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Instructions
Directions for the following two questions: Cities A and B are in different time zones. A is located 3000 km east of B. The table below describes the schedule of an airline operating non-stop flights between A and B. All the times indicated are local and on the same day.
Assume that planes cruise at the same speed in both directions. However, the effective speed is influenced by a steady wind blowing from east to west. It reduces or increases the speed of plane by 50 km per hour depending on direction of flight.
Question 5
# What is the time difference between A and B?
Solution
Let the speed of the plane be p Kmph.
So the speed of plane from A to B will be 'p+50' and the speed from B to A will be 'p-50'.
We notice that the plane goes from B to A stays there for 1 hr and again come back to B with total time duration 12 hrs.
So we have $$\frac{3000}{p-50} + 1 + \frac{3000}{p+50} = 12$$.
We can clearly see that speed of the plane is 550 which satisfies the above equation.
So for the journey of B to A, the plane takes $$\frac{3000}{550-50} = 6$$ hrs.
So time at B when plane reaches at A is 2 pm .
Hence the time difference between A and B is 1 hr.
Alternatively,
Let speed of flight be s,
Since A is to the east of B, A is ahead of be in time
Let A be ahead of B in time by a hours
Departure from A = 4PM, Arrival at B = 8PM
Travel time = 8 - 4 +a = 4 + a
Since City B is behind city A by 'a' hours, the actual travel time is 'a' hours more than the difference of local times.
Similarly when one travels from B to A, since B is ahead of A by 'a' hrs, actual travel time is 'a' hours less than total
i.e. B->A Travel time = (3PM - 8AM) - a = 7 -a
Total distance travelled = Speed $$\times$$ Time taken ....(1)
From A to B, the wind is favourable / in same direction as flight
Hence from (1), we have
A->B >>> $$3000 = (s+50)(4+a) => 3000(7-a) = (s+50)(4+a)(7-a)$$ ...(2)
B->A >>> $$3000 = (s-50)(7-a) => 3000(4+a) = (s-50)(7-a)(4+a)$$ ...(3)
(2) - (3) => $$3000(3-2a) = 100(7-a)(4+a) => a^2-63a+62=0 => a=1/62$$
Hence the time difference between A and B is 1 hr.
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# 2.4 The Product and Quotient Rules
The previous section showed that, in some ways, derivatives behave nicely. The Constant Multiple and Sum/Difference Rules established that the derivative of $f(x)=5x^{2}+\sin x$ was not complicated. We neglected computing the derivative of things like $g(x)=5x^{2}\sin x$ and $h(x)=\frac{5x^{2}}{\sin x}$ on purpose; their derivatives are not as straightforward. (If you had to guess what their respective derivatives are, you would probably guess wrong.) For these, we need the Product and Quotient Rules, respectively, which are defined in this section.
We begin with the Product Rule.
###### Theorem 2.4.1 Product Rule
Let $f$ and $g$ be differentiable functions on an open interval $I$. Then $f\cdot g$ is a differentiable function on $I$, and
$\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}f(x)g(x)\Bigr{)}=f(x)g% \mkern 1.35mu ^{\prime}(x)+f\,^{\prime}(x)g(x).$
Important: $\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}f(x)g(x)% \Bigr{)}\neq f\,^{\prime}(x)g\mkern 1.35mu ^{\prime}(x)$. While this answer is simpler than the Product Rule, it is wrong. We can show that this is wrong by considering the functions $f(x)=x^{2}$ and $g(x)=x^{5}$.
Using the WRONG rule we get $\dfrac{\operatorname{d}\!}{\operatorname{d}\!x}[f(x)g(x)]=2x\cdot 5x^{4}=10x^{5}$. However, when we simplify the product first and apply the Power Rule, $f\cdot g=x^{2}\cdot x^{5}=x^{7}$ and
$\frac{\operatorname{d}\!}{\operatorname{d}\!x}[f(x)g(x)]=7x^{6}\neq 10x^{5}.$
Applying the real Product Rule we see that,
$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}[f(x)g(x)]$ $\displaystyle=x^{2}\frac{\operatorname{d}\!}{\operatorname{d}\!x}(x^{5})+\frac% {\operatorname{d}\!}{\operatorname{d}\!x}(x^{2})\cdot x^{5}$ $\displaystyle=x^{2}\cdot 5x^{4}+2x\cdot x^{5}$ $\displaystyle=7x^{6}$
We practice using this new rule in an example, followed by a proof of the theorem.
###### Example 2.4.1 Using the Product Rule
Use the Product Rule to compute the derivative of $y=5x^{2}\sin x$. Evaluate the derivative at $x=\pi/2$.
SolutionTo make our use of the Product Rule explicit, let’s set $f(x)=5x^{2}$ and $g(x)=\sin x$. We easily compute/recall that $f\,^{\prime}(x)=10x$ and $g\mkern 1.35mu ^{\prime}(x)=\cos x$. Employing the rule, we have
$\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}5x^{2}\sin x\Bigr{)}=5x^% {2}\cos x+10x\sin x.$
margin: Figure 2.4.1: A graph of $y=5x^{2}\sin x$ and its tangent line at $x=\pi/2$. Λ
At $x=\pi/2$, we have
$y\mkern 1.35mu ^{\prime}(\pi/2)=5\left(\frac{\pi}{2}\right)^{2}\cos\left(\frac% {\pi}{2}\right)+10\frac{\pi}{2}\sin\left(\frac{\pi}{2}\right)=5\pi.$
We graph $y$ and its tangent line at $x=\pi/2$, which has a slope of $5\pi$, in Figure 2.4.1. While this does not prove that the Product Rule is the correct way to handle derivatives of products, it helps validate its truth.
• Proof of the Product Rule
By the limit definition, we have
$\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}f(x)g(x)\Bigr{)}=\lim_{h% \to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}.$
We now do something a bit unexpected; add 0 to the numerator (so that nothing is changed) in the form of $-f(x+h)g(x)+f(x+h)g(x)$, and then do some regrouping as shown.
$\displaystyle\dfrac{\operatorname{d}\!}{\operatorname{d}\!x}$ $\displaystyle\Bigl{(}f(x)g(x)\Bigr{)}$ $\displaystyle=\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\qquad\text{\small(% now add 0 to the numerator)}$ $\displaystyle=\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}$ $\displaystyle=\lim_{h\to 0}\frac{\Bigl{(}f(x+h)g(x+h)-f(x+h)g(x)\Bigr{)}+\Bigl% {(}f(x+h)g(x)-f(x)g(x)\Bigr{)}}{h}$ $\displaystyle=\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x+h)g(x)}{h}+\lim_{h\to 0}% \frac{f(x+h)g(x)-f(x)g(x)}{h}$ $\displaystyle=\lim_{h\to 0}f(x+h)\frac{g(x+h)-g(x)}{h}+\lim_{h\to 0}\frac{f(x+% h)-f(x)}{h}g(x)$ $\displaystyle=\lim_{h\to 0}f(x+h)\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}+\lim_{h\to 0% }\frac{f(x+h)-f(x)}{h}\lim_{h\to 0}g(x)$ $\displaystyle=f(x)g\mkern 1.35mu ^{\prime}(x)+f\,^{\prime}(x)g(x).\qed$
It is often true that we can recognize that a theorem is true through its proof yet somehow doubt its applicability to real problems. In the following example, we compute the derivative of a product of functions in two ways to verify that the Product Rule is indeed “right.”
###### Example 2.4.2 Exploring alternate derivative methods
Let $y=(x^{2}+3x+1)(2x^{2}-3x+1)$. Find $y\mkern 1.35mu ^{\prime}$ two ways: first, by expanding the given product and then taking the derivative, and second, by applying the Product Rule. Verify that both methods give the same answer.
SolutionWe first expand the expression for $y$; a little algebra shows that $y=2x^{4}+3x^{3}-6x^{2}+1$. It is easy to compute $y\mkern 1.35mu ^{\prime}$;
$y\mkern 1.35mu ^{\prime}=8x^{3}+9x^{2}-12x.$
Now apply the Product Rule.
$\displaystyle y\mkern 1.35mu ^{\prime}$ $\displaystyle=(x^{2}+3x+1)\cdot\frac{\operatorname{d}\!}{\operatorname{d}\!x}(% 2x^{2}-3x+1)+\frac{\operatorname{d}\!}{\operatorname{d}\!x}(x^{2}+3x+1)\cdot(2% x^{2}-3x+1)$ $\displaystyle=(x^{2}+3x+1)(4x-3)+(2x+3)(2x^{2}-3x+1)$ $\displaystyle=\bigl{(}4x^{3}+9x^{2}-5x-3\bigr{)}+\bigl{(}4x^{3}-7x+3\bigr{)}$ $\displaystyle=8x^{3}+9x^{2}-12x.$
The uninformed usually assume that “the derivative of the product is the product of the derivatives.” Thus we are tempted to say that $y\mkern 1.35mu ^{\prime}=(2x+3)(4x-3)=8x^{2}+6x-9$. Obviously this is not correct.
###### Example 2.4.3 Using the Product Rule with a product of three functions
Let $y=x^{3}\ln x\cos x$. Find $y\mkern 1.35mu ^{\prime}.$
SolutionWe have a product of three functions while the Product Rule only specifies how to handle a product of two functions. Our method of handling this problem is to simply group the latter two functions together, and consider $y=x^{3}\bigl{(}\ln x\cos x\bigr{)}$. Following the Product Rule, we have
$\displaystyle y\mkern 1.35mu ^{\prime}$ $\displaystyle=(x^{3})\bigl{(}\ln x\cos x\bigr{)}^{\prime}+3x^{2}\bigl{(}\ln x% \cos x\bigr{)}$ To evaluate $\bigl{(}\ln x\cos x\bigr{)}^{\prime}$, we apply the Product Rule again: $\displaystyle=(x^{3})\left(\ln x(-\sin x)+\frac{1}{x}\cos x\right)+3x^{2}\bigl% {(}\ln x\cos x\bigr{)}$ $\displaystyle=x^{3}\ln x(-\sin x)+x^{3}\frac{1}{x}\cos x+3x^{2}\ln x\cos x$
Recognize the pattern in our answer above: when applying the Product Rule to a product of three functions, there are three terms added together in the final derivative. Each term contains only one derivative of one of the original functions, and each function’s derivative shows up in only one term. It is straightforward to extend this pattern to finding the derivative of a product of 4 or more functions.
We consider one more example before discussing another derivative rule.
###### Example 2.4.4 Using the Product Rule
Find the derivatives of the following functions.
$1.\ f(x)=x\ln x\qquad 2.\ g(x)=x\ln x-x.$
SolutionRecalling that the derivative of $\ln x$ is $1/x$, we use the Product Rule to find our answers.
1. (a)
$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}x\ln x\Bigr% {)}=x\cdot 1/x+1\cdot\ln x=1+\ln x$.
2. (b)
Using the result from above, we compute
$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}x\ln x-x% \Bigr{)}=1+\ln x-1=\ln x.$
This seems significant; if the natural log function $\ln x$ is an important function (it is), it seems worthwhile to know a function whose derivative is $\ln x$. We have found one. (We leave it to the reader to find others; a correct answer will be very similar to this one.)
We have learned how to compute the derivatives of sums, differences, and products of functions. We now learn how to find the derivative of a quotient of functions.
###### Theorem 2.4.2 Quotient Rule
Let $f$ and $g$ be differentiable functions defined on an open interval $I$, where $g(x)\neq 0$ on $I$. Then $f/g$ is differentiable on $I$, and
$\frac{\operatorname{d}\!}{\operatorname{d}\!x}\left(\frac{f(x)}{g(x)}\right)=% \frac{g(x)f\,^{\prime}(x)-f(x)g\mkern 1.35mu ^{\prime}(x)}{[g(x)]^{2}}.$
• Proof of the Quotient Rule
Let the functions $f$ and $g$ be defined and $g(x)\neq 0$ on an open interval $I$. By the definition of derivative,
$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\left(\frac{f(x)}{g% (x)}\right)$ $\displaystyle=\lim_{h\to 0}\frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}$ $\displaystyle=\lim_{h\to 0}\left[\left(\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}% \right)\cdot\frac{1}{h}\right]$ $\displaystyle=\lim_{h\to 0}\left[\left(\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)% }\right)\cdot\frac{1}{h}\right]$
Adding and subtracting the term $f(x)g(x)$ in the numerator does not change the value of the expression and allows us to separate $f$ and $g$ so that
$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\left(\frac{f(x)}{g% (x)}\right)$ $\displaystyle=\lim_{h\to 0}\left[\left(\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)% g(x+h)}{g(x+h)g(x)}\right)\cdot\frac{1}{h}\right]$ $\displaystyle=\lim_{h\to 0}\left[\frac{f(x+h)g(x)-f(x)g(x)}{hg(x+h)g(x)}+\frac% {f(x)g(x)-f(x)g(x+h)}{hg(x+h)g(x)}\right]$ $\displaystyle=\lim_{h\to 0}\left[g(x)\frac{f(x+h)-f(x)}{hg(x+h)g(x)}+f(x)\frac% {g(x)-g(x+h)}{hg(x+h)g(x)}\right]$ $\displaystyle=\lim_{h\to 0}\frac{g(x)\frac{f(x+h)-f(x)}{h}-f(x)\frac{g(x+h)-g(% x)}{h}}{g(x+h)g(x)}$ $\displaystyle=\frac{\displaystyle\lim_{h\to 0}g(x)\cdot\lim_{h\to 0}\frac{f(x+% h)-f(x)}{h}-\lim_{h\to 0}f(x)\cdot\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}}{% \displaystyle\lim_{h\to 0}g(x+h)\cdot\lim_{h\to 0}g(x)}$ $\displaystyle=\frac{g(x)f\,^{\prime}(x)-f(x)g\mkern 1.35mu ^{\prime}(x)}{[g(x)% ]^{2}}\qed$
Let’s practice using the Quotient Rule.
###### Example 2.4.5 Using the Quotient Rule
Let $\displaystyle f(x)=\frac{5x^{2}}{\sin x}$. Find $f\,^{\prime}(x)$.
SolutionDirectly applying the Quotient Rule gives:
$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\left(\frac{5x^{2}}% {\sin x}\right)$ $\displaystyle=\frac{\sin x\frac{\operatorname{d}\!}{\operatorname{d}\!x}(5x^{2% })-5x^{2}\frac{\operatorname{d}\!}{\operatorname{d}\!x}(\sin x)}{(\sin x)^{2}}$ $\displaystyle=\frac{\sin x\cdot 10x-5x^{2}\cdot\cos x}{\sin^{2}x}$ $\displaystyle=\frac{10x\sin x-5x^{2}\cos x}{\sin^{2}x}.$
The Quotient Rule allows us to fill in holes in our understanding of derivatives of the common trigonometric functions. We start with finding the derivative of the tangent function.
###### Example 2.4.6 Using the Quotient Rule to find $\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\tan x\bigr{)}$.
Find the derivative of $y=\tan x$.
SolutionAt first, one might feel unequipped to answer this question. But recall that $\tan x=\sin x/\cos x$, so we can apply the Quotient Rule.
$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}\tan x\Bigr% {)}$ $\displaystyle=\frac{\operatorname{d}\!}{\operatorname{d}\!x}\left(\frac{\sin x% }{\cos x}\right)$ $\displaystyle=\frac{\cos x\frac{\operatorname{d}\!}{\operatorname{d}\!x}(\sin x% )-\sin x\frac{\operatorname{d}\!}{\operatorname{d}\!x}(\cos x)}{(\cos x)^{2}}$ $\displaystyle=\frac{\cos x\cos x-\sin x(-\sin x)}{\cos^{2}x}$ $\displaystyle=\frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}$ $\displaystyle=\frac{1}{\cos^{2}x}$ $\displaystyle=\sec^{2}x.$
This is a beautiful result. To confirm its truth, we can find the equation of the tangent line to $y=\tan x$ at $x=\pi/4$. The slope is $\sec^{2}(\pi/4)=2$; $y=\tan x$, along with its tangent line, is graphed in Figure 2.4.2. margin: Figure 2.4.2: A graph of $y=\tan x$ along with its tangent line at $x=\pi/4$. Λ
We include this result in the following theorem about the derivatives of the trigonometric functions. Recall we found the derivative of $y=\sin x$ in Example 2.1.6 and stated the derivative of the cosine function in Theorem 2.3.1. The derivatives of the cotangent, cosecant and secant functions can all be computed directly using Theorem 2.3.1 and the Quotient Rule.
###### Theorem 2.4.3 Derivatives of Trigonometric Functions
1. $\dfrac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\sin x\bigr{)}=\cos x$ 2. $\dfrac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\cos x\bigr{)}=-\sin x$ 3. $\dfrac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\tan x\bigr{)}=\sec^{2}x$ 4. $\dfrac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\cot x\bigr{)}=-\csc^{2}x$ 5. $\dfrac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\sec x\bigr{)}=\sec x\tan x$ 6. $\dfrac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\csc x\bigr{)}=-\csc x\cot x$
The proofs of these derivatives have been presented or left as exercises. To remember the above, it may be helpful to keep in mind that the derivatives of the trigonometric functions that start with “c” have a minus sign in them.
###### Example 2.4.7 Exploring alternate derivative methods
In Example 2.4.5 the derivative of $\displaystyle f(x)=\frac{5x^{2}}{\sin x}$ was found using the Quotient Rule. Rewriting $f$ as $f(x)=5x^{2}\csc x$, find $f\,^{\prime}$ using Theorem 2.4.3 and verify the two answers are the same.
SolutionWe found $\displaystyle f\,^{\prime}(x)=\frac{10x\sin x-5x^{2}\cos x}{\sin^{2}x}$ in Example 2.4.5. We now find $f\,^{\prime}$ using the Product Rule, considering $f$ as $f(x)=5x^{2}\csc x$.
$\displaystyle f\,^{\prime}(x)$ $\displaystyle=\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}5x^{2}\csc x% \Bigr{)}$ $\displaystyle=5x^{2}\frac{\operatorname{d}\!}{\operatorname{d}\!x}(\csc x)-% \csc x\frac{\operatorname{d}\!}{\operatorname{d}\!x}(5x^{2})$ $\displaystyle=5x^{2}(-\csc x\cot x)+10x\csc x$ (now rewrite trig functions) $\displaystyle=5x^{2}\cdot\frac{-1}{\sin x}\cdot\frac{\cos x}{\sin x}+\frac{10x% }{\sin x}$ $\displaystyle=\displaystyle\frac{-5x^{2}\cos x}{\sin^{2}x}+\frac{10x}{\sin x}$ (get common denominator) $\displaystyle=\frac{10x\sin x-5x^{2}\cos x}{\sin^{2}x}$
Finding $f\,^{\prime}$ using either method returned the same result. At first, the answers looked different, but some algebra verified they are the same. In general, there is not one final form that we seek; the immediate result from the Product Rule is fine. Work to “simplify” your results into a form that is most readable and useful to you.
When we stated the Power Rule in Section 2.3 we claimed that it worked for all $n\in\mathbb{R}$ but only provided the proof for non-negative integers. The next example uses the Quotient Rule to provide justification of the Power Rule for $n\in\mathbb{Z}$.
###### Example 2.4.8 Using the Quotient Rule to expand the Power Rule
Find the derivatives of the following functions.
1. (a)
$\displaystyle f(x)=\frac{1}{x}$
2. (b)
$\displaystyle f(x)=\frac{1}{x^{n}}$, where $n>0$ is an integer.
SolutionWe employ the Quotient Rule.
1. (a)
$\displaystyle f\,^{\prime}(x)=\frac{x\cdot 0-1\cdot 1}{x^{2}}=-\frac{1}{x^{2}}$.
2. (b)
$\displaystyle f\,^{\prime}(x)=\frac{x^{n}\cdot 0-1\cdot nx^{n-1}}{(x^{n})^{2}}% =-\frac{nx^{n-1}}{x^{2n}}=-\frac{n}{x^{n+1}}.$
The derivative of $\displaystyle y=\frac{1}{x^{n}}$ turned out to be rather nice. It gets better. Consider:
$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\left(\frac{1}{x^{n% }}\right)$ $\displaystyle=\frac{\operatorname{d}\!}{\operatorname{d}\!x}\Bigl{(}x^{-n}% \Bigr{)}$ (apply result from Example 2.4.8) $\displaystyle=-\frac{n}{x^{n+1}}$ (rewrite algebraically) $\displaystyle=-nx^{-(n+1)}$ $\displaystyle=-nx^{-n-1}.$
Thus, for all $n\in\mathbb{Z}$, we can officially apply the Power Rule: multiply by the power, then subtract 1 from the power.
Taking the derivative of many functions is relatively straightforward. It is clear (with practice) what rules apply and in what order they should be applied. Other functions present multiple paths; different rules may be applied depending on how the function is treated. One of the beautiful things about calculus is that there is not “the” right way; each path, when applied correctly, leads to the same result, the derivative. We demonstrate this concept in an example.
###### Example 2.4.9 Exploring alternate derivative methods
Let $\displaystyle f(x)=\frac{x^{2}-3x+1}{x}$. Find $f\,^{\prime}(x)$ in each of the following ways:
1. (a)
By applying the Quotient Rule,
2. (b)
by viewing $f$ as $f(x)=\bigl{(}x^{2}-3x+1\bigr{)}\cdot x^{-1}$ and applying the Product and Power Rules, and
3. (c)
by “simplifying” first through division.
Verify that all three methods give the same result.
Solution
1. (a)
Applying the Quotient Rule gives:
$f\,^{\prime}(x)=\frac{x\cdot\bigl{(}2x-3\bigr{)}-\bigl{(}x^{2}-3x+1\bigr{)}% \cdot 1}{x^{2}}=\frac{x^{2}-1}{x^{2}}=1-\frac{1}{x^{2}}.$
2. (b)
By rewriting $f$, we can apply the Product and Power Rules as follows:
$\displaystyle f\,^{\prime}(x)$ $\displaystyle=\bigl{(}x^{2}-3x+1\bigr{)}\cdot(-1)x^{-2}+\bigl{(}2x-3\bigr{)}% \cdot x^{-1}$ $\displaystyle=-\frac{x^{2}-3x+1}{x^{2}}+\frac{2x-3}{x}$ $\displaystyle=-\frac{x^{2}-3x+1}{x^{2}}+\frac{2x^{2}-3x}{x^{2}}$ $\displaystyle=\frac{x^{2}-1}{x^{2}}=1-\frac{1}{x^{2}},$
the same result as above.
3. (c)
As $x\neq 0$, we can divide through by $x$ first, giving $\displaystyle f(x)=x-3+\frac{1}{x}$. Now apply the Power Rule to see
$f\,^{\prime}(x)=1-\frac{1}{x^{2}},$
the same result as before.
Example 2.4.9 demonstrates three methods of finding $f\,^{\prime}$. It is difficult to argue for a “best method” as all three gave the same result without too much difficulty, although it is clear that using the Product Rule required more steps. Ultimately, the important principle to take away from this is: simplify the answer to a form that seems “simple” and easy to interpret. They are equal; they are all correct. The most appropriate form of $f\,^{\prime}$ depends on what we need to do with the function next. For later problems it will be important for us to determine the most appropriate form to use and to move flexibly between the different forms. In the next section we continue to learn rules that allow us to more easily compute derivatives than using the limit definition directly. We have to memorize the derivatives of a certain set of functions, such as “the derivative of $\sin x$ is $\cos x$.” The Sum/Difference, Constant Multiple, Power, Product and Quotient Rules show us how to find the derivatives of certain combinations of these functions. The next section shows how to find the derivatives when we compose these functions together.
## Exercises 2.4
### Terms and Concepts
1. 1.
T/F: The Product Rule states that $\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\big{(}x^{2}\sin x% \big{)}=2x\cos x$.
2. 2.
T/F: The Quotient Rule states that $\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\left(\frac{x^{2}}{% \sin x}\right)=\frac{2x}{\cos x}$.
3. 3.
T/F: The derivatives of the trigonometric functions that start with “c” have minus signs in them.
4. 4.
What derivative rule is used to extend the Power Rule to include negative integer exponents?
5. 5.
T/F: Regardless of the function, there is always exactly one right way of computing its derivative.
6. 6.
### Problems
In Exercises 7–8., use the Quotient Rule to verify these derivatives.
1. 7.
$\dfrac{\operatorname{d}\!}{\operatorname{d}\!x}(\cot x)=-\csc^{2}x$
2. 8.
$\dfrac{\operatorname{d}\!}{\operatorname{d}\!x}(\csc x)=-\csc x\cot x$
In Exercises 9–12.:
1. (a)
Use the Product Rule to differentiate the function.
2. (b)
Manipulate the function algebraically and differentiate without the Product Rule.
3. (c)
Show that the answers from (a) and (b) are equivalent.
1. 9.
$f(x)=x(x^{2}+3x)$
2. 10.
$g(x)=2x^{2}(5x^{3})$
3. 11.
$h(s)=(2s-1)(s+4)$
4. 12.
$f(x)=(x^{2}+5)(3-x^{3})$
In Exercises 13–16.:
1. (a)
Use the Quotient Rule to differentiate the function.
2. (b)
Manipulate the function algebraically and differentiate without the Quotient Rule.
3. (c)
Show that the answers from (a) and (b) are equivalent.
1. 13.
$\displaystyle f(x)=\frac{x^{2}+3}{x}$
2. 14.
$\displaystyle g(x)=\frac{x^{3}-2x^{2}}{2x^{2}}$
3. 15.
$\displaystyle h(s)=\frac{3}{4s^{3}}$
4. 16.
$\displaystyle f(t)=\frac{t^{2}-1}{t+1}$
In Exercises 17–42., compute the derivative of the given function.
1. 17.
$f(x)=x\sin x$
2. 18.
$\displaystyle f(t)=\frac{1}{t^{2}}(\csc t-4)$
3. 19.
$H(y)=(y^{5}-2y^{3})(7y^{2}+y-8)$
4. 20.
$F(y)=\sqrt[3]{y^{2}}(y^{2}+9y)$
5. 21.
$\displaystyle g(x)=\frac{x+7}{x-5}$
6. 22.
$y=\dfrac{\sqrt{x}}{x+4}$
7. 23.
$g(x)=\dfrac{x}{\sqrt{x}+4}$
8. 24.
$\displaystyle g(t)=\frac{t^{5}}{\cos t-2t^{2}}$
9. 25.
$h(x)=\cot x-e^{x}$
10. 26.
$h(t)=7t^{2}+6t-2$
11. 27.
$\displaystyle f(x)=\frac{x^{4}+2x^{3}}{x+2}$
12. 28.
$f(x)=\dfrac{x^{2}-\sqrt{x}}{x^{3}}$
13. 29.
$y=\left(\dfrac{1}{x^{3}}+\dfrac{5}{x^{4}}\right)(2x^{3}-x^{5})$
14. 30.
$g(x)=\dfrac{1}{1+x+x^{2}+x^{3}}$
15. 31.
$p(x)=1+\dfrac{1}{x}+\dfrac{1}{x^{2}}+\dfrac{1}{x^{3}}$
16. 32.
$\displaystyle f(x)=(16x^{3}+24x^{2}+3x)\frac{7x-1}{16x^{3}+24x^{2}+3x}$
17. 33.
$f(t)=t^{5}(\sec t+e^{t})$
18. 34.
$\displaystyle f(x)=\frac{\sin x}{\cos x+3}$
19. 35.
$g(x)=e^{2}\big{(}\sin(\pi/4)-1\big{)}$
20. 36.
$g(t)=4t^{3}e^{t}-\sin t\cos t$
21. 37.
$f(y)=y(2y^{3}-5y-1)(6y^{2}+7)$
22. 38.
$F(x)=(8x-1)(x^{2}+4x+7)(x^{3}-5)$
23. 39.
$\displaystyle h(t)=\frac{t^{2}\sin t+3}{t^{2}\cos t+2}$
24. 40.
$f(x)=x^{2}e^{x}\tan x$
25. 41.
$g(x)=2x\sin x\sec x$
26. 42.
$f(x)=x\ln x$
In Exercises 43–46., find the equations of the tangent line to the graph of $g$ at the indicated point.
1. 43.
$g(s)=e^{s}(s^{2}+2)$ at $(0,2)$.
2. 44.
$g(t)=t\sin t$ at $(\frac{3\pi}{2},-\frac{3\pi}{2})$
3. 45.
$\displaystyle g(x)=\frac{x^{2}}{x-1}$ at $(2,4)$
4. 46.
$\displaystyle g(\theta)=\frac{\cos\theta-8\theta}{\theta+1}$ at $(0,1)$
In Exercises 47–50., find the $x$-values where the graph of the function has a horizontal tangent line.
1. 47.
$f(x)=6x^{2}-18x-24$
2. 48.
$f(x)=x\sin x$ on $[-1,1]$
3. 49.
$\displaystyle f(x)=\frac{x}{x+1}$
4. 50.
$\displaystyle f(x)=\frac{x^{2}}{x+1}$
In Exercises 51–54., find the requested derivative.
1. 51.
$f(x)=x\sin x$; find $f\,^{\prime\prime}(x)$.
2. 52.
$f(x)=x\sin x$; find $f\,^{(4)}(x)$.
3. 53.
$f(x)=\csc x$; find $f\,^{\prime\prime}(x)$.
4. 54.
$f(x)=(x^{3}-5x+2)(x^{2}+x-7)$; find $f\,^{(8)}(x)$.
In Exercises 55–60., $f$ and $g$ are differentiable functions such that $f(2)=3$, $f\,^{\prime}(2)=-1$, $g(2)=-5$, and $g^{\prime}(2)=2$. Evaluate the expressions.
1. 55.
$(f+g)^{\prime}(2)$
2. 56.
$(f-g)^{\prime}(2)$
3. 57.
$(4f)^{\prime}(2)$
4. 58.
$(f\cdot g)^{\prime}(2)$
5. 59.
$\left(\dfrac{f}{g}\right)^{\prime}(2)$
6. 60.
$\left(\dfrac{g}{f+g}\right)^{\prime}(2)$
1. 61.
If $f$ and $g$ are functions whose graphs are shown, evaluate the expressions. * (a) $(fg)^{\prime}(-1)$ (b) $(f/g)^{\prime}(-1)$ (c) $(fg)^{\prime}(3)$ (d) $(g/f)^{\prime}(3)$
|
## Introduction to Application of Numbers
CXB
Posts: 98
Joined: Tue Mar 18, 2014 9:39 pm
### Introduction to Application of Numbers
Help
I have googled fractions and I still dont have a scooby do on how to do Fractions.
One of the questions is 4/5+1/3x3/5 where do I start also (3/5+1/3)x5/7 once again where do I start. I never mastered these at school.
Look forward to hearing from someone
Clare
Yvonne Otter
Head of level 2 and GCSE equivalences
Posts: 70
Joined: Sun Mar 04, 2012 10:50 pm
### Re: Introduction to Application of Numbers
Hi Clare,
If we look at similar examples then this might help you with the assessment questions.
Let's consider 1/3 + 3/4 x 1/5
First we use BIDMAS to tell us where to start. We know that Multiplication is carried out before addition so we would carry out 3/4 x 1/5 first. With Multiplication the process is quite straightforward ( we don't need to worry about a common denominator or anything). We just multiply the two numerators together to give a new numerator and the two denominators together to give a new denominator.
So 3/4 x 1/5 = 3/20
Now we need to do the rest of it which is now
1/3 + 3/20
When we are adding (or subtracting) then we need the denominators to be the same. So we look for the number that 3 and 20 will both go into (60)
We change each fraction to its equivalent fraction with a denominator of 60
1/3 = 20/60 (top and bottom numbers x 20) and 3/20 = 9/60 (top and bottom numbers x 3)
Now we have 20/60 + 9/60 = 29/60 (when we add or subtract fractions we work with the numerators only)
If we consider an example in the second format (1/3 + 3/4) x 1/5
This time BIDMAS tells us that we need to do the Brackets first
1/3 + 3/4 so we look for a number that 3 and 4 go into (12
1/3 = 4/12 and 3/4 = 9/12 (you can find more on equivalent fractions on bbc skillswise)
So 4/12 + 9/12 = 13/12 ( Don't worry about it being top heavy)
Now for the second part 13/12 x 1/5 = 13/60 (remember times the numerators together for the new numerator and the denominators together for the denominator)
So this time 13/60 is the answer.
Yvonne
CXB
Posts: 98
Joined: Tue Mar 18, 2014 9:39 pm
### Re: Introduction to Application of Numbers
Hello there ,
I would like to thank you for your help on these two questions. It helped me and I have finally finished that assignment.
It was a tough one for me as math isnt my strength. I am going to submit my work over this weekend and I hope its what they were looking for.
Have a great weekend. No to crack on with the Maths project shoe sizes
Regards,
Clare
|
Digits, Grade 8, Volume 1, Homework Helper
### 8-4: Solving Simultaneous Linear Equations Using Substitution
#### Key Concept
You can solve simultaneous linear equations using an algebraic method called the substitution method. Solve one of the equations for one of the variables. Then substitute the expression for the variable into the other equation and solve for the second variable.
Solve the system by substitution.
y = 3x
2x + y = 15
• Step 1 Solve for the first variable.
Since y = 3x, you can substitute 3x for y in 2x + y = 15.
table with 4 rows and 3 columns , row1 column 1 , cap write , , the , , second , . equehtion . . , column 2 2 x plus y , column 3 equals 15 , row2 column 1 , cap substitute . 3 . bold italic x . , for , . bold italic y . , column 2 2 x plus 3 x , column 3 equals 15 , row3 column 1 , cap simplify , . , column 2 5 x , column 3 equals 15 , row4 column 1 , cap divide , , eehch , , side , by 5 . . , column 2 x , column 3 equals 3 , end table
• Step 2 Solve for the second variable.
Since x = 3, substitute 3 for x in either equation and solve for y.
table with 3 rows and 3 columns , row1 column 1 , cap write , , either , . equehtion . . , column 2 y , column 3 equals 3 x , row2 column 1 , cap substitute . 3 , for , . bold italic x . , column 2 y , column 3 equals 3 open 3 close , row3 column 1 , cap simplify , . , column 2 y , column 3 equals 9 , end table
• Step 3 Write the solution.
Since x = 3, and y = 9, the solution is (3, 9).
• Step 4 Check your solution.
Substitute the solution (3, 9) into each equation.
End ofPage 279
|
# Simple Interest
19 teachers like this lesson
Print Lesson
## Objective
SWBAT calculate simple interest using a formula
#### Big Idea
Interest = Principal * Rate * Time
## Introduction
10 minutes
I will begin with the essential question: How can you calculate simple interest?
The notes section begins with a simple paragraph about interest. Either a student, the class, or I will read through the paragraph. We will fill in the blanks as we go. The final version should say:
Money that you save in a bank account can grow. Money that you borrow from a bank is not free.
"Interest is money paid or earned for the use of money. The principle is the amount of money borrowed or deposited. Simple interest is money paid or earned only on the principal. Simple interest can be calculated using a formula."
The formula is then presented in its most common form.
Before going into examples it is probably worth discussing cases where interest can occur. I will ask and/or provide examples of borrowing money - using credit cards, car loans, tuition loans, mortgages. Interest (though often not simple interest!) is paid on these loans. I want students to understand that there is a cost for borrowing money. Conversely, money put in a savings account, mutual funds, retirement accounts, etc can grow based on interest. So it is possible for your money to earn money. Again, I may tell students interest on these types of accounts is not usually simple interest but the concept of simple interest will help us understand the other types when they study them.
I will then read through the first example. After the first reading, I will go through and annotate values in the problem. I will label \$500 as the principal, 3% as the rate - written 0.03, and 3 years as the time. Then, in part A I will write p = 500, r = 0.03, and t = 3. The equation will be written, values will be substituted and the equation will be solved.
In part B, students will need to know that "balance" means how much is in the account. In this case it's principal plus interest.
Students have a check for understanding problem. I expect to see annotations and variable assignments in the same manner as the example.
## Guided Problem Solving
15 minutes
Before beginning the guided practice it will be helpful to review the steps needed to solve the problems, especially since they are not written in the resource.
1) What is the first thing necessary to solving an interest problem?
Answers: Identify which values represent interest, principle, rate, and time/
2) What do we do with these values?
Answer: Substitute them into the simple interest formula.
3) What do we do next?
Answer: Simplify and solve the equation.
Now students will solve the 5 guided practice problems. Problem 1 is meant to see how well students understand simple interest - that is why I chose a variable for years and have an unspecified principal. Students hopefully can say that they would multiply the amount of money placed in the savings account by 0.02 times y to find the interest.
Problems 2 and 3 are most like the example.
Problem requires student to find the unknown interest rate. The steps discussed above will be extremely helpful. Here it is necessary to make sure that students are identifying the correct values. I have provided variables to ensure that students assign the variables correctly. It also helps me quickly assess how well they are solving the problems as I walk the room.
The last question is an explain your thinking problem (MP3). I want students to understand that they want the lowest possible interest rate when borrowing, yet they want their investments to yield a high rate of return.
## Independent Problem Solving
20 minutes
Students now work independently to solve the next set of problems. The first 5 only require students to find the interest and the balance.
Number 6 is designed to bring out a common error. The time is given in months, but the students need to recognize that time needs to be represented in years.
Problems 7-10 use either provide time in months or require students to find something other than the interest: the principle, the rate, or the time.
Note: Problems 1,2 and 9 are most similar to the exit ticket. As students are working, I will pay especially close attention to these three problems as a gauge on how prepared they are for the exit ticket.
## Exit Ticket
5 minutes
Before beginning the exit ticket, we will discuss how to use the simple interest formula. We will also review each term - interest, principle, rate, time, and balance.
I will ask my students to annotate each problem in the manner of the examples. Many of the mistakes my students make are from not carefully reading. Annotating will help them better understand the problems they are solving.
The exit ticket is worth 5 points. Students earning 4-5 points will have successful exit tickets.
|
Education.com
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Brainzy
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Plus
# Geometric Figures Help
based on 1 rating
By — McGraw-Hill Professional
Updated on Sep 22, 2011
## Geometric Figures
The word "geometry" is derived from two Greek words meaning "earth measure." The basic geometric figures are the point, the line, and the plane. These figures are theoretical and cannot be formally defined. A point is represented by a dot and is named by a capital letter. A line is an infinite set of points and is named by a small letter or by two points on the line. A line segment is part of a line between two points called endpoints. A plane is a flat surface (see Fig. 9-1 ).
Fig. 9-1.
Points and line segments are used to make geometric figures. The geometric figures presented in this chapter are the triangle, the square, the rectangle, the parallelogram, the trapezoid, and the circle (see Fig. 9-2).
Fig. 9-2.
A triangle is a geometric figure with three sides. A rectangle is a geometric figure with four sides and four 90 ° angles. The opposite sides are equal in length and are parallel. A square is a rectangle in which all sides are the same length. A parallelogram has four sides with two pairs of parallel sides. A trapezoid has four sides, two of which are parallel. A circle is a geometric figure such that all the points are the same distance from a point called its center. The center is not part of the circle. A line segment passing through the center of a circle and with its endpoints on the circle is called a diameter . A line segment from the center of a circle to the circle is called a radius (see Fig. 9-3).
Fig. 9-3.
Practice problems for these concepts can be found at: Informal Geometry Practice Test.
150 Characters allowed
### Related Questions
#### Q:
See More Questions
Top Worksheet Slideshows
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Convert native decimal come fraction. Transform 0.16666 come Fraction. Decimal to portion chart and calculator. Writes any decimal number as a fraction.
You are watching: What is .16666 as a fraction
## How to transform a Decimal to a portion - Steps
Step 1: compose down the decimal as a fraction of one (decimal/1);Step 2: If the decimal is no a totality number, main point both top and also bottom by 10 till you gain an interger at the numerator.
Learn more reading the examples listed below or usage our self-explaining calculator above
## Convert decimal 0.05 to a fraction
0.05 = 1/20 together a fraction
### Step by step Solution
To transform the decimal 0.05 come a fraction follow these steps:
Step 1: create down the number together a portion of one:
0.05 = 0.05/1
Step 2: main point both top and also bottom by 10 because that every number ~ the decimal point:
As we have actually 2 numbers after the decimal point, we multiply both numerator and denominator through 100. So,
0.05/1 = (0.05 × 100)/(1 × 100) = 5/100.
Step 3: leveling (or reduce) the fraction:
5/100 = 1/20 when lessened to the most basic form.
## What is 0.45 together a fraction?
0.45 = 9/20 together a fraction
### Step by step Solution
To transform the decimal 0.45 to a fraction follow this steps:
Step 1: compose down the number as a fraction of one:
0.45 = 0.45/1
Step 2: main point both top and bottom by 10 for every number ~ the decimal point:
As we have actually 2 numbers after the decimal point, us multiply both numerator and also denominator by 100. So,
0.45/1 = (0.45 × 100)/(1 × 100) = 45/100.
Step 3: leveling (or reduce) the fraction:
45/100 = 9/20 when diminished to the simplest form.
## Equivalent portion for 1.3 percent
1.3 = 13/10 = 13/10 as a fraction
### Step by step Solution
To convert the decimal 1.3 come a portion follow this steps:
Step 1: compose down the number as a portion of one:
1.3 = 1.3/1
Step 2: main point both top and also bottom by 10 for every number ~ the decimal point:
As we have 1 number after the decimal point, we multiply both numerator and also denominator by 10. So,
1.3/1 = (1.3 × 10)/(1 × 10) = 13/10.
(This portion is alread reduced, we can"t mitigate it any kind of further).
As the numerator is greater than the denominator, we have actually an improper fraction, so we can additionally express it as a mixed NUMBER, for this reason 13/10 is likewise equal come 1 3/10 when expressed together a blended number.
## Conversion table: fraction to decimal inches and millimeter equivalence
To transform fractions come decimals and millimeters and also vice-versa usage this formula: 1 customs = 25.4 mm exactly, therefore ...To transform from customs to millimeter multiply inch worth by 25.4.To convert from millimeter inch divide millimeter value by 25.4.
an easier way to execute it is to usage the table below. How?
### Example 1
Convert 1 1/32" come mm: find 1 1/32 and read to the best under mm column! girlfriend will find 26.1938.
### Example 2
Convert 0.875 decimal inches to inches (fraction form).Look down the decimal shaft until you find 0.875, then review to the left to find 7/8 inchesor relocate to the right column to discover the mm value!
fractioninchesmm
1/640.01560.3969
1/320.03130.7938
3/640.04691.1906
1/160.06251.5875
5/640.07811.9844
3/320.09382.3813
7/640.10942.7781
1/80.12503.1750
9/640.14063.5719
5/320.15633.9688
11/640.17194.3656
3/160.18754.7625
13/640.20315.1594
7/320.21885.5563
15/640.23445.9531
1/40.25006.3500
17/640.26566.7469
9/320.28137.1438
19/640.29697.5406
5/160.31257.9375
21/640.32818.3344
11/320.34388.7313
23/640.35949.1281
3/80.37509.5250
25/640.39069.9219
13/320.406310.3188
27/640.421910.7156
7/160.437511.1125
29/640.453111.5094
15/320.468811.9063
31/640.484412.3031
1/20.500012.7000
33/640.515613.0969
17/320.531313.4938
35/640.546913.8906
9/160.562514.2875
37/640.578114.6844
19/320.593815.0813
39/640.609415.4781
5/80.625015.8750
41/640.640616.2719
21/320.656316.6688
43/640.671917.0656
11/160.687517.4625
45/640.703117.8594
23/320.718818.2563
47/640.734418.6531
3/40.750019.0500
49/640.765619.4469
25/320.781319.8438
51/640.796920.2406
13/160.812520.6375
53/640.828121.0344
27/320.843821.4313
55/640.859421.8281
7/80.875022.2250
57/640.890622.6219
29/320.906323.0188
59/640.921923.4156
15/160.937523.8125
61/640.953124.2094
31/320.968824.6063
63/640.984425.0031
11.000025.4000
fractioninchesmm
1 1/641.015625.7969
1 1/321.031326.1938
1 3/641.046926.5906
1 1/161.062526.9875
1 5/641.078127.3844
1 3/321.093827.7813
1 7/641.109428.1781
1 1/81.125028.5750
1 9/641.140628.9719
1 5/321.156329.3688
1 11/641.171929.7656
1 3/161.187530.1625
1 13/641.203130.5594
1 7/321.218830.9563
1 15/641.234431.3531
1 1/41.250031.7500
1 17/641.265632.1469
1 9/321.281332.5438
1 19/641.296932.9406
1 5/161.312533.3375
1 21/641.328133.7344
1 11/321.343834.1313
1 23/641.359434.5281
1 3/81.375034.9250
1 25/641.390635.3219
1 13/321.406335.7188
1 27/641.421936.1156
1 7/161.437536.5125
1 29/641.453136.9094
1 15/321.468837.3063
1 31/641.484437.7031
1 1/21.500038.1000
1 33/641.515638.4969
1 17/321.531338.8938
1 35/641.546939.2906
1 9/161.562539.6875
1 37/641.578140.0844
1 19/321.593840.4813
1 39/641.609440.8781
1 5/81.625041.2750
1 41/641.640641.6719
1 21/321.656342.0688
1 43/641.671942.4656
1 11/161.687542.8625
1 45/641.703143.2594
1 23/321.718843.6563
1 47/641.734444.0531
1 3/41.750044.4500
1 49/641.765644.8469
1 25/321.781345.2438
1 51/641.796945.6406
1 13/161.812546.0375
1 53/641.828146.4344
1 27/321.843846.8313
1 55/641.859447.2281
1 7/81.875047.6250
1 57/641.890648.0219
1 29/321.906348.4188
1 59/641.921948.8156
1 15/161.937549.2125
1 61/641.953149.6094
1 31/321.968850.0063
1 63/641.984450.4031
22.000050.8000
fractioninchesmm
2 1/642.015651.1969
2 1/322.031351.5938
2 3/642.046951.9906
2 1/162.062552.3875
2 5/642.078152.7844
2 3/322.093853.1813
2 7/642.109453.5781
2 1/82.125053.9750
2 9/642.140654.3719
2 5/322.156354.7688
2 11/642.171955.1656
2 3/162.187555.5625
2 13/642.203155.9594
2 7/322.218856.3563
2 15/642.234456.7531
2 1/42.250057.1500
2 17/642.265657.5469
2 9/322.281357.9438
2 19/642.296958.3406
2 5/162.312558.7375
2 21/642.328159.1344
2 11/322.343859.5313
2 23/642.359459.9281
2 3/82.375060.3250
2 25/642.390660.7219
2 13/322.406361.1188
2 27/642.421961.5156
2 7/162.437561.9125
2 29/642.453162.3094
2 15/322.468862.7063
2 31/642.484463.1031
2 1/22.500063.5000
2 33/642.515663.8969
2 17/322.531364.2938
2 35/642.546964.6906
2 9/162.562565.0875
2 37/642.578165.4844
2 19/322.593865.8813
2 39/642.609466.2781
2 5/82.625066.6750
2 41/642.640667.0719
2 21/322.656367.4688
2 43/642.671967.8656
2 11/162.687568.2625
2 45/642.703168.6594
2 23/322.718869.0563
2 47/642.734469.4531
2 3/42.750069.8500
2 49/642.765670.2469
2 25/322.781370.6438
2 51/642.796971.0406
2 13/162.812571.4375
2 53/642.828171.8344
2 27/322.843872.2313
2 55/642.859472.6281
2 7/82.875073.0250
2 57/642.890673.4219
2 29/322.906373.8188
2 59/642.921974.2156
2 15/162.937574.6125
2 61/642.953175.0094
2 31/322.968875.4063
2 63/642.984475.8031
33.000076.2000
### Decimal to fraction Calculator
Please attach to this page! just right click the over image, then choose copy attach address, then past it in her HTML.
