triangle has edges of equal length. Hence the rotation about its center permutes its vertices cyclically. Thus, T permutes the set of vectors P_0, P_1, P_3 cyclically.
• Section 3.1 page 119: 2, 4, 5, 12, 14, 22, 24, 32, 38, 40, 42, 44, 48 (in 48 part c assume both that rank A is 1 and that ker(A) is orthogonal to im(A) and show that A is the projection onto im(A) in the usual sense, ignoring Excercise 2.2.33)
• Section 3.2 page 131: 2, 4, 6, 12, 20, 19, 26, 32, 34, 46, 53, 54
• Section 3.3 page 133: 4, 3, 9, 16, 22, 26, 28, 38, 40, 42, 43, 67*, 75, 78*, 82, 83 (starred problems are challenge problems).
• Section 3.4 page 159: 4, 6, 8, 17. 22, 26, 29, 33, 34, 37 (see hint below), 40 (see hint below), 41 (first interpret this plane as the plane orthogonal to some vector), 43, 46*, 55, 57, 69, 60 (Hint: use the idea of 69), 71
• Hint for problems 37, 40, and 41: Guess a basis related to the geometric problem and check that the matrix is diagonal by computing the matrix.
• Hint for problems 33 and 34: Assume only that v_1, v_2, v_3 are three unit vectors that are pairwise orthogonal (i.e., v_i dot v_j is zero, if i is different from j). You do not need to use vector product here. It follows that {v_1, v_2, v_3} is a basis for R^3 (you may assume this).
• Section 4.1 page 176: 1 to 6, 10, 16, 18, 20, 25, 27, 36, 47, 48, 50, 55
• Section 4.2 page 184: 1 to 6, 10, 14, 22, 23, 26, 27, 30, 43, 51, 52, 53, 57 (find also a basis for the kernel), 60, 64, 65*, 66.
• Check that xe^{-x} belongs to the kernel of the linear transformation in excercise 40 page 171. Use it and Theorem 4.1.7 to find a basis for the kernel of the linear transformation. Carefully justify why the solutions you found are linearly independent, and why they span the kernel.
• Section 4.3 page 195: 5, 15, 21, 22, 29, 32, 33, 34, 48, 49
• Section 6.1 page 275: 1, 2, 5, 10, 12, 16, 26, 28, 32, 36, 45, 46, 48, 56
• Section 6.2 page 289: 1, 5, 12, 15, 16, 30, *31, 37 (justify all your answers!!!), 38, 46
• Let V be an n-dimensional vector space with basis {v_1, ..., v_n}, [ ]:V -> R^n the coordinate linear transformation, and S:R^n->V its inverse, given by S(c_1, ... c_n)=c_1v_1+ ... +c_nv_n. Let T:V->V be a linear transformation. We get the composite linear transformation from R^n to R^n, mapping a vector x to [T(S(x))], i.e., to the coordinate vector in R^n of the vector T(S(x)) in V. Being linear, the above transformation is given by multiplication by a square n by n matrix B, i.e., [T(S(x))]=Bx, for all x in R^n. The matrix B is called the matrix of T in the given basis (section 4.3). Its i-th column, by definition, is b_i=Be_i=[T(S(e_i))]=[T(v_i)]. Thus, the i-th column of B is the coordinate vector of T(v_i). The determinant det(T) is defined to be det(B) (Def 6.2.11). Use the equation b_i=[T(v_i)] and the standard basis {1, x, x^2} of P_2 and the standard basis of R^{2 x 2} to solve the following problems in section 6.2 page 289:
17, 20
• Section 6.3 page 305: 1, 2, 3 (translate the triangle first so that one of its vertices is the origin),
4, 7, 11,
Let A be a 3 by 3 matrix, with det(A)=7, u, v, w three vectors in R^3, such that the parallelopiped determined by them (i.e., the one with vertices 0, u, v, w, u+v, u+w, v+w, u+v+w) has volume 5 units. Find the volume of the parallelopiped determined by Au, Av, Aw. Carefully justify your answer!
• Section 7.1 page 323: 1-6, 9, 10, 12, 15 (see Definition 2.2.2 in section 2.2), 16, 19, 38.
• Extra problem for section 7.1: Find the matrix of the reflection A of the plane about the line x=y. Find all eigenvalues and eigenvectors of A and a basis of R^2 consisting of eigenvectors of A. Find the matrix of A with respect to the basis you found.
• Section 7.2 page 336: 1-4 (see Definition 7.2.6 for the algebraic multiplicity), 8, 12, 14, 15, 17, 19 (see Fact 7.2.8), 22 (use Theorem 6.2.1 to write a careful justification), 25, 27, 28*, 29, 33
• Extra Problem for section 7.3
• Section 7.3 page 345: 1, 2, 7, 8, 9, 10, 12, 13, 16, 21, 22, 24 (Hint: Theorem 7.3.5 part c suggests that we choose a matrix similar to the one in problem 23), 27, 28 (see Definitions 7.2.6 and 7.3.2), 36, 40, 41, 42, 45, 46, 47
• Section 7.4 page 355: 1, 8, 25, 33
• Extra Problem on diagonalization (Highly Recommended!!!)
• Section 7.5 page 353: 1, 2, 7, 8, 15, 17, 21, 23, 24. |
# Eureka Math Geometry Module 2 Lesson 31 Answer Key
## Engage NY Eureka Math Geometry Module 2 Lesson 31 Answer Key
### Eureka Math Geometry Module 2 Lesson 31 Example Answer Key
Example 1.
Find the area of ∆ GHI.
Allow students the opportunity and the time to determine what they must find (the height) and how to locate it (one option is to drop an altitude from vertex H to side $$\overline{GI}$$). For students who are struggling, consider showing just the altitude and allowing them to label the newly divided segment lengths and the height.
→ How can the height be calculated?
By applying the Pythagorean theorem to both of the created right triangles to find x,
h2 = 49 – x2 h2 = 144 – (15 – x)2
49 – x2 = 144 – (15 – x)2
49 – x2 = 144- 225 + 30x – x2
130 = 30x
x = $$\frac{13}{3}$$
HJ = $$\frac{13}{3}$$, IJ = $$\frac{32}{3}$$
The value of x can then be substituted into either of the expressions equivalent to h2 to find h.
h2 = 49 – $$\left(\frac{13}{3}\right)^{2}$$
h2 = 49 – $$\frac{169}{9}$$
h = $$\frac{4 \sqrt{17}}{3}$$
→ What is the area of the triangle?
Area = $$\left(\frac{1}{2}\right)(15)\left(\frac{4 \sqrt{17}}{3}\right)$$
Area = 10√17
Example 2.
A farmer is planning how to divide his land for planting next year’s crops. A triangular plot of land is left with two known
side lengths measuring 500 m and 1,700 m.
What could the farmer do next in order to find the area of the plot?
→With just two side lengths known of the plot of land, what are the farmer’s options to determine the area of his plot of land?
He can either measure the third side length, apply the Pythagorean theorem to find the height of the triangle, and then calculate the area, or he can find the measure of the included angle between the known side lengths and use trigonometry to express the height of the triangle and then determine the area of the triangle.
→ Suppose the included angle measure between the known side lengths is 30°. What is the area of the plot of land? Sketch a diagram of the plot of land.
Area = $$\frac{1}{2}$$ (1700)(500) sin 30
Area = 212 500
The area of the plot of land is 212,500 square meters.
### Eureka Math Geometry Module 2 Lesson 31 Opening Exercise Answer Key
Three triangles are presented below. Determine the areas for each triangle, if possible. If it is not possible to find the area with the provided information, describe what Is needed in order to determine the area.
The area of ∆ ABC is $$\frac{1}{2}$$(5)(12), or 30 square units, and the area of ∆ DEF is $$\frac{1}{2}$$ (8)(2o), or 80 square units. There is not enough information to find the height of ∆ GHI and, therefore, the area of the triangle.
Is there a way to find the missing information?
Without further information, there is no way to calculate the area.
### Eureka Math Geometry Module 2 Lesson 31 Exercise Answer Key
Exercise 1.
A real estate developer and her surveyor are searching for their next piece of land to build on. They each examine a plot of land in the shape of ∆ ABC. The real estate developer measures the length of $$\overline{A B}$$ and $$\overline{A C}$$ and finds them to both be approximately 4,000 feet, and the included angle has a measure of approximately 50°. The surveyor measures the length of $$\overline{A C}$$ and $$\overline{B C}$$ and finds the lengths to be approximately 4,000 feet and 3,400 feet, respectively, and measures the angle between the two sides to be approximately 65°.
a. Draw a diagram that models the situation, labeling all lengths and angle measures.
b. The real estate developer and surveyor each calculate the area of the plot of land and both find roughly the same area. Show how each person calculated the area; round to the nearest hundred. Redraw the diagram with only the relevant labels for both the real estate agent and surveyor.
c. What could possibly explain the difference between the real estate agent’s and surveyor’s calculated areas?
The difference in the area of measurements can be accounted for by the approximations of the measurements taken instead of exact measurements.
### Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key
Find the area of each triangle. Round each answer to the nearest tenth.
Question 1.
Area = $$\frac{1}{2}$$ (12)(9)(sin 21)
Area = 54(sin 21) ≈ 19.4
The area of the triangle is approximately 19.4 square units.
Question 2.
Area = $$\frac{1}{2}$$ (2)(11)(sin 34)
Area = 11(sin 34) ≈ 6.2
The area of the triangle is approximately 6.2 square units.
Question 3.
Area = $$\frac{1}{2}$$(8) (6$$\frac{1}{2}$$) (sin 55)
Area = 26(sin 55) ≈ 21.3
The area of the triangle is approximately 21.3 square units.
Question 4.
The included angle is 60° by the angle sum oía triangle.
Area = $$\frac{1}{2}$$ (12)(6 + 6√3) sin 60
Area = 6(6 + 6√3) $$\left(\frac{\sqrt{3}}{2}\right)$$
Area = (36 + 36√3) $$\left(\frac{\sqrt{3}}{2}\right)$$
Area = 18√3 + 18(3)
Area = 18√3 + 54 ≈ 85.2
The area of the triangle is approximately 85.2 square units.
Question 5.
In ∆ DEF, EF = 15, DF = 20, and m∠F = 63°. Determine the area of the triangle. Round to the nearest tenth.
Area = $$\frac{1}{2}$$ (20) (15) sin(63) ≈ 133.7
The area of ∆ DEF is 133.7 units2.
Question 6.
A landscape designer is designing a flower garden for a triangular area that is bounded on two sides by the client’s house and driveway. The length of the edges of the garden along the house and driveway are 18 ft. and 8 ft., respectively, and the edges come together at an angle of 80°. Draw a diagram, and then find the area of the garden to the nearest square foot.
The garden is in the shape of a triangle in which the lengths of two sides and the included angle have been provided.
Area(ABC) = $$\frac{1}{2}$$ (8 ft.)(18 ft.) sin 80
Area(ABC) = (72 sin 80) ft2
Area(ABC) ≈ 71 ft2
Question 7.
A right rectangular pyramid has a square base with sides of length 5. Each lateral face of the pyramid is an isosceles triangle. The angle on each lateral face between the base of the triangle and the adjacent edge is 75°. Find the surface area of the pyramid to the nearest tenth.
Using tangent, the altitude of the triangle to the base of length 5 is equal to 2.5 tan 75.
Using tangent, the altitude of the triangle to the base of length 5 is equal to 2.5 tan 75.
Area = $$\frac{1}{2}$$ bh
Area = $$\frac{1}{2}$$ (5)(2.5 sin 75)
Area = 6. 25(sin 75)
The total surface area of the pyramid is the sum of the four lateral faces and the area of the square base:
SA = 4(6. 25(sin 75)) + 52
SA = 25 sin 75 + 25
SA ≈ 49.1
The surface area of the right rectangular pyramid is approximately 49.1 square units.
Question 8.
The Pentagon building in Washington, DC, is built in the shape of a regular pentagon. Each side of the pentagon measures 921 ft. in length. The building has a pentagonal courtyard with the same center. Each wall of the center courtyard has a length of 356 ft. What is the approximate area of the roof of the Pentagon building?
Let A1 represent the area within the outer perimeter of the Pentagon building in square feet.
The area within the outer perimeter of the Pentagon building is approximately 1,459, 379 ft2.
The area of the roof of the Pentagon building is approximately 1, 241, 333 ft2.
Question 9.
A regular hexagon is inscribed in a circle with a radius of 7. Find the perimeter and area of the hexagon.
The regular hexagon can be divided into six equilateral triangular regions with each side of the triangles having a length of 7. To find the perimeter of the hexagon, solve the following:
6 7 = 42, so the perimeter of the hexagon is 42 units.
To find the area of one equilateral triangle:
Area = $$\frac{1}{2}$$ (7)(7) sin 60
Area = $$\frac{49}{2}\left(\frac{\sqrt{3}}{2}\right)$$
Area = $$\frac{49 \sqrt{3}}{4}$$
The area of the hexagon is six times the area of the equilateral triangle.
Total Area = 6$$\left(\frac{49 \sqrt{3}}{4}\right)$$
Total Area = $$\frac{147 \sqrt{3}}{2}$$ ≈ 127.3
The total area of the regular hexagon is approximately 127.3 square units.
Question 10.
In the figure below, ∠AEB is acute. Show that Area (∆ ABC) = AC · BE · sin ∠AEB.
Let θ represent the degree measure of angle AEB, and let h represent the altitude of ∆ ABC (and ∆ ABE).