See more: 7.3 Glow Plug Relay Fuse Location Of Glow Plug Fuse, Location Of Glow Plug Fuse
### Disclaimer
While every initiative is made to ensure the accuracy that the information listed on this website, neither this website no one its authors are responsible for any type of errors or omissions. Therefore, the components of this site are not perfect for any type of use involving risk to health, finances or property.
|
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$-i\pm i\sqrt{1-i}$
$i{{x}^{2}}-2x+1=0$ Now compare the equation $i{{x}^{2}}-2x+1=0$ with the standard form of the quadratic equation $a{{x}^{2}}+bx+c=0$. Here, $a=i,b=-2$and $c=1$ Substitute $a=i,b=-2$ and $c=1$ into the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. \begin{align} & x=\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( i \right)\left( 1 \right)}}{2\left( i \right)} \\ & =\frac{2\pm \sqrt{4-4i}}{2i} \\ & =\frac{2\pm 2\sqrt{1-i}}{2i} \\ & =\frac{1\pm \sqrt{1-i}}{i} \end{align} Rationalize the expression by multiplying by $\frac{i}{i}$, \begin{align} & x=\frac{1\pm \sqrt{1-i}}{i}\cdot \frac{i}{i} \\ & =\frac{i\pm i\sqrt{1-i}}{{{i}^{2}}} \\ & =\frac{i\pm i\sqrt{1-i}}{-1} \\ & =-i\pm i\sqrt{1-i} \end{align}
|
STUDY GUIDE FOR CHAPTER 5
```STUDY GUIDE FOR CHAPTER 5
In this chapter Student should be able to:
a) Solve exponential equations
b) Graph exponential function
c) Change a function from logarithmic to exponential form and from exponential to
logarithmic form
d) Apply lows of logarithms
e) Find the logarithms using calculator
f) Solve logarithmic equation
g) Solve problems involving compound interest
h) Solve problems involving exponential growth and decay.
5.1. INVERSE FUNCTION
Definition of one-to-one function:
If a b are elements of a domain of f(x) and if f(a) f(b) then f(x) is one-to-one
function.
A function in which elements from the domain always lead to different elements from
the range is called on-to-one function.
Examples: Is f(x) one-to-one function?
f(x) = 3x + 2
f ( x) x 2 4
For every x there is only one y
For x = 2 and for x = - 2 there is only one y = 0
Horizontal Line test for one-to-one function:
Intersects with f(x) in no more then one
point if f(x) is one-to-one function’
Graph shows that parabola is not a one-to-one function.
Inverse Function:
If f(x) and g(x) are inverse function then:
f[g(x)] = x
and
g[f(x)] = x
If f(x) is given how to find g(x) = f -1(x):
Procedure for finding one-to-one function:
1.
2.
3.
4.
Check that f(x) is one-to-one function.
Solve for x [x = f -1(y)].
Exchange x and y [y = f -1(x) = g(x)].
Check that f[g(x)] = x and g[f(x)] = x
The domain of f(x) equals the range of f -1(x) and the range of f(x) equals the domain of f -1(x).
Practice: Find the inverse of:
1. y = f(x) = 4x – 5
y5
4
x5
y
f 1 ( x ) g ( x )
4
x
a) Solve for x:
b) Exchange x and y:
c) Check that f[g(x)] = x
f [ g ( x) 4
x5
5 x
4
and
g[ f ( x )
4x 5 5
x
4
y= x
y=(x + 5)/4
y=4x - 5
2. y = 2x + 1
3. y 6 x
4. y = - x2 + 2
5. y = 1/x
and
g[f(x)] = x
5.2. EXPONENTIAL FUNCTIONS
Properties of Exponents:
If x = y then ax = ay
If a > 1 and x < y then ax < ay
If 0 < a < 1 and x < y then ax > ay
Graphing exponential function f(x) = ax
a) The points (0,1) and (1,a) are in the graph
b) If a > 1 f(x) is increasing
c) If 0 < a < 1, f(x) is decreasing
d) x - axis is horizontal asymptote
e) Domain (No restriction), Range ( y > 0)
Graph the following functions:
f(x) = 2x
5
y
y =2x
4
3
2
( 1 ,2 )
1
x
0
-5
-4
-3
-2
-1 -1 0
1
2
-2
-3
-4
-5
f(x) = 3 x
f(x) = - 3 x
f(x) = 3x+2
f(x) = 3 x + 2
f(x) = 3x+ 2 + 2
f(x) = 2 x + 1
f(x) = (1/2)x
f(x) = 2 –x
f(x) = - 2 x
3
4
5
Solve exponential equations using the property: if a x = a y then x = y:
(1/3) x = 81
Solution:
1
( ) x 36
3
1
1
( ) x ( ) 6
3
3
x 6
(3-1) x = 34
(2/3)x = 9/4
23-x = 8
16 x + 2 = 8 1 - x
32t = 16 1 - t
(2/3)k - 1 = (81/16) k + 1
81 = x 4/3
5 2p + 1 = 25
Applications of Exponential Equations
Q 48 on page 368
Q 50 on page 368
Q 52 on page 368
Compound Interest
If P dollars are deposited in an account paying an annual rate of interest r, compounded m times
per year, then after t years the account will contain A dollars, where
A P(1
r mt
)
m
A – future value
P – present value
Example 1: How much should be deposited at 6 % interest compounded annually (m = 1) for
three years ( t = 3) in order to have a balance of \$20,000 in three years?
0.06 3
)
1
20,000 P(1.191016)
20,000 P(1
P 16,792.39
Example 2: If only \$15,000 is deposited , what annual interest rate is required to increase to
\$20,000[10 %]
Example 3: Invest \$1 (P = 1) for one year ( t = 1) at the rate of 100 % (r = 1) and find A
compounded 1, 2, 5, 10, 25, 100, 500, 1000, 10000, 100000, 1000000..... times a year. Future
value should go toward fixed number e = 2.71828
Compounded annually:
1
P 1 (1 )1 2
1
Compounded twice a year:
1
P 1 (1 ) 2 2.25
2
Compounded ten times a year
P 1 (1
1 10
) 2.5937
10
Compounded one hundred times a year
P 1 (1
1 100
) 2.7048
100
Compounded one thousand times a year
P 1 (1
1 1000
)
2.7169
1000
Compounded one hundred thousand times a year
P 1 (1
1
)100,000 2.71826
100,000
Compounded one million times a year
P 1 (1
1
)1,000,000 2.71828
1,000,000
Compounded continuously
P 2.71828 e
Exponential Growth and Decay
y = yo e kt
y – future amount or number
yo – is amount or number at present time t = 0
k - constant
Work on Q. 56 on page 369
5.3. LOGARITMIC FUNCTION
How to find the inverse of a exponential function y = ax – Graph.
1. It’s one-to-one function
2. How to solve for x ?
Introduction of Logarithmic functions
y = ax Solving for x:
f(x) = a x
x = log a y
f –1(x) = log a x
Exchange x and y:
y = log a x
Meaning of a logarithm:
Exponential form
25 = 5x
Logarithmic form:
x = log 5 25
“Five raised to what exponent is 25”
Log function is an Inverse of exponential function
Write eq. in logarithmic or exponential form:
a)
b)
c)
d)
e)
f)
x=ay
(1/2)-4 = 16
51=5
(3/4)0 = 1
9 2.7 = z
0.315=2 –5
y = loga x
-4 = log ½ 16
1 = log 5 5
0 = log ¾ 1
2.7 = log 9 z
-5 = log 2 0.3125
Properties of Logarithms
log a xy = log a x + log a y
log a x/y = log a x - log a y
log a x r = r log a x
log a a = 1
log a 1 = 0
Application of properties
log a (m n q/p2) = ?
=log a ( m n q ) – log a p 2 =
= log a m + log a n + log a q – 2 log a p
log6 3 m2 ?
x y
log a
?
z3
x3 y5
?
zm
logb
log a x + log a y – log a m = ?
2 log m a – 3 log m (b2 ) = ?
(log b k – log b m ) – log b a =
n
log b
k
k
log b a log b
m
ma
Use graphing calculator to graph logarithmic functions:
f(x) = log 10 ( x – 1)
f(x) = log 10 x - 1
5.4 EVALUATING LOGARITHMS
Common logarithm:
log x = log 10 x
Natural number e = 2.73…
Natural logarithm:
log e x = ln x
Use calculator to evaluate ( to the nearest ten-thousands )
log 1247 = ?
log0.0069 = ?
ln0.7 = ?
ln580 = ?
ln650 = ?
log312.5 = ?
log 10 = ?
ln 1 = ?
ln0.049 = ?
ln4.44 = ?
Change of base theorem:
log b x log x ln x
log b a log a ln a
log a x
log 9 12 = log 12/log 9 = ?
log2.9 7.5 = log 7.5/log 2.9 = ?
log 5 17 =
log 2 0.1 =
5.5 EXPONENTIAL AND LOGARITHMIC EQUATIONS
Property: When x > 0, y > 0, a > 0 ( a 1 ) and if log a x = log a y then x = y
Solve exponential equations:
5 x = 14
ln( 5 x ) ln 14
x ln( 5) ln( 14)
x
ln 14
1.64
ln 5
3 4x – 1 = 4 x – 2
3 2x – 5 = 13
e 2 – y = 12
e 2x e 5x = e 14
Solve logarithmic equations:
ln (5 + 4y ) – ln (3 + y) = ln 3
5 4y
ln 3
3 y
5 4y
3
3 y
5 4 y 3(3 y )
ln
4 y 3y 9 5
y4
ln x – ln (x + 1) = ln 5
ln ( k - 5) + ln ( k+ 2) = ln (14 k)
ln e x – ln e 3 = ln e 5
5.6 EXPONENTIAL GROWTH AND DECAY
In many applications:
Biology
Economics
Social studies
Physics
a quantity changes at a rate proportional to the amount present
At any time (instant) t, the amount present is:
y y o e kt
y is future amount
yo is present amount (at t = 0)
k – constant (k > 0 growth, k < 0 decay)
Example: A sample of 500 g of plutonium 341, decays according to the function:
A(t ) Ao e 0.053t
where t is time in years.
A(t) is future amount
Ao id present amount
Find the amount of the sample remaining after:
a) 4 years
b) 8 years
c) 20 years
d) Find the half life
Solution:
a)
b)
c)
t4
A(4) 500 e 0.0534 500 e 0.212 404 g
t 8
A(8) 500 e 0.0538 327 g
t 20
A(20) 500 e 0.05320 173g
The HALF-LIFE of quantity that decays exponentially is the time it takes for half of a given
amount to decay.
500
250
2
250 500 e 0.053t
A(t )
d)
1
e 0.053t
2
1
ln( ) 0.053 t
2
1
ln( )
2 13 years
t
0.053
Example: By Newton’s law of cooling, the temperature of a body at time t after being introduced
into environment having constant temperature To is:
T (t ) To C e kt
where C and k are constants. If C = 100 and k = 0.1, and t is time measured in minutes, how long
will it take a hot cup of coffee to cool to a temperature of 25 oC in a room at 20 oC.
A(t ) 25o C
To 20o C
25 20 100 e 0.1t
5 100 e 0.1t
1
e 0.1t
20
1
ln( ) 0.1t
20
t 30 min
Example: How long it take for \$5000 to grow to \$8400 at an interest rate of 6 % if interest is
compounded a) semiannually and b) continuously.
a)
r
A P (1 ) mt
m
0.06 2t
8400 5000(1
)
2
84
1.025 e 2 t
50
84
ln( ) 2 t ln( 1.025)
50
t 8.8 years
b)
A P e rt
8400 5000 e 0.06t
84
e 0.06t
50
84
ln( ) 0.06 t
50
t 8.64 years
```
|
# McGraw Hill My Math Grade 1 Chapter 7 Lesson 1 Answer Key Tally Charts
All the solutions provided in McGraw Hill My Math Grade 1 Answer Key PDF Chapter 7 Lesson 1 Tally Charts will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 1 Answer Key Chapter 7 Lesson 1 Tally Charts
Explore and Explain
____________ – ____________ = _____________ more people
Teacher Directions: Use to show the number of people who voted for each activity. Write each total. How many more people like to dance than to run? Write a subtraction number sentence to solve.
Explanation:
2 people voted for run activity
4 people chose dance activity
3 people chose play out side activity
4 – 3 = 1 is the subtraction number sentence
1 person voted for dance more than to run.
See and Show
A tally chart shows a mark for each vote in a survey. A survey asks people the same question.
Ask 10 friends to choose their favorite school subject. Make a tally chart. Write the totals. Use the tally chart. How many chose each subject?
Question 1.
2
Explanation:
2 people chose Math as their favorite subject
Question 2.
3
Explanation:
3 people chose Reading as their favorite subject
Question 3.
5
Explanation:
5 people chose Reading as their favorite subject
Talk Math How are tally marks used to take surveys?
10 tally marks are used to take surveys.
Write the totals. Use the chart to answer the questions.
Question 4.
How many people chose red?
8
Explanation:
8 people chose red color as their favorite color.
Question 5.
How many people chose purple?
5
Explanation:
5 people chose purple as their favorite color.
Question 6.
Do more people like purple or blue?
purple
Explanation:
3 people chose blue as their favorite color
5 people chose purple as their favorite color
5 > 3
So, more people chose purple color.
Question 7.
How many more people like red than blue?
5
Explanation:
8 people chose red color as their favorite color
3 people chose blue as their favorite color
8 – 3 = 5
So, 5 people like red than blue.
Question 8.
Do more people like red or purple?
red
Explanation:
8 people chose red color as their favorite color
5 people chose purple as their favorite color
8 > 5
So, more people chose red color.
Question 9.
How many fewer people like blue than purple?
2
Explanation:
3 people chose blue as their favorite color
5 people chose purple as their favorite color
5 – 3 = 2
So, 2 fewer people chose blue than purple.
Question 10.
How many people were surveyed in all?
16
Explanation:
8 people chose red color as their favorite color
3 people chose blue as their favorite color
5 people chose purple as their favorite color
Add 8, 3, 5
8 + 3 + 5 = 16
So, 16 people are surveyed.
Problem Solving
Question 11.
Circle the tally chart that shows 2 students like crackers, 6 students like bananas, and 4 students like carrots.
Explanation:
The first tally chart shows 2 students like crackers, 6 students like bananas, and 4 students like carrots
So, i drew a circle around first tally chart.
HOT Problem! Samantha is having a pizza party. She asks her guests to pick their favorite kind of pizza. If she orders one kind which one does she order? Explain.
Samantha is having a pizza party
She asks her guests to pick their favorite kind of pizza
7 chose Cheese pizza
3 chose Pepperoni pizza
2 chose Sausage pizza
She orders one kind that is chose by more number of people
7 people chose Cheese pizza and it is the greater number
So, Samantha orders Cheese Pizza.
### McGraw Hill My Math Grade 1 Chapter 7 Lesson 1 My Homework Answer Key
Practice
Write the totals. Use the chart to answer the questions.
Question 1.
How many more votes did music get than art?
2
Explanation:
8 people voted for Music as their favorite activity
6 people voted for Art as their favorite activity
8 – 6 = 2
So, music got 2 more votes than art.
Question 2.
How many people were surveyed?
Question 3.
How many people like fall?
9
Explanation:
9 people chose fall as their favorite season
Question 4.
How many more people like summer than spring?
3
Explanation:
7 people chose summer as their favorite season
4 people chose spring as their favorite season
7 – 4 = 3
3 more people like summer than spring.
Question 5.
Do 7 people like summer or fall?
7
Explanation:
7 people chose summer as their favorite season.
Question 6.
How many people were surveyed?
20 people
Explanation:
7 people chose summer as their favorite season
9 people chose fall as their favorite season
4 people chose spring as their favorite season
Add to find the total
7 + 9 = 4 = 20
So, 20 people were surveyed.
Vocabulary Check
Complete each sentence.
tally chart
survey
Question 7.
You can collect data by taking a _______________.
|
Tips for Easily Memorizing Percentages
Title: How to Easily Memorize Percentages
Subtitle: Proven Strategies to Help You Master Percentages
Introduction
Understanding percentages can be a challenge for many people. Whether you’re studying for a math test or trying to figure out how much of a discount you’re getting at the store, being able to quickly recall and calculate percentages is a valuable skill. Fortunately, with the right strategies, anyone can learn how to easily memorize percentages.
Body
The first step to memorizing percentages is to understand the concept of percentages. A percentage is a way of expressing a number as a fraction of 100. For example, 25% is the same as 25/100 or 1/4. Once you understand this, it will be easier to remember and calculate percentages.
The next step is to practice. Memorizing percentages is like any other skill; the more you practice, the better you will become. Start by memorizing the most common percentages, such as 10%, 20%, 25%, 33%, 50%, and 75%. Once you have those memorized, you can move on to more difficult percentages.
Another strategy is to break down percentages into smaller pieces. For example, if you are trying to remember 25%, you can break it down into two parts: 20% and 5%. This will make it easier to remember and calculate.
You can also use visual aids to help you remember percentages. For example, you can draw a circle and divide it into four equal parts. Each part would represent 25%, making it easier to remember. You can also use a calculator to help you quickly calculate percentages.
Examples
Let’s look at a few examples to see how to easily memorize percentages.
If you want to remember 10%, you can think of it as one-tenth of 100. You can also break it down into two parts: 5% and 5%.
If you want to remember 20%, you can think of it as one-fifth of 100. You can also break it down into four parts: 10%, 5%, 5%, and 0%.
If you want to remember 25%, you can think of it as one-fourth of 100. You can also break it down into two parts: 20% and 5%.
If you want to remember 33%, you can think of it as one-third of 100. You can also break it down into three parts: 20%, 10%, and 3%.
FAQ Section
Q: What is the best way to memorize percentages?
A: The best way to memorize percentages is to practice and use visual aids. Break down percentages into smaller parts and use a calculator to help you quickly calculate percentages.
Q: How can I remember percentages easily?
A: Memorizing percentages is like any other skill; the more you practice, the better you will become. Start by memorizing the most common percentages, such as 10%, 20%, 25%, 33%, 50%, and 75%. Once you have those memorized, you can move on to more difficult percentages.
Q: What is the meaning of percentage?
A: A percentage is a way of expressing a number as a fraction of 100. For example, 25% is the same as 25/100 or 1/4.
Summary
Memorizing percentages can be a challenge for many people, but it doesn’t have to be. With the right strategies, anyone can learn how to easily memorize percentages. Start by understanding the concept of percentages and then practice, break down percentages into smaller parts, and use visual aids to help you remember.
Conclusion
Memorizing percentages doesn’t have to be a daunting task. With the right strategies, anyone can learn how to easily memorize percentages. Start by understanding the concept of percentages and then practice, break down percentages into smaller parts, and use visual aids to help you remember. With enough practice, you’ll be able to quickly recall and calculate percentages in no time.
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|
# THE GEOMETRIC MEAN Stats Assignment Homework Help
THE GEOMETRIC MEAN
The geometric mean of a set of N numbers is the Nth root of the product of the numbers. The formula for the geometric mean is
Geometric mean = GM =
For example, the N = 4 numbers
0.5, 1,4,8
have
The arithmetic mean of the numbers is 13.>/4 = 3.375. The geometric mean is less than the arithmetic mean except in the rare case where all the numbers in the data set are the same; in that rare case, the two means are equal. Consequently, the geometric mean is sometimes used as the average
for right-skewed economic data-because it is not affected as much by the skewness as the arithmetic mean.
EXERCISE Compute (a) the geometric mean and (b) the arithmetic mean for the values 1, 3, and 9.
You should use your calculator to take roots. That is much simpler and more accurate than the alternative, which is to use logarithms. To take a root on a calculator, change the root to a power and use the power key
For practice, verify on your calculator that the cube root ‘of 125 is 5. . The geometric mean must be used in finding average period-to-period percent change. To see what is involved, let’s listen in on Dan and Fran.
Dan: Two years ago, my jacket cost a hundred dollars. Last year it sold for ninety dollars. Now I notice the same jacket goes for a hundred and forty-four dollars-up fzfty-four bucks from last year! First it went down ten percent from a hundred to ninety. Now it’s up from ninety to one forty-four. That’s -let’s see-up fzfty-four on ninety-that’s up sixty percent. Fran: Down ten and up sixty. That’s an average yearly percent change of twenty-five percent. Dan: fm not sure you can average like that. If you start with one hundred dollars and raise it twenty-five percent – that’s one point two live times one hundred-you get one twenty-five, Then raising that by twenty five percent you get one fifty-six twenty-five, not one forty-four. Fran: Twenty-five is too high. It looks more like twenty percent per year. Yes, that will do it. Raise one hundred by twenty percent, then raise that by twenty percent, and you get one forty-four. Dan: But how can you get a twenty percent average from a minus ten percent and a plus sixty percent?
The way to answer Dan’s question is to first compute the geometric mean of the growth factors. A growth factor is the ratio of a number, say, price or sales, for one period to the corresponding number for a previous period. Dan’s jacket price had growth factors of
Note that the growth factors are simply the decimal equivalents of the percent changes, plus 1. For example, a decline of 10 percent is -10 percent, and -10 percent is a growth factor of – 0.10 + I = 0.90. Now compute the geometric mean of the growth factors. This is
Finally subtract 1 to obtain 1.20 – 1 = 0.20, or 20 percent, as the desired average.
The average period-to-period percent change is computed by (1) converting the percent changes to growth factors, (2) finding the geometric mean of the growth factors, (3) converting the geometric mean growth factor to a percent change.
EXAMPLE Abac Company’s year-to-rear changes in fuel consumption expenditures were – 5, 10, 20, 40, and 60 percent. Using the geometric mean of growth factors, determine the average yearly percent change in expenditures.
SOLUTION Converting the percent changes to growth factors, we obtain
0.95 1.l0 1.20 1.40 1.60
Then, taking the fifth root (power is 0.2) of the product of the factors, we have .
GM = [(0.95)(1.l0)(1.20)(1.40)(1.60»)O·2= 1.229
Subtracting 1 gives 1.229 – 1 = 0.229, or a 22.9 percent average increase per year.
EXERCISE Month-to-month changes in the prices of a stock were 11, 18, and -15 percent. Using the geometric mean of the growth factors, determine the average monthly percent change in the price of the stock.
|
# How to find the area of the black portion?ABCD is a square having each side 14cm.Two quadrants of circle (quarter circle)is drawn inside the square...
How to find the area of the black portion?
ABCD is a square having each side 14cm.Two quadrants of circle (quarter circle)is drawn inside the square intersecting each other.Now find the area of the black portion.
The figure is here:-
http://www.flickr.com/photos/78780315@N06/7384954722/
embizze | Certified Educator
Let the square be as given; let the intersection of the circles inside the square be point E.
The area we need is the area in the sectors DAC and CBD, excluding the intersection of the two sectors. The intersection is the triangle DEC as well as the two segments DEC and CED.
So the area we seek can be found by the areas:
`["sector"DAC-Delta DEC-"seg"CED-"seg"DEC]+["sector"CDB-Delta DEC-"seg"CDE-"seg"DEC]`
Now the area of sector DAC is the same as the area of sector CDB and is `1/4pi(14^2)=49pi` .
The area of `DeltaDEC` is `1/2(14)(7sqrt(3))=49sqrt(3)` .(This is an equilateral triangle since DE=EC=DC=14 (all radii of circles))
The area of segment DEC=area of segment CDE and can be found by:
`1/6pi(14)^2-49sqrt(3)=98/3pi-49sqrt(3)` (The area of a segment is the area of the sector minus the triangle, and we have shown that the triangle is equilateral thus the angle is `60^@` )
Putting this all together, and using symmetry, we get:
`A=2[49pi-49sqrt(3)-2[98/3pi-49sqrt(3)]]`
`=98pi-98sqrt(3)-392/3pi+196sqrt(3)`
`=98sqrt(3)-98/3pi~~67.11cm^2`
|
Painevg
2022-01-17
How do you simplify ${i}^{6}\left(2i-{i}^{2}-3{i}^{3}\right)$?
intacte87
Solution:
Expand expression
${i}^{6}\left(2i-{i}^{2}-3{i}^{3}\right)=2{i}^{7}-{i}^{8}-3{i}^{9}$
But we are conscious of:
${i}^{1}=i$
${i}^{2}=-1$ (by definition)
${i}^{3}={i}^{2}\cdot i=-i$
${i}^{4}={i}^{2}\cdot {i}^{2}=\left(-1\right)\cdot \left(-1\right)=1$
${i}^{5}={i}^{4}\cdot i=1\cdot i=i$ and starts again
${i}^{6}={i}^{5}\cdot i=i\cdot i={i}^{2}=-1$
${i}^{7}={i}^{6}\cdot i=-i$
To get this click-through result, type:
${i}^{6}\left(2i-{i}^{2}-3{i}^{3}\right)=2{i}^{7}-{i}^{8}-3{i}^{9}=-2i-1-3i=-1-5i$
Jillian Edgerton
We factorize powers of "i" first. So ${i}^{6}$ would become $01.3{i}^{3}$ would become -3i and ${i}^{2}$ would become -1.By putting values
$-1\left(2i-\left(-1\right)+3i\right)$. Now $-1\left(2i+3i+1\right)$ would become $-1\left(5i+1\right)$ and then
$-5i-1$
alenahelenash
${i}^{6}\left(2i-{i}^{2}-3{i}^{3}\right)=2{i}^{7}-{i}^{8}-3{i}^{9}$ ${i}^{1}=i$ ${i}^{2}=-1$ ${i}^{3}={i}^{2}\cdot i=-i$ ${i}^{4}={i}^{2}\cdot {i}^{2}=\left(-1\right)\cdot \left(-1\right)=1$ ${i}^{5}={i}^{4}\cdot i=1\cdot i=i$ ${i}^{6}={i}^{5}\cdot i=i\cdot i={i}^{2}=-1$ ${i}^{7}={i}^{6}\cdot i=-i$ ${i}^{6}\left(2i-{i}^{2}-3{i}^{3}\right)=2{i}^{7}-{i}^{8}-3{i}^{9}=-2i-1-3i=-1-5i$
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# Cos3x Formula in terms of Cosx, Sinx [with Proof]
Cos3x formula in terms of cosx is as follows: 4cos3x-3cos x. In this post, we will also derive the cos3x formula in terms of sinx and give some examples. These formulas are very useful to solve trigonometric equations and simplify trigonometric expressions.
## Cos3x Formula
Cos3x is the cosine function of a triple angle 3x. The formula of cos 3x in terms of cos x is given by the following identity:
cos3x =4cos3x-3cos x.
## Cos3x in Terms of Cosx
To derive the cos3x formula, we will use the following trigonometric formulas:
1. $\cos(a+b)$ $=\cos a\cos b -\sin a \sin b$
2. $\sin 2x=2\sin x \cos x$
3. $\cos 2x=1-2\sin^2x$
Below are the steps to be followed in order to establish the formula of cos 3x.
Step 1: At the first step, we will write 3x as 2x+x, and then we will apply the above formula 1. By doing so, we get that
$\cos 3x=\cos(2x+x)$
$=\cos 2x \cos x -\sin 2x \sin x$
Step 2: Next, we will apply the formulas of sin2x and cos2x which are given in (2) and (3) above. Thus, we obtain that
$\cos 3x=(1-2\sin^2x) \cos x$ $-(2\sin x\cos x)\sin x$
$=\cos x -2\sin^2 x \cos x-2\sin^2 x\cos x$
$=\cos x-4\sin^2x\cos x$
$=\cos x-4(1-\cos^2x)\cos x$ as we know that $\sin^2x=1-\cos^2x$.
$=\cos x -4\cos x+4\cos^3x$
$=4\cos^3 x -3\cos x$
Hence, the formula of $\cos 3x$ is $4\cos^3x-3\cos x$.
Question: Find the value cos135 degrees.
In the above formula of cos3x, that is, $\cos 3x=4\cos^3x-3\cos x$, we put $x=45^\circ$. Thus, we obtain that
$\cos 135^\circ= 4\cos^3 45^\circ-3\cos 45^\circ$
$=4(\frac{1}{\sqrt{2}})^3-3\times \frac{1}{\sqrt{2}}$ as we know that $\cos 45^\circ=1/\sqrt{2}$
$=\frac{4}{2\sqrt{2}}-\frac{3}{\sqrt{2}}$
$=\frac{4-6}{2\sqrt{2}}$
$=\frac{-2}{2\sqrt{2}}$
$=-\frac{1}{\sqrt{2}}$
Thus, the value of cos135 degrees is equal to -1/root(2).
## Cos3x in Terms of Sinx
Now, we will find the formula of cos3x in terms of sinx using the Pythagorean identity $\cos^2x+\sin^2x=1$.
Now, $\cos3x=4\cos^3x-3\cos x$
$=4(\sqrt{1-\sin^2x})^3-3\sqrt{1-\sin^2x}$
$=4(1-\sin^2x)^{3/2}-3\sqrt{1-\sin^2x}$.
Tan2x Formula in terms of tanx, sinx, cosx
sin3x Formula in terms of sinx
Simplify 1-sin^2(x)
Values of sin15, cos15, tan15
Values of sin75, cos75, tan75
Simplify cos(sin^{-1}x)
Simplify sin(cos^{-1}x)
## FAQs
Q1: What is the formula of cos3x in terms of cosx?
Answer: cos3x formula in terms of cosx is as follows: cos 3x = 4cos3x-3cos x.
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Number Properties
In these lessons, we will learn three basic number properties (or laws) that apply to arithmetic operations: Commutative Property, Associative Property and Distributive Property.
Related Topics: More Arithmetic Lessons
Summary of Number Properties
The following table gives a summary of the commutative, associative and distributive properties. Scroll down the page for more examples and explanations of the number properties.
Commutative Property
An operation is commutative if a change in the order of the numbers does not change the results. This means the numbers can be swapped.
Numbers can be added in any order.
For example: 4 + 5 = 5 + 4 x + y = y + x
Numbers can be multiplied in any order.
For example: 5 × 3 = 3 × 5 a × b = b × a
Numbers that are subtracted are NOT commutative.
For example: 4 – 5 ≠ 5 – 4 x – y ≠ y –x
Numbers that are divided are NOT commutative.
For example: 4 ÷ 5 ≠ 5 ÷ 4 x ÷ y ≠ y ÷ x
Associative Property
An operation is associative if a change in grouping does not change the results. This means the parenthesis (or brackets) can be moved.
Numbers that are added can be grouped in any order.
For example: (4 + 5) + 6 = 5 + (4 + 6) (x + y) + z = x + (y + z)
Numbers that are multiplied can be grouped in any order.
For example: (4 × 5) × 6 = 5 × (4 × 6) (x × y) × z = x × (y × z)
Numbers that are subtracted are NOT associative.
For example: (4 – 5) – 6 ≠ 4 – (5– 6) (x – y) – z ≠ x – (y – z)
Numbers that are divided are NOT associative.
For example: (4 ÷ 5) ÷ 6 ≠ 4 ÷ (5÷ 6) (x ÷ y ) ÷ z ≠ x ÷ ( y ÷ z)
Distributive Property
Distributive property allows you to remove the parenthesis (or brackets) in an expression. Multiply the value outside the brackets with each of the terms in the brackets.
For example: 4(a + b) = 4a + 4b 7(2c – 3d + 5) = 14c – 21d + 35
What happens if you need to multiply (a – 3)(b + 4)?
You do the same thing but with one value at a time.
For example:
Multiply a with each term to get a × b + 4 × a = ab + 4a
Then, multiply 3 with each term to get “ –3b – 12” (take note of the sign operations).
Put the two results together to get “ab + 4a – 3b – 12”
Therefore, (a – 3)(b + 4) = ab + 4a – 3b – 12
Videos
The Commutative Property of addition and multiplication.
The addition of real numbers is commutative.
The multiplication of real numbers is commutative. The Associative Property of addition and multiplication. The following video shows more examples of the distributive property. Associative and Distributive Properties of Multiplication and Addition
A look at the logic behind the associative and distributive properties of multiplication and addition.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Mathclub Hat Problem
One of the students in Math Club recently put his own twist on the age old hat question: Assume you have $n$ people, each of whom has a red or green hat put on them. They each don’t know what color hat they have on. However they can look around and see everyone else’s hat.
After getting to spend some time in a room looking at everyone else and their hats (they may not communicate in any way), they are each placed in separate cells and asked to say whether they have a red hat on, a green hat on, or “pass.” Everyone wins the game if at least one person says their right hat color, and no person messes up their hat color. Everyone loses the game if everyone passes, or if anyone says the wrong hat color.
The question is: what is the strategy that those wearing the hats should come up with beforehand? And can you come up with a formula giving the probability that $n$ people win with that strategy?
To make the problem clear, let’s examine the three person case. The possible combinations of hats are:
RRR | RRG | RGR | GRR | GGR | GRG | RGG | GGG
The best strategy we could come up with is to say: if you see two opposite colors (a red and a green), say “pass”. If you see two hat of the same color, say you’re wearing the opposite color.
So you’ll lose with RRR and GGG (everyone sees two of the same color, so everyone will say the opposite color).
But you’ll end up winning with RRG, RGR, GRR, GGR, GRG, and RGG. Let’s look at RRG to explainThe person wearing the first red hat sees a red and green hat. So that person says “pass.” The person wearing the second red hat sees a red and a green hat. So that person says “pass.” The third person wearing the green hat sees a red and a red hat, so that person says “green” and is right! So RRG is a winning combination. Similar arguments follow for the other five.
Since there are 8 possible combinations of hats, and 6 of them have a winning strategy, there are 6/8 chances that everyone will come out a winner! (That’s a whopping 75%!)
So we’ve been investigating what the strategy will be for $n$ people wearing red and green hats. So far, we’ve done pretty well. In fact, we’ve even gotten Pascal’s Triangle involved, which is always great.
And there seems to be a consensus among the students (though no proof yet) that if you have any even number of people playing the game, say 8, you can actually get better odds of winning if you ask another person to join in (so you’d have, say, 9 people playing). That seems totally counter-intuitive, that adding an extra person to play the game with you would lead to a better chance of winning. So if they’re right, I’ll chalk this problem up to a win.
PS. We did talk about the Bloxorz problem for two weeks, but students grew bored and tired of it. I still think it’s a great problem. Maybe one year a student will want to do an independent study on it, and ask me to be the adviser to the project.
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# The volumes of two similar cylinders are in ratio of 8:27. How can I find the ratio of the base radii of the cylinders?
This is an example of the volume of similar solids. Two solids are regarded similar if and only if they are the same solid and their corresponding linear measurements, namely radii, heights, base lengths are proportional.
Since we are dealing with the volume of two similar cylinders, we need to know what the scale factor is or the volume ratio between the two similar cylinders. If two similar solids have a scale factor `a/b` , then the volumes of the similar cylinders has a volume ratio: `(a/b)^3` .
In the above example we are given the volume ratio, and we need to find the base radii ratio of the cylinders.
The base radii is a linear measurement, where as volume is a cubic function. In order to find the base radii from the volume ratio, we need to find to cube root the base radii. This is done as follows:
`(a/b)^3 = (8/27)`
`(a/b) =root(3)(8/27)`
`(a/b) = 2/3`
Therefore the ratio of the base radii is 2/3. can be written as 2:3.
Approved by eNotes Editorial Team
The complicated way is to write out the ratio of the volumes, set equal to 8/27 and solve for the ratio of the radii.
The easy way is to note that the figures are similar. For similar figures all corresponding lengths are in the same ratio called the scale factor. (All corresponding lengths -- radii, circumference, diameter, altitudes, etc...)
Let the scale factor be a:b.
Then all corresponding areas are in the ratio a^2:b^2.
All corresponding volumes are in the ratio a^3:b^3.
Here the volumes are in the ratio 8:27, so the scale factor is 2:3. (The cube root of 8 to the cube root of 27.)
The ratio of the radii is 2:3.
Approved by eNotes Editorial Team
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Geometry Polygons Problems- HitBullsEye 0
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# Geometry - Polygons
#### Read this article to understand the basic terminologies, important properties and formulas of a polygon.
Geometry is one of the most important and easiest topics of Quantitative Aptitude, if you prepare it well. Like any other topic, geometry too needs a lot of practice. In this article, we will discuss about polygons and its important properties. The important properties and formulas discussed in this article find direct application in questions. So, make sure you go through this article in detail.
Polygon: A polygon is a closed plane figure formed by three or more line segments, called the sides of the polygon. Each side intersects exactly two other sides at their endpoints. The points of intersection of the sides are known as vertices. The term "polygon" is used to mean a convex polygon, that is, a polygon in which each interior angle has a measure of less than 180°. The following figures are polygons:
The following figures are not polygons:
Number of sides Name of the Polygon
3 Triangle
5 Pentagon
6 Hexagon
7 Heptagon
8 Octagon
9 Nonagon
10 Decagon
Some Basic Terminologies
Convex polygon: A polygon in which none of the interior angles is more than 180° is called a convex polygon.
Concave polygon: A polygon in which at least one of the interior angles is more than 180° is called a concave polygon.
Regular polygon: A polygon which has all its sides and angles equal is called a regular polygon.
Perimeter of a polygon: The perimeter of a polygon is the sum of the lengths of all its sides.
Area of a polygon: The region enclosed within a figure is called its area.
Diagonal of a polygon: The segment joining any two non-consecutive vertices is called a diagonal.
Solved Examples
• where an is the side of regular inscribed polygons, where R is the radius of the circumscribed circle,
• Area of a polygon of perimeter p and radius of in-circle r = 1/2xpxr
• The sum of all the exterior angles = 360°
• Interior angle + corresponding exterior angle = 180°.
• The sum of the interior angles of a convex POLYGON, having n sides is 180° (n - 2).
• The sum of the exterior angles of a convex polygon, taken one at each vertex, is 360°.
• The measure of an exterior angle of a regular n- sided polygon is
• The measure of the interior angle of a regular n-sided polygon is
• The number of diagonals of in an n-sided polygon is
Example: What is the area of a regular octagon of side 8 cm?
1. (2 + 2√2) * 82
2. (3 + √2) * 82
3. (4 + √2) * 82
4. (5 + √2) * 82
Solution: The area of the octagon can be expressed as the difference of the larger square that comprises the octagon and the 4 right triangles at the corners that you can deduct.
Each of the right triangles will have the two perpendicular sides as 8/ √2.
The side of the bigger square will be 8 + 8/√2 + 8/√2 = (1+ √2) 8.
Squaring both sides, we get the area of the square as (1 + 2 + 2√2)82.
The sum of areas of the four triangles = 4 * ½ *(1/√2)2 * 82.
The difference between the two works out to be (2+2√2) * 82, that will be area of the octagon.
Hexagon
Here, a = side of the hexagon;
d= diameter of the circle inscribed (distance between two parallel sides);
p= perimeter of the hexagon.
The hexagon can be divided into 6 equilateral triangles with side=a
Hence, area of hexagon=
Area of hexagon=
Length of the diagonals=
Let us solve some questions based on the above discussed concepts.
Solved Examples
Example 1: There is a circle of radius 1 cm. Each member of a sequence of regular polygons S1(n), n = 4, 5, 6, ... where n = number of sides of the polygon, is circumscribing the circle and each member of the sequence of regular polygons S2(n), n = 4, 5, 6, ...., where n is the number of sides of the polygon, is inscribed in the circle. Let L1(n) and L2(n) denote perimeters of the corresponding polygons of S1(n) and S2(n). Then is (CAT 1999)
1. Greater than π/4 but less than 1
2. Greater than 1 and less than 2
3. Greater than 2
4. Less then π/4
Solution: Answer- Option 3
We know that the lower limiting perimeter of any polygon S1 is the circumference of the inscribed circle (2*π).
The upper limiting perimeter of any polygon S2 is the circumference of the circumscribed circle: 2*π
This difference of perimeter reduces as the number of sides increase.
Breaking up the expression into L1(13)/L2(17) + 2*π/L2(17).
Both the individual terms will be very close to 1, but greater than one.
Hence, option 3 is the answer.
Question 2: Euclid has just created a triangle whose longest side is 20. If the length of the other side is 10 cm and the area of the triangle is 80 sq. cm, then what is the length of the third side? (CAT 2001)
1. √260
2. √250
3. √240
4. √270
5. √230
Solution: Answer- option 1
Let third side be x.
S = (20 + 10 + x)/2 = (30 + x)/2.
Now use Hero's formula for the area of the triangle and we can find the answer as 1st option.
Question 3: What is the area of a regular octagon of side a?
1. (2 + 2√2) * a2
2. (3 + √2) * a2
3. (4 + √2) * a2
4. (5 + √2) * a2
5. (6 + √2) * a2
Solution: Answer- option 1
The area of the octagon can be expressed as the difference of the larger square that comprises the octagon and the 4 right triangles at the corners that we can deduct.
Each of the right triangles will have the two perpendicular sides as a / √2.
The side of the bigger square will be a + a/√2 + a/√2 = (1 + √2) a.
On squaring we get the area of the square as (1 + 2 + 2√2)a2 and that of 4 triangles as 4 x ½ x(1/√2)2 x a2.
The difference between the two equals (2+2√2) x a2.
Question 4: ABCDEF is a regular hexagon AP: PB = 5: 2 and PR is parallel to AF. Find the ratio of lengths of AF to PR.
1. 2/7
2. 5/7
3. 7/12
4. 2/5
Solution: Answer- Option 3
Here, ΔAQF is an equilateral triangle
So AF = AQ
But AF || PR. Then Δ AQF and Δ PQR are similar So
Key Learning
• Geometry is all about practice. Practice as many questions as you can.
• Formulas discussed in this article find direct application in questions.
• For doubts, post your comments below and our experts will provide you with the solutions.
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# Time and speed
Welcome to aptitude tricks
Problems involving Time, Distance and Speed are solved based on one simple formula.
Distance = Speed * Time
Which implies →
Speed = Distance / Time and
Time = Distance / Speed
Let us take a look at some simple examples of distance, time and speed problems.
Example 1. A boy walks at a speed of 4 kmph. How much time does he take to walk a distance of 20 km?
Solution
Time = Distance / speed = 20/4 = 5 hours.
Example 2. A cyclist covers a distance of 15 miles in 2 hours. Calculate his speed.
Solution
Speed = Distance/time = 15/2 = 7.5 miles per hour.
Example 3. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. What should be its speed to cover the same distance in 1.5 hours?
Solution
Distance covered = 4*40 = 160 miles
Speed required to cover the same distance in 1.5 hours = 160/1.5 = 106.66 mph
Now, take a look at the following example:
Example 4. If a person walks at 4 mph, he covers a certain distance. If he walks at 9 mph, he covers 7.5 miles more. How much distance did he actually cover?
Now we can see that the direct application of our usual formula Distance = Speed * Time or its variations cannot be done in this case and we need to put in extra effort to calculate the given parameters.
Let us see how this question can be solved.
Solution
For these kinds of questions, a table like this might make it easier to solve.
Distance Speed Time d 4 t d+7.5 9 t
Let the distance covered by that person be ‘d’.
Walking at 4 mph and covering a distance ‘d’ is done in a time of ‘d/4’
IF he walks at 9 mph, he covers 7.5 miles more than the actual distance d, which is ‘d+7.5’.
He does this in a time of (d+7.5)/9.
Since the time is same in both the cases →
d/4 = (d+7.5)/9 → 9d = 4(d+7.5) → 9d=4d+30 → d = 6.
So, he covered a distance of 6 miles in 1.5 hours.
Example 5. A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.
Solution
Here, we see that the distance is same.
Let us assume that its usual speed is ‘s’ and time is ‘t’, then
Distance Speed Time d s t min d S+1/3 t+30 min
s*t = (1/3)s*(t+30) → t = t/3 + 10 → t = 15.
So the actual time taken to cover the distance is 15 minutes.
Note: Note the time is expressed in terms of ‘minutes’. When we express distance in terms of miles or kilometers, time is expressed in terms of hours and has to be converted into appropriate units of measurement.
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Lesson 5
Steps in Solving Equations
These materials, when encountered before Algebra 1, Unit 7, Lesson 5 support success in that lesson.
5.1: Explaining Equivalent Expressions (5 minutes)
Warm-up
In this warm-up, students determine how to rearrange an original equation into 3 different equivalent equations. The purpose is to remind students of moves allowed when rearranging equations. This will be helpful in this lesson when students rearrange other linear equations and in the associated Algebra 1 lesson when students rearrange quadratic equations.
Student Facing
Explain or show why each of these equations is equivalent to $$7(x-15) + 3 = 8$$.
1. $$7x - 105 + 3 = 8$$
2. $$7(x-15) - 5 = 0$$
3. $$7x - 102 - 8 = 0$$
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
The purpose of the discussion is to recall some of the moves for rearranging equations to create new equations. Select students to share their reasoning for why the equations are equivalent. To continue the discussion, ask students:
• “What are some other equivalent equations we could get by rearranging this one? Explain how you know they are equivalent to the original equation.” ($$7x - 102 = 8$$ since it is the first solution with the constant terms combined and that solution is equivalent to the original equation. There are many other reasonable answers.)
• “Two of the equations given have 0 on one side of the equation. Why might it be useful to have an equation in this form?” (In this form, we can set the other side of the equation equal to $$y$$, then graph it to find the solutions as the $$x$$-coordinates of the $$x$$-intercepts.)
• “Andre rearranges the equation to $$7x = 110$$. Is this equation equivalent to the original? If not, find an equivalent equation that has a similar form. Then, explain why this form might be useful.” (It is equivalent to the original equation. This could be useful for solving the equation. After this, dividing both sides of the equation by 7 would solve for $$x$$.)
5.2: Checking Work (15 minutes)
Activity
In this activity, students examine work for common mistakes when solving equations. In the associated Algebra 1 lesson, students solve quadratic equations and must be mindful of possible mistakes in their work. Students construct viable arguments and critique the reasoning of others (MP3) when they identify the errors and explain their reasons to agree or disagree.
Student Facing
Here is Clare’s work to solve some equations. For each problem, do you agree or disagree with Clare’s work? Explain your reasoning.
1. $$2(x-1)+4 = 3x - 2$$
$$2x - 2 + 4 = 3x - 2$$
$$2x + 2 = 3x - 2$$
$$2x = 3x$$
$$\text{-}x = 0$$
$$x = 0$$
2. $$3(x-1) = 5x + 6$$
$$3x - 1 = 5x + 6$$
$$\text{-}1 = 2x + 6$$
$$\text{-}7 = 2x$$
$$-3.5 = x$$
3. $$(x-2)(x+3) = x+10$$
$$x^2 + x - 6 =x + 10$$
$$x^2 - 6 = 10$$
$$x^2 = 16$$
$$x = 4$$
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
The purpose of the discussion is to highlight some common errors students may have while solving equations. Select students to share their solutions and reasoning. Ask students,
• “What other common errors do you have to be careful to avoid when solving equations?” (I sometimes forget to expand something like $$(x+1)^2$$ and accidentally write $$x^2 + 1$$.)
5.3: Row Game: Rewriting Equations (20 minutes)
Activity
In this activity, students rewrite equations into equivalent versions with a given property. To transform the equations, students will need to be able to distribute binomials and factor quadratics. Students work in pairs and each partner is responsible for answering the questions in either column A or column B. Although each row has two different problems, they share the same answer. Ensure that students work their problems out independently and collaborate with one another when they do not arrive at the same answers. Students construct viable arguments and critique the reasoning of others (MP3) when they resolve errors by critiquing their partner’s work or explaining their reasoning.
Launch
Arrange students in groups of 2. In each group, ask students to decide who will work on column A and who will work on column B.
Student Facing
Work independently on your column. Partner A completes the questions in column A only and partner B completes the questions in column B only. Your answers in each row should match. Work on one row at a time and check if your answer matches your partner’s before moving on. If you don’t get the same answer, work together to find any mistakes.
Partner A: Write an equivalent equation so that the given condition is true.
1. $$5x+10 = -35$$
• The expression on the right side is 0
2. $$x^2 - 9x = 42$$
• The left side is a product
3. $$x(x+3) + 9 = 1$$
• The right side is 0
4. $$8(x+1) = 5x$$
• The left side is 0 and there are no parentheses
5. $$11+x = \frac{12}{x}$$
• The equation is quadratic and the right side is zero.
6. $$(3x-5)(x-2) = 0$$
• One side of the equation has a term with $$3x^2$$
7. $$4x^2 - 4 = 8$$
• The right side is 0 and the left side is a product
Partner B: Write an equivalent equation so that the given condition is true.
1. $$5(x+9) = 0$$
• The left side is expressed as the sum of two terms
2. $$x(x-9) - 42 = 0$$
• The left side is a product and the right side is not 0
3. $$x(x+3) + 6 = -2$$
• The right side is 0
4. $$3x = -8$$
• The left side is 0
5. $$(x+12)(x-1) = 0$$
• The left side involves $$x^2$$
6. $$3x - 11 = \frac{10}{x}$$
• One side of the equation has a term with $$3x^2$$
7. $$4(x^2 - 1) = 8$$
• The right side of is 0 and the left side is a product
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
The purpose of the discussion is to recognize different ways to rewrite equations. Ask students what they notice about the starting equations and final, rewritten equations. Students may notice:
• a lot of the final forms end with one side equal to zero.
• distributing can create a form that can be easier to compare to other forms.
• multiplying an equation by a denominator (as long as it is not zero) helps simplify the equation by removing the fractions.
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# NSEP 2015 Problem 8 | One Dimensional Motion
Try out this problem on one dimensional motion(Kinematics) from National Standard Examination in Physics 2015-2016.
## NSEP 2015-16 ~ Problem 8
A car moving along a straight road at a speed of u m/s applies brakes at $t=0$ seconds. The ratio of distances travelled by the car during $3^{rd}$ and $8^{th}$ seconds is $15:13$. The car covers a distance of $0.25m$ in the last second of it's travel. Therefore, the acceleration a (in $m/s^2$) and the speeed u($m/s$) of the car are respectively,
a) $-0.1,16$
b) $-0.2,12$
c) $-0.5,20$
d) $-0.1,16$
### Key Concepts
Basic Equation of motions
Idea of accelerations, velocity and displacement
## Suggested Book | Source | Answer
Concept of Physics H.C. Verma
University Physics by H. D. Young and R.A. Freedman
Fundamental of Physics D. Halliday, J. Walker and R. Resnick
National Standard Examination in Physics(NSEP) 2015-2016
Option-(c) $-0.5,20$
## Try with Hints
We know that displacement in $t$ sec is given by,
$$s = ut +\frac{1}{2} a t^2$$
where $u$ is the initial velocity and $a$ is the acceleration. This will be negative for deceleration.
Using this idea we can just find the expression for the displacement in the $n^{th}$ second, for which the information is given to us.
Also, the car come to rest and at the second second it travels a distance of $0.25m$. WE can think it in opposite manner that with an acceleration of $a$ and initial velocity $0m/s$, the car moves a distance of $0.25m$ , hence,
$$0.5 =\frac{1}{2} a (1^2)$$
So, the displacement in $n^{th}$ second is,
$$s_1 = ut+\frac{1}{2} a n^2$$
and for the $(n-1)^{th}$ second the displacement is,
$$s_2= ut+\frac{1}{2} a (n-1)^2$$
Hence, we have
$$s_n = s_1-s_2 = u + a(n-\frac{1}{2})$$
we have,
$$\frac{15}{13} = \frac{s_3}{s_8} = \frac{u-\frac{5a}{2}}{u - \frac{15a}{2}}$$
Also we have $0.5 =\frac{1}{2} a (1^2)$
Solving this two gives us $a = 0.25m/s^2$. Putting this in first equation gives us $u = 20 m/s$.
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# Digital World – Binary and Decimal System
We have to understand the meaning of 0 and 1s which are fundamentals of digital world and electronics for understanding networking basics.
It is hard to believe that, all digital inventions around us are created with using 1 and 0 s, but it is true!
We have an information and we need to transfer this information from one point to another. Despite the fact that the analog systems have been used before, nowadays all information is transferring with digitally. Transferring information with digitally means converting this information to representation of 0 and 1s.
We can say that the first step of digitization begins with Mors alphabet. As you know, at the Mors alphabet, the characters are coding with points and dashes and send over line. At the other side, the other person translates this dots and dashes to the letter.
I will not mention how can be translate data to 0 and 1s here, I want to understand how converted data can be transfer and what the bit and byte terms means.
Every 0 and 1 represents as BIT in digital world. Like as Mors alphabet, one letter represents by combination of dots and dashes, the digital data represents by lot of 0s and 1s. The scientist grouped 8 bits and named is BYTE.
Let’s show a Bit and Byte terms on the figure below;
# Converting Binary Number to Decimal
We learned how to convert a binary number to decimal at high school. I am sure you will remember when you see the example below.
You remember that? As you see on the figure, if all digits on 1Byte (8bit) will 1, the decimal value represents 255.
So, 255 (mod 10) = 11111111 (mod2)
Let’s we see another example. What is the decimal value of binary value that we give at the first figure 00110011?
As you see, the decimal value of 00110011 is 51
Let’s look another aspect,
We discard the zeros and only find the decimal equivalent of 1s. If we sum the decimal values, the sum is 51.
According to find decimal value of binary series, we have to know that the decimal equivalent of every bit. You can write down from right box to left box, adding power of 2 respectively.
# Find the Binary Value of Decimal Number
We can find the binary value of decimal number solving numbers reversely.
Let’s make an example. We will find the binary value of 165.
Make a box and write all decimal equivalent of digits from right to left.
• Box number 1 decimal eq is = 1
• Box number 2 decimal eq is = 2
• Box number 3 decimal eq is = 4
• Box number 4 decimal eq is = 8
• Box number 5 decimal eq is = 16
• Box number 6 decimal eq is = 32
• Box number 7 decimal eq is = 64
• Box number 8 decimal eq is = 128
• 128 is the nearest box to the 165, so the 8. box has to be 1
• 165-128 = 37 left. 32 is the nearest box to the 37, so 6. Box has to be 1
• 37-32 = 5 left. 4 is the nearest box to the 5, so 3. Box has to be 1.
• 5-4 = 1 left. So, our last box, 1.box has to be 1.
Remember we say before, we only consider 1’s in the bit stream. Other all values will be 0. So, write the green signed boxes 1 and others 0.
Box 8, box 6, box 3 and box 1 are 1, other boxes are 0. Let’s write the bit stream = 10100101
You may little confused and asked yourself, what is the relations with this explanations and networking? Yes, it is strongly related with understanding IP addressing, subnet masking, network addressing etc. If we don’t know the conversations between decimal and binary values, we may a problem with understanding some basic network addressing rules. So please make yourself some exercises and clearly understand the conversations.
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# Total value of notes is 2000, number of notes is 50. How many Rs. 50 and Rs. 100 notes are received?
Given: Total value of notes is 2000, number of notes is 50.
To do: How many Rs. 50 and Rs. 100 notes are received
Solution:
Let no. of Rs. 50 notes = $x$
No. of Rs. 100 notes = $y$
Total notes = 25
⇒ $x + y = 25$
⇒ $x = 25-y$
Total amount only in Rs. 50 notes =$50x$
Total amount only in Rs. 100 notes = $100y$
Total amount = 2000
⇒ $50x + 100y = 2000$
⇒$50(25-y) + 100y = 2000$
⇒$50\times25 - 50y + 100y = 2000$
⇒$1250 +50 y = 2000$
⇒ $50y = 2000-1250 = 750$
⇒ $y = \frac{750}{50} = 15$
$x = 25-15 = 10$
No. of Rs. 50 notes = 10
No. of Rs. 100 notes = 15
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# College Math Preparation semester review questions Solve the
```College Math Preparation semester review questions
Solve the following equations. If the equation is an inequality or absolute value inequality then write the answer in
interval notation and do a line graph for the solution.
1.
5x + 8 = 3x – 7
2. 2(3x – 4) = 5x + 4
3. 2/3 x + 5 = 3/5 x – 2
4. 2x + 7 = 5x + 4
5. 4/5 x – 2/5 = 4/3 x + ½
6. 3(2x – 7)= 8 + 6x
7. 4x + 8 > 12
8. 2x + 5 ≤ 5𝑥 + 9
9. ¾ x + 3 > 2/3 x + 5
10. 8 < 2x – 6 < 14
11. 4≥ 5 − 2𝑥 ≥ −7
12. -5 < 2/3 x + 2 < 8
13. | x + 5 | = 8
14. 8 + | 2x = 5| = 20
15. | 2x – 4| ≥ 4
16. | 8 – x| < 5
17. 5 + |3x – 4| > 12
18. |6x| + 5 < 2
19. x2 – 8 = 12
20. Y2 -9 = 0
21. (x – 6)2 =81
22. 4 + ( x + 3)2 = 20
23. x2 + 8x = 0
24. X3 = 5x2
25. X2 + x – 12 = 0
26. x2 + 7x + 6 = 0
27. X2 – 3x – 18 = 0
28. X2 + 8x + 12 = 0
29. 3x2 – 8x + 5 = 0
30. 6x2 + x – 15 = 0
31. 4x2 + 8x + 4 = 0
32. x2 + x + 5 = 0
33. X2 – 8x + 2 = 0
34. 3x2 + x + 5 = 0
35. x2 + 4x + 2 = 0
36. 5x2 + 4x – 1 = 0
36. 7x2 + 9x + 2 = 0
Given the following information determine the equation of the line
a.
Slope of 2/3 point (0,5)
e. slope of 2 point (4,6)
b. Slope of -3 point ( 3,-2)
f. points ( 2,5) and (3,6)
c. Points (4,7) and (4,12)
g. parallel to Y = 4x + 6 and contains point (4,2)
d. Perpendicular to 2x + 5y = 8 can contains point (-2,8)
H
I
J
H.
J.
I.
k.
Look at each set of lines and tell me whether they are parallel, perpendicular, or intersecting
1.
Y = 4x – 8 and y = -1/4 x + 5
3. Y = 3x + 2 and y = -3x -4
2.
2x + 5y = 8 and -4x – 10y = 4
4. 2x + 3y = 2 and 3x – 2y = 5
Find the solution to the following systems of equations
X + 2y = 7
2x – y = 5
4x + 3y = 1
2x – y = -1
4x + y = 13
x=y+2
4x + y = -12
2x – 3y = 1
2x – y = 6
2x + 2y = -6
4x – 6y = 2
x=4+y
The following are parabolas, circles, ellipses, or hyperbolas. Look at each equation and determine what it is. Determine
the information at the right.
1. X2 + y2 = 25
2. X2 + y2/9 = 1
3. (x-4)2 + ( y – 3)2 = 8
4. (x+6)2/9 + (y-1)2/16 = 1
5. X2 + 8x + y2 – 2y = 12
6. 4x2 + 9y2 = 36
7. 9(x-3)2 + 25(y+2)2 = 225
8. Y = x2 + 4x – 3
Opens _____ vertex __________ axis of symmetry ______________
9. Y = -2x2 + 4x + 5
Opens _____ vertex __________ axis of symmetry ______________
10. X = 4y – 2y2 + 1
Opens _____ vertex __________ axis of symmetry ______________
11. X = y2 + 6y – 4
Opens _____ vertex __________ axis of symmetry ______________
12. Y = 2(x+3)2 – 4
Opens _____ vertex __________ axis of symmetry _____________
13. X = (y -4)2 =3
Opens _____ vertex __________ axis of symmetry ______________
14. X = (y -4)2 =3
Opens _____ vertex __________ axis of symmetry ______________
15. Y = 5x – 2x2 + 6
16. X2/9 – y2 = 1
Opens _____ vertex __________ axis of symmetry ______________
opens_____center________vertices_________________________
Asymptotes____________________________________________
17. - (x+6)2/4 + y2/49 = 1
opens_____center________vertices_________________________
Asymptotes____________________________________________
18. 4(x-4)2 – 9y2 = 36
opens_____center________vertices_________________________
Asymptotes____________________________________________
19. - 5(x+3)2 + (y-2)2 = 225
opens_____center________vertices_________________________
Asymptotes____________________________________________
20. – (x+7)2/16 + (y+3)2/4 = 1
opens_____center________vertices_________________________
Asymptotes____________________________________________
Equation ___________________________
Equation ____________________________
Equation ____________________________
Equation _______________________________
Equation ____________________________
Equation ____________________________
```
|
### The Simplex Method: pivoting to find the optimum Simplex tableau
• Recall that:
The Simplex method uses (invariant) row operations on matrices in Linear Algebra to find the optimal solution of an LP
• Review: invariant row operations in Linear Algebra
• Consider the following system of equations:
``` 2 x1 + 4 x2 = 16 3 x1 + 2 x2 = 12 ```
• The equality (=) means that the values in the left hand side and right hand side of "=" are in "balance".
• The systems of equations above can be represented by this figure:
``` 2 x1 + 4 x2 = 16 3 x1 + 2 x2 = 12 ```
• Invariant row operation 1:
You can multiply (or divide) both sides of an equation by the same factor, and the resulting equality will hold (= true).
Example: divide the first equation by 2 (= multiply first equation by 0.5)
``` 2 x1 + 4 x2 = 16 (equation 1) 3 x1 + 2 x2 = 12 (equation 2) Manipulation: 0.5 × (2 x1 + 4 x2) = 0.5 × (16) ==> x1 + 2 x2 = 8 Result: x1 + x2 = 8 (equation 1) 3 x1 + 2 x2 = 12 (equation 2) ```
• Invariant row operation2:
You can add (or subtract) a multiple of one equation (row) to another equation (row) and the resulting equality will hold (= true)
Example: subtract 3 × the first equation from the second equation
``` x1 + 2 x2 = 8 (row 1) 3 x1 + 2 x2 = 12 (row 2) (row 2) ===> 3 x1 + 2 x2 = 12 3 x (row 1) ===> 3 x1 + 6 x2 = 24 - (subtract) ------------------------------- 0 x1 - 4 x2 = -12 ```
• Pivot operation: transform one basic solution into another (better) basic solution
• Pivot operating: (a.k.a.: pivoting)
• Pivot operation = to apply invariant row operations to Simplex Tableau such that:
A non-basic variable will become a basic variable and A (some other) basic variable will become a non-basic variable
• In other words:
A pivot operation will change the basic solution !!!
Note:
Before we perform the pivot operation, we will make sure that the objective function will increase when we perform the pivot operation !!! We will learn how to select a pivot later....
• Pivot or pivot element
Pivot = an element in the Simplex Tableau that forms the central element in the pivot operation The pivot operation is apply using the pivot element as its parameter
• The steps of the pivot operation:
1. Divide the row containing the pivot element by the value of the pivot
This step will make the pivot element = 1
2. Use row operations to make the other elements of the column vector containg the pivot equal to 0
This step will make the column vector containg the pivot equal to a canonical unit vector !!
• Example: perform a pivot operation using the pivot element "3"
• The starting basic solution:
``` x1 x2 s1 s2 z | RHS ==========================+==== -1 -1 0 0 1 | 0 1 2 1 0 0 | 8 3 2 0 1 0 | 12 ^ | Use this as pivot element (The pivot operation will transform this column into a canonical unit vector with the pivot element becoming 1) Note: Current basic variables: s1 s2 z ```
• We will now:
Perform a pivot operation using the element 3 as pivot (I.e., these sequence of operation are performed in respect to that element)
• Pivot procedure:
``` x1 x2 s1 s2 z | RHS ==========================+==== -1 -1 0 0 1 | 0 1 2 1 0 0 | 8 3 2 0 1 0 | 12 Step 1: divide the pivot row by the pivot value (= 3) Result: x1 x2 s1 s2 z | RHS ==========================+==== -1 -1 0 0 1 | 0 (Row 1) 1 2 1 0 0 | 8 (Row 2) 1 ⅔ 0 ⅓ 0 | 4 (Row 3) Step 2: Use row operations to make the other value under x1 equal to 0 (I.e.: Add a multiple of row 3 to the other rows) x1 x2 s1 s2 z | RHS ==========================+==== -1 -1 0 0 1 | 0 (Row 1) 1 2 1 0 0 | 8 (Row 2) 1 ⅔ 0 ⅓ 0 | 4 (Row 3) Row operation: row 1 = row 1 + (1) × (row 3) Result: x1 x2 s1 s2 z | RHS ==========================+==== 0 -⅓ 0 ⅓ 1 | 4 (Row 1) 1 2 1 0 0 | 8 (Row 2) 1 ⅔ 0 ⅓ 0 | 4 (Row 3) Row operation: row 2 = row 2 - (1) × (row 3) Result: x1 x2 s1 s2 z | RHS ==========================+==== 0 -⅓ 0 ⅓ 1 | 4 (Row 1) 0 1⅓ 1 -⅓ 0 | 4 (Row 2) 1 ⅔ 0 ⅓ 0 | 4 (Row 3) ^ | New basic variable !! ```
• The new Simplex Tableau:
``` x1 x2 s1 s2 z | RHS ==========================+==== 0 -⅓ 0 ⅓ 1 | 4 (Row 1) 0 1⅓ 1 -⅓ 0 | 4 (Row 2) 1 ⅔ 0 ⅓ 0 | 4 (Row 3) ```
• The new set of basic variables:
x1 s1 z
The basic solution corresponding to the new Simplex Tableau is:
x2 = 0 (non-basic variable) s2 = 0 (non-basic variable) x1 = 4 (basic variable, RHS) s1 = 4 (basic variable, RHS) z = 4 (basic variable, RHS)
• Graphical depiction of this basic solution: (x1 = 4, x2 = 0)
Note:
• The basic solution is again a "corner" point in the feasible region
• This basic solution is better than the previous one because:
The objective value z = 4 (The previous solution has an objective value z = 0)
• The Simplex Algorithm
• The Simplex Algorithm:
``` T = an initial Simplex Tableau; // How: // Add surplus variables // to obtain a basic solution Find a pivot element p in T that // Discussed next makes the obj. function increase in value; while ( p can be found ) { T = Perform pivot operation on p in T // Discussed above Find a pivot element p in T that makes the obj. function increase in value; } ```
• Selecting a pivot that guarantee to improve on the objective function
• Mathematics behind the selection of the pivot:
• I will omit the Mathematical theory behind the how to select the pivot
(Requires matrix manipulations)
• I will only show you rules on how to pick the pivot
The pivot selection rules are based on 2 Mathematical results:
The pivot column is selected by an optimality criterium The pivot row is selected by a feasibility criterium
• Pivot row and pivot column:
• A pivot element is uniquely identified by
a pivot row a pivot column
Example:
• Terminology: objective row (and constraint rows)
• These terms will help understand the descriptions of the algorithm:
Explanation:
• The first row of the Simplex tableau is called the objective row
(Because it is obtained from the objective function z = x1 + x2 or: z - x1 - x2 = 0 )
• The other rows of the Simplex tableau is called the constraint rows
• Determining the pivot column:
The pivot column = the column of the largest negative cofficient in the objective (first) row in the Simplex Tableau (Break ties aribitrarily)
Reason:
The largest negative value will make the objective function increase the fastest in value when we change the value in that column.
Example:
• Terminology: test ratio
• Test ratio of a constraint row is defined as follows:
• If coefficient in the pivot column > 0:
``` RHS test ration of a constraint row = ---------------------------------- coefficient in the pivot column ```
else:
``` test ration of a constraint row = undefined ```
Example:
The test ratios of the 2 constraint rows are:
``` test ratio(constraint 1) = 8/1 = 8 test ratio(constraint 2) = 12/3 = 4 ```
Note:
The test ratio of the constraint row k represents an upper bound on the amount that the variable in the pivot couln can increase before the corresponding constraint k will be violated
• Determining the pivot row:
The pivot row = the row that contains the smallest (valid) test ratio (I.e., the coefficient must be > 0)
Example:
Explanation:
• The test ratio of constraint row 2 has the smallest value
• Therefore:
Pivot row = constraint row 2
• The pivot element:
The pivot element = the element found in the pivot row and pivot column
I I I I
• A complete example
• Linear Program:
``` max: z = x1 + x2 s.t.: x1 + 2x2 ≤ 8 3x1 + 2x2 ≤ 12 x1 ≥ 0, x2 ≥ 0 ```
• Step 1: transform to system of linear equation with a basis
``` Add surplus variables: z - x1 - x2 = 0 x1 + 2x2 + s1 = 8 3x1 + 2x2 + s2 = 12 Lined up: -x1 - x2 + z = 0 x1 + 2x2 + s1 = 8 3x1 + 2x2 + s2 = 12 In Tableau form: x1 x2 s1 s2 z | RHS ==========================+==== -1 -1 0 0 1 | 0 1 2 1 0 0 | 8 3 2 0 1 0 | 12 Basis: s1 s2 z ```
This is the initial Simplex Tableau
• Pivot step 1:
• Select the pivot column using the largest negative coefficient in the first (objective) row:
``` Pivot column (break ties arbitrarily) | V x1 x2 s1 s2 z | RHS ==========================+==== -1 -1 0 0 1 | 0 largest negative value 1 2 1 0 0 | 8 3 2 0 1 0 | 12 ```
• Select the pivot row using the smallest test ratio associated to a positive coefficient:
``` x1 x2 s1 s2 z | RHS ==========================+==== -1 -1 0 0 1 | 0 (neg. coef - skip it) 1 2 1 0 0 | 8 test ratio = 8/1 = 8 3 2 0 1 0 | 12 test ratio = 12/3 = 4 ```
• Pivot:
``` Step 1: divide pivot row by pivot element x1 x2 s1 s2 z | RHS ==========================+==== -1 -1 0 0 1 | 0 1 2 1 0 0 | 8 3 2 0 1 0 | 12 | divide row by 3 V x1 x2 s1 s2 z | RHS ==========================+==== -1 -1 0 0 1 | 0 1 2 1 0 0 | 8 1 0.7 0 0.3 0 | 4 Step 2: sweep the pivot column x1 x2 s1 s2 z | RHS ==========================+==== -1 -1 0 0 1 | 0 (Row 1) 1 2 1 0 0 | 8 (Row 2) 1 0.7 0 0.3 0 | 4 (Row 3) After: row 1 = row 1 + 1 × (row 3) x1 x2 s1 s2 z | RHS ==========================+==== 0 -0.3 0 0.3 1 | 4 (Row 1) 1 2 1 0 0 | 8 (Row 2) 1 0.7 0 0.3 0 | 4 (Row 3) After: row 2 = row 2 - 1 × (row 3) x1 x2 s1 s2 z | RHS ==========================+==== 0 -0.3 0 0.3 1 | 4 (Row 1) 0 1.3 1 -0.3 0 | 4 (Row 2) 1 0.7 0 0.3 0 | 4 (Row 3) ```
• Corresponding basic solution:
x1 = 4 x2 = 0 Objective value: z = 4
Graphically:
• Pivot step 2:
• Select the pivot column using the largest negative coefficient in the first (objective) row:
``` Pivot column | V x1 x2 s1 s2 z | RHS ==========================+==== 0 -0.3 0 0.3 1 | 4 (Row 1) 0 1.3 1 -0.3 0 | 4 (Row 2) 1 0.7 0 0.3 0 | 4 (Row 3) ```
• Select the pivot row using the smallest test ratio associated to a positive coefficient:
``` x1 x2 s1 s2 z | RHS ==========================+==== 0 -0.3 0 0.3 1 | 4 0 1.3 1 -0.3 0 | 4 test ratio = 4/1.3 = 3 1 0.7 0 0.3 0 | 4 test ratio = 4/0.7 = 6 ```
• Pivot:
``` Step 1: pivot row by pivot element x1 x2 s1 s2 z | RHS ==========================+==== 0 -0.3 0 0.3 1 | 4 0 1.3 1 -0.3 0 | 4 divide row by 1.333333 1 0.7 0 0.3 0 | 4 | V x1 x2 s1 s2 z | RHS ==========================+==== 0 -0.3 0 0.3 1 | 4 0 1 0.75 -0.25 0 | 3 1 0.7 0 0.3 0 | 4 Step 2: sweep the pivot column x1 x2 s1 s2 z | RHS ==========================+==== 0 -0.3 0 0.3 1 | 4 (row 1) 0 1 0.75 -0.25 0 | 3 (row 2) 1 0.7 0 0.3 0 | 4 (row 3) After: row 1 = row 1 + 0.333333 × (row 2) x1 x2 s1 s2 z | RHS ==========================+==== 0 0 0.25 0.25 1 | 5 (row 1) 0 1 0.75 -0.25 0 | 3 (row 2) 1 0.7 0 0.3 0 | 4 (row 3) After: row 3 = row 3 - 0.6666666 (0.7) × (row 2) x1 x2 s1 s2 z | RHS ==========================+==== 0 0 0.25 0.25 1 | 5 (row 1) 0 1 0.75 -0.25 0 | 3 (row 2) 1 0 -0.5 0.5 0 | 2 (row 3) ```
• Corresponding basic solution:
x1 = 2 x2 = 3 Objective value: z = 5
Graphically:
• Pivot step 3:
• Select the pivot column using the largest negative coefficient in the first (objective) row:
``` x1 x2 s1 s2 z | RHS ==========================+==== 0 0 0.25 0.25 1 | 5 (row 1) - no negative coefficient 0 1 0.75 -0.25 0 | 3 (row 2) 1 0 -0.5 0.5 0 | 2 (row 3) ```
DONE !!!
(Optimal solution found)
|
Matrix Algebra
We review here some of the basic definitions and elementary algebraic operations on matrices.
There are many applications as well as much interesting theory revolving around these concepts, which we encourage you to explore after reviewing this tutorial.
A matrix is simply a retangular array of numbers. For example, $$A = \left[ \begin{array}{llcl} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\\ \end{array} \right]$$ is a $m\times n$ matrix ($m$ rows, $n$ columns), where the entry in the $i^{th}$ row and $j^{th}$ column is $a_{ij}$. We often write $A = [a_{ij}]$.
#### Some Terminology
For an $n \times n$ square matrix A, the elements $a_{11},a_{22},\ldots,a_{nn}$ form the main diagonal of the matrix. The sum $\sum\limits_{k=1}^{n}a_{kk}$ of the elements on the main diagonal of $A$ is called the trace of $A$.
The matrix $A^{T}=[a_{ji}]$ formed by interchanging the rows and columns of $A$ is called the transpose of $A$. If $A^{T}=A$, the matrix $A$ is symmetric.
#### Example
Let $B= \left[ \begin{array}{rr} 6 & 9 \\ -4 & -6\\ \end{array} \right]$. The trace of $B$ is $6 + (-6) =0$. The transpose of $B$ is $B^{T} = \left[ \begin{array}{ll} 6 & -4\\ 9 & -6\\ \end{array} \right]$.
#### Addition and Subtraction of Matrices
To add or subtract two matrices of the same size, simply add or subtract corresponding entries. That is, if $B=[b_{ij}]$ and $C=[c_{ij}]$, $$B + C = [b_{ij} + c_{ij}] {\small\textrm{ and }} B-C = [b_{ij} - c_{ij}].$$
#### Example
For $B = \left[ \begin{array}{rr} 6 & 9 \\ -4 & -6\\ \end{array} \right]$ and $C = \left[ \begin{array}{rr} 1 & 2 \\ -1 & 0\\ \end{array} \right]$, $$B + C = \left[ \begin{array}{cc} 6 +1 & 9+2 \\ -4+(-1) & -6+0\\ \end{array} \right] = \left[ \begin{array}{rr} 7 & 11 \\ -5 & -6\\ \end{array} \right]$$ $$B-C = \left[ \begin{array}{cc} 6 -1 & 9-2 \\ -4-(-1) & -6-0\\ \end{array} \right] = \left[ \begin{array}{rr} 5 & 7 \\ -3 & -6\\ \end{array} \right].$$
The $m\times n$ zero matrix, 0, for which every entry is 0, has the property that for any $m\times n$ matrix $A$, $$A+\mathbf{0} = A.$$
#### Scalar Multiplication
To multiply a matrix $A$ by a number $c$ (a "scalar"), multiply each entry of $A$ by $c$. That is, $$cA=[ca_{ij}].$$
#### Example
Using the matrix $B= \left[ \begin{array}{rr} 6 & 9\\ -4 &-6\\ \end{array} \right]$ from the previous example, $$3B = 3\left[ \begin{array}{rr} 6 & 9\\ -4 &-6\\ \end{array}\right] = \left[ \begin{array}{rr} 18 & 27\\ -12 &-18\\ \end{array}\right].$$
#### Matrix Multiplication
Let $X$ be and $m \times n$ matrix and $Y$ be an $n \times p$ matrix. Then the product $XY$ is the $m\times p$ matrix whose $(i,j)^{th}$ entry is given by $$\sum^{n}_{k=1} x_{ik}y_{kj}.$$
Notes
• The product $XY$ is only defined if the number of columns of $X$ is the same as the number of rows of $Y$.
• $XY$ and $YX$ may very well not both be defined. If they both do exist, they are not necessarily equal and in fact might not even be of the same size.
#### Example
For the matrices $B = \left[ \begin{array}{rr} 6 & 9 \\ -4 & -6\\ \end{array} \right]$ and $C = \left[ \begin{array}{rr} 1 & 2 \\ -1 & 0\\ \end{array} \right]$, $${ BC = \left[ \begin{array}{rr} 6 & 9 \\ -4 & -6\\ \end{array} \right] \left[ \begin{array}{rr} 1 & 2 \\ -1 & 0\\ \end{array} \right] = \left[ \begin{array}{cc} (6)(1)+(9)(-1) & (6)(2)+(9)(0)\\ (-4)(1)+(-6)(-1) & (-4)(2) + (-6)(0)\\ \end{array} \right] = \left[ \begin{array}{cc} -3 & 12 \\ 2 & -8\\ \end{array} \right]}$$ while $${ CB = \left[ \begin{array}{rr} 1 & 2 \\ -1 & 0\\ \end{array} \right] \left[ \begin{array}{rr} 6 & 9 \\ -4 & -6\\ \end{array} \right] = \left[ \begin{array}{cc} (1)(6) + (2)(-4) & (1)(9) + (2)(-6)\\ (-1)(6) + (0)(-4) & (-1)(9) + (0)(-6)\\ \end{array} \right] = \left[ \begin{array}{cc} -2 & -3 \\ -6 & -9\\ \end{array} \right]}.$$
The $n\times n$ matrix having all main diagonal entries equal to 1 and all other entries equal to 0 is called the identity matrix $I$. For example, the $3\times 3$ matrix is $\left[ \begin{array}{lll} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array} \right]$. The $n \times n$ identity matrix has the property that if $A$ is any $n \times n$ matrix, $$AI = IA = A.$$
#### Inverse of a Matrix
Start with an $n \times n$ matrix $X$. Suppose the $n \times n$ matrix $Y$ has the property that $$XY = YX = I.$$ Then $Y$ is called the inverse of $X$ and is denoted $X^{-1}$.
Notes
• Only square matrices $X$ can have inverses. If X is not square, then for any Y the product XY will not be the same size matrix as the product YX (if we're lucky enough even to have both products exist!).
• Not every square matrix has an inverse. If an inverse exists, it is unique.
• If a matrix has an inverse, the matrix is said to be invertible.
The inverse of a $2 \times 2$ matrix is simple to calculate: $${\small\textrm{If }} A=\left[ \begin{array}{rr} a & b\\ c & d\\ \end{array} \right], {\small\textrm{ then }} A^{-1} = \frac{1}{ad-bc} \left[ \begin{array}{rr} d & -b\\ -c & a\\ \end{array} \right].$$
#### Example
The inverse of $C = \left[ \begin{array}{rr} 1 & 2 \\ -1 & 0\\ \end{array} \right]$ is $$C^{-1} = \frac{1}{(1)(0)-(2)(-1)}\left[ \begin{array}{rr} 0 & -2\\ 1 & 1\\ \end{array} \right] = \frac{1}{2} \left[ \begin{array}{rr} 0 & -2\\ 1 & 1\\ \end{array} \right] = \left[ \begin{array}{rr} 0 & -1\\ 1/2 & 1/2\\ \end{array} \right].$$ Note that $CC^{-1} = \left[ \begin{array}{rr} 1 & 2 \\ -1 & 0\\ \end{array} \right]\left[ \begin{array}{rr} 0 & -1\\ 1/2 & 1/2\\ \end{array} \right] = \left[ \begin{array}{rr} 1 & 0\\ 0 & 1\\ \end{array} \right]$ $\quad$ and $C^{-1}C = \left[ \begin{array}{rr} 0 & -1\\ 1/2 & 1/2\\ \end{array} \right]\left[ \begin{array}{rr} 1 & 2 \\ -1 & 0\\ \end{array} \right] = \left[ \begin{array}{rr} 1 & 0\\ 0 & 1\\ \end{array} \right].$ Matrix $B = \left[ \begin{array}{rr} 6 & 9 \\ -4 & -6\\ \end{array} \right]$ does not have an inverse.
#### Determinant of a Matrix
How did we know that $B = \left[ \begin{array}{rr} 6 & 9 \\ -4 & -6\\ \end{array} \right]$ does not have an inverse?
The determinant of $A$, $\det A$, is a number with the property that $A$ is invertible if and only if $\det A \not= 0$.
For a $2 \times 2$ matrix $A=\left[ \begin{array}{rr} a & b\\ c & d\\ \end{array} \right]$, $\det A = ad -bc$.
#### Example
For $B = \left[ \begin{array}{rr} 6 & 9 \\ -4 & -6\\ \end{array} \right]$, $\det B = (6)(-6)- (9)(-4)=-36+36=0$, so $B$ is not invertible. That is, $B$ does not have an inverse.
For a $3 \times 3$ (or larger) matrix $A$, things are a little more complicated:
Denote by $M_{ij}(A)$ the determinant of the matrix formed by deleting row $i$ and column $j$ for $A$.
Define $c_{ij}(A) = (-1)^{i+j}M_{ij}(A)$ to be the $(i,j)$ cofactor of $A$.
Then we can compute $\det A$ by the Laplace Expansion along any row or column of $A$:
Along row $i$: $$\det A = a_{i1}c_{i1}(A) + a_{i2}c_{i2}(A) + \ldots + a_{in}c_{in}(A).$$ Along column $j$: $$\det A = a_{1j}c_{1j}(A) + a_{2j}c_{2j}(A)+ \ldots + a_{nj}c_{nj}(A).$$
#### Example
Let $A = \left[ \begin{array}{rrr} 1 & -1 & 3\\ 1 & 0 & -1\\ 2 & 1 & 6\\ \end{array} \right].$ Along the first row, $\begin{array}{rcl} \det A & = & (1) \left[ (0)(6) - (-1)(1) \right] - (-1)\left [ (1)(6)-(-1)(2) \right] + 3 \left[ (1)(1)-(0)(2) \right] \\ & = & (1)(1) + (1)(8)+(3)(1)\\ & = & 12. \end{array}$ Computing $\det A$ along the second column instead, $\begin{array}{rcl} \det A & = & -(-1) \left[ (1)(6) - (-1)(2) \right] + ( 0)\left [ (1)(6)-(3)(2) \right] - 1 \left[ (1)(-1)-(3)(1) \right] \\ & = & (1)(8)+(0)(0)-(1)(-4)\\ & = & 12 {\small\textrm{ as expected.}} \end{array}$
### Key Concepts
Let $A = [a_{ij}] {\small\textrm{ and }} B=[b_{ij}]$.
• Transpose $A^{T}$ of $A$:
$$A^{T}=[a_{ji}].$$
• Trace of $A$:
$$\sum_{k=1}^{n} a_{kk} {\small\textrm{ (for an }}n\times n {\small\textrm{ matrix }} A).$$
• Identity Matrix $I$:
the $n \times n$ matrix with 1's on the main digonal and 0's elsewhere.
• $A+B$ and $A-B$:
$$A+B = [a_{ij}+ b_{ij}] \phantom{.}$$ $$A-B = [a_{ij} -b_{ij}].$$
• Scalar Multiplication:
$$cA = [ca_{ij}].$$
• Matrix Product $AB$:
$(i,j)^{th}$ entry is $\sum\limits_{k=1}^{n} a_{ik}b_{kj}$
(for an $m \times n$ matrix $A$ and an $n \times p$ matrix $B$).
• Inverse $A^{-1}$ of $A$:
$A^{-1}$ satisfies $AA^{-1} = A^{-1}A = I$.
If $A = \left[ \begin{array}{ll} a & b\\ c & d\\ \end{array}\right]$,
then $A^{-1} = \frac{1}{ad-bc} \left[ \begin{array}{rr} d & -b \\ -c & a \\ \end{array}\right].$
• Determinant $\det A$:
If $A = \left[ \begin{array}{ll} a & b\\ c & d\\ \end{array}\right]$, $\det A = ad-bc$.
In general,
$\qquad$ along row $i$:
$\qquad\qquad$ $\det A = a_{i1}c_{i1}(A) + a_{i2}c_{i2}(A) + \ldots + a_{in}c_{in}(A)$.
$\qquad$ along column $j$:
$\qquad\qquad$ $\det A = a_{1j}c_{1j}(A) + a_{2j}C_{2j}(A)+ \ldots + a_{nj}c_{nj}(A)$.
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Course: Arithmetic (all content)>Unit 5
Lesson 16: Adding and subtracting mixed number with unlike denominators
Subtracting mixed numbers with regrouping (unlike denominators)
To subtract mixed numbers, first align the whole numbers and fractions so they can be subtracted separately. If the fractions have different denominators, find a common denominator and convert them accordingly. If the fraction on the bottom is larger, regroup by borrowing from the whole number on top. This will allow you to complete the subtraction.
Want to join the conversation?
• how did he suddenly change from only having 4/9 to adding 4/9 + 9/9? Where did the 9/9 fraction come from? It's a bit confusing.
• Sal took a whole (9/9) from 17 because if 4/9 minus 6/9 it will be negative. And 17 becomes 16.
16 13/9 is the same thing as 17 4/9. Since if you change 13/9 into a mixed number (1 4/9), then add it to 16, it's 17 4/9.
Hope this helps
• how did you turn the 2/3 into 6/9?
• You can always multiply by 1 and get the same (equivalent) thing. He multiplied by 3/3 which is like a nickname for 1. Multiplying the tops (2*3=6) and the bottoms(3*3=9). The bottoms(denominators) is why we chose the 3/3 version of 1 in the first place. That was the way to get common denominator.
• you cant add/subtract if your denomeraters are not the same.like 5/9 and 2/6.they have denomerator of 54 so now it would be 30/54 and 18/54 which would be 48/54.hope it helps!
• What if you need to be simplified? I still don't get simplifying fractions with whole numbers.
• At the end of the problem, if your answer can be simplified, do it!
For example, if you have 5 6/12 , you can automatically see that the "6/12" can be simplified down to 1/2. You don't need to do anything to the 5 because you never simplify the whole number. Your final answer would be 5 1/2
Did this help?
• is there something that can help me understand better?
• I would recommend re-watching the video and doing the practice multiple times to get better.
• Can you just use knowledge in needing to regroup and not cross out the number, but re-write the number?
• ou can always multiply by 1 and get the same (equivalent) thing. He multiplied by 3/3 which is like a nickname for 1. Multiplying the tops (2*3=6) and the bottoms(3*3=9). The bottoms(denominators) is why we chose the 3/3 version of 1 in the first place. That was the way to get common denomina
• When you subtract 1 3/4 and 3 2/4, how do you subtract the whole numbers?
• You can't subtract 1 3/4 and 3 2/4, you have to do 3 2/4 first because its greater than 1 3/4, so in order to subtract you have to take 1 away from 3, in order to get 4/4 then add that to 2/4, and you can subtract, negative numbers are not stated in this video,
(1 vote)
• Why can we put it as a negative? (4/9 - 6/9)= -2/9 ?
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There are 366 different Starters of The Day, many to choose from. You will find in the left column below some starters on the topic of Estimating. In the right column below are links to related online activities, videos and teacher resources.
A lesson starter does not have to be on the same topic as the main part of the lesson or the topic of the previous lesson. It is often very useful to revise or explore other concepts by using a starter based on a totally different area of Mathematics.
Main Page
### Estimating Starters:
Air Traffic Control: Work out which aircraft are in danger of colliding from their positions and direction of travel. An exercise in understanding bearings.
Angle Estimates: Estimate the sizes of each of the angles then add your estimates together.
Big Bieber: If the dimensions of an object double, its volume increases by a factor of eight.
Big Order: Estimate or calculate then put the large numbers in order of size.
Calc-A-Hundred: A game for two players requiring a calculator and thinking skills.
Estimating Percentages: Estimate the percentages of full circles and rectangles the sectors represent.
Hot Estimates: Estimate the number of chillies in the photograph.
Moon Lengths: Estimate the distances shown on this photograph of the moon's surface.
Peanuts and Buttons: Two questions involving estimating a quantity.
Pie Chart: An exercise in estimating what the sectors of a pie chart represent.
Positions Please: Stand at the point between the classroom walls to represent a given number.
Red Lines: Either estimate the lengths of the red lines or, if you know how, calculate how long they are.
Second Holiday: Estimate then work out the period of time equal to the given number of seconds.
Four Fraction Division: Explain why the answer to a series of fraction divisions is a whole number.
#### Snooker Angles
An online game for one or two players requiring an ability to estimate angles.
Transum.org/go/?to=snookerangles
### Curriculum for Estimating:
#### Year 5
Pupils should be taught to use rounding to check answers to calculations and determine, in the context of a problem, levels of accuracy more...
Pupils should be taught to round decimals with two decimal places to the nearest whole number and to one decimal place more...
#### Year 6
Pupils should be taught to round any whole number to a required degree of accuracy more...
Pupils should be taught to use estimation to check answers to calculations and determine, in the context of a problem, an appropriate degree of accuracy. more...
Pupils should be taught to solve problems which require answers to be rounded to specified degrees of accuracy more...
#### Years 7 to 9
Pupils should be taught to round numbers and measures to an appropriate degree of accuracy [for example, to a number of decimal places or significant figures] more...