Area(∆ ABC) = $$\frac{1}{2}$$ · AC · h
sin θ = $$\frac{h}{B E}$$ which implies that h = BE · sin θ.
Therefore, by substitution:
Area(∆ ABC) = $$\frac{1}{2}$$ AC · BE · sin ∠AEB.
Question 11.
Let ABCD be a quadrilateral. Let w be the measure of the acute angle formed by diagonals $$\overline{AC}$$ and $$\overline{BD}$$. Show that
Area(ABCD) = $$\frac{1}{2}$$ AC · BD · sin w.
(Hint: Apply the result from Problem 10 to ∆ ABC and ∆ ACD.)
Let the intersection of $$\overline{AC}$$ and $$\overline{BD}$$ be called point P.
Using the results from Problem 10, solve the following:
Area(∆ ABC) = $$\frac{1}{2}$$AC · BP · sin w and
Area(∆ ADC) = $$\frac{1}{2}$$AC · PD · sin w
Area(ABCD) = ($$\frac{1}{2}$$AC · BP · sin w) + ($$\frac{1}{2}$$Ac · PD sin w) Area is additive.
Area(ABCD) = ($$\frac{1}{2}$$Ac. sin w) · (BP + PD) Distributive property
Area(ABCD) = ($$\frac{1}{2}$$Ac. sin w) · (BD) Distance is additive
And commutative addition gives us Area(ABCD) = $$\frac{1}{2}$$ · AC · BD · sin w.
### Eureka Math Geometry Module 2 Lesson 31 Exit Ticket Answer Key
Question 1.
Given two sides of the triangle shown, having lengths of 3 and 7 and their included angle of 49°, find the area of the triangle to the nearest tenth.
Area = $$\frac{1}{2}$$ (3)(7)(sin 49)
Area = 10. 5(sin 49) ≈ 7.9
The area of the triangle is approximately 7.9 square units.
Question 2.
In isosceles triangle PQR, the base QR = 11, and the base angles have measures of 71.45°. Find the area of ∆ PQR to the nearest tenth.
Drawing an altitude from P to midpoint M on $$\overline{Q R}$$ cuts the isosceles triangle into two right triangles with
tan 71.45 = $$\frac{P M}{5.5}$$
Area = $$\frac{1}{2}$$ bh
Area = $$\frac{1}{2}$$ (11)(5. 5(tan 71.45)) |
# Line intersecting circle.xls
(4/1)
### Description
Purpose of calculation
Find intersection between a line and a circle with centre at origin.
Calculation Reference
Geometry for Enjoyment and Challenge, Rhoad
Calculation Validation
Check graphically in chart.
XLC addin used to verify cell formulas.
Calculation Reference
Geometry for Enjoyment and Challenge, Rhoad
Geometry
Coordinate geometry
To find the intersection between a line and a circle with center at the origin, we can use the general equation of a circle:
x^2 + y^2 = r^2
where (0,0) is the center of the circle, and r is the radius.
Let the equation of the line be y = mx + b, where m is the slope and b is the y-intercept.
Substituting y = mx + b into the equation of the circle, we get:
x^2 + (mx + b)^2 = r^2
Expanding the squares, we get:
x^2 + m^2x^2 + 2bmx + b^2 = r^2
Rearranging, we get:
(m^2 + 1) x^2 + 2bm x + (b^2 - r^2) = 0
This is a quadratic equation in x. Solving for x using the quadratic formula, we get:
x = [-bm ± sqrt((bm)^2 - (m^2 + 1)(b^2 - r^2))] / (m^2 + 1)
Substituting this value of x into the equation of the line, we get the corresponding values of y:
y = mx + b
So the points of intersection between the line and the circle are:
(x1, y1) = ([-bm + sqrt((bm)^2 - (m^2 + 1)(b^2 - r^2))] / (m^2 + 1), m([-bm + sqrt((bm)^2 - (m^2 + 1)(b^2 - r^2))] / (m^2 + 1)) + b)
(x2, y2) = ([-bm - sqrt((bm)^2 - (m^2 + 1)(b^2 - r^2))] / (m^2 + 1), m([-bm - sqrt((bm)^2 - (m^2 + 1)(b^2 - r^2))] / (m^2 + 1)) + b)
These are the two points of intersection between the line and the circle.
### Calculation Preview
28 Apr 2023
File Size: 115.50 Kb |
Assignment 9
"No employment can be managed without arithmetic, no mechanical invention without geometry." -- (Benjamin Franklin) The effects of time have destroyed many great mathematical works of significance. One survivor is the Rhind Papyrus, from approximately 1800 B.C., which provides much information about ancient Egyptian mathematics. It is about 1 foot high and 18 feet long, and rests in a British museum. The Rhind Papyrus contains 85 problems written in hieratic form. Many of the problems involve the concepts of counting and measuring. Egyptian arithmetic operations didn't include fractions that contained numerators other than 1. The Rhind Papyrus contains a table that allows the reader to represent fractions as a sum of fractions with a numerator of 1. For instance, 2/97 = 1/56 + 1/679 + 1/776. When Herkimer was a clergyman, what did he say to the parishioner whose name he couldn't remember? Answer: "I can't remember your name, but your faith is familiar." Things Herky would like to know: How did a fool and his money get together in the first place? Why is the word abbreviation so long? Reading: Review Section 2.3, as necessary, Written: Pages 87-88/58-63 (Write these up neatly.)
Mathematical word analysis:FRACTION: From the Latin word frangere (to break). In this sense, a fraction is a broken portion of some whole.
To determine a line, one needs two basic things: The slope and a point on the line. It is true that if you are given two points on the line, you can "determine" the line, but to describe it with an equation you need to calculate the slope.
Example: What is the equation of the line with slope 3 that contains the point (2,15).
Response: y - 15 = 3(x - 2) ==> y = 3x +9.
Example: What is the equation of the line containing (3,6) and (-1,14).
Response: The slope of the line is (6-14)/[3-(-1)] = -8/4 = -2. The equation of the line is
y - 6 = -2(x - 3) ==> y = -2x +12. |
# Lesson video
In progress...
Hi, I'm Miss.
Kidd-Rossiter, and I'm going to be taking today's lesson on density as rate.
Before we get started, please make sure you're in a nice, quiet place, free from any distractions, and that you've got something to write with and something to write on.
If you need to pause the video now to get any of that sorted, then please do.
If not, let's get going.
So, on your screen, you've got a graph that's telling us something about honey.
On the X-axis, we've got litres.
On the Y-axis, we've got kilogrammes.
Pause the video now and think about what is the graph telling us about honey? And what rate can you see here? Pause now.
Excellent, what did you think? What does kilogrammes measure? What does litres measure? What is the graph telling us? Let's talk about it together.
Well, the first thing that we need to understand is what kilogrammes and litres are representing.
So, kilogrammes is mass, okay? So, kilogrammes is mass, and when people talk about how much they weigh, what they're talking about is their mass there, okay? And litres is a measure of capacity.
So that's how much of something can fit somewhere.
So, a water bottle might have a capacity of one litre or two litres.
From the graph, we can see that five litres of honey has a mass of seven kilogrammes.
So, five litres of honey has a mass of seven kilogrammes.
What does that mean? What's our rate here then? So, can we work out what one litre would be? So, five litres has a mass of seven kilogrammes.
What would one litre have a mass of? Pause the video now and think about that.
Excellent, it would be seven divided by five, wouldn't it? Kilogrammes.
So, one litre is 7/5 of a kilogramme, or we could say one litre is 1.
4 kilogrammes.
So, this is what we call the density.
So we use the word density to describe the mass of a substance per, for example, litre.
So that's what density is.
So, we can see here, that the density is the constant of proportionality, and the density is the rate here.
So, the rate of kilogrammes per litre.
So, our rate is 1.
4 kilogrammes per litre.
So, pause the video here, navigate to the independent task, and when you're ready to go through the answers, resume the video.
Good luck! Well done on that independent task.
Let's go through some of the answers.
So, what is the mass of two litres of fruit juice? So, if we read up from our X-axis and then across, we can see that the mass is eight kilogrammes.
So, let's draw ourselves a little table where we have litres and then kilogrammes.
We know that eight kilogrammes is two litres, so that must mean that one litre is how many kilogrammes? Excellent, four.
So, we can see here that our constant of proportionality is four, so that means that our density must be four kilogrammes per litre.
How many litres of fruit juice have a mass of nine kilogrammes? So, we know if we're going this way, we multiply by four.
So if we're going in the opposite way, we can either times by the reciprocal, which is 1/4, or divide by four.
They're equivalent calculations, aren't they? So, that gives us 9/4, which is 2.
25 litres.
If you left it as 9/4, that's absolutely fine as well.
And then finally, how many litres of fruit juice have a mass of 50 kilogrammes? So again, we multiply by the quarter or divide by four, so 50/4 gives us 12.
5 litres.
It's fun then.
10 litres of sparkling water have a mass of nine kilogrammes.
So, let's draw our table again, litres and kilogrammes.
We've got 10 and nine, so that means that this way, our constant of proportionality or our density is 9/10, so the density of sparkling water will be 0.
9 or 9/10 kilogrammes per litre.
How many litres of sparkling water have a mass of five kilogrammes? So, if we're going back the other way, we'd be multiplying by 10/9, which will give us 50/9, which is equivalent to 5.
5 recurring litres.
And then finally, what is the mass of 200 litres of sparkling water? Well, we're multiplying by 9/10, so that gives us 180 kilogrammes.
Use the previous two graphs to help with this section.
So, 10 litres of sparkling water are mixed with one litre of fruit juice.
So all together, we've got how many litres then? Excellent, 11 litres.
And if the mass of 10 litres of sparkling water is nine kilogrammes, and the mass of one litre of fruit juice is four kilogrammes, what's the mass of the total mixture? Excellent, 13 kilogrammes.
So, we've just worked out that one, haven't we? So that's the answer to that.
What's the density of the mixture? So, I'm just going to do this one up here, where I've got a little bit more room.
Density.
Then we know, we draw our table, litres and kilogrammes.
11 litres is 13 kilogrammes, so that means that we must multiply by 13/11, so that means that our density must be 13/11 kilogrammes per litre, which is equivalent to 1.
18 recurring kilogrammes per litre.
So we know that for every 11 litres, we have 13 kilogrammes.
So, hopefully you've managed to do this really accurately on your graph.
So, I know that for 11 litres, I've got 13 kilogrammes.
That means for 22 litres, I must have 26 kilogrammes.
Which means for 44 litres, I must have 52 kilogrammes.
For 66 litres, I must have 78 kilogrammes.
And so on.
Then you will have drawn with a ruler, a line that goes through these points, a straight line that goes through these points all the way through to zero.
But like I say, yours will be done with a ruler.
How many different sentences can you write to compare densities of vegetable oil, water, and honey? If you're happy to give this a go, pause the video now and get going.
If you need a little bit of support, there's some sentence starters here to help you out.
So, pause the video now and give this task a go.
Excellent.
There were so many different responses here, so I'm sure that you've written loads down.
If you're really interested in this topic, it might be good for you to go and research and draw some of your own graphs to represent density.
That's the end of today's lesson, so thank you very much for all your hard work.
I hope you've enjoyed it.
Don't forget to go and take the end of the lesson quiz so that you can show me what you've learned, and hopefully I'll see you again soon.
Bye!. |
## Solve Lesson 17 Page 30 Math 10 SBT – Kite>
Topic
a) Representation of the solution domain of the system of inequalities $$\left\{ {\begin{array}{*{20}{c}}{x + y \le 5}\\{3x + 2y \le 12}\\{x \ge 1}\\{ y \ge 0}\end{array}} \right.\left( {III} \right)$$
b) Find x, y are solutions of the system of inequalities (III) such that $$F = 3x + 7y$$ maximum and minimum values.
Solution method – See details
a)
• Step 1: Draw a line $$d:x – 2y = 4$$.
• Step 2: Take a point $$M\left( {{x_o};{y_o}} \right)$$ that is not on d (we usually take the origin O if $$c \ne 0$$. (a{x_o} + b{y_o}\) and compare with c
• Step 3: Conclusion
• If $$a{x_o} + b{y_o} < c$$ then the half-plane (excluding line d) containing the point M is the solution domain of the inequality $$ax + by < c$$
• If $$a{x_o} + b{y_o} > c$$ then the half-plane (excluding d) containing no point M is the solution domain of the inequality $$ax + by > c$$
b)
– Representation of the solution domain of the system of inequalities on the coordinate system
$$F\left( {x;y} \right)$$ reaches max or min at one of the vertices so we only need to compute the value of $$F\left( {x;y} \right)$$ at those vertices
Detailed explanation
a) We draw four lines:
dfirst: x + y = 5 is a straight line passing through two points with coordinates (0; 5) and (5; 0);
d2: 3x + 2y = 12 is a line passing through two points with coordinates (4; 0) and (0; 6);
d3: x = 1 is a line parallel to the vertical axis and passing through the point (1; 0);
d4: y = 0 is the horizontal axis.
We define each solution domain of each inequality in the system, crossing out the parts that are not in the solution domain of each inequality.
The solution domain of the system of inequalities is the region in quadrilateral ABCD with A(1; 0), B(1; 4), C(2; 3) and D(4; 0) as shown in the following figure:
b) We have the expression F = 3x + 7y reaching the maximum value, the smallest value at one of the vertices of the quadrilateral ABCD.