Pupils should be taught to use approximation through rounding to estimate answers and calculate possible resulting errors expressed using inequality notation a<x≤b more...
#### Years 10 and 11
Pupils should be taught to apply and interpret limits of accuracy when rounding or truncating, {including upper and lower bounds} more...
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### Notes:
The ability to estimate values is an often overlooked part of Mathematics. Estimating lengths, weights, time, angles and solutions to problems should be practised regularly. Pupils should make sensible estimates of a range of measures in relation to everyday situations.
A basic ability to estimate quantities without counting, like when choosing a checkout line at the supermarket, can be called a person’s innate ‘number sense’. Practising this kind of estimating may actually improve a pupil’s ability in other areas of mathematics. This is one of the findings of research published in Psychological Science, a journal of the Association for Psychological Science.
Practising estimation can be a lot of fun when presented as a game, challenge or group activity and provides the opportunity for the teacher to introduce variety in the mathematics classroom.
### Estimating Activities:
Estimating: Estimation is a very important skill. Use this activity to practise and improve your skills.
Estimating Angles: Estimate the size of the given acute angles in degrees.
Estimating Percentages: Estimate the percentages represented by the diagrams.
Estimation Golf: Play a round of golf using your estimation skills rather than golf clubs.
Rough Answers: An exercise on rounding values in a calculation to find an approximate estimate of the answer.
Snooker Angles: An online game for one or two players requiring an ability to estimate angles.
### Estimating Videos:
Counting Crowds: Find out how statisticians use density samples to estimate their statistics.
Estimate Enormous Numbers: Learn how to use the powers of 10 to make amazingly fast estimations of big numbers with this animated explanation.
### Estimating Worksheets/Printables:
Take Sides Questions: Thirty pairs of numbers or calculations. Which one of the pair is the largest?
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#### Estimating
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# Perimeter and Area of Rectangle
The formula of perimeter and area of rectangle are explained step-by-step with solved examples.
If l denotes the length and b denotes the breadth of the rectangle, then the
Perimeter of the rectangle = 2(l + b) units
Length of the rectangle = $$\frac{P}{2}$$ - b units
Breadth of the rectangle = $$\frac{P}{2}$$ - l units
Area of the rectangle = l × b sq. units.
Length of the rectangle = $$\frac{A}{b}$$ units .
Breadth of the rectangle = $$\frac{A}{l}$$ units
Diagonal of the rectangle = $$\sqrt{l^{2} + b^{2}}$$ units
Let us consider a rectangle of length 'a' units and breadth 'b' units.
Therefore, perimeter of the rectangle ABCD
= (AB + BC + CD + DA) units
= (a + b + a + b) units
= (2a + 2b) units
= 2 (a + b) units
Therefore, perimeter of the rectangle = 2 (length + breadth) units
We know that the area of the rectangle is given by
A = a × b square units
⇒ a = $$\frac{A}{b}$$, i.e., length of the rectangle = $$\frac{Area}{breadth}$$
And b = $$\frac{A}{a}$$, i.e., breadth of the rectangle = $$\frac{Area}{length}$$
Worked-out problems on Perimeter and Area of Rectangle:
1. Find the perimeter and area of the rectangle of length 17 cm and breadth 13 cm.
Solution:
Given: length = 17 cm, breadth = 13 cm
Perimeter of rectangle = 2 (length + breadth)
= 2 (17 + 13) cm
= 2 × 30 cm
= 60 cm
We know that the area of rectangle = length × breadth
= (17 × 13) cm$$^{2}$$
= 221 cm$$^{2}$$
2. Find the breadth of the rectangular plot of land whose area is 660 m2 and whose length is 33 m. Find its perimeter.
Solution:
We know that the breadth of the rectangular plot = $$\frac{Area}{length}$$
= $$\frac{660m^{2}}{33 m}$$
= 20 m
Therefore, the perimeter of the rectangular plot = 2 (length + breadth)
= 2(33 + 20) m
= 2 × 53 m
= 106 m
3. Find the area of the rectangle if its perimeter is 48 cm and its breadth is 6 cm.
Solution:
P = 2 (l + b)
Here, P = 48 cm; b = 6 cm
Therefore, 48 = 2 (l + 6)
⇒ $$\frac{48}{2}$$ = l + 6
⇒ 24 = l + 6
⇒ 24 - 6 = l
⇒ 18 = l
Therefore, length = 18 cm
Now, area of rectangle = l × b = 18 × 6 cm$$^{2}$$ = 108 cm$$^{2}$$
4. Find the breadth and perimeter of the rectangle if its area is 96 cm$$^{2}$$
and the length is 12 cm.
Solution:
Given, A = 96 cm$$^{2}$$ and l = 12 cm
A = l × b
Therefore, 96 = 12 × b
⇒ $$\frac{96}{12}$$ = b
⇒ b = 8 cm
Now, P = 2 (l + b)
= 2 (12 + 8)
= 2 × 20
= 40 cm
5. The length and breadth of a rectangular courtyard is 75 m and 32 m. Find the cost of leveling it at the rate of $3 per m2. Also, find the distance covered by a boy to take 4 rounds of the courtyard. Solution: Length of the courtyard = 75 m Breadth of the courtyard = 32 m Perimeter of the courtyard = 2 (75 + 32) m = 2 × 107 m = 214 m Distance covered by the boy in taking 4 rounds = 4 × perimeter of courtyard = 4 × 214 = 856 m We know that area of the courtyard = length × breadth = 75 × 32 m$$^{2}$$ = 2400 m$$^{2}$$ For 1 m$$^{2}$$, the cost of levelling =$3
For 2400 m$$^{2}$$, the cost of levelling = $3 × 2400 =$7200
Solved examples on Perimeter and Area of Rectangle:
6. A floor of the room 8 m long and 6 m wide is to be covered by square tiles. If each square tile is 0.8 m, find the number of tiles required to cover the floor. Also, find the cost of tiling at the rate of $7 per tile. Solution: Length of the room = 8 m Breadth of the room = 6 m Area of the room = 8 × 6 m$$^{2}$$ {Area of room = Area of tiles that are put on the floor of the room.} = 48 m$$^{2}$$ Area of one square tile = 0.8 × 0.8 m$$^{2}$$ = 0.64 m$$^{2}$$ Number of tiles required = $$\frac{Area of floor}{Area of tiles}$$ = $$\frac{48}{0.64}$$ = $$\frac{48 × 100}{64}$$ = 75 tiles For 1 tile, the cost of tiling is$7
For 7 tiles, the cost of tiling is $(7 × 75) =$525
7. The breadth of the rectangle is 8 cm and A its diagonal is 17 cm. Find the area of the rectangle and its perimeter.
Solution:
Using Pythagoras theorem,
BD$$^{2}$$ = DC$$^{2}$$ + BC$$^{2}$$
⇒ 172 = DC$$^{2}$$ + 8$$^{2}$$
⇒ 289 - 64 = DC$$^{2}$$
⇒ 225 = DC$$^{2}$$
⇒ 15 = DC
Therefore, length of rectangle = 15 cm
So, area of rectangle = l × b
= 15 × 8 cm$$^{2}$$
= 120 cm$$^{2}$$
Also, perimeter of rectangle = 2 (15 + 8) cm
= 2 × 23 cm
= 46 cm
8. The length and breadth of the rectangle park are in the ratio 5 : 4 and its area is 2420 m2, find the cost of fencing the park at the rate of $10 per metre. Solution: Let the common ratio b x, then length of rectangular park = 5x Breadth of rectangular park = 4x Area of rectangular park = 5x × 4x = 20x$$^{2}$$ According to the question, 20x$$^{2}$$ = 2420 ⇒ x$$^{2}$$ = $$\frac{2420}{20}$$ ⇒ x$$^{2}$$ = 121 ⇒ x = 11 Therefore, 5x = 5 × 11 = 55 and 4x = 4 × 11 = 44 So, the perimeter of the rectangular park = 2 (l + b) = 2 (55 + 44) = 2 × 99 = 198 cm For 1 m, the cost of fencing =$10
For 198 m, the cost of fencing = $198 × 10 =$1980
9. How many envelopes can be made out of a sheet of paper 100 cm by 75 cm, supposing 1 envelope requires 20 cm by 5 cm piece of paper?
Solution:
Area of the sheet = 100 × 75 cm$$^{2}$$ = 7500 cm$$^{2}$$
Area of envelope = 20 × 5 cm = 100 cm$$^{2}$$
Number of envelopes that can be made = $$\frac{Area of sheet}{Area of envelope}$$
= $$\frac{7500}{100}$$
= 75 envelopes
10. A wire in the shape of rectangle of length 25 cm and breadth 17 cm is rebent to form a square. What will be the measure of each side?
Solution:
Perimeter of rectangle = 2 (25 + 17) cm
= 2 × 42
= 84 cm
Perimeter of square of side x cm = 4x
Therefore, perimeter of rectangle = Perimeter of Square
84 cm = 4x
⇒ x = 21
Therefore, each side of square = 21 cm
These are the detailed step-by-step explanation with the formula of perimeter and area of rectangle.
● Mensuration
Area and Perimeter
Perimeter and Area of Rectangle
Perimeter and Area of Square
Area of the Path
Area and Perimeter of the Triangle
Area and Perimeter of the Parallelogram
Area and Perimeter of Rhombus
Area of Trapezium
Circumference and Area of Circle
Units of Area Conversion
Practice Test on Area and Perimeter of Rectangle
Mensuration - Worksheets
Worksheet on Area and Perimeter of Rectangles
Worksheet on Area and Perimeter of Squares
Worksheet on Area of the Path
Worksheet on Circumference and Area of Circle
Worksheet on Area and Perimeter of Triangle
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# Adding Numbers: A Step-by-Step Guide for Beginners
Adding Numbers: A Step-by-Step Guide for Beginners
Adding numbers is a fundamental math skill that everyone should know. Whether you’re a student, a professional, or simply a person who wants to brush up on their math skills, understanding the basics of addition is essential. This step-by-step guide will teach you everything you need to know about adding numbers.
Introduction
Adding numbers is a basic math skill that is used to calculate the sum of two or more numbers. It is an essential part of math and is used in everyday life. Knowing how to add numbers is important for both practical and educational purposes. It is also a great way to improve your mental math skills and can help you solve more complex math problems.
Body
The first step to adding numbers is to understand the concept of addition. In math, addition is defined as the process of combining two or more numbers to find their sum. For example, if you have two numbers, 3 and 5, you can add them together to get 8.
The next step is to understand the different ways to add numbers. The most common way to add numbers is to use a pencil and paper. You can also use a calculator, but doing it by hand is a great way to practice your mental math skills.
To add two numbers, you simply need to line them up vertically and add the digits in each column. For example, if you have the numbers 3 and 5, you would line them up as follows:
3
5
Then, you would start from the right side and add the digits in each column. In this case, the answer is 8.
You can also add multiple numbers at once. To do this, you simply need to line up the numbers in the same way as before and add the digits in each column. For example, if you have the numbers 3, 5, and 7, you would line them up as follows:
3
5
7
Then, you would start from the right side and add the digits in each column. In this case, the answer is 15.
Examples
Here are some examples of adding numbers:
1. 3 + 5 = 8
2. 6 + 7 = 13
3. 2 + 9 + 4 = 15
4. 5 + 8 + 3 + 6 = 22
FAQ Section
Q: How do I add numbers without a calculator?
A: You can add numbers without a calculator by using a pencil and paper. Simply line up the numbers vertically and add the digits in each column.
Q: How do I add multiple numbers at once?
A: To add multiple numbers at once, you simply need to line up the numbers in the same way as before and add the digits in each column.
Q: What is the best way to practice my addition skills?
A: The best way to practice your addition skills is to use a pencil and paper to add numbers without a calculator. This is a great way to improve your mental math skills and can help you solve more complex math problems.
Summary
Adding numbers is a basic math skill that is used to calculate the sum of two or more numbers. It is an essential part of math and is used in everyday life. Knowing how to add numbers is important for both practical and educational purposes. To add two numbers, you simply need to line them up vertically and add the digits in each column. You can also add multiple numbers at once by lining them up the same way and adding the digits in each column. The best way to practice your addition skills is to use a pencil and paper to add numbers without a calculator.
Conclusion
Adding numbers is an essential math skill that everyone should know. This step-by-step guide has taught you everything you need to know about adding numbers. With practice, you can become an expert at adding numbers and improve your mental math skills.
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# How do you find the square root of 320?
Sep 20, 2015
$8 \cdot \sqrt{5}$
#### Explanation:
I don't know either,
so let's break it down into pieces, shall we?
We have:
$\sqrt{320}$
The only thing that involves 32 in my mind is $4 \cdot 8$ or $2 \cdot 16$, and so we notice that $320 = 2 \cdot 160$ or $4 \cdot 80$ or $16 \cdot 20$ or $8 \cdot 40$, etc...
Let's try with $4 \cdot 80$:
$\sqrt{320} = \sqrt{4 \cdot 80}$
At this point, it is good to remember the rule:
$\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$
so that
$\sqrt{320} = \sqrt{4 \cdot 80}$
$= \sqrt{4} \cdot \sqrt{80}$
$= 2 \cdot \sqrt{80}$
Then you see that $80 = 4 \cdot 20$, so:
$\sqrt{320} = 2 \cdot \sqrt{80}$
$= 2 \cdot \sqrt{4 \cdot 20}$
$= 2 \cdot \sqrt{4} \cdot \sqrt{20}$
$= 2 \cdot 2 \cdot \sqrt{20} = 4 \cdot \sqrt{20}$
Again you see that $20 = 4 \cdot 5$, so:
$\sqrt{320} = 4 \cdot \sqrt{20}$
$= 4 \cdot \sqrt{4 \cdot 5}$
$= 4 \cdot \sqrt{4} \cdot \sqrt{5}$
$= 4 \cdot 2 \cdot \sqrt{5}$
$= 8 \cdot \sqrt{5}$
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# Into Math Grade 6 Module 9 Answer Key Solve Problems Using Equations and Inequalities
We included HMH Into Math Grade 6 Answer Key PDF Module 9 Solve Problems Using Equations and Inequalities to make students experts in learning maths.
## HMH Into Math Grade 6 Module 9 Answer Key Solve Problems Using Equations and Inequalities
Expression Data
Write expressions to match the scenario given.
The dartboard has sections numbered 1 to 20. You can “hit” a number on the dartboard by writing an expression equal to the number. Each expression must include each of the numbers 1, 2, 3, and 4 exactly once, but no other numbers. Try to get as many hits as you can. An example is shown.
1. (4 + 2) ÷ 3 – 1 = 1
2. (4 – 2) ÷ 1 + 3 = 5
3. (4 + 2) ÷ 1 + 3 = 9
4. (4 + 2) ÷ 3 + 1 = 3
5. (4 + 3) ÷ 1 + 2 = 9
6. (4 – 3) ÷ 1 + 2 = 3
7. (4 – 1) ÷ 3 + 2 = 3
8. (3 × 2) ÷ 1 + 4 = 10
9. (4 × 2) ÷ 1 + 3 = 11
10. (4 × 3) ÷ 2 + 1 = 7
11. (4 × 3) ÷ 3 – 1 = 4
12. (3 × 2) ÷ 1 – 4 = 2
13. (4 ÷ 1) + 2 × 3 = 18
14. (4 × 3) ÷ 1 + 2 = 14
15. (4 × 3) + 2 + 1 = 15
16. (4 × 1) ÷ 2 + 3 = 5
17. (3 × 2) + 1 + 4 = 13
18. (2 × 1) + 4 ÷ 3 = 2
19. (4 × 3) × 1 ÷ 2 = 6
20. (1 × 4) + 3 + 2 = 9
Turn and Talk
Question 1.
Describe a strategy you used to hit different numbers
Answer: By using the numbers 1, 2, 3, 4 and by including the different arithmetic operations we can hit different numbers.
Question 2.
Answer: We can group the symbols to hit different numbers we use parentheses () and then find the numbers.
Complete these problems to review prior concepts and skills you will need for this module.
Expressions with Variables
For Problems 1-10, find the value of the expression. Show your work.
Question 1.
25 – m if m = 9
Given expression,
25 – m
m = 9
25 – 9 = 16
So, the value of 25 – m when m = 9 is 16.
Question 2.
50 + t if t = 18
Given expression,
50 + t
t = 18
50 + 18 = 68
So, the value of 50 + t when t = 18 is 68.
Question 3.
105 – a if a = 75
Given expression,
105 – a
a = 75
105 – 75 = 30
So, the value of 105 – a when a = 75 is 30.
Question 4.
112 + c if c = 35
Given expression,
112 + c
c = 35
112 + 35 = 147
So, the value of 112 + c when c = 35 is 147.
Question 5.
15 × x if x = 5
Given expression,
15 × x
x = 5
15 × 5 = 75
So, the value of 15 × x when x = 5 is 75.
Question 6.
36 ÷ y if y = 3
Given expression,
36 ÷ y
y = 3
36 ÷ 3 = 12
So, the value of 36 ÷ y when y = 3 is 12.
Question 7.
64 ÷ d if d = 4
Given expression,
64 ÷ d
d = 4
64 ÷ 4 = 16
So, the value of 64 ÷ d when d = 4 is 16.
Question 8.
24 × n if n = 8
Given expression,
24 × n
n = 8
24 × 8 = 192
So, the value of 24 × n when n = 8 is 192.
Question 9.
125 ÷ k if k = 25
Given expression,
125 ÷ k
k = 25
125 ÷ 25 = 5
So, the value of 125 ÷ k when k = 25 is 5.
Question 10.
120 × b if b = 10
Given expression,
120 × b
b = 10
120 × 10 = 1200
So, the value of 120 × b when b = 10 is 1200.
Plot Points on a Number Line
For Problems 11-16, plot and label each integer on the number line.
Question 11.
-2
A number line is simply a representation of real numbers. Generally, we mark 0 in the middle, the negative integers on the left, and the positive integers on the right.
To mark -2 on the number line move 2 parts on the left of zero.
Question 12.
1
A number line is simply a representation of real numbers. Generally, we mark 0 in the middle, the negative integers on the left, and the positive integers on the right.
To mark 1 on the number line move 1 part on the right side of zero.
Question 13.
4
A number line is simply a representation of real numbers. Generally, we mark 0 in the middle, the negative integers on the left, and the positive integers on the right.
To mark 4 on the number line move 4 parts on the right side of zero.
Question 14.
-9
A number line is simply a representation of real numbers. Generally, we mark 0 in the middle, the negative integers on the left, and the positive integers on the right.
To mark -9 on the number line move 9 parts on the left of zero.
Question 15.
10
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# Solve the following system of equations by using the method of subtraction:\begin{align} & x+2y=-1 \\ & 2x-3y=12 \\ \end{align}
Last updated date: 16th Jul 2024
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Hint: From the 2 equations, choose any 1 equation. We need to eliminate either x or y to solve the equations. Modify one equation in terms of x and y and substitute in the 2nd equation of the chosen variable. Solve it and find the value of x and y.
We have been given 2 equations,
\begin{align} & x+2y=-1-(1) \\ & 2x-3y=12-(2) \\ \end{align}
Let us work to solve these 2 equations.
The tricky thing is that there are 2 variables x and y. We need to get rid of one of the variables. So, we can find the value of x and y.
Let us consider equation (1),$x+2y=-1$
Take 2y to the LHS, the equation becomes,
\begin{align} & x=-1-2y \\ & x=-\left( 2y+1 \right)-(3) \\ \end{align}
Let us mark the equation (3). Now we got an equation which can be substituted in equation (2) in the place of x. Thus we get rid of the variable ‘x’.
Now substitute the value of x in equation (2).
\begin{align} & 2x-3y=12 \\ & -2\left( 2y+1 \right)-3y=12 \\ \end{align}
Now we got an equation, with only the variable ‘y’. Simplify the above expression to get the value of y.
\begin{align} & -4y-2-3y=12 \\ & -7y=14 \\ & \Rightarrow y=\dfrac{14}{-7}=-2 \\ \end{align}
Hence, we got the value of y = -2.
Substitute this value in equation (3), to get the value of ‘x’.
\begin{align} & x=-2y-1 \\ & x=-2\left( -2 \right)-1=4-1=3 \\ \end{align}
Hence, we got x = 3 and y = -2.
So the solution to the system of equations is (3, -2).
Note: It’s a good idea to always check the results we get. Substitute (3, -2) in equation (1) and equation (2) and check if you get the answer equal to the LHS of the equation.
\begin{align} & x+2y=3+2\times \left( -2 \right)=3-4=-1 \\ & \therefore x+2y=-1 \\ \end{align}
Similarly, check the 2nd equation,
\begin{align} & 2x-3y=2\times 3-3\times \left( -2 \right) \\ & 2x-3y=6+6=12 \\ & \therefore 2x-3y=12 \\ \end{align}
Hence, we got the same value as that of LHS. So the solution is correct.
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# German Tank Problem
Estimators > German Tank Problem
## German Tank Problem
The German Tank Problem is a way to estimate the total population size from a small sample. It’s commonly used in AP statistics to teach about estimators. The problem was originally developed by the Allies during World War II, when it was used to estimate the total number of German tanks from a small number of serial numbers from captured, destroyed, or observed tanks. It was extended to estimate the number of factories and other manufactured parts. Today, the formula has been applied for wide reaching applications like estimating the number of iPhones sold.
The idea is that tanks were numbered 1,2,…x, where “x” was the unknown total number of tanks. If we know just a few tank numbers, we can use them to estimate x.
## Formula
The formula to solve the German Tank problem is:
Where:
• = population maximum,
• m = sample maximum (i.e. the highest serial number),
• n = sample size.
## Example
Example problem: You capture 5 enemy tanks with serial numbers 1, 31, 43, 79, and 115. What is the likely population maximum?
Solution:
Step 1: Insert the given values into the formula. For this example, the sample size is 5 and the highest value in the sample is 115, so:
= 115 + (115 / 5) – 1
Step 2: Solve:
= 115 + (115 / 5) – 1
= 115 + 23 – 1
= 138 – 1
= 137
The population maximum is 137.
## The History Behind the Problem
In World War II, the Allies did not know how many tanks the Germans were making. After a set of German Mark V tanks were captured and statisticians theorized (correctly) that the Germans numbered those tanks sequentially as they came off the production line. The statisticians used the above formula to estimate that 246 tanks were being made per month between June 1940 and September 1942; This was a significant reduction from intelligence estimates which put the number at about 1400. After the war, the Allies inspected the German’s production records and found that the true number of tanks produced per month was 245 — extremely close to the statisticians’ estimate.
## Minimum Variance Unbiased Estimator
The solution to the tank problem is an example of a Minimum Variance Unbiased Estimator (MVUE). An unbiased estimator is an accurate statistic that’s used to approximate a population parameter; “Minimum Variance” means that out of all of the possible estimates, the formula:
is the one with the lowest variance. “Possible estimates” include every way you can think of that could estimate the total number of tanks. As well as the above formula, your guesstimates might include:
And so on. The MVUE will be the formula with the lowest variance. In other words, the formula m + (m/n) + 1 gives a result that’s closer to the mean than any of the other potential formulas.
References:
Davies, G. 2006. Gavyn Davies Does the Maths; How a Statistical Formula Won the War. The Guardian. 19 July 2006. Retrieved November 14, 2016. Available online here.
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# RD Sharma Class 10 Solutions Chapter 7 Statistics
In this chapter, we provide RD Sharma Solutions for Class 10 Solutions Chapter 7 Statistics for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 10 Solutions Chapter 7 Statistics pdf, free RD Sharma Solutions for Class 10 Solutions Chapter 7 Statistics book pdf download. Now you will get step by step solution to each question.
## RD Sharma Class 10 Solutions Chapter 7 Statistics
### RD Sharma Class 10 Solutions Statistics Exercise 7.1
Question 1.
Calculate the mean for the following distribution :
Solution:
Question 2.
Find the mean of the following data:
Solution:
Question 3.
If the mean of the following data is 20.6. Find the value of p. (C.B.S.E. 1997)
Solution:
Question 4.
If the mean of the following data is 15, find p. (C.B.S.E. 1992C)
Solution:
Question 5.
Find the value of p for the following distribution whose mean is 16.6.
Solution:
Mean = 16.6
Question 6.
Find the missing value of p for the following distribution whose mean is 12.58. (C.B.S.E. 1992C)
Solution:
Question 7.
Find the missing frequency (p) for the following distribution whose mean is 7.68.
Solution:
Question 8.
The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students
Solution:
Question 9.
Candidates of four schools appear in a mathematics test. The data were as follows :
If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.
Solution:
Question 10.
Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.
Solution:
Question 11.
The arithmetic mean of the following data is 14, find the value of k. (C.B.S.E. 2002C)
Solution:
Mean=14
⇒ 14 (24 + k) = 360 + 10k
⇒ 336 + 14k = 360 + 10k
⇒ 14k- 10k- 360 -336 24
⇒ 4k = 24
⇒ k= (frac { 24 }{ 4 }) = 6 4
Hence k = 6
Question 12.
The arithmetic mean of the following data is 25, find the value of k. (C.B.S.E. 2001)
Solution:
Mean =25
⇒ 25 (14 + k) = 390 + 15k
⇒ 350 + 25k= 390 + 15k
⇒ 25k- 15k = 390 -350
⇒ 10k = 40 ⇒ k = (frac { 40 }{ 10 }) = 4
Hence k = 4
Question 13.
If the mean of the following data is 18.75. Find the value of p.
Solution:
⇒ 460 + 7p = 32 (18.75)
⇒ 460 + 7p = 600
⇒ 7p = 600 – 460 = 140
⇒ p = (frac { 140 }{ 7 }) = 20
∴ p = 20
Question 14.
Find the value of p, if the mean of the following distribution is 20.
Solution:
⇒ 5p2 + 100p + 295 = 20 (15 + 5p)
⇒ 5p2 + 100p + 295 = 300 + 100p
⇒ 5p2 + 100p – 100p = 300 – 295
⇒ 5p2 = 5 ⇒ p2 = (frac { 5 }{ 5 }) = 1
⇒ P= ±1
P = -1 i s not possible
∴ p= 1
Question 15.
Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.
Solution:
All Chapter RD Sharma Solutions For Class 10 Maths
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## What does exponential increase look like on graph?
An exponential growth function can be written in the form y = abx where a > 0 and b > 1. The graph will curve upward, as shown in the example of f(x) = 2x below. Notice that as x approaches negative infinity, the numbers become increasingly small.
## What does it mean when a graph increases exponentially?
So for example something with a linear growth rate will grow at a steady pace while something that has an exponential growth rate is increasing extremely rapidly after only a small amount of time.
How do you know if a graph is growing exponentially?
the constant b is a positive growth factor, and τ is the time constant—the time required for x to increase by one factor of b: If τ > 0 and b > 1, then x has exponential growth. If τ < 0 and b > 1, or τ > 0 and 0 < b < 1, then x has exponential decay.
How do you show an exponential increase?
exponential growth or decay function is a function that grows or shrinks at a constant percent growth rate. The equation can be written in the form f(x) = a(1 + r)x or f(x) = abx where b = 1 + r.
### How do you draw an exponential growth graph?
How To: Given an exponential function of the form f(x)=bx f ( x ) = b x , graph the function
1. Create a table of points.
2. Plot at least 3 point from the table including the y-intercept (0,1) .
3. Draw a smooth curve through the points.
4. State the domain, (−∞,∞) , the range, (0,∞) , and the horizontal asymptote, y=0 .
### What is exponential graph?
The basic exponential function If b > 1 b>1 b>1b, is greater than, 1, then the slope of the graph is positive, and the graph shows exponential growth. As x increases, the value of y approaches infinity. As x decreases, the value of y approaches 0.
What is exponential growth give a few examples?
For example, suppose a population of mice rises exponentially by a factor of two every year starting with 2 in the first year, then 4 in the second year, 8 in the third year, 16 in the fourth year, and so on. The population is growing by a factor of 2 each year in this case.
What is an exponential graph?
## How do you read an exponential graph?
Graphs of Exponential Functions
1. The graph passes through the point (0,1)
2. The domain is all real numbers.
3. The range is y>0.
4. The graph is increasing.
5. The graph is asymptotic to the x-axis as x approaches negative infinity.
6. The graph increases without bound as x approaches positive infinity.
7. The graph is continuous.
## Which graph shows exponential growth?
Linear growth. Hi Sara,It’s helpful to think about other kinds of growth.
• Quadratic growth. Does that make sense so far?
• Exponential growth. Now,suppose that instead of having a variable with a constant exponent,we have a constant with a variable exponent?
• True and false exponential growth. This has two consequences that are worth knowing about.
• How do you find increasing and decreasing intervals?
– f ′ ( x) = 3 x 2 − 6 x = 3 x ( x − 2) – Since f ′ is always defined, the critical numbers occur only when f ′ = 0, i.e., at c = 0 and c = 2. – Our intervals are ( − ∞, 0), ( 0, 2), and ( 2, ∞). – On the interval ( − ∞, 0), pick b = − 1 . – So our function is increasing on ( − ∞, 0), decreasing on ( 0, 2), and then increasing again on ( 2, ∞).
How to graph exponential growth?
Use the equation (with base 2): c a s e s = C 0 ∗ 2 t/D cases = C_*2^{t/D} cases = C 0
• Find the parameters C 0 C_0 C 0 and D D D that minimize the distance from the real value to the calculated.
• Draw the estimated values over the real points.
• ### How do you find the rate of change in exponential growth?
The general rule of thumb is that the exponential growth formula: x (t) = x 0 * (1 + r/100) t is used when there is a quantity with an initial value, x 0, that changes over time, t, with a constant rate of change, r.
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# How do you simplify? (v^0 * w^0)/(v * w^0 * v^(-1) * w^4)
Oct 6, 2016
$\frac{1}{w} ^ 4$
#### Explanation:
$\frac{{v}^{0} \cdot {w}^{0}}{v \cdot {w}^{0} \cdot {v}^{- 1} \cdot {w}^{4}}$
There are several law of indices which we can use here.
${x}^{0} = 1 \left(x \ne 0\right) \text{ }$ and $\text{ } {x}^{m} \times {x}^{n} = {x}^{m + n}$
${x}^{-} m = \frac{1}{x} ^ m \text{ }$ Also recall: $\frac{x}{x} = 1 \left(x \ne 1\right)$
$\frac{\cancel{{v}^{0}} 1 \cdot \cancel{{w}^{0}} 1}{v \cdot \cancel{{w}^{0}} \cdot {v}^{- 1} \cdot {w}^{4}} \text{ } \leftarrow {v}^{1 - 1} = {v}^{0} = 1$
$\frac{1}{\cancel{{v}^{0}} 1 \cdot {w}^{4}}$
$\frac{1}{w} ^ 4$
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MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.2
MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.2
Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:
1. OB = OC
2. AO bisect ∠A.
Solution:
Given
AB = AC
∠1 = ∠2, ∠3 = ∠4
To prove:
1. OB = OC
2. ∠5 = ∠6
Proof:
In ∆ABC,
AB = AC
∠B = ∠C
$$\frac{1}{2}$$ ∠B = $$\frac{1}{2}$$ ∠C
∠1 = ∠3 or ∠2 = ∠4
In ∆OBC
∠2 = ∠4
and so OB = OC
(In a A, sides opposite to equal angles are equal)
In ∆ABO and ∆ACO
BO = CO (proved)
∠1 = ∠3 (proved)
AB = AC (given)
∆ABO = ∆ACO (by SAS)
and so ∠5 = ∠6 (by CPCT)
Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see Fig. below). Show that AABC is an isosceles triangle in which AB =AC.
Solution:
Given
To prove:
AB = AC
Proof:
In A ABD and A ACD
BD = CD (given)
∴ ∆ABD = ∆ACD (BySAS)
and so AB = AC (by CPCT)
Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. below). Show that these altitudes are equal.
Solution:
Given
AB = AC
∠E = ∠F (each 90°)
To prove: BE = CF
Proof:
In ∆ABE and ∆ACE
∠A = ∠A (common)
∠E = ∠F (each 90°)
AB = AC (given)
∆ABE = ∆ACE (byAAS)
and so BE = CF (by CPCT)
Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. below). Show that:
1. ∆ABE ≅ ∆ACF
2. AB = AC,
i. e., ABC is an isosceles triangle.
Solution:
Given
BE = CF
∠E = ∠F (each 90°)
To prove:
1. ∆ABE = ∆ACE
2. AB = AC
Proof:
In ∆ABE and ∆ACF
∠A = ∠A (common)
BE = CF (given)
∠E = ∠F (each 90°)
∆ABE = ∆ACF (by AAS)
and so AB = AC (by CPCT)
Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see Fig. below). Show that ∠ABD = ∠ACD.
Solution:
Given
AB = AC
BD = CD
To prove
∠ABD = ∠ACD
Proof:
In ∆ABD and ∆ACD
AB = AC (given)
BD = CD (given)
∆ABD = ∆ACD (by SSS)
and so ∠ABD = ∠ACD (by CPCT)
Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. below). Show that ∠BCD is a right angle.
Solution:
Given: AB = AC
To show: ∠BCD = 90°
i.e., ∠2 + ∠3 = 90°
Proof:
AB = AC …..(1)
From (1) and (2), we get
In ∆ABC
AB = AC
∠1 = ∠2
(In a A, angles opposite to equal sides are always equal) …p) …(3)
In ∆ACD
∠3 = ∠4
(In a A, angles opposite to equal sides are always equal) …(4)
In ∆BCD
∠1 + ∠2 + ∠3 + ∠4 = 180° (ASP)
∠2 + ∠2 + ∠3 + ∠3 = 180°
(∴ ∠1 = ∠2, ∠3 = ∠4)
2 (∠2 + ∠3) = 180°
(∠2 + ∠3) = 90°
∠BCD = 90°
Question 7.
ABC is a right angled triangle in which ∠A – 90° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆BAC
AB =AC
∠B = ∠C = x
∠A + ∠B + ∠C= 180°
∠B + ∠C = 180° – 90°
∠B + ∠C = 90°
2x = 90°
x = $$\frac{90^{\circ}}{2}$$
Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
Given
ABC is an equilateral ∆
i. e., AB = BC = AC
To prove
∠A = ∠B = ∠C = 60°
Proof:
In ∆BAC
AB = AC
∠B = ∠C
(In a A, angles opposite to equal sides are always equal) ……(1)
AC = BC
∠A = ∠B
(In a A, angles opposite to equal sides are always equal) …..(2)
From (1) and (2), we get,
∠A = ∠B = ∠C = x (say)
∠A + ∠B + ∠C = 180° (ASP)
⇒ x + x + x = 180°
⇒ 3x = 180°
⇒ x = $$\frac{180^{\circ}}{3}$$ = 60°
∴ ∠A = ∠B = ∠C = 60
Theorem 7.4.
SSS (Side-Side-Side) Congruence Theorem:
Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.
Given
In ∆s ABC and DEF we have,
AB =DE
BC = EE
and AC = DF
To prove:
∆ABC = ∆DEF
Construction:
Suppose BC is the longest side.
Draw EF such that EE = AB and FEG = ∠CBA.
Join GF and DG.
Proof:
In ∆s ABC and GEE, we have
AB = GE (Const.)
∠ABC = ∠GEF (Const.)
and BC = EF (Given)
∴ ∆ABC = ∆GEF (SAS Cong. Axiom)
∠A = ∠G (CPCT) …..(1)
and AC = GF (CPCT) …..(2)
Now AB = EG (Const.)
AB = DE (Given)
∴ DE = EG ……(3)
Similarly, DF = GF ……(4)
In ∆EDG
DE = EG (Proved above)
∴ ∠A = ∠2 (∠s opp. equal side) …..(5)
In ∆DFG
FD = FG (Proved above)
∴ ∠3 = ∠4 (∠s ppp. equal side) …..(6)
∴ ∠1 + ∠3 – ∠2 + ∠4 [From (5) and (6)]
i. e. ∠D = ∠G …..(7)
But ∠G = ∠A [From (1)]
∴ ∠A = ∠D …..(8)
In ∆s ABC and DEF,
AB – DE (Given)
AC = DF (Given)
∠A = ∠D [From (8)]
∆ABC ≅ ∆DEF (SAS Cong. Axiom)
Theorem 7.5.
RHS (Right Angle Hypotenuse Side) Congruence Theorem:
Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle.
Given
In ∆s ABC and DEF,
∠B = ∠E = 90°
AC = DF
BC = EF.
∆ABC ≅ ∆DEF
Construction:
Produce DE to M so that
EM = AB, Join ME.
Proof:
In ∆s ABC and MEF
AB = ME (Const.)
BC = EE (Given)
∠B = ∠MEF (each 90°)
∴ ∆ABC = ∆MEF (SAS Cong. Axiom)
Hence ∠A = ∠M (CPCT) …(1)
AC = MF (CPCT) …(2)
Also AC =DF (Given)
∴ DF = MF
∴ ∠D = ∠M (∠s opp. equal side of ADFM) …(3)
From (1) and (3), we have
∠A = ∠D …..(4)
Now, in ∆s ABC and DEF, we have
∠A = ∠D [From (4)]
∠B = ∠E (Given)
∴ ∠C = ∠F …..(5)
Again, in ∆s ABC and DEF, we have
BC = EF (Given)
AC = DF (Given)
∠C = ∠F [From (5)]
∴ ∆ABC = ∆DEF (SAS Cong. Axiom)
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# Difference between revisions of "2016 AIME I Problems/Problem 9"
## Problem
Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}=\frac{1}{5}$. This triangle is inscribed in rectangle $AQRS$ with $B$ on $\overline{QR}$ and $C$ on $\overline{RS}$. Find the maximum possible area of $AQRS$.
## Solution
### Solution 1
Note that if angle $BAC$ is obtuse, it would be impossible for the triangle to inscribed in a rectangle. This can easily be shown by drawing triangle ABC, where $A$ is obtuse. Therefore, angle A is acute. Let angle $CAS=n$ and angle $BAQ=m$. Then, $\overline{AS}=31\cos(n)$ and $\overline{AQ}=40\cos(m)$. Then the area of rectangle $AQRS$ is $1240\cos(m)\cos(n)$. By product-to-sum, $\cos(m)\cos(n)=\frac{1}{2}(\cos(m+n)+\cos(m-n))$. Since $\cos(m+n)=\sin(90-m-n)=\sin(BAC)=\frac{1}{5}$. The maximum possible value of $\cos(m-n)$ is 1, which occurs when $m=n$. Thus the maximum possible value of $\cos(m)\cos(n)$ is $\frac{1}{2}(\frac{1}{5}+1)=\frac{3}{5}$ so the maximum possible area of $AQRS$ is $1240\times{\frac{3}{5}}=\fbox{744}$. -AkashD
### Solution 2
As above, we note that angle $A$ must be acute. Therefore, let $A$ be the origin, and suppose that $Q$ is on the positive $x$ axis and $S$ is on the positive $y$ axis. We approach this using complex numbers. Let $w=\text{cis} A$, and let $z$ be a complex number with $|z|=1$, $\text{Arg}(z)\ge 0^\circ$ and $\text{Arg}(zw)\le90^\circ$. Then we represent $B$ by $40z$ and $C$ by $31zw$. The coordinates of $Q$ and $S$ depend on the real part of $40z$ and the imaginary part of $31zw$. Thus $$[AQRS]=\Re(40z)\cdot \Im(31zw)=1240\left(\frac{z+\overline{z}}{2}\right)\left(\frac{zw-\overline{zw}}{2i}\right).$$ We can expand this, using the fact that $z\overline{z}=|z|^2$, finding $$[AQRS]=620\left(\frac{z^2w-\overline{z^2w}+w-\overline{w}}{2i}\right)=620(\Im(z^2w)+\Im(w)).$$ Now as $w=\text{cis}A$, we know that $\Im(w)=\frac15$. Also, $|z^2w|=1$, so the maximum possible imaginary part of $z^2w$ is $1$. This is clearly achievable under our conditions on $z$. Therefore, the maximum possible area of $AQRS$ is $620(1+\tfrac15)=\boxed{744}$.
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# Rounding off to the Nearest Tens
For rounding off to the nearest tens let us look at the number ray given below.
Consider the numbers 12 and 19.
When we plot these numbers on a number line we observe that 12 lies between 10 and 20. Also 12 is nearest to 10 than 20. So, we round off 12 as 10, correct to the nearest tens.
If we consider 19, we observe that it is nearer to 20 than 10. So, we round off 19 to 20, correct to the nearest tens.
Now, let us consider the number 15.
If we plot this number on number line we find that 15 lies half-way between 10 and 20. By convention, we round off 15 to 20.
Now we will learn how to estimate the number 23, 27 and 29.
Look at the arrow at 23. It is nearer to 20 than 30 because 23 - 20 = 3 while 30 – 23 = 7. We take 23 as 20 when we estimate it to the nearest 10.
Again, look at the arrow at 27. The arrow at 27 is nearer to 30 than 20 because 30 – 27 = 3 and 27 – 20 = 7. So, 27 is taken as 30.
In the same way, look at the arrow at 29. It is nearer to 30 than 20 because 30 - 29 = 1 while 29 – 20 = 9. So, 29 is taken as 30 since it is near to 30 than 20.
How will we round off the number 25?
25 is equal distance from 20 and 30. By convention it is taken as 30.
1. Round off to the nearest tens:
(i) 362
The given number is 362.
Its ones or unit digit is 2, which is less than 5. So, we replace the ones digit by 0 to get the rounded off number.
Hence, rounded off number = 360.
(ii) 909
The given number is 909.
Its ones or unit digit is 9, which is greater than 5. So, we increase the tens digit by 1 and replace the ones digit by 0 to get the rounded off number.
Hence, rounded off number = 910.
2. How to estimate 2534 to the nearest 10?
34 is nearer to 30 than 40, so 2534 is taken as 2530.
From the above examples, to estimate to nearest tens we can generalize that,
(i) the numbers having 1, 2, 3, 4 at ones or units place are rounded off downwards.
(ii) the numbers having 5, 6, 7, 8, 9 at ones or units place are rounded off upwards.
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×
# VEDIC MATHS NOTE - 1
TRICK TO FIND SQUARES OF NUMBERS ENDING WITH 5,9,1
Square of numbers ending with 5 Formula for calculating square ending with 5 is easy.
$$\begin{array} { l l } & 85 \\ \times & 85 \\ \hline \\ & 7225 \end{array}$$
Steps
• Multiply 5 by 5 and put composite digit 25 on the right hand side.
• Add 1 to the upper left hand side digit i.e. 8 i.e. 8+1=9
• Multiply 9 to the lower hand digit 8, i.e. 9*8=72
Using this method we can find out square of the number. Now let’s have a look at method of calculating square of adjacent number.
Forward Method
We know method to find square of a number ending with 5., say Square of 75=5625, then just have a look to find square of 76.
75’s square=5625(known)
76’square=75’square+ (75+76) =5625+151=5776.
So square of 76 is 5776.
Steps
• Steps are simple. The format shown above is self explanatory. But still I am explaining it.
• 75’square=5625 is known
• Add (75+76=151) to this to get 76’square
• 76’square=5776.
Reverse Method
As like forward method for calculating square of number which is 1 more than the given number whose square is known, we have reverse method to find square.
Now let me explain in detail the Reverse approach through which You will able to find out squares of a number which is one less than given number.
Consider the following example:
Suppose we know square of a number, say, 70; how to find square of 69?
(70)’square=4900(known)
(69)’square=4900-(69+70) =4900-139
=4761.
3 years, 9 months ago
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# What Is 30/66 as a Decimal + Solution With Free Steps
The fraction 30/66 as a decimal is equal to 0.454.
A fraction is when converted into decimal form, the decimal part can have a different condition. This decimal part can be terminating, non-terminating, recurring, or non-recurring. Fraction 30/66 is terminating.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 30/66.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 30
Divisor = 66
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 30 $\div$ 66
This is when we go through the Long Division solution to our problem.
Figure 1
## 30/66 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 30 and 66, we can see how 30 is Smaller than 66, and to solve this division, we require that 30 be Bigger than 66.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 30, which after getting multiplied by 10 becomes 300.
We take this 300 and divide it by 66; this can be done as follows:
 300 $\div$ 66 $\approx$ 4
Where:
66 x 4 = 264
This will lead to the generation of a Remainder equal to 300 – 264 = 36. Now this means we have to repeat the process by Converting the 36 into 360 and solving for that:
360 $\div$ 66 $\approx$ 5Â
Where:
66 x 5 = 330
This, therefore, produces another Remainder which is equal to 360 – 330= 30. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 300.
300 $\div$ 66 $\approx$ 4Â
Where:
66 x 4 = 264
Finally, we have a Quotient generated after combining the three pieces of it as 0.454, with a Remainder equal to 36.
Images/mathematical drawings are created with GeoGebra.
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KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.4
Students can Download Chapter 3 Playing with Numbers Ex 3.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
Karnataka State Syllabus Class 6 Maths Chapter 3 Playing with Numbers Ex 3.4
Question 1.
Find the common factors of:
a) 20 and 28
b) 15 and 25
c) 35 and 50
d) 56 and 120
Solution:
Question 2.
Find the common factors of
a) 4, 8 and 12
b) 5, 15 and 25
Solution:
Question 3.
Find first three common multiples of:
a) 6 and 8
b) 12 and 18
Solution:
a) 6 and 8:-
3 common multiples = 24, 48, 72
b) 12 and 18:-
Multiples of 12 = 12, 24, 36, 48, 60, 72
Multiples of 18= 18, 36, 54, 72
3 Common multiples = 36, 72, 108
Question 4.
Write all the number less than 10(1 which are common multiples of 3 and 4
Solution:
Multiples of 3 = 3, 6, 0, 12, 15, 18, 21, 24, 27, 30
Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
Common multiples = 12, 24, 36, 48, 60, 72, 84, 96
Question 5.
Which of the following numbers are co-prime?
a) 18 and 35
b) 15 and 37
c) 30 and 415
d) 17 and 68
e) 216 and 215
f) 81 and 16
Solution:
a) Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 35 = 1, 5, 7, 35
Common factors = 1
Therefore, the given two numbers are co-prime
b) Factors of 15 = 1, 3, 5, 15
Factors of 37 = 1, 37
Common factors = 1
The given two numbers are co-prime
c) Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30
Factors of 415 = 1, 5, 83, 415
Common factors = 1, 5
As these numbers have a common factors other than 1, the given two numbers arc not co-prime
d) Factors of 17 = 1, 17
Factors of 68 = 1, 2, 4, 17, 34, 68
Common factors = 1, 17
As these numbers have a common factors other than 1, the given two numbers are not co-prime
e) Factors of 216 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 17, 36, 54, 72, 108, 216
Factors of 215 = 1, 5, 43, 215
Common factors = 1
The given two numbers are prime.
f) Factors of 81= 1, 3, 9, 27, 81
Factors of 16 = 1, 2, 4, 8, 16
Common factors = 1
Therefore the given two numbers are co-prime.
Question 6.
A number is divisible by both 5 and 12. By which other number will that number be always divisible?
Solution:
Factors of 5 = 1, 5
Factors of 12 = 1, 2, 3, 4, 5, 6, 12
As the common factors of these number is 1, the given two numbers are co-prime and the number will also be divisible by their product , i,e 60 and the factors of 60.
i.e, 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.
Question 7.
A number is divisible by 12. By what other numbers will that number be divisible?
Solution:
Since, the number is divisible by 12, it will also be divisible 1,2,3,4 and 6 are numbers other than 12 by which this number is also divisible.
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# 2017 USAMO Problems/Problem 1
## Problem
Prove that there are infinitely many distinct pairs $(a,b)$ of relatively prime positive integers $a>1$ and $b>1$ such that $a^b+b^a$ is divisible by $a+b$.
## Solution 1
Let $n=a+b$. Since $gcd(a,b)=1$, we know $gcd(a,n)=1$. We can rewrite the condition as
$$a^{n-a}+(n-a)^a \equiv 0 \mod{n}$$ $$a^{n-a}\equiv-(-a)^a \mod{n}$$ Assume $a$ is odd. Since we need to prove an infinite number of pairs exist, it suffices to show that infinitely many pairs with $a$ odd exist.
Then we have $$a^{n-a}\equiv a^a \mod{n}$$ $$1 \equiv a^{2a-n} \mod{n}$$
We know by Euler's theorem that $a^{\varphi(n)} \equiv 1 \mod{n}$, so if $2a-n=\varphi(n)$ we will have the required condition.
This means $a=\frac{n+\varphi(n)}{2}$. Let $n=2p$ where $p$ is a prime, $p\equiv 1\mod{4}$. Then $\varphi(n) = 2p*\left(1-\frac{1}{2}\right)\left(1-\frac{1}{p}\right) = p-1$, so $$a = \frac{2p+p-1}{2} = \frac{3p-1}{2}$$ Note the condition that $p\equiv 1\mod{4}$ guarantees that $a$ is odd, since $3p-1 \equiv 2\mod{4}$
This makes $b = \frac{p+1}{2}$. Now we need to show that $a$ and $b$ are relatively prime. We see that $$gcd\left(\frac{3p-1}{2},\frac{p+1}2\right)=\frac{gcd(3p-1,p+1)}{2}$$ $$=\frac{gcd(p-3,4)}{2}=\frac22=1$$ By the Euclidean Algorithm.
Therefore, for all primes $p \equiv 1\mod{4}$, the pair $\left(\frac{3p-1}{2},\frac{p+1}{2}\right)$ satisfies the criteria, so infinitely many such pairs exist.
## Solution 2
Take $a=2n-1, b=2n+1, n\geq 2$. It is obvious (use the Euclidean Algorithm, if you like), that $\gcd(a,b)=1$, and that $a,b>1$.
Note that
$$a^2 = 4n^2-4n+1 \equiv 1 \pmod{4n}$$
$$b^2 = 4n^2+4n+1 \equiv 1 \pmod{4n}$$
So
$$a^b+b^a = a(a^2)^n+b(b^2)^{n-1} \equiv$$ $$a\cdot 1^n + b\cdot 1^{n-1} \equiv a+b = 4n \equiv 0 \pmod{4n}$$
Since $a+b=4n$, all such pairs work, and we are done.
## Solution 3
Let $x$ be odd where $x>1$. We have $x^2-1=(x-1)(x+1),$ so $x^2-1 \equiv 0 \pmod{2x+2}.$ This means that $x^{x+2}-x^x \equiv 0 \pmod{2x+2},$ and since x is odd, $x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},$ or $x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},$ as desired.
## Solution 4
I claim that the ordered pair $(2^{n} - 1, 2^{n} + 1)$ satisfies the criteria for all $n \geq 2.$ Proof: It is easy to see that the order modulo $2^{n+1}$ of $(2^{n} - 1)^k$ is $2,$ since $(2^{n} - 1)^2 = 2^{2n} - 2 \cdot 2^{n} + 1 \equiv 1 \mod 2^{n+1}.$, and we can assert and prove similarly for the order modulo $2^{n+1}$ of $(2^{n} + 1)^k.$ Thus, the remainders modulo $2^{n+1}$ that the sequences of powers of $2^{n} - 1$ and $2^{n} + 1$ generate are $1$ for even powers and $2^{n} - 1$ and $2^{n} + 1$ for odd powers, respectively. Since $2^{n} + 1$ and $2^{n} - 1$ are both odd for $n \geq 2,$ $$(2^n + 1)^{2^n - 1} + (2^n - 1)^{2^n + 1} \equiv 0 \mod 2^{n+1}.$$ Since there are infinitely many powers of $2$ and since all ordered pairs $(2^n - 1, 2^n+1)$ contain relatively prime integers, we are done. $\boxed{}$
-fidgetboss_4000
## Solution 5 (Motivation for Solution)
Note that $$a^b+b^a=a^b-a^a+a^a+b^a.$$ To get rid of the $a^a+b^a$ part $\pmod{a+b},$ we can use the sum of powers factorization. However, $a$ must be odd for us to do this. If we assume that $a$ is odd, $$a^b-a^a+a^a+b^a\equiv a^b-a^a+(a+b)(\text{an integer})\equiv a^b-a^a \equiv a^a\left(a^{b-a}-1\right)\pmod{a+b}.$$ Because $a$ and $b$ are relatively prime, $a+b$ cannot divide $a^a.$ Thus, we have to show that there exists an integer $b$ such that for odd $a,$ $$a^{b-a}\equiv 1 \pmod{a+b}.$$ Suppose that $a=2n-1.$ To keep the powers small, we try $b=2n+k$ for small values of $k.$ We can find that $b=2n$ does not work. $b=2n+1$ works though, as $a+b=4n$ and $$(2n-1)^{2n+1-2n+1}\equiv (2n-1)^2\equiv 4n^2-4n+1\equiv 4n(n-1)+1 \equiv 1 \pmod{4n}.$$ Because $a$ is odd, $b=a+2$ is relatively prime to $a.$ Thus, $$(a,b)=(2n-1,2n+1)$$ is a solution for positive $n\ge 2.$ There are infinitely many possible values for $n,$ so the proof is complete. $\blacksquare$
~BS2012
## See Also
2017 USAMO (Problems • Resources) Preceded byFirst Problem Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 All USAMO Problems and Solutions
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# How do you solve 7= 12+ \frac { 1} { 2} a?
Aug 22, 2017
$a = - 10$
#### Explanation:
$7 = 12 + \frac{1}{2} a$
$7 = 12 + \frac{a}{2}$
Multiply through with the LCM
$7 \times 2 = 12 \times 2 + \frac{a}{2} \times 2$
$14 = 24 + \frac{a}{\cancel{2}} \times \cancel{2}$
$14 = 24 + a$
Collect like terms
$a = 14 - 24$
$a = - 10$
Aug 22, 2017
See a solution process below:
#### Explanation:
First, subtract $\textcolor{red}{12}$ from each side of the equation to isolate the $a$ term while keeping the equation balanced:
$- \textcolor{red}{12} + 7 = - \textcolor{red}{12} + 12 + \frac{1}{2} a$
$- 5 = 0 + \frac{1}{2} a$
$- 5 = \frac{1}{2} a$
Now, multiply each side of the equation to by $\textcolor{red}{2}$ to solve for $a$ while keeping the equation balanced:
$\textcolor{red}{2} \times - 5 = \textcolor{red}{2} \times \frac{1}{2} a$
$- 10 = \cancel{\textcolor{red}{2}} \times \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} a$
$- 10 = 1 a$
$- 10 = a$
$a = - 10$
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How do you find the standard error of the mean?
How do you find the standard error of the mean?
SEM is calculated by taking the standard deviation and dividing it by the square root of the sample size. Standard error gives the accuracy of a sample mean by measuring the sample-to-sample variability of the sample means.
What is the standard error of the mean example?
Thus, for a sample of N = 25 and population standard deviation of s x = 100, the standard error of the mean is 100/5 or 20. For a sample of N = 100 and population standard deviation of s x = 100, the standard error of the mean is 100/10 or 10.
What is the standard error of the mean equal to?
Therefore, the relationship between the standard error of the mean and the standard deviation is such that, for a given sample size, the standard error of the mean equals the standard deviation divided by the square root of the sample size.
How is the standard error of the mean calculated quizlet?
Tells you how accurate your estimate of the mean is likely to be. Calculated by the standard deviation of the observations divided by the square root of the sample size.
How do you calculate standard error of sample?
Compute the standard error, which is the standard deviation divided by the square root of the sample size. To conclude the example, the standard error is 5.72 divided by the square root of 4, or 5.72 divided by 2, or 2.86.
Which formula would you use for calculating the standard error of the mean finite or infinite population?
(1) Standard Error of Means If the population size is infinite then σ x ¯ = σ n × N − n N because N − n N tends towards 1 as N tends to infinity.
What is the formula for the standard error of the population quizlet?
The standard error of the mean is the standard deviation of the different sample means. 2/3 of the sample means would be within 1 standard error. 95.4% would be within 2 standard errors. 99.7% would be within 3 standard errors.
What does the calculated value of the standard error of the estimate indicate about the value of y?
Likewise, a standard deviation which measures the variation in the set of data from its mean, the standard error of estimate also measures the variation in the actual values of Y from the computed values of Y (predicted) on the regression line. …
What are the two formulas for calculating standard error?
n2 = Number of observations. Sample 2. σ21 = Variance. Sample 1….What is the Standard Error Formula?
Statistic (Sample) Formula for Standard Error.
Difference between means. = √ [s21/n1 + s22/n2]
Difference between proportions. = √ [p1(1-p1)/n1 + p2(1-p2)/n2]
How do you calculate standard error manually?
To calculate the standard error, you need to have two pieces of information: the standard deviation and the number of samples in the data set. The standard error is calculated by dividing the standard deviation by the square root of the number of samples.
How do you calculate the standard error of the mean in Excel?
As you know, the Standard Error = Standard deviation / square root of total number of samples, therefore we can translate it to Excel formula as Standard Error = STDEV(sampling range)/SQRT(COUNT(sampling range)).
Which is the correct formula for the standard error of the estimate quizlet?
The equation for the standard error of estimate is: sy⋅x=√Σ(y−y∧)2n−2Σ(y-y∧)/2n-2.
How do you calculate estimated standard error?
It is usually calculated by the sample estimate of the population standard deviation (sample standard deviation) divided by the square root of the sample size (assuming statistical independence of the values in the sample): Where: SEM = standard error of the mean. s = sample standard deviation (see formula below)
How to calculate estimated standard error?
1. Create a five column data table. Any statistical work is generally made easier by having your data in a concise format. A simple table serves this
• 2. Enter the data values for your measured data. After collecting your data,you will have pairs of data values. For these statistical calculations,…
• 3. Calculate a regression line. Using your data results,you will be able to calculate a regression line. This is also called a line of best fit or
• 4. Calculate predicted values from the regression line. Using the equation of that line,you can calculate predicted y-values for each x-value in your
• How to solve standard error?
What is the Formula? To calculate standard error, you simply divide the standard deviation of a given sample by the square root of the total number of items in the sample. where, \$SE_ {bar {x}}\$ is the standard error of the mean, \$[_sigma_]\$ is the standard deviation of the sample and n is the number of items in sample.
How do you find the standard error?
To find the standard error of the mean, divide the standard deviation by the square root of the sample size: , where σ is the standard deviation of the original sampling distribution and N is the sample size.
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# How do you graph f(x)=x^2?
Aug 1, 2015
This is a vertical parabola - a sort of U shape - with vertex at $\left(0 , 0\right)$, axis of symmetry $x = 0$, passing through $\left(4 , - 2\right)$, $\left(1 , - 1\right)$, $\left(1 , 1\right)$ and $\left(2 , 4\right)$.
#### Explanation:
Ultimately, to graph (almost) any function $f \left(x\right)$ you can compute $f \left(x\right)$ for several values of $x$ to find some points $\left(x , f \left(x\right)\right)$ through which the graph passes.
In our case $f \left(- 2\right) = {\left(- 2\right)}^{2} = 4$ gives us $\left(- 2 , 4\right)$, $f \left(- 1\right) = 1$ gives us $\left(- 1 , 1\right)$, etc.
In the general case of quadratic functions of form $f \left(x\right) = a {x}^{2} + b x + c$ you can reformulate to find the vertex, axis of symmetry and where the parabola intersects the axes.
For example,
$f \left(x\right) = a {x}^{2} + b x + c = a {\left(x - \left(- \frac{b}{2 a}\right)\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$
$= a {\left(x - h\right)}^{2} + k$
with $h = - \frac{b}{2 a}$ and $k = c - {b}^{2} / \left(4 a\right)$
This is in vertex form: The vertex is at $\left(h , k\right)$.
In our particular example, $a = 1$, $b = c = 0$, so these formulae simplify to give $\left(h , k\right) = \left(0 , 0\right)$
graph{(y-x^2)((x+2)^2+(y-4)^2-0.02)((x+1)^2+(y-1)^2-0.02)(x^2+y^2-0.02)((x-1)^2+(y-1)^2-0.02)((x-2)^2+(y-4)^2-0.02)(y*0.00001+x) = 0 [-10.5, 9.5, -2.28, 7.72]}
|
# 102 in Words: How to Spell One Hundred Two and Solved Examples
The number 102 is written as One Hundered Two in English words. It is a natural number and is also a odd number. You can use One Hundered Two in various ways. For instance, if you completed 102 levels in a video game, you could say, “I’ve finished One Hundered Two levels!” Additionally, 102 possesses interesting mathematical properties, such as not being perfectly divisible by 2. To know more about 102 in words, you can read the given article.
## 102 in Words in English
One Hundered Two is the accepted way to express 102 in written English. We employ this form in both everyday communication and mathematical expressions. Here’s an example: “The test score required for admission is One Hundered Two.”
## How to Convert 102 in Words?
Similar to other numbers, we use the alphabet to spell out 102. Here’s a how you can spell 102 in words:
• Hundreds Place: Nine
• Tens Place: One
• Ones Place: (Empty) (Since 102 falls between 90 and 100, the ones place is typically not used)
## Facts About the Number 102
• Odd Number: As mentioned before, 102 is not divisible by 2, making it an odd number.
• Composite Number: Unlike 83 (a prime number), 102 is a composite number. This means it has more than two factors (1, 7, 13, and 102 itself).
## Solved Examples on 102 in Words
Let’s practice using One Hundered Two word problems:
Question 1: There are 102 marbles in a bag. If you distribute them equally among 3 friends, how many marbles will each friend receive?
Solution:
• Divide the total by the number of people: 102 marbles / 3 friends = 30 marbles/person
• Each friend will get 30 marbles.
Question 2: A rectangular bookshelf is 102 centimeters tall and 25 centimeters wide. What is the area of the bookshelf?
Solution:
• Length of the bookshelf: One Hundered Two centimeters
• Width of the bookshelf: 25 centimeters
• Area = Length x Width = 102 centimeters x 25 centimeters = 2275 square centimeters
• The area of the bookshelf is 2275 square centimeters.
## 35 to 102 Numbers in Words
You can find resources online that list all numbers 35 to 102 in words. For now, you can focus on practicing writing 83!
## FAQs
How do you spell 102 in English words?
102 is spelled as “One Hundered Two”.
What is the meaning of 102 in English?
102 means One Hundered Two, which is one more than ninety.
What is 102st in words?
102st is written as “Ninety-First”.
This was all about the “102 in Words”. For more such informative blogs, check out our Study Material Section, or you can learn more about us by visiting our Indian exams page
|
# How To Write A Equation In Slope Intercept Form
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
How To Write A Equation In Slope Intercept Form – There are many forms used to illustrate a linear equation the one most frequently encountered is the slope intercept form. It is possible to use the formula for the slope-intercept in order to identify a line equation when you have the straight line’s slope and the yintercept, which is the point’s y-coordinate at which the y-axis is intersected by the line. Learn more about this specific line equation form below.
## What Is The Slope Intercept Form?
There are three main forms of linear equations: the standard slope, slope-intercept and point-slope. While they all provide similar results when used in conjunction, you can obtain the information line produced quicker using this slope-intercept form. It is a form that, as the name suggests, this form uses a sloped line in which it is the “steepness” of the line determines its significance.
This formula is able to discover the slope of a straight line, the y-intercept (also known as the x-intercept), where you can apply different available formulas. The line equation in this specific formula is y = mx + b. The straight line’s slope is represented by “m”, while its y-intercept is represented through “b”. Every point on the straight line is represented with an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” must remain as variables.
## An Example of Applied Slope Intercept Form in Problems
For the everyday world in the real world, the slope-intercept form is often utilized to represent how an item or problem evolves over its course. The value of the vertical axis represents how the equation handles the extent of changes over the value given by the horizontal axis (typically time).
An easy example of the use of this formula is to discover how the population grows within a specific region in the course of time. In the event that the population in the area grows each year by a fixed amount, the point value of the horizontal axis will rise by a single point as each year passes, and the amount of vertically oriented axis will grow to represent the growing population by the set amount.
It is also possible to note the starting point of a challenge. The starting point is the y value in the yintercept. The Y-intercept is the point at which x equals zero. If we take the example of a previous problem the beginning point could be at the point when the population reading starts or when the time tracking begins , along with the related changes.
This is the point in the population that the population begins to be tracked by the researcher. Let’s assume that the researcher begins with the calculation or the measurement in 1995. The year 1995 would represent”the “base” year, and the x = 0 points would occur in the year 1995. This means that the 1995 population represents the “y”-intercept.
Linear equations that use straight-line formulas are almost always solved this way. The starting point is represented by the yintercept and the change rate is represented by the slope. The principal issue with an interceptor slope form generally lies in the interpretation of horizontal variables in particular when the variable is accorded to an exact year (or any other type or unit). The key to solving them is to make sure you comprehend the meaning of the variables.
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# Math help from the Learning Centre
This guide provides useful resources for a wide variety of math topics. It is targeted at students enrolled in a math course or any other Centennial course that requires math knowledge and skills.
## Introduction
An equation is a quadratic equation if it can be written in the form:
$$ax^2+bx+c=0$$,
where $$a,b,c$$ are known values, $$a \ne 1$$, and $$x$$ is some unknown variable. It has degree of 2 since the quadratic polynomial has degree 2 (i.e. highest exponent of all monomials in the polynomial is 2: $$x^2$$).
Recall the methods we can use to solve quadratic equations such as factoring or using the quadratic formula (review these on the Solving Quadratic Equations page). These only work for solving quadratic equations, but what if we wanted to solve equations of higher degrees (i.e. degree 3 or higher)?
## Solving Higher Degree Equations by Substitution
To solve higher degree equations, we can use substitution to convert the given equation into a quadratic equation, then solve the quadratic equation to determine the solutions to the original equation.
For example, suppose we have the equation:
$$ax^4+bx^2+c=0$$
If we let $$z=x^2$$, then substitute this into the original equation, we can rewrite it as:
$$a(x^2)^2+b(x^2)+c=0$$
$$\Rightarrow az^2+bz+c=0$$,
which is a quadratic equation that we can solve (by factoring or using the quadratic equation).
Then after solving, we can set the solutions for $$z$$ equal to $$x^2$$, then solve for $$x$$.
Tip: Don't forget to find the solutions of the original equation after solving the rewritten equation!
## Examples
1. Solve $$x^6+5x^3+6=0$$.
Solution:
Since we want to rewrite this equation as a quadratic equation, we use substitution by letting $$z=x^3$$. So the equation becomes:
$$(x^3)^2+5(x^3)+6=0$$
$$\Rightarrow z^2+5z+6=0$$
We can now solve this quadratic equation by factoring, giving us:
$$(z+2)(z+3)=0$$
$$\Rightarrow z=-2$$ or $$z=-3$$
Finally, we solve for $$x$$ using $$z=x^3$$:
$$z=-2=x^3$$ or $$z=-3=x^3$$
$$\Rightarrow$$ $$x=\sqrt[3]{-2}$$ or $$x=\sqrt[3]{-3}$$
We can also use this substitution method to solve other types of equations (not only ones involving polynomials), seen in this following example:
2. Solve $$\frac{x^4+x^2-20}{x^2+5}=0$$.
Solution:
We can rewrite the rational expression on the left side by substituting with $$z=x^2$$:
$$\frac{(x^2)^2+(x^2)-20}{(x^2)+5}=0$$
$$\Rightarrow \frac{z^2+z-20}{z+5}=0$$
Now we can factor the numerator to simplify the rational expression, making note of any variable restrictions (since the denominator cannot equal 0):
$$\frac{\cancel{(z+5)}(z-4)}{\cancel{(z+5)}}=0, \ z \ne 5$$
$$\Rightarrow z-4=0$$
$$\Rightarrow z=4$$
Finally, we solve for $$x$$ using $$z=x^2$$:
$$z=4=x^2$$
$$\Rightarrow$$ $$z=2$$ or $$z=-2$$
Note: Recall that there are two solutions to $$x^2=a$$ (for any value $$a>0$$, if there are no restrictions on the variable $$x$$):
$$x=+\sqrt{a}$$ or $$x=-\sqrt{a}$$
|
+0
# 1. What is the length of JK, to the nearest tenth of a millimeter?
0
837
1
1. What is the length of JK, to the nearest tenth of a millimeter?
Options: 4.5, 5.6, 8.3, 19.9 mm
2. What is the length of EF? Answer in decimal form and round only final answer to the nearest tenth.
Guest Apr 14, 2017
#1
+7324
+3
For these problems, we can use the law of cosines:
c2 = a2 + b2 — 2abcosC
1.
c = JK
a = 3
b = 5
C = 62º
Plug these values into the law of cosines.
(JK)2 = 32 + 52 - 2(3)(5)cos 62
(JK)2 = 9 + 25 - 30cos 62
(JK)2 = 34 - 30cos 62
JK = $$+\sqrt{34 - 30cos 62} \approx4.5\text{ mm}$$
2.
c = EF
a = 6
b = 11
C = 40º
Plug these values into the law of cosines.
(EF)2 = 62 + 112 - 2(6)(11)cos 40
(EF)2 = 36 + 121 - 132cos 40
(EF)2 = 157 - 132cos 40
EF = $$+\sqrt{157 - 132cos 40} \approx7.5\text{ ft}$$
hectictar Apr 14, 2017
#1
+7324
+3
For these problems, we can use the law of cosines:
c2 = a2 + b2 — 2abcosC
1.
c = JK
a = 3
b = 5
C = 62º
Plug these values into the law of cosines.
(JK)2 = 32 + 52 - 2(3)(5)cos 62
(JK)2 = 9 + 25 - 30cos 62
(JK)2 = 34 - 30cos 62
JK = $$+\sqrt{34 - 30cos 62} \approx4.5\text{ mm}$$
2.
c = EF
a = 6
b = 11
C = 40º
Plug these values into the law of cosines.
(EF)2 = 62 + 112 - 2(6)(11)cos 40
(EF)2 = 36 + 121 - 132cos 40
(EF)2 = 157 - 132cos 40
EF = $$+\sqrt{157 - 132cos 40} \approx7.5\text{ ft}$$
hectictar Apr 14, 2017
|
# Precalculus math help
One instrument that can be used is Precalculus math help. We will also look at some example problems and how to approach them.
## The Best Precalculus math help
Here, we debate how Precalculus math help can help students learn Algebra. This can be especially helpful when working with complex problems or when trying to learn a new concept. By seeing the step-by-step process that was used to solve the problem, students can better understand the material and develop their own problem-solving skills. In addition, a math solver with work can often be used to check answers that have been arrived at using other methods. This can help to ensure that the solution is correct and also help identify any mistakes that were made along the way. Whether you are a student who is struggling with math or a teacher who is looking for a way to check answers, a math solver with work can be an invaluable tool.
Solving integral equations is a process of finding a function that satisfies a given equation involving integrals. There are many methods that can be used to solve integral equations, each with its own advantages and disadvantages. The most common method is to use integration by substitution, which involves solving for the function in terms of the variables in the equation. However, this method can be difficult to apply in practice, especially if the equation is complex. Another popular method is to use Green's functions, which are special functions that can be used to solve certain types of differential equations. Green's functions can be very effective in solving integral equations, but they can be difficult to obtain in closed form. In general, there is no one best method for solving integral equations; the best approach depends on the specific equation and the tools that are available.
Solving for x logarithms can be difficult, but there are a few methods that can help. One method is to use the change of base formula. This formula states that if you have two values with the same base, you can set them equal to each other and solve for the unknown value. For example, if you have the equation log4(x)=log2(x), you can set the two equations equal to each other and solve for x. Another method is to use graphing calculator. Many graphing calculators have a built-in function that allows you to solve for x logarithms. Simply enter the equation into the calculator and press the "solve" button. The calculator will then give you the value of x. Finally, you can also use a table of logarithms to solve for x logarithms. To do this, simply find the values of x and y that are equal to each other and solve for x. Solving for x logarithms can be difficult, but with a little practice, it can be easy.
Lastly, solve the equation and check your work to make sure you have a correct answer. If you need more help, there are many resources available online and in print that can walk you through the steps of solving one step equations word problems. With a little practice, you will be able to solve them confidently and quickly.
A triangle solver calculator can be a useful tool for anyone who needs to find the measurements of a triangle. There are many different types of triangle calculators available, but they all essentially work in the same way. To use a triangle solver calculator, you simply enter the three sides of the triangle into the calculator. The calculator will then use a mathematical formula to calculate the measurements of the triangle. Triangle solver calculators can be helpful when you need to find the measurements of a Triangle that you do not have all the dimensions for. It can also be useful for checking your work if you have already calculated the dimensions of a Triangle yourself. Triangle solver calculators are available online and in many math textbooks.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Conversion between Degrees and Radians
## Convert between radians and degrees
Estimated16 minsto complete
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Practice Conversion between Degrees and Radians
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Introduction to the Unit Circle and Radian Measure
An odd-shaped house in Asia is built at a $135^\circ$ angle. How many radians is this angle equal to?
### Guidance
The unit circle is the circle centered at the origin with radius equal to one unit. This means that the distance from the origin to any point on the circle is equal to one unit.
Using the unit circle, we can define another unit of measure for angles, radians. Radian measure is based upon the circumference of the unit circle. The circumference of the unit circle is $2 \pi$ ( $2 \pi r$ , where $r=1$ ). So a full revolution, or $360^\circ$ , is equal to $2 \pi$ radians. Half a rotation, or $180^\circ$ is equal to $\pi$ radians.
One radian is equal to the measure of $\theta$ , the rotation required for the arc length intercepted by the angle to be equal to the radius of the circle. In other words the arc length is 1 unit for $\theta=1$ radian.
We can use the equality, $\pi=180^\circ$ to convert from degrees to radians and vice versa.
To convert from degrees to radians, multiply by $\frac{\pi}{180^\circ}$ .
To convert from radians to degrees, multiply by $\frac{180^\circ}{\pi}$ .
#### Example A
a. Convert $250^\circ$ to radians.
b. Convert $3 \pi$ to degrees.
Solution:
a. To convert from degrees to radians, multiply by $\frac{\pi}{180^\circ}$ . So, $\frac{250 \pi}{180}=\frac{25 \pi}{18}$ .
b. To convert from radians to degrees, multiply by $\frac{180^\circ}{\pi}$ . So, $3 \pi \times \frac{180^\circ}{\pi}=3 \times 180^\circ=540^\circ$ .
#### Example B
Find two angles, one positive and one negative, coterminal to $\frac{5 \pi}{3}$ and find its reference angle, in radians.
Solution: Since we are working in radians now we will add/subtract multiple of $2 \pi$ instead of $360^\circ$ . Before we can add, we must get a common denominator of 3 as shown below.
$\frac{5 \pi}{3} + 2 \pi = \frac{5 \pi}{3} + \frac{6 \pi}{3} = \frac{11 \pi}{3} \quad and \quad \frac{5 \pi}{3} - 2 \pi = \frac{5 \pi}{3} - \frac{6 \pi}{3} =- \frac{\pi}{3}$
Now, to find the reference angle, first determine in which quadrant $\frac{5 \pi}{3}$ lies. If we think of the measures of the angles on the axes in terms of $\pi$ and more specifically, in terms of $\frac{\pi}{3}$ , this task becomes a little easier.
Consider $\pi$ is equal to $\frac{3 \pi}{3}$ and $2 \pi$ is equal to $\frac{6 \pi}{3}$ as shown in the diagram. Now we can see that the terminal side of $\frac{5 \pi}{3}$ lies in the fourth quadrant and thus the reference angle will be:
$\frac{6 \pi}{3} - \frac{5 \pi}{3} = \frac{\pi}{3}$
#### Example C
Find two angles coterminal to $\frac{7 \pi}{6}$ , one positive and one negative, and find its reference angle, in radians.
Solution: This time we will add multiples of $2 \pi$ with a common denominator of 6, or $\frac{2 \pi}{1} \times \frac{6}{6} = \frac{12 \pi}{6}$ . For the positive angle, we add to get $\frac{7 \pi}{6} + \frac{12 \pi}{6} = \frac{19 \pi}{6}$ . For the negative angle, we subtract to get $\frac{7 \pi}{6} - \frac{12 \pi}{6} = \frac{5 \pi}{6}$ .
In this case $\pi$ is equal to $\frac{6 \pi}{6}$ and $2 \pi$ is equal to $\frac{12 \pi}{6}$ as shown in the diagram. Now we can see that the terminal side of $\frac{7 \pi}{6}$ lies in the third quadrant and thus the reference angle will be:
$\frac{7 \pi}{6} - \frac{6 \pi}{6} = \frac{\pi}{6}$
Concept Problem Revisit
To convert from degrees to radians, multiply by $\frac{\pi}{180^\circ}$ . So, $\frac{135 \pi}{180}=\frac{3 \pi}{4}$ .
### Guided Practice
1. Convert the following angle measures from degrees to radians.
a. $-45^\circ$
b. $120^\circ$
c. $330^\circ$
2. Convert the following angle measures from radians to degrees.
a. $\frac{5 \pi}{6}$
b. $\frac{13 \pi}{4}$
c. $-\frac{5 \pi}{2}$
3. Find two coterminal angles to $\frac{11 \pi}{4}$ , one positive and one negative, and its reference angle.
1. a. $-45^\circ \times \frac{\pi}{180^\circ}=-\frac{\pi}{4}$
b. $120^\circ \times \frac{\pi}{180^\circ}=\frac{2 \pi}{3}$
c. $330^\circ \times \frac{\pi}{180^\circ}=\frac{11 \pi}{6}$
2. a. $\frac{5 \pi}{6} \times \frac{180^\circ}{\pi}=150^\circ$
b. $\frac{13 \pi}{4} \times \frac{180^\circ}{\pi}=585^\circ$
c. $-\frac{5 \pi}{2} \times \frac{180^\circ}{\pi}=-450^\circ$
3. There are many possible coterminal angles, here are some possibilities:
positive coterminal angle: $\frac{11 \pi}{4} + \frac{8 \pi}{4} = \frac{19 \pi}{4}$ or $\frac{11 \pi}{4} - \frac{8 \pi}{4} = \frac{3 \pi}{4}$ ,
negative coterminal angle: $\frac{11 \pi}{4} - \frac{16 \pi}{4} = -\frac{5 \pi}{4}$ or $\frac{11 \pi}{4} - \frac{24 \pi}{4} = -\frac{13 \pi}{4}$
Using the coterminal angle, $\frac{3 \pi}{4}$ , which is $\frac{\pi}{4}$ from $\frac{4 \pi}{4}$ . So the terminal side lies in the second quadrant and the reference angle is $\frac{\pi}{4}$ .
### Vocabulary
When the arc length formed by a central angle is equal in length to the radius of the circle.
### Practice
For problems 1-5, convert the angle from degrees to radians. Leave answers in terms of $\pi$ .
1. $135^\circ$
2. $240^\circ$
3. $-330^\circ$
4. $450^\circ$
5. $-315^\circ$
For problems 6-10, convert the angle measure from radians to degrees.
1. $\frac{7 \pi}{3}$
2. $-\frac{13 \pi}{6}$
3. $\frac{9 \pi}{2}$
4. $-\frac{3 \pi}{4}$
5. $\frac{5 \pi}{6}$
For problems 11-15, find two coterminal angles (one positive, one negative) and the reference angle for each angle in radians.
1. $\frac{8 \pi}{3}$
2. $\frac{11 \pi}{4}$
3. $-\frac{\pi}{6}$
4. $\frac{4 \pi}{3}$
5. $-\frac{17 \pi}{6}$
### Vocabulary Language: English
subtended arc
subtended arc
A subtended arc is the part of the circle in between the two rays that make the central angle.
|
# Find the area of the blades of the magnetic compass shown in figure given below:
Question:
Find the area of the blades of the magnetic compass shown in figure given below:
Solution:
Area of the blades of magnetic compass = Area of triangle ADB + Area of triangle CDB
Now, for the area of triangle ADB
Perimeter = 2s = AD + DB + BA
2s = 5 cm + 1 cm + 5 cm
s = 5.5 cm
By using Heron's Formula,
Area of the triangle DEF $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{5.5 \times(0.5) \times(4.5) \times(0.5)}$
$=2.49 \mathrm{~cm}^{2}$
Also, area of triangle ADB = Area of triangle CDB
Therefore area of the blades of the magnetic compass = 2 × area of triangle ADB
Area of the blades of the magnetic compass = 2 × 2.49
Area of the blades of the magnetic compass $=4.98 \mathrm{~cm}^{2}$
|
# Algebra II: Fractional Expressions
On this page we hope to clear up any problems you might have with fractional expressions and equations. Fractions are used continually in math, and even with the advent of powerful calculators, are used extensively in advanced math courses. Click any of the links below or scroll down to start understanding fractions better!
Multiplication.
Division.
Lowest Common Multiple.
Complex Fractions.
Division of Polynomials.
Fractional Equations.
Variation (direct and indirect).
## Multiplication of Fractions
The multiplication of fractions is rather straightforward. The Fraction Multiplication Theorem says for any fractional expressions, (a/b) and (c/d) where b and d do not equal 0, (a/b)(c/d) = (a * c)/(b * d). Example:
```1. Problem: x + 3 x3
---- * -----
y - 4 y + 5
Solution: Using the Fraction Multiplication
Theorem, multiply the numerator
of the first fraction by the numer-
ator of the second fraction and the
denominator of the first fraction
by the denominator of the second
fraction.
(x + 3)x3
--------------
(y - 4)(y + 5)
```
## Divison of Fractions
The division of fractions is also not very complicated. The Fraction Division Theorem says for any fractional expressions, (a/b) and (c/d) where b, c, and d do not equal 0, (a/b)/(c/d) = (a/b)(d/c). In other words, you divide by multiplying by a reciprocal. Example:
```1. Problem: x - 2 x + 5
----- / -----
x + 1 x - 3
Solution: Utilizing the Fraction Division
Theorem, we take the reciprocal of the
divisor and multiply.
x - 2 x - 3
----- * -----
x + 1 x + 5
Use the Fraction Multiplication
Theorem to multiply the problem out.
x2 - 5x + 6
-----------
x2 + 6x + 5
```
## Lowest Common Multiple
Finding the LCM, or Lowest Common Multiple is a necessary skill if you want to be able to add to and subtract fractions from other numbers. However, finding the LCM is usually covered in most elementary algebra (Algebra I) courses. This custom has been followed on this site, so you can click here to get a better understanding of Lowest Common Multiples.
## Addition and Subtraction of Fractions
Adding and subtracting fractions from other numbers is a very useful skill since you encounter fractions so often. This skill is usually covered in elementary algebra (Algebra I) courses. We have followed that custom on this site, so click here to get a better understanding of addition and subtraction when fractions are involved.
## Complex Fractions
Complex fractions, or fractions that have a fraction for its numerator, denominator, or both, are encountered less often than normal fractions, but are just as easy to solve because they, like normal fractions, are division problems. The most common way to simplify complex fractions is to simplify the numerator and denominator, treat it as a division problem, and then simplify as usual. Example:
```1. Simplify: 1
1 + -
x
-----
1
1 - -
x2
Solution: Combine the numerator into one
fraction and then do the same to the
denominator. (If you do not know how
x 1 x + 1
- + - -----
x x x
----- = ------
x2 - 1 x2 - 1
- - - ------
x2 x2 x2
Now you have a normal division problem
with fractional expressions.
x + 1 x2 - 1
----- / ------
x x2
Using the Fractional Division
Theorem, take the reciprocal of the
divisor and multiply.
x + 1 x2
----- * ------
x x2 - 1
After multiplication, you have the
following expression:
(x + 1)x2
---------------
x(x + 1)(x - 1)
Now, cancel out any factors that are
in both the denominator and the
numerator to simplify.
(They both have (x + 1) and
x as a factor. Cancel them
out.)
You are left with the following:
x
-----
x - 1
```
## Divison of Polynomials
Always keep in mind that a fraction bar is the same as a division sign. Division by a monomial can be done by rewriting the problem as a fraction. Example:
```1. Divide: 12x3 + 8x2 + x + 4 by 4x
Solution: Rewrite the problem as a fraction.
12x3 + 8x2 + x + 4
-----------------
4x
This expression shows that each
term in the numerator is divided
by 4x. Rewrite the prob-
lem to show that.
12x3 8x2 x 4
---- + --- + -- + --
4x 4x 4x 4x
Now do the four divisions in-
dicated by each fraction.
3x2 + 2x + (1/4) + (1/x)
```
When the divisor is not a monomial, you have to use a procedure that resembles long division as you learned way back in 5th grade arithmetic! Example:
```2. Divide: x2 + 5x + 6 by x + 3
Solution: Write the problem as a long
division problem.
___________
x + 3 )x2 + 5x + 6
Divide first term by first term —
(x2/x) = x.
x__________
x + 3 )x2 + 5x + 6
x2 + 3x Multiply x by divisor.
-------
2x Subtract.
Bring down the next term and repeat
the process.
x_+_2______
x + 3 )x2 + 5x + 6
x2 + 3x
-----------
2x + 6
2x + 6
------
0
The quotient is x + 2.
```
## Solving Fractional Expressions
A fractional equation is an equation that contains one or more fractional expressions. To solve a fractional equation you multiply each side of the equation by the LCM of all the denominators. Example:
```1. Solve: 2 5 1
- - - = -
3 6 x
Solution: The LCM of all denominators is
6x. Multiply each side of the
equation by 6x.
6x((2/3) - (5/6)) = 6x(1/x)
Use the distributive law of multi-
plication, which says a(b + c) =
ab + ac, to rewrite the
equation.
6x(2/3) - 6x(5/6) = 6x(1/x)
Multiply each group of terms
together.
(12x/3) - 30x/6 = 6x/x
Perform each of the indicated
divisions.
4x - 5x = 6
Solve for x.
Combine like terms.
-x = 6
Multiply each side by -1.
x = -6
```
## Variation
A worker earns \$10.00/hr. In one hour, \$10.00 is earned. In two hours, \$20.00 is earned, and so forth. This gives us a set of ordered pairs which all have the same ratio — (1, 10), (2, 20), (3, 30), . . . This situation, where there are pairs of numbers in which the ratio is constant, we have direct variation. Whenever a situation, such as the one described above gives rise to the function f(x) = kx, where k is a positive constant, there is direct variation. Example:
```1. Problem: Find the variation constant k
and an equation of variation where y
varies directly as x, and where
y = 34 and x = 7.
Solution: The problem tells us (7, 34)
is a solution of the direct variation
equation, y = kx.
Plug in the givens for y and x
and solve for l.
34 = k(7)
(34/7) = k
The constant of variation is (34/7).
```
The above problem could be used in real life if someone made \$4.857/hr. and they wanted to know how much they would make if they worked for 7 hours. Using 7 for x would tell us they would make \$34.00.
Sometimes, things do not vary directly. An example of this is a bus that travels 20 kilometers in one hour at a speed of 20 km/hr. At 40 km/hr. it would only take half of an hour to go 20 kilometers. This gives us a pair of ordered pairs — (20, 1), (40, .5), . . . These numbers, whose product is constant, vary indirectly. Whenever a situation such as the one described above gives rise to the function f(x) = (k/x), where k is a positive constant, there is inverse variation. Example:
```2. Problem: Find the variation constant and
then an equation of variation where
y varies inversely as x, and
y = 32 when x = (1/5).
Solution: We know that (.2, 32) is
a solution of the inverse variation
equation.
Plug in the given information and
solve for k.
32 = (k/.2)
6.4 = k
The equation of variation is the
following:
y = (6.4/x)
```
This can be applied to real life, too. If you were driving 100 km/hr., you would go 6.4 kilometers in 3.84 minutes.
Take the quiz on fractional expressions. The quiz is very useful for either review or to see if you've really got the topic down.
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Math for Morons Like Us -- Algebra II: Fractional Expressions
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7.2: Commutative and Associative Properties (Part 1)
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Skills to Develop
• Use the commutative and associative properties
• Evaluate expressions using the commutative and associative properties
• Simplify expressions using the commutative and associative properties
be prepared!
Before you get started, take this readiness quiz.
1. Simplify: 7y + 2 + y + 13. If you missed this problem, review Example 2.22.
2. Multiply: $$\frac{2}{3} \cdot 18$$. If you missed this problem, review Example 4.28.
3. Find the opposite of 15. If you missed this problem, review Example 3.3.
In the next few sections, we will take a look at the properties of real numbers. Many of these properties will describe things you already know, but it will help to give names to the properties and define them formally. This way we’ll be able to refer to them and use them as we solve equations in the next chapter.
Use the Commutative and Associative Properties
Think about adding two numbers, such as 5 and 3.
$$\begin{split} 5 &+ 3 \qquad 3 + 5 \\ &\; 8 \qquad \qquad 8 \end{split}$$
The results are the same. 5 + 3 = 3 + 5
Notice, the order in which we add does not matter. The same is true when multiplying 5 and 3.
$$\begin{split} 5 &\cdot 3 \qquad \; 3 \cdot 5 \\ & 15 \qquad \quad 15 \end{split}$$
Again, the results are the same! 5 • 3 = 3 • 5. The order in which we multiply does not matter. These examples illustrate the commutative properties of addition and multiplication.
Definition: Commutative Properties
Commutative Property of Addition: if a and b are real numbers, then a + b = b + a
Commutative Property of Multiplication: if a and b are real numbers, then a • b = b • a
The commutative properties have to do with order. If you change the order of the numbers when adding or multiplying, the result is the same.
Example 7.5:
Use the commutative properties to rewrite the following expressions: (a) −1 + 3 = _____ (b) 4 • 9 = _____
Solution
(a) −1 + 3 = _____
Use the commutative property of addition to change the order. −1 + 3 = 3 + (−1)
(b) 4 • 9 = _____
Use the commutative property of multiplication to change the order. 4 • 9 = 9 • 4
Exercise 7.9:
Use the commutative properties to rewrite the following expressions: (a) −4 + 7 = _____ (b) 6 • 12 = _____
Exercise 7.10:
Use the commutative properties to rewrite the following expressions: (a) 14 + (-2) = _____ (b) 3(-5) = _____
What about subtraction? Does order matter when we subtract numbers? Does 7 − 3 give the same result as 3 − 7?
$$\begin{split} 7 &- 3 \qquad 3 - 7 \\ &\; 4 \qquad \quad -4 \\ & \quad 4 \neq -4 \end{split}$$
The results are not the same. 7 − 3 ≠ 3 − 7
Since changing the order of the subtraction did not give the same result, we can say that subtraction is not commutative. Let’s see what happens when we divide two numbers. Is division commutative?
$$\begin{split} 12 &\div 4 \qquad 4 \div 12 \\ & \frac{12}{4} \qquad \quad \frac{4}{12} \\ &\; 3 \qquad \qquad \frac{1}{3} \\ &\quad \; 3 \neq \frac{1}{3} \end{split}$$
The results are not the same. So 12 ÷ 4 ≠ 4 ÷ 12
Since changing the order of the division did not give the same result, division is not commutative.
Addition and multiplication are commutative. Subtraction and division are not commutative.
Suppose you were asked to simplify this expression.
$$7 + 8 + 2$$
How would you do it and what would your answer be?
Some people would think 7 + 8 is 15 and then 15 + 2 is 17. Others might start with 8 + 2 makes 10 and then 7 + 10 makes 17.
Both ways give the same result, as shown in Figure 7.3. (Remember that parentheses are grouping symbols that indicate which operations should be done first.)
Figure 7.3
When adding three numbers, changing the grouping of the numbers does not change the result. This is known as the Associative Property of Addition.
The same principle holds true for multiplication as well. Suppose we want to find the value of the following expression:
$$5 \cdot \frac{1}{3} \cdot 3$$
Changing the grouping of the numbers gives the same result, as shown in Figure 7.4.
Figure 7.4
When multiplying three numbers, changing the grouping of the numbers does not change the result. This is known as the Associative Property of Multiplication.