At A(1; 0) with x = 1 and y = 0, then F = 3.1 + 7.0 = 3;
At B(1; 4) with x = 1 and y = 4, then F = 3.1 + 7.4 = 31;
At C(2; 3) with x = 2 and y = 3, then F = 3.2 + 7.3 = 27;
At D(4; 0) with x = 4 and y = 0, then F = 3.4 + 7.0 = 12.
So the maximum value of F is 31 at x = 1 and y = 4, the minimum value of F is 3 at x = 1 and y = 0 |
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# 8.22: Problem Solving Plan Proportions
Difficulty Level: At Grade Created by: CK-12
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### Let’s Think About It
Takeru Kobayashi is a competitive eater famous for eating hot dogs among other things. Once, he ate 110 hot dogs in 10 minutes at the New York State Fair. How many hot dogs can Kobayashi eat in 1 minute?
In this concept, you will learn the problem solving strategy: use a proportion.
### Guidance
You can solve ratio problems using proportions. A proportion is an equation that shows two equivalent ratios. In order to use a proportion, the ratios in the proportion must also compare the same things.
Let’s look at a problem.
A car travels 55 miles in two hours. A bus travels 85 kilometers in two hours. Which vehicle traveled a farther distance?
This problem compares speed per hour; however, it is comparing miles to kilometers. The units are not the same and you cannot use a proportion to solve this problem without converting the units first.
Let’s look at another problem.
A cheetah can run 75 miles per hour. If you could run three times as fast as a cheetah, how fast would you be able to run?
The first ratio compares the cheetah’s speed per hour. The second ratio compares a person’s speed per hour. Both ratios can be written as miles per hour. You can use a proportion to solve this problem.
First, write a proportion showing the comparison.
cheetah's speednumber of hours=person's speednumber of hours\begin{align*}\frac{\text{cheetah's speed}}{\text{number of hours}}= \frac{\text{person's speed}}{\text{number of hours}}\end{align*}
Next, take the data and fill it into the proportion.
751=x3\begin{align*}\frac{75}{1}=\frac{x}{3}\end{align*}
The cheetah runs 75 miles per hour. Per means “divided by” and “hour” refers to 1 hour. The numerator is 75 and the denominator is 1. The person runs three times as fast, so he or she would go as far in 1 hour as a cheetah would in 3 hours. The denominator is 3. The person’s speed is unknown so it is represented with the variable x\begin{align*}x\end{align*}.
Then, solve the proportion using cross products.
xx==75(3)225\begin{align*}\begin{array}{rcl} x & = & 75(3)\\ x & = & 225 \end{array}\end{align*}
If a person ran three times as fast as a cheetah, he or she would run 225 mph.
### Guided Practice
Use a proportion to solve the following problem.
If a person can run 3 miles in 20 minutes, how long will it take the same person to run 12 miles at the same rate?
First, set up the proportion.
milesminutes=milesminutes\begin{align*}\frac{\text{miles}}{\text{minutes}}= \frac{\text{miles}}{\text{minutes}}\end{align*}
Next, fill in the given information.
320=12x\begin{align*}\frac{3}{20} = \frac{12}{x}\end{align*}
Now we cross multiply and solve for x\begin{align*}x\end{align*}.
3x3xx===20(12)24080\begin{align*}\begin{array}{rcl} 3x & = & 20(12)\\ 3x & = & 240\\ x & = & 80 \end{array}\end{align*}
The person would run 12 miles in 80 minutes or 1 hour and 20 minutes.
### Examples
Use the information below to answer the following questions.
A cheetah runs 75 miles per hour.
#### Example 1
If you could run twice as fast as a cheetah, how fast could you run?
First, set up the proportion.
mileshour=mileshour\begin{align*}\frac{\text{miles}}{\text{hour}}= \frac{\text{miles}}{\text{hour}}\end{align*}
Next, fill in the given information.
751=x2\begin{align*}\frac{75}{1} = \frac{x}{2}\end{align*}
Then, cross multiply and solve for x\begin{align*}x\end{align*}.
xx==75(2)150\begin{align*}\begin{array}{rcl} x & = & 75(2)\\ x & = & 150 \end{array}\end{align*}
You could run 150 miles per hour.
#### Example 2
If you could run half as fast as a cheetah, how fast could you run?
First, set up the proportion.
mileshour=mileshour\begin{align*}\frac{\text{miles}}{\text{hour}}= \frac{\text{miles}}{\text{hour}}\end{align*}
Next, fill in the given information.
751=x0.5\begin{align*}\frac{75}{1} = \frac{x}{0.5}\end{align*}
Then, cross multiply and solve for x\begin{align*}x\end{align*}.
xx==75(0.5)37.5\begin{align*}\begin{array}{rcl} x & = & 75(0.5)\\ x & = & 37.5 \end{array}\end{align*}
You could run 37.5 miles per hour.
#### Example 3
If you could run four times as fast as a cheetah, how fast could you run?
First, set up the proportion.
mileshour=mileshour\begin{align*}\frac{\text{miles}}{\text{hour}}= \frac{\text{miles}}{\text{hour}}\end{align*}
Next, fill in the given information.
751=x4\begin{align*}\frac{75}{1} = \frac{x}{4}\end{align*}
Then, cross multiply and solve for x\begin{align*}x\end{align*}.
xx==75(4)300\begin{align*}\begin{array}{rcl} x & = & 75(4)\\ x & = & 300 \end{array}\end{align*}
You could run 300 miles per hour.
Remember the competitive eater Takeru Kobayashi?
He once ate 110 hot dogs in 10 minutes. To find out how many hot dogs he can eat in 1 minute, write a proportion and solve for x\begin{align*}x\end{align*}.
First, set up the proportion.
hot dogsminutes=hot dogsminutes\begin{align*}\frac{\text{hot dogs}}{\text{minutes}}= \frac{\text{hot dogs}}{\text{minutes}}\end{align*}
Next, fill in the given information.
11010=x1\begin{align*}\frac{110}{10} = \frac{x}{1}\end{align*}
Then, cross multiply and solve for x\begin{align*}x\end{align*}.
10x10xx===110(1)11011\begin{align*}\begin{array}{rcl} 10x & = & 110(1)\\ 10x & = & 110\\ x & = & 11 \end{array}\end{align*}
Takeru Kobayashi can eat 11 hot dogs in 1 minute.
### Explore More
Solve each word problem by using a proportion.
1. In a diagram for the new garden, one inch is equal to 3 feet. If this is the case, how many feet is the actual garden edge if the measurement on the diagram is 5 inches?
2. If two inches on a map are equal to three miles, how many miles are represented by four inches?
3. If eight inches on a map are equal to ten miles, how many miles are 16 inches equal to?
4. Casey drew a design for bedroom. On the picture, she used one inch to represent five feet. If her bedroom wall is ten feet long, how many inches will Casey draw on her diagram to represent this measurement?
5. If two inches are equal to twelve feet, how many inches would be equal to 36 feet?
6. If four inches are equal to sixteen feet, how many feet are two inches equal to?
7. The carpenter chose a scale of 6" for every twelve feet. Given this measurement, how many feet would be represented by 3"?
8. If 9 inches are equal to 27 feet, how many feet are equal to three inches?
9. If four inches are equal to 8 feet, how many feet are equal to two inches?
10. If six inches are equal to ten feet, how many inches are five feet equal to?
11. If four inches are equal to twelve feet, how many inches are equal to six feet?
12. For every 20 feet of fence, John drew 10 inches on his plan. If the real fence is only 5 feet long, how many inches will John draw on his plan?
13. If eight inches are equal to twelve feet, how many inches are equal to six feet?
14. How many inches are equal to 20 feet if 4 inches are equal to 10 feet?
15. How many inches are equal to 8 feet if six inches are equal to 16 feet?
16. Nine inches are equal to twelve feet, so how many inches are equal to 4 feet?
17. If a person runs two miles in twelve minutes, how long will it take them to run 4 miles at the same rate?
18. A person runs 1 mile in 16 minutes. Given this information, how long will it take him/her to run 3 miles?
19. If a person runs two miles in twenty minutes, at what rate does he/she run one mile?
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 8.22.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
Problem Solving
Problem solving is using key words and operations to solve mathematical dilemmas written in verbal language.
Proportion
A proportion is an equation that shows two equivalent ratios.
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# AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers Ex 4
AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers Ex 4 Textbook Questions and Answers.
## AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Exercise 4
Question 1.
Find the reciprocal of each of the following fractions.
(i) $$\frac { 5 }{ 8 }$$
(ii) $$\frac { 8 }{ 7 }$$
(iii) $$\frac { 13 }{ 7 }$$
(iv) $$\frac { 3 }{ 4 }$$
Solution:
i) Reciprocal of $$\frac { 5 }{ 8 }$$ = $$\frac { 8 }{ 5 }$$
ii) Reciprocal of $$\frac { 8 }{ 7 }$$ = $$\frac { 7 }{ 8 }$$
iii) Reciprocal of $$\frac { 13 }{ 7 }$$ = $$\frac { 7 }{ 13 }$$
iv) Reciprocal of $$\frac { 3 }{ 4 }$$ = $$\frac { 4 }{ 3 }$$
Question 2.
Find
(i) 18 ÷ $$\frac { 3 }{ 4 }$$
(ii) 8 ÷ $$\frac { 7 }{ 3 }$$
(iii) 3 ÷ 2$$\frac { 1 }{ 3 }$$
(iv) 5 ÷ 3$$\frac { 4 }{ 7 }$$
Solution:
Question 3.
Find
(i) $$\frac { 2 }{ 5 }$$ ÷ 3
(ii) $$\frac { 7 }{ 8 }$$ ÷ 5
(iii) $$\frac { 4 }{ 9 }$$ ÷ $$\frac { 4 }{ 5 }$$
Solution:
Question 4.
Deepak can paint $$\frac { 2 }{ 5 }$$ of a house in one day. lfhe continues working at this rate, how many days will he take to paint the whole house?
Amount of work done by deepak in 1 day = $$\frac { 2 }{ 5 }$$ th
Total work can be done in = 1 ÷ $$\frac { 2 }{ 5 }$$ days
= $$\frac { 2 }{ 5 }$$ = $$1 \times \frac{5}{2}=\frac{1 \times 5}{2}=\frac{5}{2}=2 \frac{1}{2}$$ days. |
# Base of Exponential form
## Definition
A number which considered as a parameter to split any number as its multiplicative factors for expressing it in exponential notation is called base of the exponential notation.
Any number can be expressed as multiplicative factors on the basis of another number. The number which considered as a parameter and its number multiplicative factors are expressed in exponential notation to represent the respective number symbolically.
Due to splitting a number as multiplicative factors on the basis of a particular number to express it in exponential form, the number which is taken as a parameter to split the number as its multiplicative factors is known as base of the exponential notation. It is also called as the base of the exponential form.
### Example
$125$ is a number and express it as multiplicative factors of another number $5$.
$125 = 5 \times 5 \times 5$
Express this expansion in exponential notation.
$125 = 5^{\, \displaystyle 3}$
On the basis of number $5$, the number $125$ is expressed as three multiplicative factors of $5$. Hence, the number $5$ is called the base of the exponential notation for the number $125$.
Observe the following two examples for better understanding.
$2401 = 7 \times 7 \times 7 \times 7$
$\implies 2401 = 7^{\displaystyle \, 4}$
The number $7$ is called the base of the exponential form.
$59049 = 9 \times 9 \times 9 \times 9 \times 9$
$\implies 59049 = 9^{\displaystyle \, 5}$
The number $9$ is called the base of the exponential notation.
#### Algebraic form
Assume, $m$ is a number and it is split into $n$ number of multiplicative factors on the basis of another number $b$.
$m = \underbrace{b \times b \times b \times … \times b}_{\displaystyle n factors}$
Express relation between three of them in exponential notation.
$m = b^{\displaystyle n}$
On the basis of number $b$, the number $m$ is split into $n$ number of multiplicative factors. Therefore, the number $b$ is called base of the exponential form.
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# Eureka Math Algebra 2 Module 4 Lesson 14 Answer Key
## Engage NY Eureka Math Algebra 2 Module 4 Lesson 14 Answer Key
### Eureka Math Algebra 2 Module 4 Lesson 14 Example Answer Key
Example 1: Polls
A recent poll stated that 40% of Americans pay ‘a great deal” or a ‘fair amount” of attention to the nutritional information that restaurants provide. This poll was based on a random sample of 2,027 adults living in the United States.
The 40% corresponds to a proportion of 0.40, and 0.40 is called a sample proportion. It is an estimate of the proportion of all adults who would say they pay “a great deal” or a “fair amount” of attention to the nutritional information that restaurants provide.
If you were to take a random sample of 20 Americans, how many would you predict would say that they pay attention to nutritional information? In this lesson, you will investigate this question by generating distributions of sample proportions and investigating patterns in these distributions.
Your teacher will give your group a container of dried beans. Some of the beans in the container are black. With your classmates, you are going see what happens when you take a sample of beans from the container and use the proportion of black beans in the sample to estimate the proportion of black beans in the container (a population proportion).
Exploratory Challenge 1/Exercises 1-9:
Exercise 1.
Each person in the group should randomly select a sample of 20 beans from the container by carefully mixing all the beans and then selecting one bean and recording its color. Replace the bean, mix the bag, and continue to select one bean at a time until 20 beans have been selected. Be sure to replace each bean and mix the bag before selecting the next bean. Count the number of black beans in your sample of 20.
Answers will vary, but the number of black beans will center around 8.
Exercise 2.
What is the proportion of black beans in your sample of 20? (Round your answer to 2 decimal places.) This value is called the sample proportion of black beans.
Answers will vary, but the sample proportions will center around 0.4.
Exercise 3.
Write your sample propomon on a sticky note, and place the note on the number line that your teacher has drawn on the board. Place your note above the value on the number line that corresponds to your sample proportion.
Class data will vary. One possible sampling distribution is shown below.
The graph of all the students’ sample proportions is called the sampling distribution of the samples’ proportions. This sampling distribution is an approximation of the actual sampling distribution of all possible samples of size 20.
Exercise 4.
Describe the shape of the distribution.
Answers might vary, but the shape is generally mound-shaped.
Exercise 5.
What was the smallest sample proportion observed?
Answers will vary. Based on the sample graph: 0.15
Exercise 6.
What was the largest sample proportion observed?
Answers will vary. Based on the sample graph: 0.65
Exercise 7.
What sample proportion occurred most often?
Answers will vary, but the sample proportion should be around 0.4. Based on the sample graph: 0.35
Exercise 8.
Using technology, find the mean and standard deviation of the sample proportions used to construct the sampling distribution created by the class.
Answers will vary, but the mean will be approximately 0.4, and the standard deviation will be approximately 0.11.
Exercise 9.
How does the mean of the sampling distribution compare with the population proportion of 0.40?
Answers will vary, but the two values should be about the same. In theory, the mean of the sampling distribution of sample proportions is equal to the population proportion.
Example 2: Sampling Variability
What do you think would happen to the sampling distribution if everyone in the class took a random sample of 40 beans from the container? To help answer this question, you will repeat the process described in Example 1, but this time you will draw a random sample of 40 beans instead of 20.
Exploratory Challenge 2/Exercises 10 – 21:
Exercise 10.
Take a random sample with the replacement of 40 beans from the container. Count the number of black beans in your sample of 40 beans.
Answers will vary, but the number of black beans will center around 16.
Exercise 11.
What is the proportion of black beans in your sample of 40? (Round your answer to 2 decimal places.)
Answers will vary, but the sample proportions will center around 0.40.
Exercise 12.
Write your sample proportion on a sticky note, and place It on the number line that your teacher has drawn on the board. Place your note above the value on the number line that corresponds to your sample proportion.
Class data will vary. One possible sampling distribution is shown below.
Exercise 13.
Describe the shape of the distribution.
Answers may vary, but the shape is generally mound-shaped.
Exercise 14.
What was the smallest sample proportion observed?
Answers will vary. Based on the sample graph: 0.30
Exercise 15.
What was the largest sample proportion observed?
Answers will vary. Based on the sample graph: 0. 55
Exercise 16.
What sample proportion occurred most often?
Answer will vary but will be approximately 0.4.
Exercise 17.
Using technology, find the mean and standard deviation of the sample proportions used to construct the sampling distribution created by the class.
Answers will vary, but the mean will be approximately 0.4 and the standard deviation approximately 0.08.
Exercise 18.
How does the mean of the sampling distribution compare with the population proportion of 0.40?
Answers will vary, but the two values should be about the same. In theory, the mean of the sampling distribution of sample proportions is equal to the population proportion.
Exercise 19.
How does the mean of the sampling distribution based on random samples of size 20 compare to the mean of the sampling distribution based on random samples of size 40?
The two means are approximately the same, about 0.4.
Exercise 20.
As the sample size increased from 20 to 40, describe what happened to the sampling variability (standard deviation of the distribution of sample proportions)?
The standard deviation of the distribution of the sample proportions based on a sample size of 40 is less than the standard deviation of the distribution of the sample proportions based on a sample size of 20.
Exercise 21.
What do you think would happen to the variability (standard deviation) of the distribution of sample proportions if the sample size for each sample was 80 instead of 40? Explain.
Because the standard deviation decreased as sample size increased from 20 to 40, I expect that the standard deviation will decrease further when the sample size is 80.
### Eureka Math Algebra 2 Module 4 Lesson 14 Problem Set Answer Key
Question 1.
A class of 28 eleventh graders wanted to estimate the proportion of all juniors and seniors at their high school with part-time jobs after school. Each eleventh grader took a random sample of 30 juniors and seniors and then calculated the proportion with part-time jobs. Following are the 28 sample proportions.
0.7, 0.8, 0.57, 0.63, 0.7, 0.47, 0.67, 0.67, 0.8, 0.77, 0.4, 0.73, 0.63, 0.67, 0.6, 0.77, 0.77, 0.77, 0.53, 0.57, 0.73, 0.7, 0.67, 0.7, 0.77, 0.57, 0.77, 0.67
a. Construct a dot plot of the sample proportions.
b. Describe the shape of the distribution.
Skewed to the left
c. Using technology, find the mean and standard deviation of the sample proportions.
Mean= 0.67
Standard deviation = 0. 1
d. Do you think that the proportion of all juniors and seniors at the school with part-time jobs could be 0.7? Do you think it could be 0.5? Justify your answers based on your dot plot.
It is likely that the proportion of all juniors and seniors with part-time jobs could be 0.70 since 0.70 is near the center of the dot plot. It is unlikely that the proportion of all juniors and seniors Is 0.5 since there are very few samples with a sample proportion of 0.5 or less.
e. Suppose the eleventh graders had taken random samples of size 60. How would the distribution of sample proportions based on samples of size 60 differ from the distribution for samples of size 30?
The sampling distribution would be mound-shaped with approximately the same mean of the sampling distribution based on size 30, but the standard deviation of the sampling distribution based on size 60 would be smaller than one based on samples of size 30.
Question 2.
A group of eleventh graders wanted to estimate the proportion of all students at their high school who suffer from allergies. Each student in one group of eleventh graders took a random sample of 20 students, while each student in another group of eleventh graders each took a random sample of 40 students. Below are the two sampling distributions (shown as histograms) of the sample proportions of high school students who said that they suffer from allergies. Which histogram is based on random samples of size 40? Explain.
Histogram A is based on random samples of size 40 because it has less variability than Histogram B.
Question 3.
The nurse in your school district would like to study the proportion of all high school students in the district who usually get at least eight hours of sleep on school nights. Suppose each student in your class takes a random sample of 20 high school students in the district and each calculates their sample proportion of students who said that they usually get at least eight hours of sleep on school nights. Below is a histogram of the sampling distribution.
a. Do you think that the proportion of all high school students who usually get at least eight hours of sleep on school nights could have been 0.4? Do you think It could have been 0.55? Could It have been 0.75? Justify your answers based on the histogram.
The proportion of all high school students who usually get at least eight hours of sleep is likely around 0.4 since that is near the center of the sampling distribution. The proportion could be 0.55 since that is still close to the center of the distribution. It is unlikely that the proportion of all high school students is 0.75 since none of the samples produced sample proportions as large as 0.75.
b. Suppose students had taken random samples of size 60. How would the distribution of sample proportions based on samples of size 60 differ from those of size 20?
The means of the two distributions would be relatively close, but the standard deviation of the distribution based on samples of size 60 would be smaller than the standard deviation of the distribution based on sample sizes of 20.
### Eureka Math Algebra 2 Module 4 Lesson 14 Exit Ticket Answer Key
A group of eleventh graders wanted to estimate the population proportion of students in their high school who drink at least one soda per day. Each student selected a different random sample of 30 students from the high school and calculated the proportion that drinks at least one soda per day. The dot plot below shows the sampling distribution. This distribution has a mean of 0.51 and a standard deviation of 0.09.
Question 1.
Describe the shape of the distribution. |
# Factor Theorem
In mathematics, factor theorem is used when factoring the polynomials completely. It is a theorem that links factors and zeros of the polynomial.
According to factor theorem, if f(x) is a polynomial of degree n ≥ 1 and ‘a’ is any real number, then, (x-a) is a factor of f(x), if f(a)=0.
Also, we can say, if (x-a) is a factor of polynomial f(x), then f(a) = 0. This proves the converse of the theorem. Let us see the proof of this theorem along with examples.
## What is a Factor Theorem?
Factor theorem is commonly used for factoring a polynomial and finding the roots of the polynomial. It is a special case of a polynomial remainder theorem.
As discussed in the introduction, a polynomial f(x) has a factor (x-a), if and only if, f(a) = 0. It is one of the methods to do the factorisation of a polynomial.
## Proof
Here we will prove the factor theorem, according to which we can factorise the polynomial.
Consider a polynomial f(x) which is divided by (x-c), then f(c)=0.
Using remainder theorem,
f(x)= (x-c)q(x)+f(c)
Where f(x) is the target polynomial and q(x) is the quotient polynomial.
Since, f(c) = 0, hence,
f(x)= (x-c)q(x)+f(c)
f(x) = (x-c)q(x)+0
f(x) = (x-c)q(x)
Therefore, (x-c) is a factor of the polynomial f(x).
### Another Method
f(x)= (x-c)q(x)+f(c)
If (x-c) is a factor of f(x), then the remainder must be zero.
(x-c) exactly divides f(x)
Therefore, f(c)=0.
The following statements are equivalent for any polynomial f(x)
• The remainder is zero when f(x) is exactly divided by (x-c)
• (x-c) is a factor of f(x)
• c is the solution to f(x)
• c is a zero of the function f(x), or f(c) =0
## How to Use Factor Theorem
The steps are given below to find the factors of a polynomial using factor theorem:
Step 1 : If f(-c)=0, then (x+ c) is a factor of the polynomial f(x).
Step 2 : If p(d/c)= 0, then (cx-d) is a factor of the polynomial f(x).
Step 3 : If p(-d/c)= 0, then (cx+d) is a factor of the polynomial f(x).
Step 4 : If p(c)=0 and p(d) =0, then (x-c) and (x-d) are factors of the polynomial p(x).
Rather than finding the factors by using polynomial long division method, the best way to find the factors are factor theorem and synthetic division method. This theorem is used primarily to remove the known zeros from polynomials leaving all unknown zeros unimpaired, thus by finding the zeros easily to produce the lower degree polynomial.
There is another way to define the factor theorem. Usually, when a polynomial is divided by a binomial, we will get a reminder. The quotient obtained is called as depressed polynomial when the polynomial is divided by one of its binomial factors. If you get the remainder as zero, the factor theorem is illustrated as follows:
The polynomial, say f(x) has a factor (x-c) if f(c)= 0, where f(x) is a polynomial of degree n, where n is greater than or equal to 1 for any real number, c.
### Other Methods to Find Factors
Apart from factor theorem, there are other methods to find the factors, such as:
## Problems and Solutions
Factor theorem example and solution are given below. Go through once and get a clear understanding of this theorem. Factor theorem class 9 maths polynomial enables the children to get a knowledge of finding the roots of quadratic expressions and the polynomial equations, which is used for solving complex problems in your higher studies.
Consider the polynomial function f(x)= x2 +2x -15
The values of x for which f(x)=0 are called the roots of the function.
Solving the equation, assume f(x)=0, we get:
x2 +2x -15 =0
x2 +5x – 3x -15 =0
(x+5)(x-3)=0
(x+5)=0 or (x-3)=0
x = -5 or x = 3
Because (x+5) and (x-3) are factors of x2 +2x -15, -5 and 3 are the solutions to the equation x2 +2x -15=0, we can also check these as follows:
If x = -5 is the solution, then
f(x)= x2 +2x -15
f(-5) = (-5)2 + 2(-5) – 15
f(-5) = 25-10-15
f(-5)=25-25
f(-5)=0
If x=3 is the solution, then;
f(x)= x2 +2x -15
f(3)= 32 +2(3) – 15
f(3) = 9 +6 -15
f(3) = 15-15
f(3)= 0
If the remainder is zero, (x-c) is a polynomial of f(x).
### Alternate Method – Synthetic Division Method
We can also use the synthetic division method to find the remainder.
Consider the same polynomial equation
f(x)= x2 +2x -15
We use 3 on the left in the synthetic division method along with the coefficients 1,2 and -15 from the given polynomial equation.
Since the remainder is zero, 3 is the root or solution of the given polynomial.
The techniques used for solving the polynomial equation of degree 3 or higher are not as straightforward. So linear and quadratic equations are used to solve the polynomial equation.
Keep visiting BYJU’S for more information on polynomials and try to solve factor theorem questions from worksheets and also watch the videos to clarify the doubts.
## Frequently Asked Questions – FAQs
Q1
### What is the factor theorem?
According to factor theorem, if f(x) is a polynomial of degree n ≥ 1 and ‘a’ is any real number then, (x-a) is a factor of f(x), if f(a)=0.
Q2
### Why do we use factor theorem?
Factor theorem is commonly used for factoring a polynomial and finding the roots of the polynomial.
Q3
### How to find if x-a is a factor of a polynomial f(x)?
If f(x) is a polynomial, then x-a is the factor of f(x), if and only if, f(a) = 0, where a is the root.
Q4
### Is x-1 a factor of 2x4+3x2-5x+7?
Using factor theorem, if x-1 is a factor of 2x4+3x2-5x+7, then by putting x=1, the given polynomial should equal to zero.
Hence, substituting x = 1 in 2x4+3x2-5x+7, we get:
2x4+3x2-5x+7 = 2(1) + 3(1) – 5 + 7 = 2+3-5+7 = 7
Since the polynomial is not equal to zero, x-1 is not a factor of 2x4+3x2-5x+7.
Q5
### What are the other methods to find the factors of polynomials?
Apart from the factor theorem, we can use polynomial long division method and synthetic division method to find the factors of the polynomial.