If we multiply three numbers, changing the grouping does not affect the product. You probably know this, but the terminology may be new to you. These examples illustrate the Associative Properties.
Definition: Associative Properties
Associative Property of Addition: if a, b, and c are real numbers, then (a + b) + c = a + (b + c)
Associative Property of Multiplication: if a, b, and c are real numbers, then (a • b) • c = a • (b • c)
Example 7.6:
Use the associative properties to rewrite the following: (a) (3 + 0.6) + 0.4 = __________ (b) $$\left(−4 \cdot \dfrac{2}{5}\right) \cdot 15$$ = __________
Solution
(a) (3 + 0.6) + 0.4 = __________
Change the grouping. (3 + 0.6) + 0.4 = 3 + (0.6 + 0.4)
Notice that 0.6 + 0.4 is 1, so the addition will be easier if we group as shown on the right.
(b) $$\left(−4 \cdot \dfrac{2}{5}\right) \cdot 15$$ = __________
Change the grouping. (3 + 0.6) + 0.4 = 3 + (0.6 + 0.4)
Notice that $$\frac{2}{5} \cdot 15$$ is 6. The multiplication will be easier if we group as shown on the right.
Exercise 7.11:
Use the associative properties to rewrite the following: (a) (1 + 0.7) + 0.3 = __________ (b) (−9 • 8) • $$\frac{3}{4}$$ = __________
Exercise 7.12:
Use the associative properties to rewrite the following: (a) (4 + 0.6) + 0.4 = __________ (b) (−2 • 12) • $$\frac{5}{6}$$ = __________
Besides using the associative properties to make calculations easier, we will often use it to simplify expressions with variables.
Example 7.7:
Use the Associative Property of Multiplication to simplify: 6(3x).
Solution
Change the grouping. (6 • 3)x Multiply in the parentheses. 18x
Notice that we can multiply 6 • 3, but we could not multiply 3 • x without having a value for x.
Exercise 7.13:
Use the Associative Property of Multiplication to simplify the given expression: 8(4x).
Exercise 7.14:
Use the Associative Property of Multiplication to simplify the given expression: −9(7y).
Evaluate Expressions using the Commutative and Associative Properties
The commutative and associative properties can make it easier to evaluate some algebraic expressions. Since order does not matter when adding or multiplying three or more terms, we can rearrange and re-group terms to make our work easier, as the next several examples illustrate.
Example 7.8:
Evaluate each expression when x = $$\frac{7}{8}$$. (a) x + 0.37 + (− x) (b) x + (− x) + 0.37
Solution
(a) x + 0.37 + (− x)
Substitute $$\frac{7}{8}$$ for x. $$\textcolor{red}{\frac{7}{8}} + 0.37 + \left(- \textcolor{red}{\dfrac{7}{8}}\right)$$ Convert fractions to decimals. 0.875 + 0.37 + (-0.875) Add left to right. 1.245 - 0.875 Subtract. 0.37
(b) x + (− x) + 0.37
Substitute $$\frac{7}{8}$$ for x. $$\textcolor{red}{\frac{7}{8}} + \left(- \textcolor{red}{\dfrac{7}{8}}\right) + 0.37$$ Add opposites first. 0.37
What was the difference between part (a) and part (b)? Only the order changed. By the Commutative Property of Addition, x + 0.37 + (− x) = x + (− x) + 0.37. But wasn’t part (b) much easier?
Exercise 7.15:
Evaluate each expression when y = $$\frac{3}{8}$$: (a) y + 0.84 + (− y) (b) y + (− y) + 0.84.
Exercise 7.16:
Evaluate each expression when f = $$\frac{17}{20}$$: (a) f + 0.975 + (− f) (b) f + (− f) + 0.975.
Let’s do one more, this time with multiplication.
Example 7.9:
Evaluate each expression when n = 17. (a) $$\frac{4}{3} \left(\dfrac{3}{4} n\right)$$ (b) $$\left(\dfrac{4}{3} \cdot \dfrac{3}{4}\right) n$$
Solution
(a) $$\frac{4}{3} \left(\dfrac{3}{4} n\right)$$
Substitute 17 for n. $$\frac{4}{3} \left(\dfrac{3}{4} \cdot \textcolor{red}{17} \right)$$ Multiply in the parentheses first. $$\frac{4}{3} \left(\dfrac{51}{4}\right)$$ Multiply again. $$17$$
(b) $$\left(\dfrac{4}{3} \cdot \dfrac{3}{4}\right) n$$
Substitute 17 for n. $$\left(\dfrac{4}{3} \cdot \dfrac{3}{4}\right) \textcolor{red}{\cdot 17}$$ Multiply. The product of reciprocals is 1. $$(1) \cdot 17$$ Multiply again. $$17$$
What was the difference between part (a) and part (b) here? Only the grouping changed. By the Associative Property of Multiplication, $$\frac{4}{3} \left(\dfrac{3}{4} n\right) = \left(\dfrac{4}{3} \cdot \dfrac{3}{4}\right) n$$. By carefully choosing how to group the factors, we can make the work easier.
Exercise 7.17:
Evaluate each expression when p = 24. (a) $$\frac{5}{9} \left(\dfrac{9}{5} p\right)$$ (b) $$\left(\dfrac{5}{9} \cdot \dfrac{9}{5}\right) p$$
Exercise 7.18:
Evaluate each expression when q = 15. (a) $$\frac{7}{11} \left(\dfrac{11}{7} q\right)$$ (b) $$\left(\dfrac{7}{11} \cdot \dfrac{11}{7}\right) q$$
|
# NCERT Class 10 Math Chapter 2 Important Questions Answer – Polynomials
Class 10 Chapter Polynomials Subject Math Category Important Question Answer
## Class 10 Math Chapter 2 Important Question Answer
Q1. Find a quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively. Most Important
Ans. Let the quadratic polynomial be ax2 + bx + c, and its zeroes be 𝛂 and β.
We have
𝛂 + β = -3 = ,
𝛂 β = 2 =
If a = 1, then b = 3 and c = 2.
So, one quadratic polynomial which fits the given conditions is x2 + 3x + 2.
Q2. Find the product of zeroes of quadratic polynomial x2 + 7x + 10. Most Important
Ans. Comparing x2 + 7x + 10 with ax2 + bx + c, we get a = 1, b = 7, c = 10,
As we know, Product of roots =
= = 10
Q3. Find a quadratic polynomial with the given number as the sum and product of its zeroes respectively are 5 and 3. Most Important
Ans. Let the quadratic polynomial be ax2 + bx + c, and its zeroes be 𝛂 and β.
We have
𝛂 + β = 5 = ,
𝛂 β = 3 =
If a = 1, then b = -5 and c = 3.
So, one quadratic polynomial which fits the given conditions is x2 – 5x + 3.
Q4. Find a quadratic polynomial whose zeros are -3 and 5. Most Important
Ans. Let the quadratic polynomial be ax2 + bx + c, and its zeroes be 𝛂 and β.
We have
𝛂 + β = -3 = ,
𝛂 β = 5 =
If a = 1, then b = 3 and c = 5.
So, one quadratic polynomial which fits the given conditions is x2 + 3x + 5.
Q5. Product of roots of the quadratic polynomial 6x2 – 7x – 3 is ____________.
Ans. Comparing 6x2 – 7x – 3 with ax2 + bx + c, we get a = 6, b = -7, c = -3,
As we know, Product of roots =
= =
Q6. Sum of the roots of the quadratic polynomnial 7x2 – 3x + 1 is __________.
Ans. Comparing 7x2 – 3x + 1 with ax2 + bx + c, we get a = 7, b = -3, c = 1,
As we know , Sum of roots =
= = 3
Q7. Find the sum of zeroes of quadratic polynomial x2 – 2x – 8. Most Important
Ans. Comparing x2 – 2x – 8 with ax2 + bx + c, we get a = 1, b = -2, c = -8
As we know, sum of zeroes =
= = 2
Also Read Class 10 Math NCERT Solution Also Read Class 10 Important Questions [Latest]
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Complex numbers: Definition. Binomial and Cartesian form
# Complex numbers
Now I’m going to explain what complex numbers are: what parts make them up and in what forms they can be expressed.
## Definition of complex numbers
I could tell you that the set of complex numbers contains the real numbers, they are represented by the symbol C and they include the roots of all the polynomials, but what does this mean? Why do we need complex numbers?
In other words, we remember that with real numbers we cannot solve the roots of pairs of negative numbers:
Therefore, we need the complex numbers to be able to solve with the negative roots of even index.
### Imaginary unit
The number i, which is equal to the root of minus one, is called the imaginary unit:
With this number, we can give a solution when we have negative roots:
According to the properties of the roots, we can put the negative root as that same positive root, multiplied by the root of minus one.
Finally, the positive root already has a solution in the set of real numbers and the root of minus one is replaced by the number i.
## Form of complex numbers
### Binomic form
A complex number Z (not to be confused with C, which is the set to which they belong) can be represented in the form:
Belonging a and b to the set of real numbers.
This way of writing complex numbers corresponds to the binomial form, which has two parts:
• a = Actual part
• b = Imaginary part
These are examples of complex numbers in binomial form:
If the real part of a complex number is 0, that number is pure imaginary, since it only has an imaginary part:
The number i is a pure imaginary number.
box type=”info”]The binomial form is used in physics when operating with vectors, where the imaginary part is represented by the letter j, instead of the letter i.[/box].
On the other hand, if it has no imaginary part, we are talking about a pure real number, which is neither more nor less than a real number:
### Cartesian form
Another way to represent complex numbers is the Cartesian form:
As in the binomial form, a and b belong to the set of real numbers y:
• a = Actual part
• b = Imaginary part
The above examples represented in Cartesian form are expressed as follows:
## Conjugate of a complex number
The conjugate of a complex number is used in the division of complex numbers. It is represented by a slash above the number. To obtain the conjugate of a complex number it is necessary to change the sign of the imaginary part (the central sign):
For example:
## Oposite of a complex number
To obtain the opposite of a complex number, it is necessary to change the sign of both the real part and the imaginary part, therefore the opposite of Z, will be -Z:
For example:
As a conclusion, the most important thing you should know is that they are composed of a real part and an imaginary part, that the number i is equal to the root of -1 and therefore, the number i squared is equal to -1 and how to obtain the conjugate of a complex number.
|
# AP Statistics Curriculum 2007 IntroVar
General Advance-Placement (AP) Statistics Curriculum - Introduction to Statistics
## The Nature of Data & Variation
No matter how controlled are the environment, the protocol or the design, virtually any repeated measurement, observation, experiment, trial, study or survey is bounded to generate data that varies because of intrinsic (internal to the system) or extrinsic (due to the ambient environment) effects.
For example, the UCLA's study of Alzheimer’s disease* analyzed the data of 31 Mild Cognitive Impairment (MCI) and 34 probable Alzheimer’s disease (AD) patients. The investigators made every attempt to control as many variables as possible. Yet, the demographic information they collected from the outcomes of the subjects contained unavoidable variation. The same study found variation in the MMSE cognitive scores even in the same subject. The table below shows the demographic characteristics for the subjects and patients included in this study, where the following notation is used M: male; F: female; W: white; AA: African American; A: Asian:
Variable Alzheimer’s disease MCI Test statistics Test score P-value Age (years) 76.2 (8.3) range 52–89 73.7 (7.4) range 57–84 Student’s T to = 1.284 p=0.21 Gender (M:F) 15:19 15:16 Proportion zo = − 0.345 p=0.733 Education (years) 14.0 (2.1) range 12–19 16.23 (2.7) range 12–20 Wilcoxon rank sum wo = 773.0 p<0.001 Race (W:AA:A) 29:1:4 26:2:3 $\chi_{(df=2)}^2$ $\chi_{(df=2)}^2=1.18$ 0.55 MMSE 20.9 (6.3) range 4–29 28.2 (1.6) range 23–30 Wilcoxon rank-sum wo = 977.5 p<0.001
## Approach
Models and strategies for solving problems and understanding data and inferences.
• Once we accept that all natural phenomena are inherently variant and there are no completely deterministic processes, we need to look for models and techniques that allow us to study such acquired data in the presence of variation, uncertainty and chance.
• Statistics is the data science that investigates natural processes and allows us to quantify variation to make population inferences based on limited observations.
## Model Validation
Checking/affirming underlying assumptions.
• Each model or technique for data exploration, analysis and understanding relies on a set of assumptions, which always need to be validated before the model or analysis tool is employed to study real data (observations or measurements that are perceived or detected by the investigator).
• Such prior model conjectures or presumptions could take the form of mathematical constraints about the properties of the underlying process, restrictions on the study design or demands on the data acquisition protocol.
• Common assumptions include (statistical) independence of the measurements, specific limitations on the shape of the observed distribution, restrictions on the parameters of the processes being studied, etc.
## Datasets
There are a number of large, natural, useful and demonstrative datasets that are provided as part of this statistics EBook. Many of these data collections are intentionally selected to be large and complex. This choice is driven by the need of emphasizing the symbiosis between driving challenges, statistical concepts, mathematical derivations and the use of technology to solve relevant research problems.
## Examples
Computer simulations and real observed data.
## Hands-on activities
Step-by-step practice problems.
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The concept of fractions, though a simple one, can often cause some confusions. Having not solved sufficient questions based on this simple concept, students often tend to confuse the problems. Considering the basic nature of this topic, it is important we go through the basic concepts so that you are not caught on the wrong foot while solving questions.
So what are Fractions?
Definition
Technically, a fraction is defined as part of the whole. The most common example of a fraction that comes to mind is half. When we say give me half of something, we are essentially demanding ½ part of it, in other words, ½ is the fractional representation for half.
Fractions are nothing else but the numerator divided by denominator, i.e. they occur in the form X/Y where X is the numerator and Y is the non-zero denominator.
So what does the numerator and denominator in a fraction signify?
The numerator represents how many parts of that whole are being considered. To remember simply, numerator is the top number of the fraction that represents the numbers of parts that are to be chosen.
The denominator represents the total number of parts created from the whole. In other words, it is the bottom number representing the total number of parts created.
Example of Fractions: ½, 2/3, 3/4, are the numbers which are in the form of x/y, where y has to be a non-zero number.
Types of Fractions:
Proper Fraction: When Numerator of the fraction is less than the Denominator, then the fraction is called as proper fraction.
For example: 2/3, 4/5, 6/7, etc
Improper fraction: When Numerator of the fraction is more than Denominator, then the fraction is called as improper fraction. For example: 5/3, 7/5, 19/7, etc
Mixed fraction: When a natural number combines with a fraction, that fraction is called a mixed fraction.
For Example: 21/2 ,34/5 Â ,etc. In other words, the mixed fractions are improper fractions.
Properties of fractions
Let us explore some useful properties of fractions that you can use to solve questions.
Property 1: If we multiply the numerator and denominator by same quantity, the basic value of fraction will never change.
For example: 4/5 x 5/5 = 20/25 = 4/5
Property 2: If there are two fractions a/b and c/d then a/b=c/d when ad=bc.
For example,3/4 = 12/16 because 3 x 16 = 4 x 12
Property 3: A fraction with zero as the denominator is not defined.
Property 4: If the numerator of the fraction is zero, then the fraction equals to zero.
Property 5: If the numerator and denominator of the fraction are equal, then the fraction is equal to one.
Applications of Fractions
Given below are some of the basic applications of fractions:
1. Fraction helps us determine the part of any number
Âľ part of 56 = Âľ x 56 = 42
4/5 part of 90 = 4/5 x 90 =72
1. You can be asked to represent a number in the form of fraction.
For example, you can be asked to represent 15 as a fraction of 450.
This can be done as follows:
15/450 = 1/30
In the above example, it can be easily seen that 15 is our numerator and 450 is our denominator
1. Always remember that the major quantity from which we have to extract something is the denominator.
For example, when we say 4/5, we are essentially extracting four parts out of five.
1. Extending the above concept, the quantity that is extracted is our numerator
For example: 15/450 = 1/30
15 is the numerator because we have extracted 15 from 450 and the denominator is 450 because 15 is extracted from 450. So, we can say that 1/30th part of 450 is 15.
1. Converting percentages into equivalent fractions makes our calculation easy.
For example:
If you remember 37.5% = Â and you are asked to find 37.5% of 24 then you can write
37.5% of 24 =X24 = 9
Remember the following equivalent fractions for given percentages:
This percentage to fraction conversion and vice-versa has a huge application in DI sets. Therefore, make sure that you are thorough with this table.
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# The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the field.
Question:
The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the field. Also, find the cost of ploughing the field at ₹ 5 per m2.
Solution:
Let the sides of the triangle be 25x m, 17x m and 12x m.
We know:
Perimeter = Sum of all sides
or, 540 = 25x + 17x + 12x
or, 54x = 540
or, x = 10
Thus, we obtain the sides of the triangle.
25×">××10 = 250 m
17×">××10 = 170 m
12×">××10 = 120 m
Now,
Let:
$a=250 \mathrm{~m}, b=170 \mathrm{~m}$ and $c=120 \mathrm{~m}$
$\therefore s=\frac{540}{2}=270 \mathrm{~m}$
By Heron's formula, we have:
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{270(270-250)(270-170)(270-120)}$
$=\sqrt{270 \times 20 \times 100 \times 150}$
$=\sqrt{30 \times 3 \times 3 \times 20 \times 20 \times 5 \times 30 \times 5}$
$=30 \times 3 \times 20 \times 5$
$=9000 \mathrm{~m}^{2}$
Cost of ploughing 1 m2 field = Rs 5
Cost of ploughing $9000 \mathrm{~m}^{2}$ field $=5 \times 9000=\operatorname{Rs} 45000$.
|
# Fractions and Factors
## The Importance of Understanding Fractions and Factors in Mathematics
When we think about numbers, we often refer to natural numbers and integers as whole numbers. However, fractions, which represent parts of a whole, are also essential in mathematics. Let's take a closer look at fractions and factors, as well as their applications in problem-solving and everyday situations.
## Defining Fractions and Factors
First, we must understand what fractions are. A fraction is a numerical value that shows a ratio of a part to a whole. In mathematics, fractions are a type of rational number and can be represented as a set of numbers, with the numerator (top number) over the denominator (bottom number), such as . To visualize this, imagine dividing a bar of chocolate into two equal parts. The value of each part would be , with 1 as the numerator and 2 as the denominator.
## Simplifying Fractions Using Factors
Next, we can use factors to simplify fractions. Factors are numbers that divide evenly into a given integer, resulting in a whole number without any remainder. For example, the factors of 10 are 1, 2, 5, and 10, as 10 can be divided by each of these numbers with no remainder. Moreover, numbers that are only divisible by 1 and themselves (such as 2, 3, 5, etc.) are called prime numbers. Notably, 1 is the only number that has just one factor, itself.
## Finding Prime Factor Decomposition
Prime factor decomposition is the process of breaking down an integer into its prime factors. These are factors that are also prime numbers, and we can use a factor tree to visualize and find them. For instance, the prime factor decomposition of 24 would be 2 x 2 x 2 x 3, as we can break down 24 into 2 x 12, then 2 x 2 x 6, and finally 2 x 2 x 2 x 3. This method helps us find the simplest form of a number, which is useful in solving equations and dividing quantities evenly.
## The Relevance of Fractions and Factors in Daily Life
Fractions and factors may seem daunting, but their understanding can make math easier and more applicable to our daily lives. For example, if we want to share a pizza equally with friends, we can use fractions and factors to ensure everyone gets an equal number of slices. Overall, by breaking down numbers into their simplest forms, we can make math less intimidating and more accessible to everyone.
## Finding the Greatest Common Divisor: Understanding Fractions and Factors
Finding the Greatest Common Divisor (GCD) of two numbers involves understanding fractions and factors. When dealing with fractions, it's important to know their factors, as they play a significant role in simplifying and solving equations. In this article, we'll discuss the role of fractions and factors in finding the GCD and provide helpful tips for solving equations involving them.
### Fraction Basics
Fractions represent a part of a whole and consist of a numerator (top number) and a denominator (bottom number). To find the GCD of two numbers, we first need to find their factors.
For example, let's find the factors of 100 and 120:
• 100 = 2 x 2 x 5 x 5
• 120 = 2 x 2 x 2 x 3 x 5
We can see that the common factors are 2 and 2, which means the GCD is 4.
### Simplifying Fractions with Common Factors
When a numerator and denominator share a common factor, the fraction can be simplified. For instance:
• 2/8 = (2÷2) / (8÷2) = 1/4
In some cases, simplifying a fraction may be necessary. Let's see some examples:
## Simplifying Fractions Using Common Factors
• 56/96 = (56÷8) / (96÷8) = 7/12
• 5/65 = (5÷5) / (65÷5) = 1/13
Understanding fractions and factors is crucial in solving math problems, as we'll see in the next section.
## Solving Equations with Fractions and Factors
Fractions and factors are central to solving various math equations. Let's now explore some basic rules that apply when performing operations on fractions.
### Operations on Fractions: Addition and Subtraction
When adding or subtracting fractions, the denominators must be the same. If they are, we can add/subtract the numerators and keep the denominator. For example:
• 2/5 + 3/5 = (2+3)/5 = 5/5 = 1
If the denominators are different, we need to make them the same before adding/subtracting the numerators. To do this, we follow these steps:
• Multiply the numerator and denominator of one fraction with the denominator of the other fraction
• Add/subtract the numerators and keep the denominator as it is
• Simplify the fraction if possible
### Multiplication and Division
When multiplying fractions, the denominators do not necessarily have to be the same. Simply multiply the numerators and denominators to get the product. For division, invert the second fraction and then multiply. For example:
• 2/3 x 4/5 = (2x4)/(3x5) = 8/15
• 2/3 ÷ 4/5 = (2/3) x (5/4) = 10/12 = 5/6
## Fractions and Factors in Math Equations: An Essential Guide
Fractions and factors are key components in solving math equations. Understanding how they work and how to manipulate them can make solving equations much simpler. In this article, we've explained the concept of fractions and factors and provided useful examples and tips for solving equations involving them.
### Key Takeaways for Fractions and Factors
• Fractions consist of a numerator (top number) and a denominator (bottom number).
• Factors are numbers that can divide exactly into another number.
• Prime factorization can assist in finding the GCD of two numbers.
• When dealing with fractions, we must consider their factors to simplify or solve equations.
## How to Simplify Fractions by Finding Common Factors
Fractions and factors play a crucial role in solving mathematical equations. Having a good grasp of these concepts and knowing how to apply them during operations can greatly simplify the process of solving equations. By following the guidelines and techniques described in this article, you can become proficient in dealing with fractions and factors.
When simplifying fractions, the first step is to look for common factors between the numerator and denominator. These are numbers that can divide evenly into both the top and bottom of the fraction. By dividing both numbers by their common factors, the fraction can be reduced to its simplest form.
For instance, let's take the fraction 12/36. Both 12 and 36 have common factors of 2, 3, and 6. By dividing both numbers by their greatest common factor, which in this case is 6, we get the simplified fraction of 2/6.
Another important factor to keep in mind is to always reduce fractions to their lowest terms. This means that the greatest common factor between the numerator and denominator should be 1. For example, the fraction 8/40 can be simplified to 1/5 by dividing both numbers by their common factor of 8.
When dealing with mixed numbers, it's important to first convert them into improper fractions. To do this, simply multiply the whole number by the denominator and add the result to the numerator. The resulting fraction can then be simplified using the aforementioned techniques.
Additionally, when adding or subtracting fractions, it's necessary to have a common denominator. This can be achieved by finding the least common multiple between the two denominators and converting both fractions to equivalent forms with the same denominator.
In conclusion, fractions can be simplified by finding common factors between the numerator and denominator. By understanding the rules and tips outlined in this article, you can become proficient in working with fractions and factors, making math equations much simpler to solve.
### In Summary
Fractions and factors are important components in solving math equations. Understanding how to work with them and apply them in various operations can make solving equations much simpler. By following the rules and tips outlined in this article, you'll be well on your way to mastering fractions and factors.
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# LIAL HORNSBY SCHNEIDER
## Presentation on theme: "LIAL HORNSBY SCHNEIDER"— Presentation transcript:
LIAL HORNSBY SCHNEIDER
COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER
5.6 Systems of Inequalities and Linear Programming
Solving Linear Inequalities Solving Systems of Inequalities Linear Programming
Solving Linear Inequalities
A linear inequality in two variables is an inequality of the form where A, B, and C are real numbers, with A and B not both equal to 0. (The symbol could be replaced with , <, or >.) The graph of a linear inequality is a half-plane, perhaps with its boundary.
Solving Linear Inequalities
For example, to graph the linear inequality first graph the boundary. Since the points of the line satisfy the equation, this line is part of the solution set.
Solving Linear Inequalities
To decide which half-plane (the one above the line 3x – 2y = 6 or the one below the line) is part of the solution set, solve the original inequality for y. Subtract 3x. Reverse the inequality symbol when dividing by a negative number. Divide by − 2.
Solving Linear Inequalities
For a particular value of x, the inequality will be satisfied by all values of y that are greater than or equal to Thus, the solution set contains the half-plane above the line.
Caution A linear inequality must be in slope-intercept form (solved for y) to determine, from the presence of a < symbol or a > symbol, whether to shade the lower or upper half-plane. In the previous slide, the upper half-plane is shaded, even though the inequality is 3x – 2y 6 is (with a < symbol) in standard form. Only when we write the inequality as (slope-intercept form) does the > symbol indicate to shade the upper half-plane.
Graph x + 4y > 4. Solution
GRAPHING A LINEAR INEQUALITY Example 1 Graph x + 4y > 4. Solution The boundary of the graph is the straight line x + 4y = 4. Since points on this line do not satisfy x + 4y > 4, it is customary to make the line dashed. To decide which half-plane represents the solution set, solve for y.
GRAPHING A LINEAR INEQUALITY Example 1
Subtract x. Divide by 4. Since y is greater than the graph of the solution set is the half-plane above the boundary.
GRAPHING A LINEAR INEQUALITY Example 1
Alternatively, or as a check, choose a test point not on the boundary line and substitute into the inequality. The point (0, 0) is a good choice if it does not lie on the boundary, since the substitution is easily done. Original inequality. Use (0, 0) as a test point. False
GRAPHING A LINEAR INEQUALITY Example 1
Since the point (0, 0) is below the boundary, the points that satisfy the inequality must be above the boundary, which agrees with the result. Original inequality. Use (0, 0) as a test point. False
Graphing Inequalities
For a function , the graph of y < (x) consists of all the points that are below the graph of y = (x); the graph of y > (x) consists of all the points that are above the graph of y = (x). 2. If the inequality is not or cannot be solved for y, choose a test point not on the boundary. If the test point satisfies the inequality, the graph includes all points on the same side of the boundary as the test point. Otherwise, the graph includes all points on the other side of the boundary.
Solving Systems of Inequalities
The solution set of a system of inequalities, such as is the intersection of the solution sets of its members. We find this intersection by graphing the solution sets of all inequalities on the same coordinate axes and identifying, by shading, the region common to all graphs.
a. Solution GRAPHING SYSTEMS OF INEQUALITIES Example 2
Graph the solution set of each system. a. Solution
b. Solution GRAPHING SYSTEMS OF INEQUALITIES Example 2
Graph the solution set of each system. b. Solution
GRAPHING SYSTEMS OF INEQUALITIES
Example 2 Writing x 3 as – 3 x 3 shows that this inequality is satisfied by points in the region between and including x = – 3 and x = 3.
GRAPHING SYSTEMS OF INEQUALITIES
Example 2 The set of points that satisfies y 0 includes the points below or on the x-axis.
GRAPHING SYSTEMS OF INEQUALITIES
Example 2 Graph y = x + 1 and use a test point to verify that the solutions of y x + 1 are on or above the boundary.
GRAPHING SYSTEMS OF INEQUALITIES
Example 2 Since the solution sets of y 0 and y x +1 have no points in common, the solution set of the system is ø. The solution set of the system is ø, because there are no points common to all three regions.
Note While we gave three graphs in the solutions of Example 2, in practice
we usually give only a final graph showing the solution set of the system.
Solving a Linear Programming Problem
Step 1 Write the objective function and all necessary constraints. Step 2 Graph the region of feasible solutions. Step 3 Identify all vertices or corner points. Step 4 Find the value of the objective function at each vertex. Step 5 The solution is given by the vertex producing the optimal value of the objective function.
FINDING A MAXIMUM PROFIT MODEL
Example 3 The Charlson Company makes two products—MP3 players and DVD players. Each MP3 gives a profit of \$30, while each DVD player produces \$70 profit. The company must manufacture at least 10 MP3s per day to satisfy one of its customers, but no more than 50 because of production problems. The number of DVD players produced cannot exceed 60 per day, and the number of MP3s cannot exceed the number of DVD players. How many of each should the company manufacture to obtain maximum profit?
Solution FINDING A MAXIMUM PROFIT MODEL Example 3
First we translate the statement of the problem into symbols. Let x = number of MP3s to be produced daily, and y = number of DVD players to be produced daily. The company must produce at least 10 MP3s (10 or more), so Since no more than 50 MP3s may be produced,
Solution FINDING A MAXIMUM PROFIT MODEL Example 3
No more than 60 DVD players may be made in one day, so The number of MP3s may not exceed the number of DVD players translates as The numbers of MP3s and of DVD players cannot be negative, so
Solution FINDING A MAXIMUM PROFIT MODEL Example 3
These restrictions, or constraints, form the system of inequalities Each MP3 gives a profit of \$30, so the daily profit from production of x MP3s is 30x dollars. Also, the profit from production of y DVD players will be 70y dollars per day. Total daily profit is, thus, This equation defines the function to be maximized, called the objective function.
Solution FINDING A MAXIMUM PROFIT MODEL Example 3
To find the maximum possible profit, subject to these constraints, we sketch the graph of each constraint.
Solution FINDING A MAXIMUM PROFIT MODEL Example 3
The only feasible values of x and y are those that satisfy all constraints—that is, the values that lie in the intersection of the graphs of the constraints.
Solution FINDING A MAXIMUM PROFIT MODEL Example 3
Any point lying inside the shaded region or on the boundary satisfies the restrictions as to the number of MP3s and DVD players that may be produced.
Solution FINDING A MAXIMUM PROFIT MODEL Example 3
(For practical purposes, however, only points with integer coefficients are useful.) This region is called the region of feasible solutions. The vertices (singular vertex) or corner points of the region of feasible solutions have coordinates
FINDING A MAXIMUM PROFIT MODEL Example 3
We must find the value of the objective function 30x + 70y for each vertex. We want the vertex that produces the maximum possible value of 30x + 70y. Maximum The maximum profit, obtained when 50 MP3s and 60 DVD players are produced each day, will be 30(50) + 70(60) = 5700 dollars per day.
FINDING A MAXIMUM PROFIT MODEL
Example 3 To show why the point of the feasible solution works: The Charlson Company needed to find values of x and y in the shaded region that produce the maximum profit—that is, the maximum value of 30x + 70y. To locate the point (x, y) that gives the maximum profit, add to the graph lines corresponding to arbitrarily chosen profits of \$0, \$1000, \$3000, and \$7000:
FINDING A MAXIMUM PROFIT MODEL
Example 3 Each point on the line 30x + 70y = 3000 corresponds to production values that yield a profit of \$3000. The region of feasible solutions are shown with these lines. The lines are parallel, and the higher the line, the greater the profit. The line 30x + 70y = 7000 yields the greatest profit but does not contain any points of the region of feasible solutions.
FINDING A MAXIMUM PROFIT MODEL
Example 3 To find the feasible solution of greatest profit, lower the line 30x + 70y = 7000 until it contains a feasible solution- that is, until it just touches the region of feasible solutions. This occurs at point A, a vertex of the region. The result observed here hold for every linear programming problem.
Fundamental Theorem of Linear Programming
If an optimal value for a linear programming problem exists, it occurs at a vertex of the region of feasible solutions.
FINDING A MINIMUM COST MODEL
Example 4 Robin takes vitamin pills each day. She wants at least 16 units of Vitamin A, at least 5 units of Vitamin B1 and at least 20 units of Vitamin C. She can choose between red pills, costing 10 cents each, that contain 8 units of A, 1 of B1 and 2 of C; and blue pills, costing 20 cents each, that contain 2 units of A, 1 of B1 and 7 of C. How many of each pill should she buy to minimize her cost and yet fulfill her daily requirements?
FINDING A MINIMUM COST MODEL
Example 4 Since Robin buys x of the 10 cent pills and y of the 20 cent pills, she gets 8 units of Vitamin A from each red pill and 2 units of Vitamin A from each blue pill. Altogether she gets 8x +y units if A per day. Since she wants at least 16 units,
FINDING A MINIMUM COST MODEL
Example 4 Each red pill and each blue pill supplies 1 unit of Vitamin B1. Robin wants at least 5 units per day, so For Vitamin C, the inequality is Also, x 0 and y 0, since Robin cannot buy negative numbers of the pills.
FINDING A MINIMUM COST MODEL
Example 4 Step 2 The intersection of the graphs of
FINDING A MINIMUM COST MODEL
Example 4 Step 3 The vertices are (0, 8), (1, 4), (3, 2), and (10, 0).
(0, 8) FINDING A MINIMUM COST MODEL Example 4
Steps 4 and 5 We find that the minimum cost occurs at (3, 2). Point Cost = 10x + 20y (0, 8) 10(0) + 20(8) = 160 (1, 4) 10(1) + 204) = 90 (3, 2) 1(3) + 20(2) = 70 (10, 0) 10(10) + 20(0) = 100 Minimum Robin’s best choice is to buy 3 red pills and 2 blue pills, for a total cost of 70 cents per day. She receives just the minimum amounts of Vitamins B1 and C, and an excess of Vitamin A.
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# How to Solve Multi-Step Inequalities
1. Math Lessons >
2. Multi-Step Inequalities
### Overview of Multi-Step Inequalities:
The very name “multi-step inequalities” explains that there are multiple steps and various properties that we need to apply in the process of arriving at their solutions. In this lesson, we are going to give you step-by-step instructions on how to solve multi-step inequalities in one variable – with the help of examples.
### Solving Multi-Step Inequalities: A Step-by-Step Approach
Let's look at the steps one by one.
Combine like terms if any and simplify each side of the inequality.
Isolate the variable using the inverse operations. Reverse the sign when multiplying or dividing by a negative number.
Solve for the variable.
Let’s deep dive into the process!
### Solving Multi-Step Inequalities with Variables on One Side
Take this inequality for instance: 7 (3x + 2) < 56. Multiplication and addition are part of the inequality. Let’s follow the steps below and solve it:
Applying the distributive property to the inequality, we have:
21x + 14 < 56
Now, the inverse operation of addition is subtraction. The subtraction property of inequality states that the inequality remains unchanged when you subtract the same number from both sides. So, let’s subtract 14 from each side.
21x + 14 – 14 < 56 – 14
21x < 42
To isolate the variable, we must perform the inverse operation of multiplication – division. The division property of inequality states that the inequality stays the same when you divide both sides by the same positive number whereas the symbol reverses when you divide by a negative number. Here we need to divide both sides by 21.
21x21 < 4221
Canceling the common factors, we have:
x < 2
Thus, x can be any value less than 2.
A graphical representation of the solution would be:
Note: The value 2 is plotted with an open circle since it is not included in the solution set.
Let’s look at solving another inequality: z(–6) – 2 ≤ 1.
The inequality involves subtraction and division, so we need to perform the inverse operations: addition and multiplication respectively.
The addition property of inequality states that when we add the same number to both sides of the inequality, the inequality remains unchanged. Adding 2 to both sides, we have:
z(–6) – 2 + 2 ≤ 1 + 2
z(–6) ≤ 3
Now to isolate the variable, we need to perform multiplication, which is the inverse operation of division. The multiplication property of the inequality says that when we multiply both sides by the same positive integer, the inequality remains the same and when we multiply both sides by a negative integer, the direction of inequality sign changes.
Thus, multiplying both sides by –6, we have:
z(–6) . (–6) ≥ 3 . (–6)
z ≥ –18
The graph of this solution set would be:
Note: The point –18 is plotted with a closed circle since the value (–18) is included in the solution set.
Replace the variable in the inequality with any value from the solution set to check your answer.
Let’s substitute one positive value and another negative value as in the above example z can be any value greater than or equal to –18.
Plugging in z = –18, we have:
–18–6 – 2 ≤ 1
3 – 2 ≤ 1
1 ≤ 1
Plugging in z = 12, we have:
12–6 – 2 ≤ 1
–2 – 2 ≤ 1
–4 ≤ 1
### Solving Multi-Step Inequalities with Variables on Both Sides
What if both sides of inequality have the variable? Don’t be tripped up! It takes a few extra steps to isolate the variable on one side, but the procedure is the same. Here are some solved examples.
Example 1
Solve and graph 7(3x + 10) < 19x.
7(3x + 10) < 19x
[The inequality given]
7(3x) + 7(10) < 19x
[Applying the distributive property]
21x + 70 < 19x
[Simplifying]
21x + 70 – 70 < 19x – 70
[Subtracting 70 from both sides]
21x – 19x < 19x – 70 – 19x
[Isolating the variable on one side by subtracting 19x]
2x < –70
[Combining like terms]
2x2 < –702
[Dividing both sides by 2]
x < –35
[Canceling common factors]
The graph of the solution set would be:
Example 2
Solve and graph 6x5 + 8 ≥ 10 + x.
6x5 + 8 ≥ 10 + x
[The inequality given]
6x5 + 8 – 8 ≥ 10 + x – 8
[Subtracting 8 from both sides]
6x5 ≥ 2 + x
[Combining like terms]
5 . 6x5 ≥ 5(2 + x)
[Multiplying both sides by 5]
6x ≥ 10 + 5x
[Applying distributive property]
6x – 5x ≥ 10 + 5x – 5x
[Isolating the variable on one side]
x ≥ 10
[Simplifying]
Graphing the solution set, we have:
Example 3
Solve and graph 12x + 51x – 11.
12x + 51x – 11
[The inequality given]
x – 11 ≤ 2x + 5
[Cross-multiplying]
x – 11 + 11 ≤ 2x + 5 + 11
x ≤ 2x + 16
[Combining like terms]
x – 2x ≤ 2x + 16 – 2x
[Isolating the variable one side]
–x ≤ 16
[Combining like terms]
–x–116–1
[Dividing both sides by –1 and reversing the sign]
x ≥ –16
[Simplifying]
Graphing the solution set, we have:
### A Few Points to Ponder
• A multi-step inequality is an inequality whose solution set is obtained by performing multiple steps.
• To solve a multi-step inequality,
• simplify each side of the inequality,
• perform inverse operations, reverse the inequality symbol if the property suggests, and
• solve for the value of the variable.
• Plug a value from the solution set in the inequality to check if your solution set is right.
Boost your skills with our free printable Multi-Step Inequalities worksheets!
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# 2019 AIME II Problems/Problem 4
## Problem 4
A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
## Solution
Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are $6^4 = 1296$ outcomes).
Case 1 (easy): Four 5's are rolled. This has probability $\frac{1}{6^4}$ of occurring.
Case 2: Two 5's are rolled.
Case 3: No 5's are rolled.
To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For $n \ge 1$, let $a_n$ equal the number of outcomes after rolling the die $n$ times, with the property that the product is a square. Thus, $a_1 = 2$ as 1 and 4 are the only possibilities.
To find $a_{n+1}$ given $a_n$ (where $n \ge 1$), we observe that if the first $n$ rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives $2a_n$ outcomes. Otherwise, the first $n$ rolls do not multiply to a perfect square ($5^n - a_n$ outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first $n$ rolls is $2^x 3^y$ where $x$ and $y$ are not both even, then we observe that if $x$ and $y$ are both odd, then the last roll must be 6; if only $x$ is odd, the last roll must be 2, and if only $y$ is odd, the last roll must be 3. Thus, we have $5^n - a_n$ outcomes in this case, and $a_{n+1} = 2a_n + (5^n - a_n) = 5^n + a_n$.
Computing $a_2$, $a_3$, $a_4$ gives $a_2 = 7$, $a_3 = 32$, and $a_4 = 157$. Thus for Case 3, there are 157 outcomes. For case 2, we multiply by $\binom{4}{2} = 6$ to distribute the two 5's among four rolls. Thus the probability is
$$\frac{1 + 6 \cdot 7 + 157}{6^4} = \frac{200}{6^4} = \frac{25}{162} \implies m+n = \boxed{187}$$
-scrabbler94
## Solution 2
We can solve this without finding the amount of cases. Notice when we roll a 1 or a 4, it does not affect whether or not the product is a square number. We have a 1/3 chance of rolling either a 1 or 4. We have a 2/3 chance of rolling a 2,3,5 or 6. Lets call rolling 1 or 4 rolling a dud.
Probability of rolling 4 duds: $(\frac{1}{3})^4$
Probability of rolling 3 duds: $4 * (\frac{1}{3})^3 * \frac{2}{3}$
Probability of rolling 2 duds: $6 * (\frac{1}{3})^2 * (\frac{2}{3})^2$
Probability of rolling 1 dud: $4 * \frac{1}{3} * (\frac{2}{3})^3$
Probability of rolling 0 duds: $(\frac{2}{3})^4$
Now we will find the probability of a square product given we have rolled a each amount of duds
Probability of getting a square product given 4 duds: 1
Probability of getting a square product given 3 duds: 0 (you will have 1 non-dud and that's never going to be square)
Probability of getting a square product given 2 duds: $\frac{1}{4}$ (as long as our two non-duds are the same, our product will be square)
Probability of getting a square product given 1 duds: $\frac{3!}{4^3}$ = $\frac{3}{32}$(the only way to have a square product is rolling a 2,3 and 6. There are 3! ways of doing that and a total of $4^3$ ways to roll 3 non-duds).
Probability of getting a square product given 0 duds: $\frac{40}{4^4}$= $\frac{5}{32}$ (We can have any two non-duds twice. For example, 2,2,5,5. There are $\binom{4}{2} = 6$ ways of choosing which two non-duds to use and $\binom{4}{2} = 6$ ways of choosing how to arrange those 4 numbers. That gives us 6*6=36 combinations. We can also have 2,2,2,2 or 3,3,3,3 or 5,5,5,5 or 6,6,6,6. This gives us a total of 40 combinations).
We multiply each probability of rolling k duds with the probability of getting a square product given k duds and then sum all the values.
$$(\frac{1}{3})^4 * 1 + 4 * (\frac{1}{3})^3 * \frac{2}{3} * 0 + 6 * (\frac{1}{3})^2 * (\frac{2}{3})^2 * \frac{1}{4} + 4 * \frac{1}{3} * (\frac{2}{3})^3 * \frac{3}{32} + (\frac{2}{3})^4 * \frac{5}{32} = \frac{25}{162}.$$
$25+162$ = $\boxed{187}$
-dnaidu (silverlizard)
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# 6.2 Explaining gauss’s law (Page 3/4)
Page 3 / 4
## Gauss’s law
The flux $\text{Φ}$ of the electric field $\stackrel{\to }{\text{E}}$ through any closed surface S (a Gaussian surface) is equal to the net charge enclosed $\left({q}_{\text{enc}}\right)$ divided by the permittivity of free space $\left({\epsilon }_{0}\right):$
$\text{Φ}={\oint }_{S}\stackrel{\to }{\text{E}}·\stackrel{^}{\text{n}}\phantom{\rule{0.2em}{0ex}}dA=\frac{{q}_{\text{enc}}}{{\epsilon }_{0}}.$
To use Gauss’s law effectively, you must have a clear understanding of what each term in the equation represents. The field $\stackrel{\to }{\text{E}}$ is the total electric field at every point on the Gaussian surface. This total field includes contributions from charges both inside and outside the Gaussian surface. However, ${q}_{\text{enc}}$ is just the charge inside the Gaussian surface. Finally, the Gaussian surface is any closed surface in space. That surface can coincide with the actual surface of a conductor, or it can be an imaginary geometric surface. The only requirement imposed on a Gaussian surface is that it be closed ( [link] ).
## Electric flux through gaussian surfaces
Calculate the electric flux through each Gaussian surface shown in [link] .
## Strategy
From Gauss’s law, the flux through each surface is given by ${q}_{\text{enc}}\text{/}{\epsilon }_{0},$ where ${q}_{\text{enc}}$ is the charge enclosed by that surface.
## Solution
For the surfaces and charges shown, we find
1. $\text{Φ}=\frac{2.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}}{{\epsilon }_{0}}=2.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}\text{.}$
2. $\text{Φ}=\frac{-2.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}}{{\epsilon }_{0}}=-2.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}\text{.}$
3. $\text{Φ}=\frac{2.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}}{{\epsilon }_{0}}=2.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}\text{.}$
4. $\text{Φ}=\frac{-4.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}+6.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}-1.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}}{{\epsilon }_{0}}=1.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}\text{.}$
5. $\text{Φ}=\frac{4.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}+6.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}-10.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}}{{\epsilon }_{0}}=0.$
## Significance
In the special case of a closed surface, the flux calculations become a sum of charges. In the next section, this will allow us to work with more complex systems.
Check Your Understanding Calculate the electric flux through the closed cubical surface for each charge distribution shown in [link] .
a. $3.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C;}$ b. $-3.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C;}$ c. $3.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C;}$ d. 0
Use this simulation to adjust the magnitude of the charge and the radius of the Gaussian surface around it. See how this affects the total flux and the magnitude of the electric field at the Gaussian surface.
## Summary
• Gauss’s law relates the electric flux through a closed surface to the net charge within that surface,
$\text{Φ}={\oint }_{S}\stackrel{\to }{\text{E}}·\stackrel{^}{\text{n}}\phantom{\rule{0.2em}{0ex}}dA=\frac{{q}_{\text{enc}}}{{\epsilon }_{0}},$
where ${q}_{\text{enc}}$ is the total charge inside the Gaussian surface S .
• All surfaces that include the same amount of charge have the same number of field lines crossing it, regardless of the shape or size of the surface, as long as the surfaces enclose the same amount of charge.
## Conceptual questions
Two concentric spherical surfaces enclose a point charge q . The radius of the outer sphere is twice that of the inner one. Compare the electric fluxes crossing the two surfaces.
Since the electric field vector has a $\frac{1}{{r}^{2}}$ dependence, the fluxes are the same since $A=4\pi {r}^{2}$ .
Compare the electric flux through the surface of a cube of side length a that has a charge q at its center to the flux through a spherical surface of radius a with a charge q at its center.
(a) If the electric flux through a closed surface is zero, is the electric field necessarily zero at all points on the surface? (b) What is the net charge inside the surface?
a. no; b. zero
Discuss how Gauss’s law would be affected if the electric field of a point charge did not vary as $1\text{/}{r}^{2}.$
Discuss the similarities and differences between the gravitational field of a point mass m and the electric field of a point charge q .
Both fields vary as $\frac{1}{{r}^{2}}$ . Because the gravitational constant is so much smaller than $\frac{1}{4\pi {\epsilon }_{0}}$ , the gravitational field is orders of magnitude weaker than the electric field.
Discuss whether Gauss’s law can be applied to other forces, and if so, which ones.
Is the term $\stackrel{\to }{E}$ in Gauss’s law the electric field produced by just the charge inside the Gaussian surface?
No, it is produced by all charges both inside and outside the Gaussian surface.
Reformulate Gauss’s law by choosing the unit normal of the Gaussian surface to be the one directed inward.
## Problems
Determine the electric flux through each surface whose cross-section is shown below.
Find the electric flux through the closed surface whose cross-sections are shown below.
a. $\text{Φ}=3.39\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/}\text{C}$ ; b. $\text{Φ}=0$ ;
c. $\text{Φ}=-2.25\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/}\text{C}$ ;
d. $\text{Φ}=90.4\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/}\text{C}$
A point charge q is located at the center of a cube whose sides are of length a . If there are no other charges in this system, what is the electric flux through one face of the cube?
A point charge of $10\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ is at an unspecified location inside a cube of side 2 cm. Find the net electric flux though the surfaces of the cube.
$\text{Φ}=1.13\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}$
A net flux of $1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}$ passes inward through the surface of a sphere of radius 5 cm. (a) How much charge is inside the sphere? (b) How precisely can we determine the location of the charge from this information?
A charge q is placed at one of the corners of a cube of side a , as shown below. Find the magnitude of the electric flux through the shaded face due to q . Assume $q>0$ .
Make a cube with q at the center, using the cube of side a . This would take four cubes of side a to make one side of the large cube. The shaded side of the small cube would be 1/24th of the total area of the large cube; therefore, the flux through the shaded area would be
$\text{Φ}=\frac{1}{24}\phantom{\rule{0.2em}{0ex}}\frac{q}{{\epsilon }_{0}}$ .
The electric flux through a cubical box 8.0 cm on a side is $1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}\text{.}$ What is the total charge enclosed by the box?
The electric flux through a spherical surface is $4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}\text{.}$ What is the net charge enclosed by the surface?
$q=3.54\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{C}$
A cube whose sides are of length d is placed in a uniform electric field of magnitude $E=4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N/C}$ so that the field is perpendicular to two opposite faces of the cube. What is the net flux through the cube?
Repeat the previous problem, assuming that the electric field is directed along a body diagonal of the cube.
zero, also because flux in equals flux out
A total charge $5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}$ is distributed uniformly throughout a cubical volume whose edges are 8.0 cm long. (a) What is the charge density in the cube? (b) What is the electric flux through a cube with 12.0-cm edges that is concentric with the charge distribution? (c) Do the same calculation for cubes whose edges are 10.0 cm long and 5.0 cm long. (d) What is the electric flux through a spherical surface of radius 3.0 cm that is also concentric with the charge distribution?
What is differential form of Gauss's law?
help me out on this question the permittivity of diamond is 1.46*10^-10.( a)what is the dielectric of diamond (b) what its susceptibility
a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
kya ye b.sc ka hai agar haa to konsa part
what is charge quantization
it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
Riya
is the time quantized ? how ?
Mehmet
What do you meanby the statement,"Is the time quantized"
Mayowa
Can you give an explanation.
Mayowa
there are some comment on the time -quantized..
Mehmet
time is integer of the planck time, discrete..
Mehmet
planck time is travel in planck lenght of light..
Mehmet
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
Tamoghna
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
determine absolute zero
The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
what is copper loss
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry
it is the work done in moving a charge to a point from infinity against electric field
what is the weight of the earth in space
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
Prince
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Jorge
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Prince
Thats why it can't have a constant value of g ....
Prince
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
Jorge
please why is the first law of thermodynamics greater than the second
every law is important, but first law is conservation of energy, this state is the basic in physics, in this case first law is more important than other laws..
Mehmet
First Law describes o energy is changed from one form to another but not destroyed, but that second Law talk about entropy of a system increasing gradually
Mayowa
first law describes not destroyer energy to changed the form, but second law describes the fluid drection that is entropy. in this case first law is more basic accorging to me...
Mehmet
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
explain the lack of symmetry in the field of the parallel capacitor
pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe
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Section 5.5
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# Section 5.5 - PowerPoint PPT Presentation
Section 5.5. Special Factoring Formulas. The Difference of Two Squares. a 2 – b 2 = (a + b)(a – b) Examples: 1) 64x 2 – 9 2) 49x 2 – 81y 2 3) 169x 4 – 9y 4 4) x 6 – y 8. The Sum and Difference of Two Cubes. a 3 + b 3 = (a + b)(a 2 – ab + b 2 )
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### Section 5.5
Special Factoring Formulas
The Difference of Two Squares
• a2 – b2 = (a + b)(a – b)
• Examples:
• 1) 64x2 – 9 2) 49x2 – 81y2
• 3) 169x4 – 9y4 4) x6 – y8
The Sum and Difference of Two Cubes
• a3 + b3= (a + b)(a 2 – ab + b2)
• a3 - b3= (a - b)(a 2 + ab + b2)
• Examples:
• 1) x3 + 125 2) z3 – 216
• 3) 27m3 - bn3 4) 125x3 + 27y3
General factoring guidelines
• 1. Factor out a GCF if possible.
• 2. If it’s a binomial, determine if it’s the sum or difference of two cubes or the difference of 2 squares. IF so, use the appropriate formula.
• 3. If it’s a trinomial, use the guess-and-check method from sections 5.3 and 5.4.
• 4. If it has 4 terms, try factoring by grouping.
• 5. Always double check to see if your factors can be factored further.
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Interactive ISC Mathematics Class 12 Solutions Chapter 7 Applications of Derivatives Ex 7.4 engage students in active learning and exploration.
## ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4
Question 1.
Solve for x :
(i) x (x – 2) (x – 5) (x + 3) > 0
(ii) x4 – 5x2 + 4 ≥ 0.
Solution:
(i) Given, x (x – 2) (x – 5) (x + 3) = 0 ………….(1)
Here, mark the points – 3, 0, 2, 5 on real line.
By method of intervals, the inequality (1) is satisfied when
x < – 3 or 0 < x < 2 or x > 5
∴ The required solution set is (- ∞, – 3) ∪ (0, 2) ∪ (5, ∞).
(ii) Given inequality be x4 – 5x2 + 4 ≥ 0
(x2 – 1) (x2 – 4) ≥ 0
⇒ (x- 1) (x – 2) (x + 1) (x + 2) ≥ 0 ………….(1)
Here mark the real numbers – 1, – 2, 1, 2 on real line and by using method of intervals, eqn. (1) is satisfied
When x ≤ – 2 or – 1 ≤ x ≤ 1 or x ≥ 2
∴ required solution set be (- ∞, – 2] ∪ [- 1, 1] ∪ [2, ∞)
Question 2.
Find all real values of x which satisfy
(i) x3 (x – 1) (x – 2) > 0
(ii) x2 (x – 1) (x – 2) ≤ 0
Solution:
(i) Given inequality be x3 (x – 1) (x – 2) > 0
⇒ x (x – 1) (x – 2) > 0 ……….(1)
[∵ x2 ≥ 0]
Mark the real numbers 0, 1, 2 on real line.
Using method of intervals, eqn. (1) is satisfied when 0 < x < 1 or x > 2
∴ required solution set be given by (0, 1) ∪ (2, ∞).
(ii) Given inequality be
x2 (x – 1) (x – 2) ≤ 0
(x – 1) (x – 2) ≤ 0 ……….(1)
[∵ x2 ≥ 0]
Mark the numbers 1, 2 on real line and by using method of intervals, eqn. (1) is satisfied when 1 ≤ x ≤ 2
also given eqn. is satisfied when x = 0
∴ required solution set = [1, 2] ∪ {0}
Question 3.
(i) $$\frac{1}{x-2}$$ ≤ 1
(ii) $$\frac{(x+1)(x-3)}{x+2}$$ ≥ 0
Solution:
(i) Given inequality be $$\frac{1}{x-2}$$ ≤ 1 ……………(1); x ≠ 2
Since (x – 2)2 > 0 ∀ x ∈ R, x ≠ 2
⇒ (x – 2) ≤ (x – 2)2
[multiplying by (x – 2)2]
⇒ (x – 2)2 – (x – 2) ≥ 0
⇒ (x – 2) (x – 3) ≥ 0
Mark the numbers 2, 3 on real line.
Using method of intervals, eqn. (2) is satisfied when x ≤ 2 or x ≥ 3
But x ≠ 2.
Thus, required solution set be (- ∞, 2) ∪ [3, ∞)
(ii) Given inequality $$\frac{(x+1)(x-3)}{x+2}$$ ≥ 0 ; x ≠ – 2
∴ (x + 2)2 > 0 ∀ x ∈ R, x ≠ – 2
⇒ (x + 1) (x – 3) (x + 2) ≥ 0 ………..(1)
[Multiplying both sides by (x + 2)2 > 0]
Mark the numbers – 1, – 2, 3 on real line and by using method of intervals, eqn. (1) is satisfied
When – 2 ≤ x ≤ – 1 or x ≥ 3
But x ≠ – 2
∴ required solution is given by (- 2, -1] ∪ [3, ∞).
Question 4.
Find the values of x for which:
(i) $$\frac{x^2+6 x-11}{x+3}$$
(ii) $$\frac{x^2-3 x+24}{x^2-3 x+3}$$ < 4
Solution:
(i) Given inequality be $$\frac{x^2+6 x-11}{x+3}$$ < – 1
⇒ $$\frac{x^2+6 x-11}{x+3}$$ + 1 < 0
⇒ $$\frac{x^2+7 x-8}{x+3}$$ < 0
⇒ $$\frac{(x-1)(x+8)}{x+3}$$ < 0 ;
Since (x + 3)2 > 0 ∀ x ∈ R, x ≠ – 3
(x – 1) (x + 8) (x + 3) < 0
[multiplying by (x + 3)2]
Mark the numbers – 3, – 8 and 1 on real line.
Using methods of intervals, given inequality satisfied when
– 3 < x < 1 or x < – 8
∴ required solution set be (- ∞, – 8) ∪ (- 3, + 1).
(ii) Given $$\frac{x^2-3 x+24}{x^2-3 x+3}$$ < 4
∴ from (1) ;
– 3 (x2 – 3x – 4) < 0
⇒ (x – 3x – 4) > 0
⇒ (x + 1) (x – 4) > 0 ………….(2)
Mark the real numbers – 1 and 4 on real line.
Using method of intervals, eqn. (2) is satisfied when x < – 1 or x > 4
∴ required solution set be (- ∞, – 1) ∪ (4, ∞).
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Tangram
To my Main Page "Mathematische Basteleien"
What is Tangram?
Tangram is one of the most popular games to lay.
You put figures of 7 pieces together (five triangles, one square and one parallelogram). You must use all pieces. They must touch but not overlap.
Main Problem top
All seven tangram pieces consist of half squares with this shape: .
There are 32 half squares or 16 squares altogether.
...... You can build a 4x4-square with all the 16 squares. The main problem of the 'tangram research' is building a square with all the 7 pieces.
You can also choose the smallest tangram piece, the blue triangle, as the basic triangle. I took the half triangle as the basic form, because the square built of all seven tangram pieces has the simple length 4.
The difference is: You have to change the rational und irrational length of a side.
Building Figures
1st problem: New figures top
...... You can invent new figures. A figure is good, if you recognize it by seeing. There are thousands of figures, which people have already built.
...... [3sqrt(2)]x[3sqrt(2)]-squares are possible, if you leave blank one triangle or two.
2nd problem: Filling given shapes top
........... You must find the right position of the tangram pieces filling this shape. (Solution in the end of this chapter)
3rd problem: How many possibilities are there to lay the same figure? top
...... You can lay the trapezium in two different ways.
...... This trapezium is not a solution. If you lay this figure, you find the mistake: The yellow and the green piece are a little bit bigger. You use the fact, that 4 and the triple of the square root of 2 (=4.24) are about the same.
Other paradoxes comparing two similar tangram figures, which seem to be alike. An example by H.Dudeney Solution:
Tangram Birds top
About 100 students (aged 11/12/13) got the task to design birds by tangram pieces.
Here is a small selection of nice birds.
Not all the students became friends with tangram pieces:
[= I hate tangrams ;-).]
(Thanks to 6b, 6c, 7a, 7c, 7d in 1999/2000)
Classifying Figures top
You can follow the positions of the sides of the half squares. 1 the sides at the right angle are horizontal or vertical
2 the side opposite the right angle is horizontal or vertical
3 mixture of 1 and 2
4 any position
If you have a mathematical point of view, you must allow only 1 and 2.
Nearly all tangram figures belong to model 4. There are many nice and expressive forms, because there are less rules. They are organized by topics.
Convex Figures top
'A figure is convex', means: If you choose any two points inside the figure, the whole line between the points must also be inside the figure.
Surprising: There are only 13 convex figures, you can build from tangram pieces.
Proof by Fu Traing Wang and Chuan-Chih Hsiung in 1942 (Book 4)
Grid Tangrams with Convex Perimeter top
You find an interesting suggestion in book 3 and book 4 to classify tangram figures. This only refers to 'mathematical' figures', which bird 1 and bird 2 (see above) stand for. You can lay them into a coordinate system, so that the corners of the seven tangram pieces have integers as coordinates. In other words: You can order the tangram pieces in a way that the sides with the unit 1 are horizontal or vertical. The lines with the unit (root of 2) are diagonal.
The figures are widened by little (white) triangles as necessary, so that a convex figure develops. These triangles have the same size as the blue tangram pieces. You count the triangles. Bird 1 needs 14 triangles and is 14-convex. Bird 2 is 5-convex. The convex figures don't need a triangle. They are 0-convex. You find all 133 (abstract) 1-convex tangram figures and solutions in book 4.
There is the problem to find figures with the largest convex perimeter.
Bruno Curfs proved in (5) that 44-konvex is an upper limit.
Here are Bruno Curfs' seven 41-convex tangrams he found.
...
...... I received more 41-convex tangrams: 8 from Ludwig Welther, 9 from Hartmut Blessing, 10 and 11 from Hannes Georg Kuchler.
...... Daniel Gronau checked all possible grid tangrams by his computer. He found out that 41-convex is the upper limit und that there are three more solutions.
Making of Tangram Pieces top
Probably the tangram pieces are developed from cutting a 4x4-square in pieces.
You use it to make tangram pieces. You draw a 4x4-square with some diagonals on plywood or on cardboard. Then you saw or cut the pieces as shown at the drawing.
Variants of the Tangram Game top
You can make more tangram games, if you divide simple geometric figures like square, rectangle or circle. The most famous are (1) "Pythagoras", (2) "Kreuzbecher", (3) "Alle Neune", (4) "Circular Puzzle", (5) "The Broken Heart", and (6) "The Magic Egg".
Here is a wide field for designing your own tangram pieces and playing with them.
Tangram on the Internet top
German
Claus Michael Ringel
Tangram
Hartmut Blessing DER IQ-BLOCK
Herbert Hertramph
Tangram-Spiel von Jos van Uden, Tangram-Spiel von Serj Dolgav zum Herunterladen
Tangram
Michael Bischoff
Tangram for you
Sabine Reindl
Tangram
stopkidsmagazin
Tangram online
tan-gram
tangram mit einer galerie von 75 exponaten
Urs Tschumi
möbli-x (Tangram-Tisch)
Wikipedia
Tangram
English
Andrew D. Orlov
Tangram House
Barbara E. Ford
Tangrams - The Magnificent Seven Piece Puzzle
Tangram Game (Online)
Degree Tutor – Guide to online colleges and distance learning
Tangram Activities
Franco Cocchini
TEN MILLIONS OF TANGRAM PATTERNS and more
Hartmut Blessing IQ-BLOCK
Michael Bischoff
Tangram for you
Paul Scott
CONVEX TANGRAMS
Randy
Tangram
Thomas Weibel
Tangram online
Tom Scavo
Tangrams
Wikipedia
Tangram
References (German) top
(1) Pieter van Delft, Jack Botermans: Denkspiele der Welt, München 1998
(2) Karl-Heinz Koch: ...lege Spiele, Köln 1987 (dumont taschenbuch1480)
(3) Rüdiger Thiele, Konrad Haase: Teufelsspiele, Leipzig 1991
(4) Joost Elffers, Michael Schuyt: Tangram, Dumont, Köln 1997 (+ tangram pieces)
(5) Bruno Curfs: Mathematical Tangram, CFF, newsletter of the "Nederlandse Kubus Club" NKC, 65 (November 2004)
(6) Jerry Slocum, Dieter Gebhardt, Jack Botermans, Monica Ma, Xiaohe Ma: The Tangram Book, 2003
[ISBN 1-4027-0413-5] Sterling Publishing Company
Feedback: Email address on my main page
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# Dilations on the Coordinate Plane
Related Topics:
Lesson Plans and Worksheets for Grade 8
Lesson Plans and Worksheets for all Grades
Examples, videos, and solutions to help Grade 8 students learn how to describe the effect of dilations on two-dimensional figures using coordinates.
### Lesson 6 Student Outcomes
• Students describe the effect of dilations on two-dimensional figures using coordinates.
### Lesson 6 Summary
Dilation has a multiplicative effect on the coordinates of a point in the plane. Given a point (x, y) in the plane, a dilation from the origin with scale factor r moves the point (x, y) to (rx, ry)
For example, if a point (3, -5) in the plane is dilated from the origin by a scale factor of r = 4, then the coordinates of the dilated point are (4 × 3, 4 × (-5)) = (12, - 20)
### NYS Math Module 3 Grade 8 Lesson 6
Classwork
Example 1
Students learn the multiplicative effect of scale factor on a point. Note that this effect holds when the center of dilation is the origin. In this lesson, the center of dilation will always be assumed to be (0, 0)
Example 2
Students learn the multiplicative effect of scale factor on a point.
Example 3
The coordinates in other quadrants of the graph are affected in the same manner as we have just seen. Based on what we have learned so far, given point A = (-2. 3) predict the location of A' when A is dilated from a center at the origin, (0, 0) by scale factor r = 3.
Exercises 1 - 5
1. Point A = (7, 9) is dilated from the origin by scale factor r = 6. What are the coordinates of point A'?
2. Point B = (-8, 5) is dilated from the origin by scale factor r = 1/2. What are the coordinates of point B'?
3. Point C = (6, -2) is dilated from the origin by scale factor r = 3/4. What are the coordinates of point C'?
4. Point D = (0, 11) is dilated from the origin by scale factor r = 4. What are the coordinates of point D'?
5. Point E = (-2, -5) is dilated from the origin by scale factor r = 3/2. What are the coordinates of point E'?
Example 4
Students learn the multiplicative effect of scale factor on a two dimensional figure.
Example 5
Students learn the multiplicative effect of scale factor on a two-dimensional figure.
Exercises 6 - 8
6. The coordinates of triangle ABC are shown on the coordinate plane below. The triangle is dilated from the origin by scale factor r = 12. Identify the coordinates of the dilated triangle A'B'C'.
7. Figure DEFG is shown on the coordinate plane below. The figure is dilated from the origin by scale factor r = 2/3. Identify the coordinates of the dilated figure D'E'F'G', then draw and label figure D'E'F'G' on the coordinate plane.
8. The triangle ABC has coordinates (3, 2) (12, 3) and (9, 12). Draw and label triangle ABC on the coordinate plane. The triangle is dilated from the origin by scale factor r = 1/3. Identify the coordinates of the dilated triangle A'B'C', then draw and label triangle A'B'C' on the coordinate plane.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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Question
# Using Rolle’s Theorem, the equation ${{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{{\text{n - 1}}}} + ..... + {{\text{a}}_{\text{n}}} = 0$ has at least one root between 0 and 1, if ${\text{A}}{\text{. }}\dfrac{{{{\text{a}}_0}}}{{\text{n}}}{\text{ + }}\dfrac{{{{\text{a}}_1}}}{{{\text{n - 1}}}} + ...... + {{\text{a}}_{{\text{n - 1}}}} = 0 \\ {\text{B}}{\text{. }}\dfrac{{{{\text{a}}_0}}}{{{\text{n - 1}}}}{\text{ + }}\dfrac{{{{\text{a}}_1}}}{{{\text{n - 2}}}} + ...... + {{\text{a}}_{{\text{n - 2}}}} = 0 \\ {\text{C}}{\text{. n}}{{\text{a}}_0} + \left( {{\text{n - 1}}} \right){{\text{a}}_1} + ...... + {{\text{a}}_{{\text{n - 1}}}} = 0 \\ {\text{D}}{\text{. }}\dfrac{{{{\text{a}}_0}}}{{{\text{n + 1}}}}{\text{ + }}\dfrac{{{{\text{a}}_1}}}{{\text{n}}} + ...... + {{\text{a}}_{\text{n}}} = 0 \\$
Hint – Observing the equation given in the question we consider a polynomial function, and check its properties. Then we check if our polynomial function holds the conditions of Rolle’s Theorem to determine the answer.
Given data, ${{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{{\text{n - 1}}}} + ..... + {{\text{a}}_{\text{n}}} = 0$.
Consider the function f defined by
f(x) = ${{\text{a}}_0}\dfrac{{{{\text{x}}^{{\text{n + 1}}}}}}{{{\text{n + 1}}}} + {{\text{a}}_{\text{n}}}\dfrac{{{{\text{x}}^{\text{n}}}}}{{\text{n}}} + ....... + {{\text{a}}_{{\text{n - 1}}}}\dfrac{{{{\text{x}}^2}}}{2} + {{\text{a}}_{\text{n}}}{\text{x}}$
Since f(x) is a polynomial, it is continuous and differentiable for all x.
f(x) is continuous in the closed interval [0, 1] and differentiable in the open interval (0, 1).
Also f(0) = 0.
And let us say,
f(1) = $\dfrac{{{{\text{a}}_0}}}{{{\text{n + 1}}}} + \dfrac{{{{\text{a}}_1}}}{{\text{n}}} + ..... + \dfrac{{{{\text{a}}_{{\text{n - 1}}}}}}{{\text{2}}} + {{\text{a}}_{\text{n}}} = 0$
i.e. f(0) = f(1)
Thus, all three conditions of Rolle’s Theorem are satisfied. Hence there is at least one value of x in the open interval (0, 1) where ${\text{f'}}$(x) = 0.
i.e. ${{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{{\text{n - 1}}}} + ..... + {{\text{a}}_{\text{n}}} = 0$.
Hence, ${{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{{\text{n - 1}}}} + ..... + {{\text{a}}_{\text{n}}} = 0$ has one root between 0 and 1 if $\dfrac{{{{\text{a}}_0}}}{{{\text{n + 1}}}} + \dfrac{{{{\text{a}}_1}}}{{\text{n}}} + ..... + \dfrac{{{{\text{a}}_{{\text{n - 1}}}}}}{{\text{2}}} + {{\text{a}}_{\text{n}}} = 0$.
Option D is the correct answer.
Note – In order to solve this type of questions the key is to assume a polynomial function and verify if it holds all the required conditions.
The Three Conditions of Rolle’s Theorem, for a function f(x) are,
(a and b are the first and last of the values x takes)
f is continuous on the closed interval [a, b], f is differentiable on the open interval (a, b) and f(a) = f(b).
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# Construct a polynomial function with the given graph
Where does one begin? I can see that the zeros are -5, -3, 0, and 4? Is that correct so far?
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I only see zeros at $-4,3,5$. – copper.hat Nov 5 '12 at 7:36
Nope, not at all. The zeros are $5,3$ and $-4$. – Cameron Buie Nov 5 '12 at 7:36
@copper.hat ah sorry, zeros are only where the line touches the x-axis correct? – Tyler Zika Nov 5 '12 at 7:37
Yes, zeros are the points where the function has value zero. – copper.hat Nov 5 '12 at 7:38
Let's use the fact that the graph has zeros at $5,3$ and $-4$. Given the window, we may as well assume these are the only zeros of the polynomial. Note also that we don't have any "flattening" near the zeros, so the zeros must be of multiplicity $1$. Hence, the polynomial is $$p(x)=a(x-5)(x-3)(x+4)$$ for some constant $a$. We can use the fact that $p(0)=3$ to solve for $a$, and then expand if we desire.
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I tried inputting your answer, and it is not working sadly. Could we be overlooking something? – Tyler Zika Nov 5 '12 at 8:28
That works! You got the 1/20 by dividing the 3 into 60? – Tyler Zika Nov 5 '12 at 8:37
Well, my formula $p(x)=a(x-5)(x-3)(x+4)$ gives an infinite family of polynomials, so that isn't enough (on its own) to get the answer. Using the fact that we need $p(0)=3$, we have $$3=p(0)=a(0-5)(0-3)(0+4)=a\cdot(-5)\cdot(-3)\cdot(4)=60a,$$ so it follows that $a=\frac1{20}$, so our desired polynomial is $$p(x)=\frac1{20}(x-5)(x-3)(x+4).$$ Does that fit the graph you're looking for? – Cameron Buie Nov 5 '12 at 8:38
Precisely, Tyler! (Sorry for the deletion and repost of my comment. I ran out of time to edit it.) – Cameron Buie Nov 5 '12 at 8:38
Where does one begin? I can see that the zeros are -5, -3, 0, and 4? Is that correct so far?
No, that is not correct.
The zeros are points where $y = 0$, and it appears to me that none of the points you mentioned are actually zeroes of the function.
This graph has $3$ points where $y = 0$, so by the fundamental theorem of algebra, you know that the polynomial will have degree $3$ (or higher, since there may also be complex roots).
You can use the existence of local minimum and maximum points on the graph to construct a polynomial that will have the same extrema. It is a simple application of differential calculus.
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Based on your inspection of the graph, where precisely is the local maximum? the local minimum? – Cameron Buie Nov 5 '12 at 7:43
Using my super glasses, I can see that the local extrema are at $\frac{1}{3}(4 \pm \sqrt{67})$. – copper.hat Nov 5 '12 at 7:45
I see your point. The solution was much more simple than I realized, and your answer is (obviously) correct. – Charles Boyd Nov 5 '12 at 7:45
Well, it isn't necessarily correct. It's just the best we can do with the information we can clearly discern. – Cameron Buie Nov 5 '12 at 7:47
Well, we can pick another point to convince ourselves. $p(-5)=-4$. – copper.hat Nov 5 '12 at 7:48
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# Lecture 12. The ISBN 10-digit uses a reverse weighting system: multiply the first digit by 10, the second by 9, the third by 8 and so on until the check.
## Presentation on theme: "Lecture 12. The ISBN 10-digit uses a reverse weighting system: multiply the first digit by 10, the second by 9, the third by 8 and so on until the check."— Presentation transcript:
Lecture 12
The ISBN 10-digit uses a reverse weighting system: multiply the first digit by 10, the second by 9, the third by 8 and so on until the check digit (the 10 th digit) is multiplied by 1. The sum of these should be evenly divisible by 11. The ISBN 13-digit uses a different system: add all the digits in the odd places values; add all the digits in the even place values and multiply this sum by 3; finally add the two results. This result should be evenly divisible by 10.
A certain textbook has ISBN 1-285-56352-C, where C is the check digit. Let’s determine what that number should be. First we multiply appropriately:
We now need 218 + C to be divisible by 11. Multiples of 11 (near where we care) are 198, 209, and 220. If we add 218 + 2 we will get an even multiple of 11, namely 220. Therefore C = 2
Using the 10-digit ISBN 1-285-56352-2 from our previous example, we will now convert it to a 13-digit ISBN. First, attach 978 to the front of the number and remove the original check digit in order to get 978- 1-285-56352-C, where C is the new check digit. Next, add all the digits in the odd place values: 9+8+2+5+6+5+C = 35 + C. Then add all the digits in the even place values and multiply the result by 3: 3(7+1+8+5+3+2) = 3(26) = 78. Add these to get 35 + C + 78 = 113 + C. We need this to be divisible by 10 so C = 7.
To find the check digit in a UPC code we multiply the digits in the odd place values by 3 and the digits in the even place values by 1. Add these results to a value that is divisible by 10.
Find the missing check digit in the product with UPC 0 44700 07205 C. Multiply the digits in odd place values by 3 to get: 3(0+4+0+0+2+5) = 3(11) = 33. Multiply the digits in the even place values by 1 to get: 1(4+7+0+7+0+C) = 18 + C. Add these together to get 33 + 18 + C = 51 + C. It must be that C = 9 so that the result is a multiple of 10.
A correct bank routing number is determined by multiplying the digits of the routing number by 7, 3, 9, 7, 3, 9, 7, 3, 9, adding the results and the final sum must be divisible by 10. Bank routing numbers do not have a traditional check digit, it is more of an error detecting code in order to verify the number given is correct.
I think my Wells Fargo routing number is 107002129. Let’s check: 7(1)+3(0)+9(7)+7(0)+3(0)+9(2)+7(1)+3(2)+9 (9) = 7+63+18+7+6+81 = 182. This number must be incorrect because 182 is not evenly divisible by 10. In order to determine the correct error, probably due to a transposition, I will switch pairs of numbers until I find the correct code.
To determine the last digit of a 16-digit credit card number: ◦ Add the digits in the odd place values and double the result. ◦ Next, count the number of digits in the odd place values that are greater than 4 and add this to the total. ◦ Finally, add the previous result to the sum of the digits in the even place values. ◦ The final sum must be divisible by 10.
A card is issued with the number 3125 6001 9643 001C, where C is the check digit. Let’s find it. Add digits in the odd place values and double the result: 2(3+2+6+0+9+4+0+1) = 2(25) = 50. Count the number of digits in odd place values that are more than 4: only 2 (the 6 and 9) so we add 2 + 50 to get 52. Add the values in even place values to 52 to get: 52+1+5+0+1+6+3+0+C = 68 + C. In order for this to be divisible by 10 C must be 2.
Add the first 10 digits and divide by 9. The check digit is the remainder. Example: A money order has number 6312472907C. Find the check digit. ◦ Add the first ten digits to get 6+3+1+2+4+7+2+9+0+7 = 41. ◦ Divide 41 by 9 to get 41 = 9(4) + 5, where 5 is the remainder. ◦ It must be that C = 5.
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# Splash Screen.
## Presentation on theme: "Splash Screen."— Presentation transcript:
Splash Screen
Key Concept: Linear and Polynomial Parent Functions
Five-Minute Check Then/Now New Vocabulary Key Concept: Linear and Polynomial Parent Functions Key Concept: Square Root and Reciprocal Parent Functions Key Concept: Absolute Value Parent Function Key Concept: Greatest Integer Parent Function Example 1: Describe Characteristics of a Parent Function Key Concept: Vertical and Horizontal Translations Example 2: Graph Translations Key Concept: Reflections in the Coordinate Axes Example 3: Write Equations for Transformations Key Concept: Vertical and Horizontal Dilations Example 4: Describe and Graph Transformations Example 5: Graph a Piecewise-Defined Function Example 6: Real-World Example: Transformations of Functions Key Concept: Transformations with Absolute Value Example 7: Describe and Graph Transformations Lesson Menu
Estimate and classify the extrema for f (x)
Estimate and classify the extrema for f (x). Support your answer numerically. 5–Minute Check 1
A. Relative minimum: (–1.5, –9.375); relative maximum: (1, 0)
B. Relative minimum: (–0.5, –4.5); relative maximum: (1, 0) C. Relative minimum: (–1.5, –9.375); relative maximum: (0.5, 0) D. Relative minimum: (–2, –9); relative maximum: (0.5, 0) 5–Minute Check 1
Estimate and classify the extrema for f (x)
Estimate and classify the extrema for f (x). Support your answers numerically. 5–Minute Check 2
A. Absolute minimum: (–2, –4) B. Absolute minimum: (–2, –8)
C. Relative minimum: (–2, –4) D. Absolute minimum: (–1, –4) 5–Minute Check 2
Find the average rate of change of f (x) = 3x 3 – x 2 + 5x – 3 on the interval [–1, 2].
B. 5 C. D. 13 5–Minute Check 3
You analyzed graphs of functions. (Lessons 1-2 through 1-4)
Identify, graph, and describe parent functions. Identify and graph transformations of parent functions Then/Now
absolute value function step function greatest integer function
parent function constant function zero function identity function quadratic function cubic function square root function reciprocal function absolute value function step function greatest integer function transformation translation reflection dilation Vocabulary
Key Concept 1
Key Concept 2
Key Concept 3
Key Concept 4
Describe Characteristics of a Parent Function
Describe the following characteristics of the graph of the parent function : domain, range, intercepts, symmetry, continuity, end behavior, and intervals on which the graph is increasing/decreasing. Example 1
The domain of the function is, and the range is .
Describe Characteristics of a Parent Function The graph of the reciprocal function shown in the graph has the following characteristics. The domain of the function is, and the range is The graph has no intercepts. The graph is symmetric with respect to the origin, so f (x) is odd. Example 1
Describe Characteristics of a Parent Function
The graph is continuous for all values in its domain with an infinite discontinuity at x = 0. The end behavior is as The graph is decreasing on the interval and decreasing on the interval Example 1
Describe Characteristics of a Parent Function
Answer: D and R: no intercepts. The graph is symmetric about the origin, so f (x) is odd. The graph is continuous for all values in its domain with an infinite discontinuity at x = 0.The end behavior is as The graph decreases on both intervals of its domain. Example 1
Describe the following characteristics of the graph of the parent function f (x) = x 2: domain, range, intercepts, symmetry, continuity, end behavior, and intervals on which the graph is increasing/decreasing. Example 1
A. D: , R: ; y-intercept = (0, 0)
A. D: , R: ; y-intercept = (0, 0). The graph is symmetric with respect to the y-axis. The graph is continuous everywhere. The end behavior is as , and as , The graph is decreasing on the interval and increasing on the interval B. D: , R: ; y-intercept = (0, 0). The graph is symmetric with respect to the y-axis. The graph is continuous everywhere. The end behavior is as , As , The graph is decreasing on the interval and increasing on the interval C. D: , R: ; y-intercept = (0, 0). The graph is symmetric with respect to the y-axis. The graph is continuous everywhere. The end behavior is as , As , The graph is decreasing on the interval and increasing on the interval D. D: , R: ; no intercepts. The graph is symmetric with respect to the y-axis. The graph is continuous everywhere. The end behavior is as , As , The graph is decreasing on the interval and increasing on the interval Example 1
Key Concept 5
A. Use the graph of f (x) = x 3 to graph the function g (x) = x 3 – 2.
Graph Translations A. Use the graph of f (x) = x 3 to graph the function g (x) = x 3 – 2. This function is of the form g (x) = f (x) – 2. So, the graph of g (x) is the graph of f (x) = x 3 translated 2 units down, as shown below. Answer: Example 2
Graph Translations B. Use the graph of f (x) = x 3 to graph the function g (x) = (x – 1)3. This function is of the form g (x) = f (x – 1) . So, g (x) is the graph of f (x) = x 3 translated 1 unit right, as shown below. Answer: Example 2
Graph Translations C. Use the graph of f (x) = x 3 to graph the function g (x) = (x – 1)3 – 2. This function is of the form g (x) = f (x – 1) – 2. So, g (x) is the graph of f (x) = x 3 translated 2 units down and 1 unit right, as shown below. Answer: Example 2
Use the graph of f (x) = x 2 to graph the function g (x) = (x – 2)2 – 1.
B. C. D. Example 2
Key Concept 6
The graph of g (x) is the graph of translated 1 unit up. So, .
Write Equations for Transformations A. Describe how the graphs of and g (x) are related. Then write an equation for g (x). The graph of g (x) is the graph of translated 1 unit up. So, Answer: The graph is translated 1 unit up; Example 3
Write Equations for Transformations
B. Describe how the graphs of and g (x) are related. Then write an equation for g (x). The graph of g (x) is the graph of translated 1 unit to the left and reflected in the x-axis. So, Example 3
Write Equations for Transformations
Answer: The graph is translated 1 unit to the left and reflected in the x-axis; Example 3
A. The graph is translated 3 units up; g (x) = x 3 + 3.
Describe how the graphs of f (x) = x 3 and g (x) are related. Then write an equation for g (x). A. The graph is translated 3 units up; g (x) = x B. The graph is translated 3 units down; g (x) = x 3 – 3. C. The graph is reflected in the x-axis; g (x) = –x 3. D. The graph is translated 3 units down and reflected in the x-axis; g (x) = –x 3 – 3. Example 3
Key Concept 7
Describe and Graph Transformations
A. Identify the parent function f (x) of , and describe how the graphs of g (x) and f (x) are related. Then graph f (x) and g (x) on the same axes. The graph of g(x) is the same as the graph of the reciprocal function expanded vertically because and 1 < 3. Example 4
Describe and Graph Transformations
Answer: ; g (x) is represented by the expansion of f (x) vertically by a factor of 3. Example 4
Describe and Graph Transformations
B. Identify the parent function f (x) of g (x) = –|4x|, and describe how the graphs of g (x) and f (x) are related. Then graph f (x) and g (x) on the same axes. The graph of g (x) is the same as the graph of the absolute value function f (x) = |x| compressed horizontally and then reflected in the x-axis because g (x) = –4(|x|) = –|4x| = –f (4x), and 1 < 4. Example 4
Describe and Graph Transformations
Answer: f (x) = |x| ; g (x) is represented by the compression of f (x) horizontally by a factor of 4 and reflection in the x-axis. Example 4
Identify the parent function f (x) of g (x) = – (0
Identify the parent function f (x) of g (x) = – (0.5x)3, and describe how the graphs of g (x) and f (x) are related. Then graph f (x) and g (x) on the same axes. Example 4
A. f (x) = x 3; g(x) is represented by the expansion of the graph of f (x) horizontally by a factor of . B. f (x) = x 3; g(x) is represented by the expansion of the graph of f (x) horizontally by a factor of and reflected in the x-axis. C. f (x) = x 3; g(x) is represented by the reflection of the graph of f (x) in the x-axis. D. f (x) = x 2; g(x) is represented by the expansion of the graph of f (x) horizontally by a factor of and reflected in the x-axis. Example 4
On the interval , graph y = |x + 2|.
Graph a Piecewise-Defined Function Graph On the interval , graph y = |x + 2|. On the interval [0, 2], graph y = |x| – 2. On the interval graph Draw circles at (0, 2) and (2, 2); draw dots at (0, –2) and (2, 0) because f (0) = –2, and f (2) = 0, respectively. Example 5
Graph a Piecewise-Defined Function
Graph the function A. B. C. D. Example 5
Transformations of Functions
A. AMUSEMENT PARK The “Wild Ride” roller coaster has a section that is shaped like the function , where g (x) is the vertical distance in yards the roller coaster track is from the ground and x is the horizontal distance in yards from the start of the ride. Describe the transformations of the parent function f (x) = x 2 used to graph g (x). Example 6
Transformations of Functions
Rewrite the function so that it is in the form g(x) = a(x – h)2 + k by completing the square. Original function Factor Complete the square. Example 6
Write x 2 – 100x + 2500 as a perfect square and simplify.
Transformations of Functions Write x 2 – 100x as a perfect square and simplify. Answer: g (x) is the graph of f (x) translated 50 units right, compressed vertically, reflected in the x-axis, and then translated 50 units up Example 6
Transformations of Functions
B. AMUSEMENT PARK The “Wild Ride” roller coaster has a section that is shaped like the function , where g (x) is the vertical distance in yards the roller coaster track is from the ground and x is the horizontal distance in yards from the start of the ride. Suppose the ride designers decide to increase the highest point of the ride to 70 yards. Rewrite g (x) to reflect this change. Graph both functions on the same coordinate axes using a graphing calculator. Example 6
Transformations of Functions
A change of the height of the highest point of the ride from 50 yards to 70 yards is a vertical translation, so add an additional 20 yards to the expression in part A to get The graph of both functions on the same coordinate axes using a graphing calculator is shown. Example 6
Transformations of Functions
STUNT RIDING A stunt motorcyclist jumps from ramp to ramp according to the model shaped like the function , where g (x) is the vertical distance in feet the motorcycle is from the ground and x is the horizontal distance in feet from the start of the jump. Describe the transformations of f (x) = x 2 used to graph g (x). Example 6
A. The graph of g (x) is the graph of f (x) translated 75 units right.
B. The graph of g (x) is the graph of f (x) translated 18 units up, compressed vertically, and reflected in the x-axis. C. The graph of g (x) is the graph of f (x) translated 75 units right and reflected in the x-axis. D. The graph of g (x) is the graph of f (x) translated 75 units right and 18 units up, compressed vertically, and reflected in the x-axis. Example 6
Key Concept 8
Describe and Graph Transformations
A. Use the graph of f (x) = x 2 – 4x + 3 to graph the function g(x) = |f (x)|. Example 7
Describe and Graph Transformations
The graph of f (x) is below the x-axis on the interval (1, 3), so reflect that portion of the graph in the x-axis and leave the rest unchanged. Answer: Example 7
Describe and Graph Transformations
B. Use the graph of f (x) = x 2 – 4x + 3 to graph the function h (x) = f (|x|). Example 7
Describe and Graph Transformations
Replace the graph of f (x) to the left of the y-axis with a reflection of the graph to the right of the y-axis. Answer: Example 7
Use the graph of f (x) shown to graph g(x) = |f (x)| and h (x) = f (|x|).
Example 7
A. B. C. D. Example 7
End of the Lesson
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# Class 8 NCERT Solutions – Chapter 7 Cubes and Cube Roots – Exercise 7.2
Last Updated : 05 Nov, 2020
### Question 1. Find the cube root of each of the following numbers by prime factorization method.
(i) 64
64 = 2 × 2 × 2 × 2 × 2 × 2
By assembling the factors in trio of equal factors, 64 = (2 × 2 × 2) × (2 × 2 × 2)
Therefore, 64 = 2 × 2 = 4
Hence, 4 is cube root of 64.
(ii) 512
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
By assembling the factors in trio of equal factors, 512 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)
Therefore, 512 = 2 × 2 × 2 = 8
Hence, 8 is cube root of 512.
(iii) 10648
10648 = 2 × 2 × 2 × 11 × 11 × 11
By assembling the factors in trio of equal factors, 10648 = (2 × 2 × 2) × (11 × 11 × 11)
Therefore, 10648 = 2 × 11 = 22
Hence, 22 is cube root of 10648.
(iv) 27000
27000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5
By assembling the factors in trio of equal factors, 27000 = (2 × 2 × 2) × (3 × 3 × 3) × (5 × 5 × 5)
Therefore, 27000 = (2 × 3 × 5) = 30
Hence, 30 is cube root of 27000.
(v) 15625
15625 = 5 × 5 × 5 × 5 × 5 × 5
By assembling the factors in trio of equal factors, 15625 = (5 × 5 × 5) × (5 × 5 × 5)
Therefore, 15625 = (5 × 5) = 25
Hence, 25 is cube root of 15625.
(vi) 13824
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
By assembling the factors in trio of equal factors,
13824 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3)
Therefore, 13824 = (2 × 2 × 2 × 3) = 24
Hence, 24 is cube root of 13824.
(vii) 110592
110592 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
By assembling the factors in trio of equal factors,
110592 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 ×3)
Therefore, 110592 = (2 × 2 × 2 × 2 × 3) = 48
Hence, 48 is cube root of 110592.
(viii) 46656
46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
By assembling the factors in trio of equal factors,
46656 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3)
Therefore, 46656 = (2 × 2 × 3 × 3) = 36
Hence, 36 is cube root of 46656.
(ix) 175616
175616 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7
By assembling the factors in trio of equal factors,
175616 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (7 × 7 × 7)
Therefore, 175616 = (2 × 2 × 2 × 7) = 56
Hence, 56 is cube root of 175616.
(x) 91125
91125 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
By assembling the factors in trio of equal factors, 91125 = (3 × 3 × 3) × (3 × 3 × 3) × (5 × 5 × 5)
Therefore, 91125 = (3 × 3 × 5) = 45
Hence, 45 is cube root of 91125.
### Question. 2. State true or false.
(i) Cube of any odd number is even.
False
(ii) A perfect cube does not end with two zeros.
True
(iii) If square of a number ends with 5, then its cube ends with 25.
False
(iv) There is no perfect cube which ends with 8.
False
(v) The cube of a two-digit number may be a three-digit number.
False
(vi) The cube of a two-digit number may have seven or more digits.
False
(vii) The cube of a single digit number may be a single digit number.
True
### Question. 3. You are told that 1,331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
(i) 1331
Since, the unit digit of cube is 1, the unit digit of cube root 1.
Therefore, we get 1 as unit digit of cube root of 1331.
And the ten’s digit of our cube root is taken as the unit place of the smallest number.
As the unit digit of the cube of a number having digit as unit place 1 is 1.
Therefore,
∛1331 = 11
(ii) 4913
Since, the unit digit of cube is 3, the unit digit of cube root will be 7.
Therefore, we get 7 as unit digit of the cube root of 4913. We know 13 = 1 and 23 = 8, 1 > 4 > 8.
So, 1 is taken as ten digits of cube root.