Quiz on Factor Theorem |
# How do you use synthetic division to divide 3x^3+4x^2-7x+1 by 3x-2?
Jul 30, 2015
$\textcolor{red}{\frac{3 {x}^{3} + 4 {x}^{2} - 7 x + 6}{3 x - 2} = {x}^{2} + 2 x - 1 - \frac{1}{3 x - 2}}$
#### Explanation:
We use a slightly modified table when the coefficient of $x$ does not equal $1$. Note the extra lines.
Step 1. Write only the coefficients of $x$ in the dividend inside an upside-down division symbol.
$| 3 \text{ "4" "-7" " " " } 1$
$| \textcolor{w h i t e}{1}$
stackrel("——————————————)
Step 2. Put the divisor at the left.
$\text{ "" "|3" "4" "-7" " " " } 1$
$\text{ } \textcolor{red}{2} \textcolor{w h i t e}{1} |$
" "stackrel("——————————————)
Step 3. Write the coefficient of $x$ below the division line
$\text{ "" "|3" "4" "-7" " " " } 1$
$\text{ "2|" "color(white)(1)2 " "" "4" } - 2$
" "stackrel("——————————————)
$\text{ } \textcolor{w h i t e}{1} |$
$\textcolor{red}{/ 3} \textcolor{w h i t e}{1} |$
Step 4. Drop the first coefficient of the dividend below the division symbol.
$\text{ "" "|3" "4" "-7" " " " } 1$
$\text{ "2|" "color(white)(1)2 " "" "4" } - 2$
" "stackrel("——————————————)
$\text{ } \textcolor{w h i t e}{1} | \textcolor{red}{3}$
$/ 3 \textcolor{w h i t e}{1} |$
Step 5. Divide the dropped value by the coefficient of $x$ and place the result in the row below.
$\text{ "" "|3" "4" "-7" " " " } 1$
$\text{ "2|" "color(white)(1)2 " "" "4" } - 2$
" "stackrel("——————————————)
$\text{ "" } | 3$
$/ 3 \textcolor{w h i t e}{1} | \textcolor{red}{1}$
Step 6. Multiply the result by the constant, and put the product in the next column.
$\text{ "" "|3" "4" "-7" " " " } 1$
$\text{ "2|" } \textcolor{w h i t e}{1} \textcolor{red}{2}$
" "stackrel("——————————————)
$\text{ "" } | 3$
$/ 3 \textcolor{w h i t e}{1} | 1$
Step 7. Add down the column.
$\text{ "" "|3" "4" "-7" " " " } 1$
$\text{ "2|" } \textcolor{w h i t e}{1} 2$
" "stackrel("——————————————)
$\text{ "" "|3" } \textcolor{red}{6}$
$/ 3 \textcolor{w h i t e}{1} | 1$
Step 8. Repeat Steps 5, 6, and 7 until you can go no farther.
$\text{ "" "|3" "4" "-7" " " " } 1$
$\text{ "2|" "color(white)(1)2 " "" "4" } - 2$
" "stackrel("——————————————)
$\text{ "" "|3" "6" "-3" } \textcolor{red}{- 1}$
$/ 3 \textcolor{w h i t e}{1} | 1 \text{ "2" } - 1$
$\frac{3 {x}^{3} + 4 {x}^{2} - 7 x + 6}{3 x - 2} = {x}^{2} + 2 x - 1 - \frac{1}{3 x - 2}$
Check:
$\left(3 x - 2\right) \left({x}^{2} + 2 x - 1 - \frac{1}{3 x - 2}\right) = \left(3 x - 2\right) \left({x}^{2} + 2 x - 1\right) - 1$
$= 3 {x}^{3} + 6 {x}^{2} - 3 x - 2 {x}^{2} - 4 x + 2 - 1 = 3 {x}^{3} + 4 {x}^{2} - 7 x + 1$
Jul 30, 2015
$\left(3 {x}^{3} + 4 {x}^{2} - 7 x + 1\right) \div \left(3 x - 2\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$= {x}^{2} + 2 x - 1$$\textcolor{w h i t e}{\text{XXXX}}$Remainder: -1
#### Explanation:
Note that this is simply an alternate approach to Ernest Z's answer. Some people may find one approach easier to understand than the other.
Set up as standard long division:
$3 x$ " goes into " $3 {x}^{3}$$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow {x}^{2}$ times:
Multiply $3 x - 2$ by ${x}^{2}$ and write the product below the line:
Subtract:
"Bring down" the $- 7 x$
$3 x$ " goes into " $6 {x}^{2}$$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow 2 x$ times
...and so on...
The Remainder of $\left(- 1\right)$ may be simply noted as a remainder or written as a fraction (-1/(3x-2))_ |
ma008 ยป
Contents
## All The Numbers
This last lesson before the midterm, sort of stands out on its own, and w're going to do a review section to make sure you've got a grasp on this complex new material that we covered. Just as we started out the course with building up our knowledge of sets of numbers, we're going to round up this section of the class by revisiting this and acknowledging how we've built off of it. As a review of everything that we've learned about sets of numbers, here is a blank version of the full world of numbers diagram. And I would like you to type the name of each set into the proper position on the chart.
## All The Numbers
All of the numbers that we've worked with fall under the category of complex numbers. Complex numbers can have both real and imaginary components. So that means that real numbers and imaginary numbers are each subsets of the complex numbers. Within the real numbers, we have all of the same categories as we did before.
## Everything in its Place
Now that we've filled in the proper sets in the proper places in the world of numbers diagram, let's put some numbers in it. As you did early in the course, please place each of these numbers in the place it belongs on the diagram. Label the diagram with each letter in the appropriate box.
## Everything in its Place
I filled in the numbers in the proper spots. Irrational, integer, whole, natural, rational, complex, and imaginary.
## Complex Number Manipulation
Now that we've looked at complex numbers in our world of numbers diagram, let's manipulate them a bit. Please simplify each of these expressions so that it is written in the standard form for a complex number, a plus bi.
## Complex Number Manipulation
Remember, that when we add and subtract complex numbers, the imaginary parts are like terms and the real parts are like terms. So we end up with 1 plus 3i plus the quanity negative 6 plus 2i equals negative 5 plus 5i. 1 plus 3i minus the quanity negative 6 plus 2i gives us 7 plus i. When we multiply complex numbers, this needs to happen by distributing in the same way that we did with binomials involving, involving variables. The quantity 1 plus 3i times the quantity negative 6 plus 2i, gives us negative 6, plus 2i, minus 18i, plus 6i squared. Which when simplified, gives us negative 6, minus 16i, minus 6 or negative 12 minus 16i. And lastly, when we divided complex numbers, the final thing we need to do is to make the denominator into a real number, so that we can just let it modify the coefficients of the real and imaginary parts of the denominator. To change the denominator in this way, we multiply both the denominator and the numerator by the complex conjugate of the original denominator. So, negative 6 minus 2i in this case. Now let's look at the problem. 1 plus 3i, divided by negative 6 plus 2i, gives us 1 plus 3i, times the conjugate negative 6 minus 2i, divided by negative 6 plus 2i, times negative 6 minus 2i. Simplifying further, we got negative 6 minus 2i minus 18i minus 6i squared, divided by 36 minus 4i squared. Which gives us negative six minus 20i plus 6, divided by 36 plus 4 or negative 20i over 40, and finally negative i over 2.
## Solve For x
Now that we've worked with expressions for a while, let's move on to some equations. Please solve for x in the following equation. 3x squared plus 18 equals 0. Although there are a couple of different methods you could use for solving this, I'd like you to try to do it without using the quadratic formula this time. Please write the solutions in as concise a form as you can.
## Solve For x
So, when we simplify this, we subtract 18 from both sides and we get 3x squared equals negative 18. We then divide by 3 on both sides, we get x squared equals negative 6, giving us our solutions of x equals plus or minus the square root of negative 6. And when we pull out the negative 1 as an i, we get x equals plus or minus i times the square root of 6. This is two solutions, a plus i square root of 6 and a negative i square root of 6.
## Equation of a Parabola
And now for another equation, this time of a parabola. Here is the graph of a parabola for you. Now, first of all, what is the equation of this parabola in vertex form?
## Equation of a Parabola
Your solution is y plus 3 equals negative one third times the quantity x minus 2 squared.
## Roots of the Parabola
What are the roots of this parabola? List them with commas separating them.
## Roots of the Parabola
We can use a quadratic formula to solve this problem. First we put our answer in standard form, plug in 0 for y, and then substitute into the quadratic formula to obtain our two solutions of 2 plus 3i and 2 minus 3i. |
# Solving a One-Step Linear Equation Problem Type 2 Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to Solving a One-Step Linear Equation Problem Type 2. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Solve the following linear equation.
16 – 4y = 12
A - 1
B - 2
C - 3
D - 4
### Explanation
Step 1:
16 – 4y = 12
Subtracting 16 from both sides
Step 2:
16 – 4y – 16 = 12 – 16 = − 4
− 4y = − 4
Step 3:
Dividing both sides by – 4
So, y = 1
Q 2 - Solve the following linear equation.
4w + 6 = 18
A - 1
B - 2
C - 3
D - 4
### Explanation
Step 1:
4w + 6 = 18
Subtracting 6 from both sides
Step 2:
4w + 6 – 6 = 18 – 6 = 12
4w = 12
Step 3:
Dividing both sides by 4;
$\frac{4w}{4}$ = $\frac{12}{4}$ = 3
So, w = 3
Q 3 - Solve the following linear equation.
1 + 6p = 13
A - 1
B - 2
C - 3
D - 4
### Explanation
Step 1:
1 + 6p = 13
Subtracting 1 from both sides
Step 2:
1 + 6p – 1 = 13 – 1 = 12
6p = 12
Step 3:
Dividing both sides by 6 we get
$\frac{6p}{6}$ = $\frac{12}{6}$ = 2
So, p = 2
Q 4 - Solve the following linear equation.
7 + 2z = 19
A - 3
B - 4
C - 5
D - 6
### Explanation
Step 1:
7 + 2z = 19
Subtracting 7 from both sides
Step 2:
7 + 2z – 7 = 19 – 7 = 12
2z = 12
Step 3:
Dividing both sides by 2;
$\frac{2z}{2}$ = $\frac{12}{2}$ = 6
So, z = 6
Q 5 - Solve the following linear equation.
20 – 5m = 5
A - 1
B - 2
C - 3
D - 4
### Explanation
Step 1:
20 – 5m = 5
Subtracting 20 from both sides
Step 2:
20 – 5m – 20 = 5 – 20 = − 15
− 5m = − 15;
Step 3:
Dividing both sides by −5,
$\frac{-5m}{-5}$ = $\frac{-15}{-5}$ = 3
So, m = 3
Q 6 - Solve the following linear equation.
3t + 1 = 16
A - 4
B - 5
C - 6
D - 7
### Explanation
Step 1:
3t + 1 = 16
Subtracting 1 from both sides
Step 2:
3t + 1 – 1 = 16 – 1 = 15
3t = 15;
Step 3:
Dividing both sides by 3
$\frac{3t}{3}$ = $\frac{15}{3}$ = 5
So, t = 5
Q 7 - Solve the following linear equation.
7 = 3k − 5
A - 4
B - 5
C - 6
D - 7
### Explanation
Step 1:
7 = 3k − 5
Adding 5 to both sides
Step 2:
7 + 5 = 3k – 5 + 5 = 3k
3k = 12
Step 3:
Dividing both sides with 3
$\frac{3k}{3}$ = $\frac{12}{3}$ = 4
So, k = 4
Q 8 - Solve the following linear equation.
3x + 4 = 13
A - 0
B - 1
C - 2
D - 3
### Explanation
Step 1:
3x + 4 = 13
Subtracting 4 from both sides
Step 2:
3x + 4 – 4 = 13 – 4 = 9
3x = 9;
Step 3:
Dividing both sides with 3
$\frac{3x}{3}$ = $\frac{9}{3}$ = 3
So, x = 3
Q 9 - Solve the following linear equation.
6 = 2q − 4
A - 3
B - 4
C - 5
D - 6
### Explanation
Step 1:
6 = 2q − 4
Adding 4 to both sides
Step 2:
6 + 4 = 2q – 4 + 4 = 2q
2q = 10
Step 3:
Dividing both sides by 2
$\frac{2q}{2}$ = $\frac{10}{2}$ = 5
So, q = 5
Q 10 - Solve the following linear equation.
8 + 2x = 12
A - 1
B - 2
C - 3
D - 4
### Explanation
Step 1:
8 + 2x = 12
Subtracting 8 from both sides
Step 2:
8 + 2x – 8 = 12 – 8 = 4
2x = 4;
Step 3:
Dividing both sides by 2
$\frac{2x}{2}$ = $\frac{4}{2}$ = 2
So, x = 2
solving_one_step_linear_equation_problem_type2.htm |
Addition and Subtraction of Polynomials Worksheet | Problems & Solutions
# Addition and Subtraction of Polynomials Worksheet
Addition and Subtraction of Polynomials Worksheet
• Page 1
1.
Subtract the polynomial 7$y$ - 4$y$2 - 6 from 2$y$ + 3 - $y$2.
a. 3$y$2 - 5$y$ + 9 b. - 3$y$2 + 5$y$ + 9 c. 5$y$2 - 9$y$ + 9 d. 3$y$2 + 5$y$ - 3
#### Solution:
7y - 4y2 - 6 , 2y + 3 - y2
[Original Polynomials.]
- 4y2 + 7y - 6 , - y2 + 2y + 3
[Write each expression in standard form.]
- y2 + 2y + 3
(-) - 4y2 + 7y - 6
.................................
3y2 - 5y + 9
.................................
[Line up like terms vertically and subtract.]
The result is 3y2 - 5y + 9.
2.
What is the sum of the polynomials 5$y$2 + $y$ - 6 and 6$y$2 - 5$y$ + 9?
a. 11$y$2 + 4$y$ + 3 b. 12$y$2 - 4$y$ + 5 c. 11$y$2 - 4$y$ + 3 d. 13$y$2 - 4$y$ + 6
#### Solution:
(5y2 + y - 6) + (6y2 - 5y + 9)
[Write the expression for the sum of the polynomials.]
= (5y2 + 6y2) + (y - 5y) + (-6 + 9)
[Group like terms.]
= (5 + 6)y2 + (1 - 5)y + (-6 + 9)
[Use distributive property.]
= 11y2 - 4y + 3
[Simplify.]
3.
Subtract the polynomial 24$y$2 from 31$y$2.
a. 4$y$2 b. 55$y$2 c. 7$y$2 d. -7$y$2
#### Solution:
31y2 - 24y2
[Subtraction of the polynomials using horizontal format.]
= (31 - 24)y2
[Use distributive property.]
= 7y2
[As the exponents are same, subtract the coefficients.]
4.
Add the polynomials 3$a$ + 11$c$ and 9$b$ - 8$c$ and get the simplified value.
a. 3$a$ + 9$b$ + 3$c$ b. 3$a$ + 8$b$ - 3$c$ c. 3$a$ + 9$b$ - 3$c$ d. None of the above
#### Solution:
(3a + 11c) + (9b - 8c)
[Write the expression for adding the polynomials.]
= 3a + 9b + 11c - 8c
[Use Commutative and Associative properties of addition and to group like terms.]
= 3a + 9b + 3c
[Simplify the like terms.]
5.
Find the sum of the polynomials 2$y$2 + 7$y$ and 4$y$2 - 2$y$.
a. 2$y$2 - $y$ b. 6$y$2 - 5$y$ c. 6$y$2 + 5$y$ d. $y$2 - 5$y$
#### Solution:
2y2 + 7y + 4y2 - 2y
= 2y2 + 4y2 + 7y - 2y
[Group like terms.]
= 6y2 + 5y
[Simplify.]
6.
Subtract the polynomial 5$y$2 + 1 from 2$y$2 + 2.
a. -3$y$2 + 1 b. 2$y$2 - 1 c. 2$y$2 + 1 d. -2$y$2 - 1
#### Solution:
2y2 + 2 - (5y2 + 1)
[Subtract.]
= 2y2 -5y2 + (2 - 1)
[Group like terms]
= -3y2 + 1
[Simplify.]
7.
What is the sum of the polynomials 4$y$3 + 4$y$2 + 5$y$ and 3$y$2 + 3 ?
a. $y$3 - $y$2 - 5$y$ - 3 b. 4$y$3 - 4$y$2 - 5$y$ + 3 c. 4$y$3 + 7$y$2 + 5$y$ + 3 d. None of the above
#### Solution:
4y3 + 4y2 + 5y + 3y2 + 3
= 4y3 + 4y2 + 3y2 + 5y + 3
[Group like terms]
= 4y3 + 7y2 + 5y + 3
[Combine like terms]
8.
What is the value of (5$y$2 + 1) - (-2$y$3 + 10$y$2 - 3)?
a. $y$3 - 5$y$2 - 4 b. 2$y$3 + 5$y$2 + 1 c. 2$y$3 - 5$y$2 + 4 d. None of the above
#### Solution:
(5y2 + 1) - (-2y3 + 10y2 - 3)
[Original polynomial]
= 5y2 + 1 + 2y3 - 10y2 + 3
[Apply distributive property.]
= 2y3 + 5y2 - 10y2 + 1 + 3
[Group like terms]
= 2y3 - 5y2 + 4
[Combine like terms]
9.
What is the simplified form of the polynomial obtained by subtracting -5$x$ + 5$y$ - 3$z$ from 4$x$ + 5$y$ - 4$z$?
a. 9$x$ + $z$ b. $x$ - $y$ - 4$z$ c. 9$x$ - $z$ d. $x$ + $y$ + $z$
#### Solution:
-5x + 5y - 3z and 4x + 5y - 4z
[Original polynomials]
4x + 5y - 4z - (-5x + 5y - 3z) = 4x + 5y - 4z + 5x - 5y + 3z
[Apply distributive property.]
= 4x + 5x + 5y - 5y - 4z + 3z
[Group like terms]
= 9x - z
[Combine like terms]
10.
The degree of the polynomial 7$y$4 - 10$y$ is _______.
a. 6 b. 3 c. 4 d. 5
#### Solution:
7y4 - 10y
[Original Polynomial.]
The highest exponent of y will be the degree of the polynomial.
The highest exponent of y in the polynomial is 4.
The degree of the polynomial is 4. |
One sample Z-tests
In this vignette, we work through an example Z-test, and point out a number of points where you might get stuck along the way.
Problem setup
Let’s suppose that a student is interesting in estimating how many memes their professors know and love. So they go to class, and every time a professor uses a new meme, they write it down. After a year of classes, the student has recorded the following meme counts, where each count corresponds to a single class they took:
$3, 7, 11, 0, 7, 0, 4, 5, 6, 2$
The student talks to some other students who’ve done similar studies and determines that $$\sigma = 2$$ is a reasonable value for the standard deviation of this distribution.
Assumption checking
Before we can do a Z-test, we need to make check if we can reasonably treat the mean of this sample as normally distributed. This happens is the case of either of following hold:
1. The data comes from a normal distribution.
2. We have lots of data. How much? Many textbooks use 30 data points as a rule of thumb.
Since we have a small sample, we should check if the data comes from a normal distribution using a normal quantile-quantile plot.
# read in the data
x <- c(3, 7, 11, 0, 7, 0, 4, 5, 6, 2)
# make the qqplot
qqnorm(x)
qqline(x)
Since the data lies close the line $$y = x$$, and has no notable systematic deviations from line, it’s safe to treat the sample as coming from a normal distribution. We can proceed with our hypothesis test.
Null hypothesis and test statistic
Let’s test the null hypothesis that, on average, professors know 3 memes. That is
$H_0: \mu = 3 \qquad H_A: \mu \neq 3$
First we need to calculate our Z-statistic. Let’s do this with R. Remember that the Z-statistic is defined as
$Z = \frac{\bar x - \mu_0}{\sigma / \sqrt{n}} \sim \mathrm{Normal}(0, 1)$
Calculating p-values
In R this looks like:
n <- length(x)
# calculate the z-statistic
z_stat <- (mean(x) - 3) / (2 / sqrt(n))
z_stat
#> [1] 2.371708
To calculate a two-sided p-value, we need to find
\begin{align} P(|Z| \ge |2.37|) &= P(Z \ge 2.37) + P(Z \le -2.37) \\ &= 1 - P(Z \le 2.37) + P(Z \le -2.37) \\ &= 1 - \Phi(2.37) + \Phi(-2.37) \end{align}
To do this we need to c.d.f. of a standard normal
library(distributions3)
Z <- Normal(0, 1) # make a standard normal r.v.
1 - cdf(Z, 2.37) + cdf(Z, -2.37)
#> [1] 0.01778809
Note that we saved z_stat above so we could have also done
1 - cdf(Z, abs(z_stat)) + cdf(Z, -abs(z_stat))
#> [1] 0.01770607
which is slightly more accurate since there is no rounding error.
So our p-value is about 0.0177. You should verify this with a Z-table. Note that you should get the same value from cdf(Z, 2.37) and looking up 2.37 on a Z-table.
You may also have seen a different formula for the p-value of a two-sided Z-test, which makes use of the fact that the normal distribution is symmetric:
\begin{align} P(|Z| \ge |2.37|) &= 2 \cdot P(Z \le -|2.37|) \\ &= 2 \cdot \Phi(-2.37) \end{align}
Using this formula we get the same result:
2 * cdf(Z, -2.37)
#> [1] 0.01778809
Finally, sometimes we are interest in one sided Z-tests. For the test
\begin{align} H_0: \mu \le 3 \qquad H_A: \mu > 3 \end{align}
the p-value is given by
$P(Z > 2.37)$
which we calculate with
1 - cdf(Z, 2.37)
#> [1] 0.008894043
For the test
$H_0: \mu \ge 3 \qquad H_A: \mu < 3$
the p-value is given by
$P(Z < 2.37)$
which we calculate with
cdf(Z, 2.37)
#> [1] 0.991106
Rejection regions
Preface: I am strongly opposed to make a dichotomous “reject/fail to reject” decision for hypothesis tests. If you do a hypothesis test, you should report the p-value, period. Picking an arbitrary $$\alpha$$ level rejection threshold and treating it as a gold standard is ridiculous, as evidenced by 60 years of statistical literature laden with warnings about hypothesis testing. That said, sometimes it can be useful to think about when you reject a hypothesis test.
We can think about three different rejection regions for a Z-test:
1. The rejection region in terms of the p-value
2. The rejection region in terms of the test statistic
3. The rejection region in terms of the sample mean
For a given $$\alpha$$ level threshold, all of these rejection regions are equivalent. We’ll start by thinking about the rejection of a two-sided test. That is
\begin{align} H_0 : \mu = \mu_0 \qquad H_A : \mu \neq \mu_0 \end{align}
We then calculate a test statistic $$Z_\text{obs}$$, and a p-value $$P(|Z| > |Z_\text{obs}|)$$ and reject when $$P(|Z| > |Z_\text{obs}|) < \alpha$$. This defines our first rejection region. Using our observation from before, this is exactly equivalent to rejecting when
\begin{align} P(|Z| > |Z_\text{obs}|) < \alpha & \Longleftrightarrow 2 \cdot P(Z < -|Z_\text{obs}|) < \alpha \\ & \Longleftrightarrow P(Z < -|Z_\text{obs}|) < \alpha / 2 \end{align}
and this last statement is exactly the same as when $$Z_\text{obs} < z_{\alpha / 2}$$ or $$z_{1 - \alpha / 2} < Z_\text{obs}$$. This is our second region region, in terms of the test statistic. Finally, recall that
$Z_\text{obs} = \frac{\bar x - \mu_0}{\sigma / \sqrt{n}}$
So we take the conditions and rearrange to in terms of $$\bar x$$
\begin{align} \frac{\bar x - \mu_0}{\sigma / \sqrt{n}} > z_{1 - \alpha / 2} & \Longleftrightarrow \bar x > \mu_0 + z_{1 - \alpha / 2} \frac{\sigma}{\sqrt n} \end{align}
and
\begin{align} \frac{\bar x - \mu_0}{\sigma / \sqrt{n}} < z_{\alpha / 2} & \Longleftrightarrow \bar x < \mu_0 + z_{\alpha / 2} \frac{\sigma}{\sqrt n} \end{align}
You can also think about this in terms of $$\mu_0$$. We will reject the test when $$\mu_0$$ is not in
\begin{align} \left( \bar x + z_{\alpha / 2} \frac{\sigma}{\sqrt n}, \bar x + z_{1 - \alpha / 2} \frac{\sigma}{\sqrt n} \right) \end{align}
which you may recognize as the confidence interval for $$\mu$$! So the confidence interval contains all the values of $$\mu_0$$ that we cannot reject at the $$\alpha$$ level. You can perform a similar calculation for a one sided test, resulting in a one-sided confidence bound, where one end of the interval is either $$\infty$$ or $$-\infty$$.
Power and sample size calculations
Formulas for power
We want to make sure that we actually reject our null hypothesis in the case that it is false. That is, we would like to make sure that our test has high power. Mathematically, this means that $$P(\text{reject } H_0 | H_0 \text{ is false})$$. The problem here is that $$H_0$$ can be wrong in many different ways: it could be that the true $$\mu$$ is $$2$$, it could be $$7$$, it could be $$4.26$$. So to calculate power as formulated above is not really possible. However, we can calculate power for specific versions of “$$H_0$$ is false”.
Let’s consider the case that $$H_0$$ is false, and in particular the true value of $$\mu$$ is $$\mu_A$$. In this case, the power of our test is $$P(\text{reject } H_0 | \mu = \mu_A)$$. Recall that we reject $$H_0$$ when $$\bar{x} > \mu_0 + z_{1 - \alpha / 2} \frac{\sigma}{\sqrt n}$$ or $$\bar{x} < \mu_0 + z_{\alpha / 2} \frac{\sigma}{\sqrt n}$$, so the power of our test when $$\mu = \mu_A$$ is
$P(\text{reject } H_0 \left | \mu = \mu_A \right .) = P\left( \left . \bar x > \mu_0 + z_{1 - \alpha / 2} \frac{\sigma}{\sqrt n} \right | \mu = \mu_A \right) + P\left( \left . \bar x < \mu_0 + z_{\alpha / 2} \frac{\sigma}{\sqrt n} \right | \mu = \mu_A \right).$
Remember that $$\bar X \sim \mathrm{Normal} \left(\mu, \frac{\sigma^2}{n} \right)$$. This means that, given $$\mu = \mu_A$$, $$\bar{x} \sim \mathrm{Normal} \left(\mu, \frac{\sigma^2}{n} \right)$$, which let’s us calculate the probabilities we need to find the power:
\begin{align} P\left( \left . \bar{x} > \mu_0 + z_{1 - \alpha / 2} \frac{\sigma}{\sqrt n} \right | \mu = \mu_A \right) &= P\left( \frac{\bar{x} - \mu_A}{\sigma/\sqrt{n}} > \frac{\mu_0 + z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}} - \mu_A}{\sigma/\sqrt{n}} \right) \\ &= P\left( Z > \frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} + z_{1-\alpha/2} \right), \end{align}
and similarly
$P\left( \left . \bar{x} < \mu_0 + z_{\alpha / 2} \frac{\sigma}{\sqrt n} \right | \mu = \mu_A\right) = P\left( Z < \frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} + z_{\alpha/2} \right).$
So, the power of our test, if the true population mean is $$\mu_A$$, is
\begin{align} \text{Power} &= P\left( Z > \frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} + z_{1-\alpha/2} \right) + P\left( Z < \frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} + z_{\alpha/2} \right) \\ &= \left[1 - P\left( Z < \frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} + z_{1-\alpha/2} \right)\right] + P\left( Z < \frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} + z_{\alpha/2} \right) \end{align}
Let’s calculate this if $$\mu_A = 5$$.
power_lower <- (3 - 5) / (2 / sqrt(10)) + quantile(Z, 0.025)
power_upper <- (3 - 5) / (2 / sqrt(10)) + quantile(Z, 1 - 0.025)
cdf(Z, power_lower) + (1 - cdf(Z, power_upper))
#> [1] 0.8853791
This means that the probability that we reject the null hypothesis ($$H_0: \mu = 3$$) if the true mean is $$5$$ is about 0.89.
Formulas for sample size calculations
Often times researchers like to go in the other direction: aim for a specific level of power, and calculate how many observations are needed to reach that level. To achieve a power of $$1 - \beta$$ for a one sample Z-test with $$H_0 : \mu = \mu_0$$, you need
$n \approx \left( { \sigma \cdot (z_{\alpha / 2} + z_\beta) \over \mu_0 - \mu_A } \right)^2$
samples. If $$n$$ is not an integer, round up. Often, the denominator is thought of as the detectable difference. So, the question becomes how many samples are required to have sufficient power to detect a difference of some particular size.
This equation is simply a rewrite of the equation presented above for power. Recall, the power for a two sided test is
$\text{Power} = P\left( Z > \frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} + z_{1-\alpha/2} \right) + P\left( Z < \frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} + z_{\alpha/2} \right).$
Usually, only one of these terms is contributing while the other is very close to zero. Let’s say the first term is the one clearly different from zero. To determine the sample size, we want to determine $$n$$ such that $$P\left( Z > \frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} + z_{1-\alpha/2} \right) = 1-\beta$$. Or, similarly, $$P\left( Z < \frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} + z_{1-\alpha/2} \right) = \beta$$. I.e. we need $$\frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} + z_{1-\alpha/2} = z_{\beta}$$.
\begin{align} z_{\beta} = \frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} + z_{1-\alpha/2} & \Longleftrightarrow z_{\beta} - z_{1-\alpha/2} = \frac{\mu_0 - \mu_A}{\sigma/\sqrt{n}} \\ & \Longleftrightarrow \sigma (z_{\beta} - z_{1-\alpha/2}) = (\mu_0-\mu_A)\sqrt{n} \\ & \Longleftrightarrow \frac{\sigma (z_{\beta} - z_{1-\alpha/2})}{\mu_0-\mu_A} = \sqrt{n} \\ & \Longleftrightarrow \left(\frac{\sigma (z_{\beta} - z_{1-\alpha/2})}{\mu_0-\mu_A}\right)^2 = n \end{align}
Since $$z_{1-\alpha/2} = -z_{\alpha/2}$$, we have the equation above:
$\left(\frac{\sigma (z_{\beta} + z_{\alpha/2})}{\mu_0-\mu_A}\right)^2 = n.$
As an example, say the student prior to the experiment had determined that they wanted to test if the number of memes their professors know and love is 2. They want to make sure their sample size is large enough so that they are likely to reject the null hypothesis if the true number is 3. They determine that they want a probability of 0.9 of rejecting the null if the true number is 3. So, a sample size calculation looks like this:
$n \approx \left( { 2 \cdot (1.96 + 1.28) \over 2-3 } \right)^2 = 41.99$
So to make sure that they reject the null hypothesis with a probability of 0.9 if the true value is 3, they would have to ask 53 professors.
Below is this same calculation done in R. Remember, $$\beta = 1-\text{Power}$$. Note the small discrepancy. This is due to rounding error.
(2 * (quantile(Z, 0.05 / 2) + quantile(Z, 1 - 0.9)) / (3 - 2))^2
#> [1] 42.02969 |
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• Kristjan Salev
# HOW TO MEASURE DISTANCES WITH A MATCH
Updated: May 26
In an earlier post, I wrote about how to measure heights using a mirror. However, in this short post, we will look at how to measure distances using a matchstick. In fact, any stick on which you can draw millimeter lines is suitable for this. This method is only applicable if we can roughly estimate the height of the object to which we are measuring the distance. Fortunately, for many objects, this is more or less known to us. For example, the average height of a person is 1.7 meters, and a car is 1.6 meters. If the object is located in a place inaccessible to us, for example behind a river, then its distance can be estimated very successfully using this simple method. Let's see how this distance estimation method works.
## Using similar triangles
As when measuring heights with a mirror, we use the properties of similar triangles when measuring distances with a matchstick. Two triangles are similar if their corresponding angles are congruent and their corresponding sides are in equal proportion. Simply put, these triangles are the same shape but different sizes. For example, let's take a right triangle with side lengths of 3, 4, and 5 units.
By increasing the side lengths of this triangle by 1.5 times, we get a new, larger triangle with side lengths of 4.5, 6, and 7.5 units. All the corresponding angles of these triangles are congruent, but the side lengths of the larger triangle are 1.5 times longer than the corresponding sides of the smaller triangle. One way to prove that triangles are similar is to make sure that the ratios of their corresponding sides are equal. If we take the ratios of the corresponding sides of our triangles, we see that they come out equal (6/4 = 4.5/3 = 7.5/5 = 1.5).
There are other ways to prove that triangles are similar. The easiest way is to make sure, that the corresponding angles of two triangles are congruent. Similarly, if the ratios of two pairs of corresponding sides of two triangles are equal and the included angles are congruent, then the triangles are similar.
## A match as a simple range finder
Now let's make a simple distance meter. We take a match and draw lines with millimeter intervals on one side of it. Suppose you see a person across the river and want to determine the distance to him. Holding a match in an outstretched hand and looking with one eye, bring its upper end to such a position that it coincides with the upper end of a person in the distance.
After that, slowly move your thumbnail up the match to the point where it aligns with the lowest point of the human figure. Now you have determined at which millimeter mark the thumbnail stopped. In the top image, the human figure in the distance appears to be 6 mm tall. Knowing approximately the distance of the eye from the match in the outstretched hand, we have all the necessary information to find the distance of the person beyond the river.
As you can see, two triangles are formed - triangle ABC and triangle ADE. Triangles ABC and ADE are similar because their corresponding angles are congruent.
Using the property of similar triangles that the ratio of corresponding sides is equal, we can calculate the distance to the person across the river. The ratio of the average height of a person to the length measured on a match must be equal to the ratio of the unknown distance to the distance between the eye and the match.
Let's say, for example, that the distance between the eye and the match is AD=60 cm (600 mm), the height measured on the match is DE=6 mm, and the average height of a person is BC=170 cm (1700 mm).
By writing the ratios of the corresponding sides (AB/AD=BC/DE), we can derive the unknown distance (AB) from the equation, which in our example is 170 meters.
This way of measuring the distances is not very accurate because we do not know precisely the heights of many objects. But for objects whose height we can roughly estimate, it can give a pretty good result. |
# Cycle sort algorithm
In this article, we will discuss the Cycle sort Algorithm. Cycle sort is a comparison sorting algorithm that forces array to be factored into the number of cycles where each of them can be rotated to produce a sorted array. It is theoretically optimal in the sense that it reduces the number of writes to the original array.
It is an in-place and unstable sorting algorithm. Cycle sort forces an array to be factored into a number of cycles where every element can rotate in order to produce a sorted array. The time complexity of cycle sort is O(n2), even in the best case.
Now, let's see the algorithm of cycle sort.
## Algorithm
Suppose there is an array arr with n distinct elements. Given an element A, we can find its index by counting the number of elements smaller than A.
1. If the element is at its correct position, simply leave it as it is.
2. Else, we have to find the correct position of A by counting the number of elements smaller than it. Another element B is replaced to be moved to its correct position. This process continues until we get an element at the original position of A.
The above-illustrated process constitutes a cycle. Repeat this cycle for every element of the list until the list is sorted.
## Working of Cycle sort Algorithm
Now, let's see the working of Cycle sort Algorithm. To understand the working of cycle sort algorithm, let's take an unsorted array.
Let the elements of array are - {30, 20, 10, 40, 60, 50}
Now, the given array is completely sorted.
## Cycle sort complexity
Now, let's see the time complexity of cycle sort in the best case, average case, and worst case. We will also see the space complexity of cycle sort.
### 1. Time Complexity
Case Time Complexity
Best Case O(2n)
Average Case O(2n)
Worst Case O(2n)
The time complexity of cycle sort in all three cases in O(n2). Even in the best case, it takes O(n2) time to sort the array elements. In Cycle sort, there is always the traversing of the entire subarray from the current position in order to count the number of elements less than the current element.
So, in cycle sort, it doesn't matter whether the given array is already sorted or not. It has no consequence on the running time of the algorithm. So, the cycle sort must run in the quadratic time.
### 2. Space Complexity
Space Complexity O(1) Stable No
• The space complexity of cycle sort is O(1).
## Implementation of Cycle sort
Now, let's see the programs of cycle sort in different programming languages.
Program: Write a program to implement cycle sort in C language.
Output
Program: Write a program to implement cycle sort in C++.
Output
Program: Write a program to implement cycle sort in C#.
Output
After the execution of the above code, the output will be -
Program: Write a program to implement cycle sort in Java.
Output
After the execution of the above code, the output will be -
So, that's all about the article. Hope, the article will be helpful and informative to you.
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# Similar Figures and Indirect Measurement
### Popular Tutorials in Similar Figures and Indirect Measurement
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Trying to find a missing measurement on similar figures? Make ratios from corresponding sides and set up a proportion! Solve the proportion to get your missing measurement. Figure out how to do all that by watching this tutorial!
#### What are Similar Figures?
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Word problems allow you to see the real world uses of math! This tutorial shows you how to take a word problem and use indirect measurement to turn it into a proportion. Then see how to use the mean extremes property of proportions to cross multiply and solve for the answer. Take a look!
#### What are Corresponding Parts in Similar Figures?
Identifying corresponding parts in similar figures isn't so bad, but you have to know what you're looking for. This tutorial does a great job of explaining the corresponding parts of similar figures! Take a look!
#### How do you Figure Out if Two Figures are Similar?
Follow along with this tutorial to see an example of determining if two given figures are similar.
#### How Do You Find Missing Measurements of Similar Figures Using a Scale Factor?
Looking at similar figures? Want to find a missing measurement on one of the figures? You could use a scale factor to solve! In this tutorial, learn how to create a ratio of corresponding sides with known length and use the ratio to find the scale factor. Then, write an equation using the scale factor to find your missing measurement!
#### How Do You Find a Scale Factor in Similar Figures?
Have similar figures? Want to find the scale factor? Then check out this tutorial! You'll see how to use measurements from similar figures to create a ratio and find the scale factor.
#### How Do You Solve a Scale Model Problem Using a Scale Factor?
This tutorial provides a great real world application of math! You'll see how to use the scale on a house blueprint to find the scale factor. Then, see how to use the scale factor and a measurement from the blueprint to find the measurement on the actual house! Check out this tutorial and see the usefulness blueprints and scale factor!
#### How Do You Find the Scale of a Model?
Want some practice with scale? Then check out this tutorial and you'll see how to find the scale of a model given the lengths of the model and the actual object. Take a look!
#### How Do You Use the Scale on a Map to Find an Actual Distance?
Maps help us get from one place to another. In this tutorial, you'll learn how to use a map to find an actual distance.
#### How Do You Figure Out What Size a Model Should Be if You Have a Scale?
Before tall sky scrapers are build, a scale model of the building is made, but how does the architect know what size the model should be? Follow along with this tutorial to find out!
#### What is a Scale?
In math, the term scale is used to represent the relationship between a measurement on a model and the corresponding measurement on the actual object. Without scales, maps and blueprints would be pretty useless. Check out this tutorial and learn about scale factor! |
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Problem 67
# Simplify (a) $$3^{-2}$$ (b) $$1 / 5^{-2}$$
Expert verified
The simplified expressions are: (a) $$3^{-2} = \frac{1}{9}$$ (b) $$\frac{1}{5^{-2}} = 25$$
See the step by step solution
## Step 1: Simplify expression (a) $$3^{-2}$$
First, we can rewrite the expression using the property of negative exponents: $3^{-2} = \frac{1}{3^2}$ Now, simply calculate the exponent: $\frac{1}{3^2} = \frac{1}{9}$ So, $$3^{-2} = \frac{1}{9}$$.
## Step 2: Simplify expression (b) $$\frac{1}{5^{-2}}$$
First, we can rewrite the expression using the property of negative exponents: $\frac{1}{5^{-2}} = \frac{1}{\frac{1}{5^2}}$ Now, we can simplify the expression further by multiplying the numerator and the denominator of the inner fraction by $$5^2$$: $\frac{1}{\frac{1}{5^2}} = \frac{1 \times 5^2}{1}$ Lastly, calculate the exponent: $\frac{1 \times 5^2}{1} = \frac{25}{1} = 25$ So, $$\frac{1}{5^{-2}} = 25$$.
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Dividing Polynomials
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# Dividing Polynomials - PowerPoint PPT Presentation
Dividing Polynomials. divisor. Dividing by a Monomial . If the divisor only has one term, split the polynomial up into a fraction for each term. Now reduce each fraction. 3 x 3. 4 x 2. x. 2. 1. 1. 1. 1. Long Division of Polynomials.
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Presentation Transcript
divisor
Dividing by a Monomial
If the divisor only has one term, split the polynomial up into a fraction for each term.
Now reduce each fraction.
3x3
4x2
x
2
1
1
1
1
Long Division of Polynomials
Now multiply by the divisor and put the answer below.
Bring down the next number or term
32 698
First divide 3 into 6 or x into x2
Now divide 3 into 5 or x into 11x
So we found the answer to the problem x2 + 8x – 5 x – 3 or the problem written another way:
If the divisor has more than one term, perform long division. You do the same steps with polynomial division as with integers. Let's do two problems, one with integers you know how to do and one with polynomials and copy the steps.
Subtract (which changes the sign of each term in the polynomial)
x
+ 11
2
1
Multiply and put below
x - 3 x2 + 8x - 5
64
x2 – 3x
subtract
5
8
11x
- 5
32
11x - 33
This is the remainder
26
28
Write out with long division including 0y for missing term
Divide y into y2
Divide y into -2y
Let's Try Another One
If any powers of terms are missing you should write them in with zeros in front to keep all of your columns straight.
Subtract (which changes the sign of each term in the polynomial)
y
- 2
Multiply and put below
Bring down the next term
Multiply and put below
y + 2 y2 + 0y + 8
y2 + 2y
subtract
-2y
+ 8
- 2y - 4
This is the remainder
12
REMAINDER THEOREM
Let f be a polynomial function. If f (x) is divided by x – c, then the remainder is f (c).
Let’s look at an example to see how this theorem is useful.
So the remainder we get in synthetic division is the same as the answer we’d get if we put -2 in the function.
The root of x + 2 = 0 is x = -2
using synthetic division let’s divide by x + 2
-2 2 -3 2 -1
-4 14 -32
2 -7 16 -33
the remainder
Find f(-2)
FACTOR THEOREM
Let f be a polynomial function. Then x – c is a factor of f (x) if and only if f (c) = 0
-3 -4 5 0 8
12 -51 153
-4 17 -51 161
If and only if means this will be true either way:
1. If f(c) = 0, then x - c is a factor of f(x)
2. If x - c is a factor of f(x) then f(c) = 0.
Try synthetic division and see if the remainder is 0
Opposite sign goes here
NO it’s not a factor. In fact, f(-3) = 161
We could have computed f(-3) at first to determine this.
Not = 0 so not a factor
Set divisor = 0 and solve. Put answer here.
x + 3 = 0 so x = - 3
Bring first number down below line
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Synthetic Division
There is a shortcut for long division as long as the divisor is x – k where k is some number. (Can't have any powers on x).
1
- 3
1 6 8 -2
- 3
- 9
3
1
x2 + x
3
- 1
1
This is the remainder
Put variables back in (one x was divided out in process so first number is one less power than original problem).
List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0.
Set divisor = 0 and solve. Put answer here.
x - 4 = 0 so x = 4
Bring first number down below line
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Let's try another Synthetic Division
0 x3
0 x
1
4
1 0 - 4 0 6
4
16
48
192
1
x3 + x2 + x +
4
12
48
198
This is the remainder
Now put variables back in (remember one x was divided out in process so first number is one less power than original problem so x3).
List all coefficients (numbers in front of x's) and the constant along the top. Don't forget the 0's for missing terms.
Bring first number down below line
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Let's try a problem where we factor the polynomial completely given one of its factors.
You want to divide the factor into the polynomial so set divisor = 0 and solve for first number.
- 2
4 8 -25 -50
- 8
0
50
No remainder so x + 2 IS a factor because it divided in evenly
4
x2 + x
0
- 25
0
Put variables back in (one x was divided out in process so first number is one less power than original problem).
So the answer is the divisor times the quotient:
List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0.
The second factor is the difference of squares so factor it.
Our goal in this section is to learn how we can factor higher degree polynomials. For example we want to factor:
We could randomly try some factors and use synthetic division and know by the factor theorem that if the remainder is 0 then we have a factor. We might be trying things all day and not hit a factor so in this section we’ll learn some techniques to help us narrow down the things to try.
The first of these is called Descartes Rule of Signs named after a French mathematician that worked in the 1600’s.
Rene Descartes 1596 - 1650
Descartes’ Rule of Signs
Let f denote a polynomial function written in standard form.
The number of positive real zeros of f either equals the number of sign changes of f (x) or else equals that number less an even integer.
The number of negative real zeros of f either equals the number of sign changes of f (-x) or else equals that number less an even integer.
1
2
starts Pos. changes Neg.changes Pos.
There are 2 sign changes so this means there could be 2 or 0 positive real zeros to the polynomial.
Use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros the polynomial may have.
Counting multiplicities and complex (imaginary) zeros, the total number of zeros will be the same as the degree of the polynomial.
1
starts Neg. changes Pos.
There is one sign change so there is one positive real zero.
starts Pos. Never changes
There are no negative real zeros.
Descartes rule says one positive and no negative real zeros so there must be 4 complex zeros for a total of 5. We’ll learn more complex zeros in Section 4.7.
Back to our original polynomial we want to factor:
1
We’d need to try a lot of positive or negative numbers until we found one that had 0 remainder. To help we have:
The Rational Zeros Theorem
What this tells us is that we can get a list of the POSSIBLE rational zeros that might work by taking factors of the constant divided by factors of the leading coefficient.
Both positives and negatives would work for factors
1, 2
Factors of the constant
1
1 1 1 -3 -1 2
1 2 -1 -2
1 2 -1 -2 0
So a list of possible things to try would be any number from the top divided by any from the bottom with a + or - on it. In this case that just leaves us with 1 or 2
1, 2
1
Let’s try 1
YES! It is a zero since the remainder is 0
We found a positive real zero so Descartes Rule tells us there is another one
Since 1 is a zero, we can write the factor x - 1, and use the quotient to write the polynomial factored.
1 1 2 -1 -2
1 3 2
1 3 2 0
We could try 2, the other positive possible. IMPORTANT: Just because 1 worked doesn’t mean it won’t work again since it could have a multiplicity.
1, 2
1
Let’s try 1 again, but we try it on the factored version for the remaining factor (once you have it partly factored use that to keep going---don't start over with the original).
YES! the remainder is 0
Once you can get it down to 3 numbers here, you can put the variables back in and factor or use the quadratic formula, we are done with trial and error.
Let’s take our polynomial then and write all of the factors we found:
There ended up being two positive real zeros, 1 and 1 and two negative real zeros, -2, and -1.
In this factored form we can find intercepts and left and right hand behavior and graph the polynomial
Plot intercepts
Left & right hand behavior
Touches at 1 crosses at -1 and -2.
“Rough” graph
Let’s try another one from start to finish using the theorems and rules to help us.
Using the rational zeros theorem let's find factors of the constant over factors of the leading coefficient to know what numbers to try.
1, 3, 9
factors of constant
1, 2
So possible rational zeros are all possible combinations of numbers on top with numbers on bottom:
1
2
3
4
starts Pos. changes Neg. changes Pos. Changes Neg. Changes Pos.
starts Pos. Stays positive
Let’s see if Descartes Rule helps us narrow down the choices.
No sign changes in f(x) so no positive real zeros---we just ruled out half the choices to try so that helps!
4 sign changes so 4 or 2 or 0 negative real zeros.
-1 2 13 29 27 9
-1 2 11 18 9
-2 -11 -18 -9
-2 - 9 -9
2 11 18 9 0
2 9 9 0
Let’s try -1
Yes! We found a zero. Let’s work with reduced polynomial then.
Yes! We found a zero. Let’s work with reduced polynomial then.
Let’s try -1 again
Yes! We found another one. We are done with trial and error since we can put variables back in and solve the remaining quadratic equation.
So remaining zeros found by setting these factors = 0 are -3/2 and -3. Notice these were in our list of choices.
Let’s solve the equation:
To do that let’s consider the function
If we find the zeros of the function, we would be solving the equation above since we want to know where the function = 0
By Descartes Rule: There is one sign change in f(x) so there is one positive real zero.
There are 2 sign changes in f(-x) so there are 2 or 0 negative real zeros.
Using the rational zeros theorem, the possible rational zeros are:
1, 5
1, 2
1 2 -3 -3 -5
5 2 -3 -3 -5
2 -1 -4
10 35 160
2 -1 -4 -9
2 7 32 155
1 is not a zero and f(1) = -9
Let’s try 1
5 is not a zero and f(5) = 155
Let’s try 5
On the next screen we’ll plot these points and the y intercept on the graph and think about what we can tell about this graph and its zeros.
f(5) = 155
To join these points in a smooth, continuous curve, you would have to cross the x axis somewhere between 1 and 5. This is the Intermediate Value Theorem in action. We can see that since Descartes Rule told us there was 1 positive real zero, that is must be between 1 and 5 so you wouldn’t try 1/2, but you'd try 5/2 instead.
f(0) = -5
f(1) = -9
Intermediate Value TheoremLet f denote a polynomial function. If a < b and if f(a) and f(b) are of opposite sign, then there is at least one zero of f between a and b.
f(5) = 155
In our illustration, a = 1 and b =5
So if we find function values for 2 different x’s and one is positive and the other negative, there must be a zero of the function between these two x values
In our illustration, f(a) = -9 and f(b) = 155 which are opposite signs
f(1) = -9
-1 1 8 -1 0 2
-1 -7 8 -8
1 7 -8 8 -6
We use this theorem to approximate zeros when they are irrational numbers. The function below has a zero between -1 and 0. We’ll use the Intermediate Value Theorem to approximate the zero to one decimal place.
First let’s verify that there is a zero between -1 and 0. If we find f(-1) and f(0) and they are of opposite signs, we’ll know there is a zero between them by the Intermediate Value Theorem.
So f(-1) = -6 and f(0) = 2These are opposite signs.
neg
Does the sign change occur between f(-1) and f(-0.5) or between f(-0.5) and f(0)?
pos
-1
-0.5 1 8 -1 0 2
-0.5 -3.75 2.375 -1.1875
1 7.5 -4.75 2.375 0.8125
pos
f(-1) = -6
The graph must cross the x-axis somewhere between -1 and 0
f(-0.7) = -0.9939
Let’s try half way between at x = - 0.5
Sign change
f(-0.6) = 0.0416
f(-0.5) = 0.8125
So let’s try something between -1 and - 0.5. Let’s try - 0.7. Do this with synthetic division or direct substitution.
- 0.6 is the closest to zero so this is the zero approximated to one decimal place
f(0) = 2
Notice that the sign change is between - 0.7 and - 0.6 |
# What Is Slope Intercept Form Of A Linear Equation
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
What Is Slope Intercept Form Of A Linear Equation – One of the numerous forms employed to represent a linear equation, one that is commonly used is the slope intercept form. The formula for the slope-intercept in order to identify a line equation when that you have the slope of the straight line and the y-intercept. This is the coordinate of the point’s y-axis where the y-axis intersects the line. Read more about this particular linear equation form below.
## What Is The Slope Intercept Form?
There are three fundamental forms of linear equations: the traditional slope, slope-intercept and point-slope. While they all provide identical results when utilized in conjunction, you can obtain the information line generated quicker with this slope-intercept form. Like the name implies, this form uses a sloped line in which the “steepness” of the line determines its significance.
The formula can be used to determine a straight line’s slope, the y-intercept (also known as the x-intercept), where you can apply different formulas available. The equation for this line in this particular formula is y = mx + b. The straight line’s slope is signified by “m”, while its y-intercept is indicated by “b”. Each point of the straight line can be represented using an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” have to remain as variables.
## An Example of Applied Slope Intercept Form in Problems
When it comes to the actual world, the slope intercept form is commonly used to illustrate how an item or problem evolves over the course of time. The value provided by the vertical axis represents how the equation handles the extent of changes over what is represented through the horizontal axis (typically the time).
A simple example of the use of this formula is to determine how much population growth occurs in a particular area in the course of time. In the event that the population of the area increases each year by a predetermined amount, the point value of the horizontal axis will grow by a single point with each passing year and the value of the vertical axis will increase to show the rising population by the set amount.
It is also possible to note the beginning point of a question. The starting value occurs at the y-value in the y-intercept. The Y-intercept is the place where x is zero. In the case of the above problem the beginning point could be at the time the population reading begins or when the time tracking begins , along with the related changes.
So, the y-intercept is the location at which the population begins to be tracked to the researchers. Let’s assume that the researcher begins to do the calculation or measurement in 1995. Then the year 1995 will be considered to be the “base” year, and the x = 0 point will occur in 1995. Therefore, you can say that the population of 1995 represents the “y”-intercept.
Linear equations that use straight-line formulas are almost always solved this way. The starting point is expressed by the y-intercept and the rate of change is expressed as the slope. The principal issue with the slope intercept form usually lies in the interpretation of horizontal variables especially if the variable is accorded to a specific year (or any other kind of unit). The trick to overcoming them is to ensure that you understand the definitions of variables clearly. |
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