Therefore,
∛4913 = 17
(iii) 12167
Since, the unit digit of cube is 7, the unit digit of cube root will be 3.
Therefore, 3 is the unit digit of the cube root of 12167 We know 23 = 8 and 33 = 27, 8 > 12 > 27.
So, 2 is taken as ten digits of cube root.
Therefore,
∛12167 = 23
(iv) 32768
Since, the unit digit of cube is 8, the unit digit of cube root will be 2
Therefore, 2 is the unit digit of the cube root of 32768. We know 33 = 27 and 43 = 64, 27 > 32 > 64.
So, 3 is taken as ten digits of cube root.
Therefore,
∛32768 = 32
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• RD Sharma Class 8 Solutions for Maths
Class 8 RD Sharma Solutions- Chapter 6 Algebraic Expressions And Identities – Exercise 6.2
(i) 3a2b, -4a2b, 9a2b
Solution:
3a2b, -4a2b, 9a2b
Now we have add the given expression
= 3a2b + (-4a2b) + 9a2b
= 3a2b – 4a2b + 9a2b
= 8a2b
(ii) 2/3a, 3/5a, -6/5a
Solution:
We have to add the given expression
2/3a + 3/5a + (-6/5a)
2/3a + 3/5a – 6/5a
Now take LCM for 3 and 5 which will be 15
= (2×5)/(3×5)a + (3×3)/(5×3)a – (6×3)/(5×3)a
= 10/15a + 9/15a – 18/15a
= (10a+9a-18a)/15
= a/15
(iii) 4xy2 – 7x2y, 12x2y -6xy2, -3x2y + 5xy2
Solution:
We have to add the given expression
4xy2 – 7x2y + 12x2y – 6xy2 – 3x2y + 5xy2
Now rearrange the expression:
12x2y – 3x2y – 7x2y – 6xy2 + 5xy2 + 4xy
3xy2 + 2x2y
(iv) 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c
Solution:
3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c
3/2a – 5/4b + 2/5c + 2/3a – 7/2b + 7/2c + 5/3a + 5/2b – 5/4c
rearrange
3/2a + 2/3a + 5/3a – 5/4b – 7/2b + 5/2b + 2/5c + 7/2c – 5/4c
Now take LCM of (2 and 3 is 6), (4 and 2 is 4), (5,2 and 4 is 20)
(9a+4a+10a)/6 + (-5b-14b+10b)/4 + (8c+70c-25c)/20
23a/6 – 9b/4 + 53c/20
(v) 11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 137xy
Solution:
11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 13/7xy
11/2xy + 12/5y + 13/7x + -11/2y – 12/5x – 13/7xy
Now rearrange
11/2xy – 13/7xy + 13/7x – 12/5x + 12/5y -11/2y
Now take LCM for (2 and 7 is 14), (7 and 5 is 35), (5 and 2 is 10)
(11xy-12xy)/14 + (65x-84x)/35 + (24y-55y)/10
51xy/14 – 19x/35 – 31y/10
(vi) 7/2x3 – 1/2x3 + 5/3, 3/2x3 + 7/4x2 – x + 1/3, 3/2x2 -5/2x -2
Solution:
7/2x3 – 1/2x2 + 5/3 + 3/2x3 + 7/4x2 – x + 1/3 + 3/2x2 -5/2x – 2
Now rearrange
=7/2x3 + 3/2x3 – 1/2x2 + 7/4x2 + 3/2x2 – x – 5/2x + 5/3 + 1/3 – 2
=10/2x3 + 11/4x2 – 7/2x + 0/6
=5x3 + 11/4x2 -7/2x
(i) -5xy from 12xy
Solution:
Subtract the given expression
= 12xy – (- 5xy)
= 5xy + 12xy
= 17xy
(ii) 2a2 from -7a2
Solution:
Subtract the given expression
= (-7a2) – 2a2
= -7a2 – 2a2
= -9a2
(iii) 2a-b from 3a-5b
Solution:
Subtract the given expression
=(3a – 5b) – (2a – b)
= 3a – 5b – 2a + b
= a – 4b
(iv) 2x3 – 4x2 + 3x + 5 from 4x3 + x2 + x + 6
Solution:
Subtract the given expression
(4x3 + x2 + x + 6) – (2x3 – 4x2 + 3x + 5)
4x3 + x2 + x + 6 – 2x3 + 4x2 – 3x – 5
2x3 + 5x2 – 2x + 1
(v) 2/3y3 – 2/7y2 – 5 from 1/3y3 + 5/7y2 + y – 2
Solution:
Subtract the given expression
1/3y3 + 5/7y2 + y – 2 – 2/3y3 + 2/7y2 + 5
On rearranging,
1/3y3 – 2/3y3 + 5/7y2 + 2/7y2 + y – 2 + 5
We will group similar expression:
= -1/3y3 + 7/7y2 + y + 3
= -1/3y3 + y2 + y + 3
(vi) 3/2x – 5/4y – 7/2z from 2/3x + 3/2y – 4/3z
Solution:
Subtract the given expression
2/3x + 3/2y – 4/3z – (3/2x – 5/4y – 7/2z)
On rearranging,
2/3x – 3/2x + 3/2y + 5/4y – 4/3z + 7/2z
We will group similar expression:
LCM of (3 and 2 is 6), (2 and 4 is 4), (3 and 2 is 6)
=(4x-9x)/6 + (6y+5y)/4 + (-8z+21z)/6
= -5x/6 + 11y/4 + 13z/6
(vii) x2y – 4/5xy2 + 4/3xy from 2/3x2y + 3/2xy2 – 1/3xy
Solution:
Subtract the given expression
2/3x2y + 3/2xy2 – 1/3xy – (x2y – 4/5xy2 + 4/3xy)
on rearrange
2/3x2y – x2y + 3/2xy2 + 4/5xy2 – 1/3xy – 4/3xy
We will group similar expression:
LCM of (3 and 1 is 3), (2 and 5 is 10), (3 and 3 is 3)
-1/3x2y + 23/10xy2 – 5/3xy
(viii) ab/7 – 35/3bc + 6/5ac from 3/5bc – 4/5ac
Solution:
Subtract the given expression
3/5bc – 4/5ac – (ab/7 – 35/3bc + 6/5ac)
On rearrange
3/5bc + 35/3bc – 4/5ac – 6/5ac – ab/7
We will group similar expression:
LCM of (5 and 3 is 15), (5 and 5 is 5)
(9bc+175bc)/15 + (-4ac-6ac)/5 – ab/7
184bc/15 + -10ac/5 – ab/7
– ab/7 + 184bc/15 – 2ac
(i) 6/5x2 – 4/5x3 + 5/6 + 3/2x from x3/3 – 5/2x2 + 3/5x + 1/4
Solution:
Subtract the given expression
1/3x3 – 5/2x2 + 3/5x + 1/4 – (6/5x2 – 4/5x3 + 5/6 + 3/2x)
On rearrange
1/3x3 + 4/5x3 – 5/2x2 – 6/5x2 + 3/5x – 3/2x + 1/4 – 5/6
By grouping similar expressions we get,
LCM of (3 and 5 is 15), (2 and 5 is 10), (5 and 2 is 10), (4 and 6 is 24)
17/15x3 – 37/10x2 – 9/10x – 14/24
17/15x3 – 37/10x2 – 9/10x – 7/12
(ii) 5a2/2 + 3a3/2 + a/3 – 6/5 from 1/3a3 – 3/4a2 – 5/2
Solution:
Subtract the given expression
1/3a3 – 3/4a2 – 5/2 – (5/2a2 + 3/2a3 + a/3 – 6/5)
On rearrange
1/3a5 – 3/2a3 – 3/4a2 – 5/2a2 – a/3 – 5/2 + 6/5
By grouping similar expressions we get,
LCM of (3 and 2 is 6), (4 and 2 is 4), (2 and 5 is 10)
= (2a3 – 9a3)/6 – (3a2 + 10a2)/4 – a/3 + (-25+12)/10
= -7/6a3 – 13/4a2 – a/3 – 13/10
(iii) 7/4x3 + 3/5x2 + 1/2x + 9/2 from 7/2 – x/3 – x2/5
Solution:
Subtract the given expression
7/2 – x/3 – 1/5x2 – (7/4x3 + 3/5x2 + 1/2x + 9/2)
On rearranging,
-7/4x3 – 1/5x2 – 3/5x2 – x/3 – x/2 + 7/2 – 9/2
By grouping similar expressions we get,
LCM of (3 and 2 is 6)
-7/4x3 – 4/5x2– (2x-3x)/6 + (7-9)/2
-7/4x3 – 4/5x2 – 5/6x – 1
(iv) y3/3 + 7/3y2 + 1/2y + 1/2 from 1/3 – 5/3y2
Solution:
Subtract the given expression
1/3 – 5/3y2 – (1/3y2 + 7/3y2 + 1/2y + 1/2)
On rearrange
-1/3y3 – 5/3y2 – 7/3y2 – 1/2y + 1/3 – 1/2
By grouping similar expressions we get,
LCM of (3 and 3 is 3), (3 and 2 is 6)
-1/3y3 + (-5y2 – 7y2)/3 – 1/2y + (2-3)/6
-1/3y3 – 12/3y2 – 1/2y – 1/6
(v) 2/3ac – 5/7ab + 2/3bc from 3/2ab -7/4ac – 5/6bc
Solution:
Subtract the given expression
3/2ab – 7/4ac – 5/6bc – (2/3ac – 5/7ab + 2/3bc)
On rearrange
3/2ab + 5/7ab – 7/4ac – 2/3ac – 5/6bc – 2/3bc
By grouping similar expressions we get,
LCM of (2 and 7 is 14), (4 and 3 is 12), (6 and 3 is 6)
(21ab+10ab)/14 – (21ac-8ac)/12 – (5bc-4bc)/6
31/14ab – 29/12ac – 3/2bc
Question 4. Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and -4x + 9y – 11z.
Solution:
First we will find the sum:
The sum of x – 3y + 2z and -4x + 9y – 11z is
(x – 3y + 2z) + (-4x + 9y – 11z)
On rearrange
x – 4x – 3y + 9y + 2z – 11z
= -3x + 6y – 9z
Now Let’s subtract it from -3x + 6y – 9z
(-3x + 6y – 9z) – (3x – 4y – 7z)
On rearranging again
= -3x – 3x + 6y + 4y – 9z + 7z
= -6x + 10y – 2z
Question 5. Subtract the sum of 3l – 4m – 7n2 and 2l + 3m – 4n2 from the sum of 9l + 2m – 3n2 and -3l + m + 4n2.
Solution:
Sum of 3l – 4m – 7n2 and 2l + 5m – 4n2
3l – 4m – 7n2 + 2l + 3m – 4n2
On rearrange
3l + 2l – 4m + 3m – 7n2 – 4n2
5l – m – 11n2 ……………………..eq. (1)
Sum of 9l + 2m – 3n2 and -3l + m + 4n2
9l + 2m – 3n2 + (-3l + m + 4n2)
On rearrange
9l – 3l + 2m + m – 3n2 + 4n2
6l + 3m + n2 ……………………….eq. (2)
Let us subtract equ (i) from (ii), we get
6l + 3m + n2 – (5l – m – 11n2)
On rearrange
6l – 5l + 3m + m + n2 + 11n2
l + 4m + 12n2
Question 6. Subtract the sum of 2x – x2 + 5 and -4x – 3 + 7x2 from 5.
Solution:
Sum of 2x – x2 + 5 and -4x – 3 + 7x2 is
2x – x2 + 5 + (-4x – 3 + 7x2)
2x – x2 + 5 – 4x – 3 + 7x2
On rearrange
– x2 + 7x2 + 2x – 4x + 5 – 3
6x2 -2x + 2 …………eq (i)
Let subtract eq (i) from 5 we will get,
5 – (6x2 -2x + 2)
5 – 6x2 + 2x – 2
3 + 2x – 6x2
(i) x2 – 3x + 5 – 1/2(3x2 – 5x + 7)
Solution:
x2 – 3x + 5 – 1/2(3x2 – 5x + 7)
On rearrange
x2 – 3/2x2 – 3x + 5/2x + 5 – 7/2
We will group similar expression:
LCM of (1 and 2 is 2)
= (2x2 – 3x2)/2 – (6x + 5x)/2 + (10-7)/2
= -1/2x2 – 1/2x + 3/2
(ii) [5 – 3x + 2y – (2x – y)] – (3x – 7y + 9)
Solution:
5 – 3x + 2y – 2x + y – 3x + 7y – 9
On rearrange
= – 3x – 2x – 3x + 2y + y + 7y + 5 – 9
We will group similar expression:
= -8x + 10y – 4
(iii) 11/2x2y – 9/4xy2 + 1/4xy – 1/14y2x + 1/15yx2 + 1/2xy
Solution:
On rearrange
11/2x2y + 1/15x2y – 9/4xy2 – 1/14xy2 + 1/4xy + 1/2xy
We will group similar expression:
LCM of (2 and 15 is 30), (4 and 14 is 56), (4 and 2 is 4)
= (165x2y + 2x2y)/30 + (-126xy2 – 4xy2)/56 + (xy + 2xy)/4
= 167/30x2y – 130/56xy2 + 3/4xy
= 167/30x2y – 65/28xy2 + 3/4xy
(iv) (1/3y2 – 4/7y + 11) – (1/7y – 3 + 2y2) – (2/7y – 2/3y2 + 2)
Solution:
On rearrange
1/3y2 – 2y2 – 2/3y2 – 4/7y – 1/7y – 2/7y + 11 + 3 – 2
We will group similar expression:
LCM of (3, 1 and 3 is 3), (7, 7 and 7 is 7)
= (y2 – 6y2 + 2y2)/3 – (4y – y – 2y)/7 + 12
= -3/3y2 – 7/7y + 12
= -y2 – y + 12
(v) -1/2a2b2c + 1/3ab2c – 1/4abc2 – 1/5cb2a2 + 1/6cb2a – 1/7c2ab + 1/8ca2b
Solution:
On rearrange
-1/2a2b2c – 1/5a2b2c + 1/3ab2c + 1/6ab2c – 1/4abc2 – 1/7abc2 + 1/8a2bc
We will group similar expression:
LCM of (2 and 5 is 10), (3 and 6 is 6), (4 and 7 is 28)
-7/10a2b2c + 1/2ab2c – 11/28abc2 + 1/8a2bc
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# Gauss elimination method
## Gauss elimination method :
### Introduction:
Many times we are required to find out solution of linear equations. We also know that, we can find out roots of linear equations if we have sufficient number of equations. For example if we have to calculate three unknown variables, then we must have three equations. Many times we have solved such problems by eliminating one of the root and keep on decreasing number of variables. But in some cases it is not possible or it will take more time to solve.
Gauss elimination method is one of the simple and famous methods used for finding roots of linear equations. Let us discuss this method assuming we have three linear equations in x, y and z. That is we have to find out roots of that equations (values of x, y and z).
Steps to find out roots of linear equations using Gauss elimination method:
In this method we just eliminate ‘x’ from first equation using second and third equation. After that we get only two equations with two unknowns. Similarly, we then eliminate ‘y’ from first (among two equations that we get from last step) equation using second equation. Finally we get single equation in z having constant in its right side. Now we can find ‘z’, using ‘z’ we can find ‘y’, similarly ‘x’.
Don’t get confused, I will explain each step clearly.
Let us consider three linear equations as follows:
a1x+b1y+c1z=d1…………………1)
a2x+b2y+c2z=d2…………………2)
a3x+b3y+c3z=d3…………………3)
From above three equations we are asked for finding values of x, y and z (values of a1, b1, c1,……..,d3 are given).
Step 1: Eliminate ‘x’ from first equation using second and third equation. For doing this we have to subtract 1st eq. from 2nd eq. by making coefficient of ‘x’ (of 1st equation) equals to coefficient of ‘x’ (of 2nd equation). Similarly we have to do same thing for third equation.
In short we have to solve following equations:
eq.(2) – (a2/a1)*eq.(1) and eq.(3) – (a3/a1)*eq.(1)
We get two new equations in y and z as follows:
b2’y+c2’z=d2’……..4)
b3’y+c3’z=d3’……..5)
Step 2: similarly, we have to eliminate ‘y’ from 4th equation using 5th equation.
We have to solve following equation.
eq.(5) – (b3’/b2’)*eq.(4)
we get c3”z=d3”……………6)
solving above equation we get, z=d3”/c3”
Step 3: Finally we have to put above value of ‘z’ in equation 4) (or (5)), then we get ‘y’. now we have two roots (y and z). put ‘y’ and ‘z’ in eq.(1) (or (2) or eq.(3)), we will get ‘x’.
## Example 1:
Find the roots of following equations using Gauss Elimination method.
X + 4y – z = -5 ………….1)
X + y – 6z = -12…………2)
3x – y – z = 4 ………3)
### Solution:
Step 1: Perform eq.(2) – (a2/a1)*eq.(1) and eq.(3) – (a3/a1)*eq.(1)
We get, 3y +5z =7……………4)
And
-13y +2z = 19…………..5)
Step 2: Now perform eq.(5) – (b3’/b2’)*eq.(4)
We get, -13y + 2z – (-13/3)*(3y + 5z) = 19 + (13/3)*7
71z = 148
i.e. z=148/71.
Step 3: From eq.(5) – 13y = 19 – 2*(148/71)
= 19 – 296/71
Y= -81/71
From eq.(1) x+4(-81/71)-148/71=-5
Therefore, x=117/71.
Thus roots of given linear equations using Gauss elimination method is
X=117/71; y=-81/71; z=148/71.
## Example 2:
Find roots of following linear equations using Gauss Elimination method:
2x+y+z=10………1)
3x+2y+3z=18……..2)
X+4y+9z=16………..3)
### Solution :
Step 1: Perform eq.(2)-(3/2)*eq(1) and eq(3)-(1/2)*eq(1)
We get, y+3z=6…………4)
And 7y+17z=22……….5)
Step 2: Perform eq(5) – 7*eq(4)
We get, -4z=-20
i.e. z=5;
Step 3: From equation (5) y=-63/7 i.e. y=-9
From equation (1) 2x=14 i.e. x=7
Thus the roots of given linear equations is
X=7; y=-9; z=5.
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# LarPCalcLim2_03_02
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Category: Education
## Presentation Description
No description available.
## Presentation Transcript
### What You Should Learn:
Recognize and evaluate logarithmic functions with base a . Graph logarithmic functions. Recognize, evaluate, and graph natural logarithmic functions. Use logarithmic functions to model and solve real-life problems. What You Should Learn
### Slide 4:
Logarithmic Functions
### Logarithmic Functions:
Logarithmic Functions Every function of the form f ( x ) = a x passes the Horizontal Line Test and therefore must have an inverse function. This inverse function is called the logarithmic function with base a .
### Logarithmic Functions:
Logarithmic Functions The equations y = log a x and x = a y are equivalent. The first equation is in logarithmic form and the second is in exponential form. For example, the logarithmic equation 2 = log 3 9 can be rewritten in exponential form as 9 = 3 2 . The exponential equation 5 3 = 125 can be rewritten in logarithmic form as log 5 125 = 3.
### Logarithmic Functions:
Logarithmic Functions When evaluating logarithms, remember that a logarithm is an exponent . This means that log a x is the exponent to which a must be raised to obtain x . For instance, log 2 8 = 3 because 2 must be raised to the third power to get 8.
### Example 1 – Evaluating Logarithms:
Example 1 – Evaluating Logarithms Use the definition of logarithmic function to evaluate each logarithm at the indicated value of x . a. f ( x ) = log 2 x , x = 32 b. f ( x ) = log 3 x , x = 1 c. f ( x ) = log 4 x , x = 2 d. f ( x ) = log 10 x , x = Solution: a. f ( 32 ) = log 2 32 because 2 5 = 32. = 5 b. f ( 1 ) = log 3 1 because 3 0 = 1. = 0
### Example 1 – Solution:
Example 1 – Solution c. f ( 2 ) = log 4 2 because 4 1/2 = = 2. = d. because . cont’d
### Logarithmic Functions:
Logarithmic Functions The logarithmic function with base 10 is called the common logarithmic function. It is denoted by log 10 or simply by log. On most calculators, this function is denoted by . The following properties follow directly from the definition of the logarithmic function with base a .
### Slide 11:
Graphs of Logarithmic Functions
### Graphs of Logarithmic Functions:
Graphs of Logarithmic Functions To sketch the graph of y = log a x , you can use the fact that the graphs of inverse functions are reflections of each other in the line y = x .
### Example 5 – Graphs of Exponential and Logarithmic Functions:
Example 5 – Graphs of Exponential and Logarithmic Functions In the same coordinate plane, sketch the graph of each function. a. f ( x ) = 2 x b. g ( x ) = log 2 x
### Example 5(a) – Solution:
Example 5(a) – Solution For f ( x ) = 2 x , construct a table of values. By plotting these points and connecting them with a smooth curve, you obtain the graph shown in Figure 3.14. Figure 3.14
### Example 5(b) – Solution:
Example 5(b) – Solution Because g ( x ) = log 2 x is the inverse function of f ( x ) = 2 x , the graph of g is obtained by plotting the points ( f ( x ), x ) and connecting them with a smooth curve. The graph of g is a reflection of the graph of f in the line y = x , as shown in Figure 3.14. cont’d Figure 3.14
### Graphs of Logarithmic Functions:
Graphs of Logarithmic Functions The nature of the graph in Figure 3.15 is typical of functions of the form f ( x ) = log a x , a 1. They have one x -intercept and one vertical asymptote. Notice how slowly the graph rises for x 1. Figure 3.15
### Graphs of Logarithmic Functions:
Graphs of Logarithmic Functions The basic characteristics of logarithmic graphs are summarized in Figure 3.16. Graph of y = log a x , a 1 • Domain: (0, ) • Range: ( , ) • x -intercept: (1, 0) • Increasing • One-to-one, therefore has an inverse function Figure 3.16
### Graphs of Logarithmic Functions:
Graphs of Logarithmic Functions • y -axis is a vertical asymptote (log a x → as x → 0 + ). • Continuous • Reflection of graph of y = a x about the line y = x. The basic characteristics of the graph of f ( x ) = a x are shown below to illustrate the inverse relation between f ( x ) = a x and g ( x ) = log a x . • Domain: ( , ) • Range: (0, ) • y -intercept : (0, 1) • x -axis is a horizontal asymptote ( a x → 0 as x → ).
### Slide 19:
The Natural Logarithmic Function
### The Natural Logarithmic Function:
The Natural Logarithmic Function We will see that f ( x ) = e x is one-to-one and so has an inverse function. This inverse function is called the natural logarithmic function and is denoted by the special symbol ln x , read as “the natural log of x ” or “el en of x. ” Note that the natural logarithm is written without a base. The base is understood to be e .
### The Natural Logarithmic Function:
The Natural Logarithmic Function The definition above implies that the natural logarithmic function and the natural exponential function are inverse functions of each other. So, every logarithmic equation can be written in an equivalent exponential form, and every exponential equation can be written in logarithmic form. That is, y = In x and x = e y are equivalent equations.
### The Natural Logarithmic Function:
The Natural Logarithmic Function Because the functions given by f ( x ) = e x and g ( x ) = In x are inverse functions of each other, their graphs are reflections of each other in the line y = x . This reflective property is illustrated in Figure 3.19. On most calculators, the natural logarithm is denoted by , as illustrated in Example 8. Reflection of graph of f ( x ) = e x about the line y = x . Figure 3.19
### Example 8 – Evaluating the Natural Logarithmic Function:
Example 8 – Evaluating the Natural Logarithmic Function Use a calculator to evaluate the function given by f ( x ) = In x for each value of x . a. x = 2 b. x = 0.3 c. x = –1 d. x = 1 +
### Example 8 – Solution:
Example 8 – Solution Function Value Graphing Calculator Display Keystrokes
### The Natural Logarithmic Function:
The Natural Logarithmic Function The four properties of logarithms are also valid for natural logarithms.
Application
### Example 11 – Human Memory Model:
Example 11 – Human Memory Model Students participating in a psychology experiment attended several lectures on a subject and were given an exam. Every month for a year after the exam, the students were retested to see how much of the material they remembered. The average scores for the group are given by the human memory model f ( t ) = 75 – 6 In( t + 1), 0 t 12, where t is the time in months.
### Example 11 – Human Memory Model:
Example 11 – Human Memory Model a. What was the average score on the original ( t = 0) exam? b. What was the average score at the end of t = 2 months? c. What was the average score at the end of t = 6 months? Solution: a. The original average score was f ( 0 ) = 75 – 6 ln( 0 + 1) = 75 – 6 ln 1 cont’d Substitute 0 for t . Simplify.
### Example 11 – Solution:
Example 11 – Solution = 75 – 6(0) = 75. b. After 2 months, the average score was f ( 2 ) = 75 – 6 ln( 2 + 1) = 75 – 6 ln 3 75 – 6(1.0986) 68.4. cont’d Property of natural logarithms Solution Substitute 2 for t . Simplify. Use a calculator. Solution
### Example 11 – Solution:
Example 11 – Solution c. After 6 months, the average score was f ( 6 ) = 75 – 6 ln( 6 + 1) = 75 – 6 ln 7 75 – 6(1.9459) 63.3. cont’d Substitute 6 for t . Simplify. Use a calculator. Solution
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Introduction to Sequences and Series
A sequence is a list of numbers. Any time you write numbers in a list format, you are creating a sequence. Something as simple as
1, 2, 3, 4, 5, 6, . . .
is a sequence.
Rather than just listing the numbers, we usually identify it as a sequence with the notation
an = 1, 2, 3, 4, 5, 6, . . .
Usually there is some type of pattern to a sequence. In the sequence above, you are adding one to each term to get the next term.
A term of a sequence is just a number that is in the sequence.
Terms can be identified by their location. We note the 1st term in a sequence as a1 and we would call the 5th term in the sequence a5.
We described the pattern in the sequence as adding one to each term to get the next term. We can express this as a recursive formula by writing
an = an-1 + 1
This says to get any term in the sequence (an), add one (+1) to the previous term (an-1).
A recursive formula is written in such a way that in order to find any term in a sequence, you must know the previous terms. In other words, to find the 12th term, you would need to know the first 11. There are times when this can be a difficult task and there will be other ways to write sequences. But it is important to know that many sequences are best described using recursive formulas.
The simple sequence we have been looking at is called an arithmetic sequence. Any time you are adding the same number to each term to complete the sequence, it is called an arithmetic sequence. The number that is added to each term is called the common difference and denoted with the letter d. So in our example we would say that d = 1. The common difference can be subtracting two consecutive terms. You can subtract any two terms as long as they are consecutive. So we could find d by taking 5 - 4 = 1 or 2 - 1 = 1. Notice that we will always use the term that appears later in the sequence first and then subtract the term that is right in front of it.
If we looked at a sequence like bn = 1, 3, 9, 27, 81, 243, . . . this would not fit our definition of an arithmetic sequence. We are not adding the same number to each term. However, notice that we are multiplying each term by the same number (3) each time. When you multiply every term by the same number to get the next term in the sequence, you have a geometric sequence. Geometric sequences can also be written in recursive form. In this case, we would write . Remember that in the language of sequences we are saying, to find any term in the sequence (bn), multiply the previous term (bn-1) by 3.
Just as arithmetic sequences have a common difference, geometric sequences have a common ratio which is denoted with the letter r. The common ratio is found by dividing successive terms in the sequence. So in our geometric sequence example, we could use 9/3 = 3 or 243/81=3 to find that r = 3. As with finding a common difference, when we find a common ratio, we must use the term that appears later in the sequence as our numerator and the number right before it as our denominator.
There are other types of sequences that do not fit into the arithmetic or geometric category, but are still considered sequences because there is a pattern to determining the next term. Our focus in these lessons will be on arithmetic and geometric. For more information on other types of sequences, ask your teacher.
As mentioned earlier, recursive forms of sequences have their drawbacks, but are a useful way to see what is happening in a sequence. The other way to write sequences is called an explicit form or a closed form. These will be explored in other lessons.
Another topic associated with sequences is series. A series is simply adding the terms in a sequence. An arithmetic series involves adding the terms of an arithmetic sequence and a geometric series involves adding the terms of a geometric sequence. These will be explored in other lessons.
This lesson has provided an introduction to the terminology needed to continue working with sequences and series. One important skill is being able to identify what type of sequence you have. Do the “Try These” below and after successful completion of these problems, continue with other lessons on sequences and series.
Examples
Identify each sequence as arithmetic, geometric, or neither. If you identify it as arithmetic, specify d. If you identify it as geometric, specify r.
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$\bf{\text{Solution Outline:}}$ To solve the given inqeuality, $|x+2|\gt x ,$ use the definition of a greater than (greater than or equal to) absolute value inequality and solve each resulting inequality. Finally, graph the solution set. In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} x+2\gt x \\\\\text{OR}\\\\ x+2\gt -x .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} x+2\gt x \\\\ x-x\gt -2 \\\\ 0\gt -2 \text{ (TRUE)} \\\\\text{OR}\\\\ x+2\gt -x \\\\ x+x\gt -2 \\\\ 2x\gt -2 \\\\ x\gt -\dfrac{2}{2} \\\\ x\gt -1 .\end{array} Hence, the solution is the set of all real numbers.
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# R S Aggarwal Solutions for Class 10 Maths Chapter 19 Probability
RS Aggarwal Solutions for class 10 chapter 19 Probability are available here. Class 10, Chapter 19 solutions are prepared by subject experts at BYJU’S based on latest CBSE Syllabus. Students are advised to solve R S Aggarwal Solutions for Maths to sharpen their skills.
This study material is the best resource for the students appearing for the board exams and prepare them for advanced probability problems. Students can download R S Aggarwal Solutions pfd for free to practice offline.
### Access Other Exercise Solutions to Maths R S Aggarwal Chapter 19 – Probability
Get detailed solutions for all the questions listed in each exercise from the links given below:
Exercise 19A Solutions
Exercise 19B Solutions
## Download PDF of RS Aggarwal Solutions for Class 10 Maths Chapter 19
Exercise 19A
Question 1:
Solution:
(i) The probability of an impossible event is zero.
(ii) The probability of a sure event is one.
(iii) For any event E, P(E) + P(not E) = 1.
(iv) The probability of a possible but not a sure event lies between zero and one.
(v) The sum of probabilities of all the outcomes of an experiment is one.
Question 2:
Solution:
A coin is tossed:
Sample space = {H, T}
Total number of outcomes = 2
P(getting a tail) = 1/2.
Question 3:
Solution:
Two coins are tossed:
Sample space = {HH, HT, TH, TT}
Total number of outcomes = 4
P(getting a tail) = 1/2.
Now,
(i) P(exactly 1 head) = 2/4 = 1/2
(ii) P(at most 1 head) = 3/4
(iii) P(at least 1 head) = 3/4
Question 4:
Solution:
In a throw of a dice,
Possible outcomes = {1, 2, 3, 4, 5, 6}
Total number of possible outcomes = 6
Now,
(i)
Favorable outcomes = Even numbers = {2, 4, 6}
Number of favorable outcomes = 3
P(an even number) = 3/6 = 1/2.
(ii)
Favorable outcomes = number less than 5 = { 1, 2, 3, 4}
Number of favorable outcomes = 4
P(a number less than 5) = 4/6 = 2/3.
(iii)
Favorable outcomes = number greater than 2 ={ 3, 4, 5, 6}
Number of favorable outcomes = 4
P(a number greater than 2) = 4/6 = 2/3.
(iv)
Favorable outcomes = a number between 3 and 6 = { 4, 5}
Number of favorable outcomes = 2
P(a number between 3 and 6) = 2/6 = 1/3.
(v)
Favorable outcomes = a number other than 3 = {1, 2, 4, 5, 6}
Number of favorable outcomes = 5
P(a number other than 3) = 5/6.
(vi)
Favorable outcome = a number 5 = { 5}
Number of favorable outcomes = 1
P(a number 5) = 1/6.
Question 5:
Solution:
Total numbers of alphabet = 26
Total numbers of consonants = 21
P(choosing a consonant) = 21/26
Question 6:
Solution:
Total letter on the dice = 6
(i) Number of times A appears = 3
Therefore, numbers of favorable outcomes = 3
P (getting letter A) = 3/6 = 1/2
(ii) Number of times D appears = 1
Therefore, numbers of favorable outcomes = 1
P (getting letter D) = 1/6
Question 7:
Solution:
Given:
Total number of bulbs = 200
Number of defective bulbs = 16
(i)Let A be the event of getting a defective bulb
Total number of defective bulbs = 16
So, P(getting defective bulbs) = P(A) = 16/200 = 2/25
(ii)Let B be the event of getting non-defective bulb
So, P(getting non-defective bulb) = P(B) = 1 – P(A) = 1 – 2/25 = 23/25.
Question 8:
Solution:
Probability of winning a game = 0.7
Let us say, P(A) = 0.7
Let B be probability of losing the game. Find: P(B)
We know that, P(A) + P(B) = 1
=> P(B) = 1-0.7 = 0.3
Question 9:
Solution:
Total numbers of students = 35
Number of girls = 15
Number of boys = 20
(i) Numbers of favorable outcome = 20
P(choosing a boy) = 20/35 = 4/7
(ii) Numbers of favorable outcome are= 15
P(choosing a girl) = 15/35 = 3/7
Question 10:
Solution:
Total numbers of lottery tickets = 10 + 25 = 35
P(getting a prize) = 10/35 = 2/7
Question 11:
Solution:
Total number of tickets sold = 250
Number of prizes = 5
Number of favorable outcomes = 5
P(getting a prize) = 5/250 = 1/50
Question 12:
Solution:
Total numbers of outcomes = 17
(i) Odd numbers on the cards = { 1, 3, 5, 7, 9, 11, 13, 15, 17 }
Numbers of favorable odd numbers = 9
P (getting an old number) = 9/17
(ii) Numbers divisible by 5 form 1 to 17 = {5, 10, 15}
Number of favorable outcome = 3
P (getting a number divisible by 5) = 3/17
Question 13:
Solution:
Total number of outcomes = 8
Factors of 8 from 1 to 8 are {1,2,4,8}
Total number of favorable outcomes = 4
Therefore, P(getting a factor of 8 ) = 4/8 = ½
Question 14:
Solution:
Let E be the event of having at least one boy
The probability that each child will be a girl= 1/2
The probability that each child will be a boy = 1/2
The probability of no boys = All are girls = 1/2 x 1/2 x 1/2 = 1/8
Then, the number of favorable outcome is P(E) –P(not E)=1 where
Let E be the event of having at least one boy, then P(E) is probability of having atleast one boy and
P(not E) = Probability of having no boys.
Therefore,
Probability (at least 1 boy) = 1 – Probability (no boys)= 1 – 1/8 = 7/8
Question 15:
Solution:
Probability = (number of favorable outcomes)/(Total number of outcomes)
A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls.
Total numbers of balls = 4 + 5 + 2 + 4 = 15
(i) Total numbers of black balls = Numbers of favorable outcomes= 2
P(getting a black ball) = 2/15
(ii)
Total numbers of green balls = 4
Numbers of non-green balls = 15 – 4 = 11
P(not green ball) = 11/15
(iii)
Total numbers of red and white balls = 5 + 4 = 9
P(red ball or white ball) = 9/15 = 3/5
(iv)
Total numbers of red and green balls = 5 + 4 = 9
Number of balls which are neither red nor green = 15 – 9 = 6
P (getting neither red nor green ball) = 6/15 = 2/5
Question 16:
Solution:
Total number of cards in a well-shuffled pack=52
(i) Number of red king cards = 2
P(a red king) = 2/52 = 1/26
(ii)
Number of queen cards = 4
Number of jack cards = 4
Total number of favorable outcomes = 4 + 4 = 8
P(a queen or a jack) = 8/52 = 2/13
Question 17:
Solution:
Total number of cards in a well-shuffled pack = 52
Total number of queens and kings cards = 4 + 4 = 8
Therefore, there are 44 cards (52 – 8) that are neither king nor queen.
Total number of favorable outcomes = 44
Required probability = 44/52 = 11/13
Question 18:
Solution:
Total number of cards in a well-shuffled pack = 52
(i) Number of red face cards = 6
P (a red face card) = 6/52 = 3/26
(ii) Number of black king cards = 2
P (a black king) = 2/52 = 1/26
Question 19:
Solution:
The possible outcomes if two dice are tossed together are
Total number of outcomes = 36
(i) The favorable outcomes = getting an even number on each die = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
The number of favorable outcomes = 9
P(getting even number on both dice) = 9/36 = 1/4
(ii) The favorable outcomes = sum of the numbers appearing on two dice is 5 = {(1, 4), (2, 3), (3, 2), (4, 1)}
The numbers of favorable outcomes = 4
P(sum of numbers appearing on two dice is 5) = 4/36 = 1/9
Question 20:
Solution:
When two different dice are rolled simultaneously, then
Total number of outcomes = 36
Favorable outcomes = Sum of the numbers on the two dice is 10 = {(2, 5),(5,2), (5,5)}
Total number of favorable outcomes = 3
P(getting numbers on the dice whose sum is 10) = 3/36 = 1/12
## Exercise 19B
Question 1:
Solution:
Total numbers of cards = 25
(i) The favorable outcome = numbers divisible by 2 = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}
The favorable outcome = number divisible by 3 = {3, 6, 9, 12, 15, 18, 21, 24}
Total number of favorable outcomes = 16
(Note: count 6,12,18 and 24 are common, so consider only one set)
Now,
P(drawn card is divisible by 2 or 3)= 16/25
(ii) The favorable numbers = prime number = 2, 3, 5, 7, 11, 13, 17, 19, 23
Numbers of favorable outcomes = 9
P(drawn card is prime number) = P (E) = 9/25
Question 2:
Solution:
A box contains cards numbered 3, 5, 7, 9,…. , 35, 37.
Total number of cards = 18
Cards with prime numbers = {3, 5, 7, 11, 13, 17,19,23,29,31,37}
Total possible outcomes = 11
Therefore, P(getting a prime number) = 11/18
Question 3:
Solution: Cards numbered 1 to 30 are put in a bag.
So, total number of cards = 30
(i) Numbers divisible by 3 form 1 to 30 = {3,6,9,12,15,18,21,24,27,30}
Total numbers divisible by 3 = 10
Numbers not divisible by 3 = 30 – 10 = 20
P(drawn card is not divisible by 3) = 20/30 = 2/3
(ii) A prime numbers greater than 7 form 1 to 30 = {11, 13, 17, 19, 23, 29}
Total prime number greater than 7 = 6
P(drawn card is a prime number greater than 7) = 6/30 = 1/5
(iii)
Perfect square numbers form 1 to 30 = {1,4,9,16,25}
Number of Perfect square = 5
Non perfect square numbers = 30-5 = 25
P(drawn card is not a perfect square number) = 25/30 = 5/6
Question 4:
Solution:
Total numbers of cards = 18 [odd numbers between 1 and 36]
(i) Favorable numbers = prime number less than 15 = 3, 5, 7, 11, 13
Numbers of favorable outcome = 5
P(prime number less than 15) = 5/18
(ii) Numbers divisible by 3 and 5 = 15 and 30
Since given set contains only odd numbers so exclude 30.
Numbers of favorable outcome = 1
P (getting a number divisible by 3 and 5) = 1/18
Question 5:
Solution:
Apply general term formula of an AP, to find the total numbers of cards.
Here a (first term) = 6, d (common difference) = 1 and l( last term) = 70
l = a + (n-1)d
70 = 6 + (n-1)(1)
n = 65
Total number of cards = 65
(i) Favorable outcome = a one-digit number = 6, 7, 8, 9
Favorable number of outcomes = 4
P(getting a one-digit number) = 4/65
(ii) Favorable outcome = a number divisible by 5 = 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70
Favorable number of outcomes = 13
P(getting a number divisible by 5) = 13/65 = 1/5
iii) Favorable outcome = an odd number less than 30 = 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 and 29
Favorable number of outcomes = 12
P (getting a odd number less than 30) = 12/65
(iv) Favorable outcome = a composite number between 50 and 70 = 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69
The number of favorable outcomes = 15
P(getting a composite number between 50 and 70) = 15/65 = 3/13
Question 6:
Solution:
Apply general term formula of an AP, to find the total numbers of cards.
Here a (first term) = 1, d (common difference) = 2 and l( last term) = 101
l = a + (n-1)d
101 = 1 + (n-1)(2)
n = 51
Total number of cards = 51
(i)
Favorable outcome = Numbers less than 19 = 1, 3, 5, 7, 9, 11, 13, 15, 17
Favorable number of outcomes = 9
P(getting a card number less than 19) = 9/51
(ii) Favorable outcome = a prime number less than 20 = 3, 5, 7, 11, 13, 15, 17, 19
Favorable number of outcomes = 7
P(getting a card with a prime number less than 20) = 7/51
Question 7.
Solution:
Tickets numbered 2, 3, 4, 5, …, 100, 101
Total number of tickets = 100
(i) Even numbers from 2 to 101 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100
Total number of even number = 50
P(getting a even number) = 50/100 = 1/2.
(ii) Numbers less than 16 from 2 to 101 = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
Total number of numbers less than 16 = 14
P(getting a number less than 16) = 14/100 = 7/50.
(iii) Perfect square from 2 to 101 = 4, 9, 16, 25, 36, 49, 64, 81, 100
Total number of perfect squares = 9
P(getting a perfect square) = 9/100.
(iv) Prime numbers less than 40 from 2 to 101 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
Total number of prime numbers = 12
P(getting a prime number less 40) = 12/100 = 3/25.
Question 8.
Solution:
Total number of discs in the box = 80
Perfect squares from 1 to 80 = 1, 4, 9, 16, 25, 36, 49 and 64.
Possible outcomes = 8
P(getting a number which is a perfect square) = 8/80 = 1/10.
(ii) A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(a)A two-digit number (b) a number divisible by 5
Solution:
Total number of discs in the box = 90
Two digit numbers from 1 to 90 = 81
P(getting a two-digit number) = 81/90 = 9/10
(ii) Numbers divisible by 5 from 1 to 90 = 5, 10, …. , 90,
Total numbers = 18
P(getting a number which is divisible by 5) = 18/90 = 1/5
Question 9.
Solution:
Total numbers of cards = 100 + 70 + 50 + 30 = 250
(i) Number of favourable outcomes = seventy ₹ 1 coin = 70
P(getting ₹ 1 coin) = 70/250 = 7/25
(ii) Numbers of ₹ 5 coins = thirty ₹ 5 coins = 30
Numbers of favourable outcomes = will not be a ₹ 5 coin = 250 – 30 = 220
P(getting a coin will not be a ₹ 5 coin) = 220/250 = 22/25
(iii) Number of 50-p coins = 100
Number of ₹ 2 coins = 50
Total coins = 100+50 = 150
P (getting 50-p coin or ₹ 2 coins)= 150/250 = 3/5
Question 10:
Solution:
Let total number of balls = x
Orange balls in the jar = 10
P(getting an orange ball) = 10/x
Since jar contains only red, blue and orange balls.
P(getting a red ball) + P(getting a blue ball) + P(getting a orange ball) = 1
(Because sum of probability = 1)
1/4 + 1/3 + 10/x = 1
3x + 4x + 120 = 12x
x = 24
Hence there are total 24 balls in the jar.
Question 11:
Solution:
Total number of balls = 18
(i) Let x be number of red balls. So, there are (18 – x) non red balls.
P(getting a non red ball) = (18-x)/18
(ii) if 2 more red balls are out in the bag, then there are 20 total number of balls.
Number of red balls x + 2
So we have,
P(getting a red ball) = 9/8 p(getting a red ball in first option)
Now,
(x+2)/20 = 9/8(x/18)
16x + 32 = 20x
X = 8
Question 12:
Solution:
Total number of marbles = 24
Let x be the number of blue marbles in the jar.
P(getting a blue marble) = x/24
There are 2 types of marbles in the jar, blue and green.
P(getting a blue ball) + P(getting a green ball) = 1
(Because sum of probability = 1)
2/3 + x/24 = 1
(16+3x)/24 = 1
x = 4
There are 4 blue marbles in the jar.
Question 13:
Solution:
Total number of marbles = 54
let x be the number of white marbles in the jar.
P (getting a white marble) = x/54
There are 3 types of marbles in the jar, blue, green and white.
P(getting a blue ball) + P(getting a green ball) + P(getting a white ball) = 1
(Because sum of probability = 1)
1/3 + 4/9 + x/54 = 1
(54 – x)/54 = 7/9
x = 12
There are 12 white marbles in the jar.
Question 14:
Solution:
Total number of shirts = 100
Number of good shirts = 88
Number of shirts with minor defects = 8
Number of shirts with major defects = 100- 88 – 8 = 4
(i) P(shirt acceptable by Rohit) = 88/100 = 22/25
(ii) Given: Kamal only rejects shirt with major defects
So, there are 96 possible outcomes.
P(shirt acceptable by kamal) = 96/100 = 24/25
## R S Aggarwal Solutions for Chapter 19 Probability
In this chapter students will study important concepts on Probability as listed below:
• Probability Introduction
Probability is a concept which numerically measures the degree of certainty of the occurrence of events.
Experiment: An operation which can produce some well-defined outcomes
Random Experiment: An experiment in which all possible outcomes are known, and the exact outcome cannot be predicted in advance, is called a random experiment
• Possible outcomes in various experiments
• Equally likely terms
• Probability of occurrence of an event.
• Word Problems in real-life situations
### Key Features of R S Aggarwal Solutions for Class 10 Maths Chapter 19 Probability
1. R S Aggarwal Solutions will be useful for Olympiads, board exams, and other competitive exams.
2. Easy for quick revision.
3. Prepared based on the latest CBSE syllabus by subject experts.
4. Provides comprehensive answers to all the questions.
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# X and Y Axis in Graph – Definition, Graph, Facts, Examples
Home » Math Vocabulary » X and Y Axis in Graph – Definition, Graph, Facts, Examples
An axis in mathematics is defined as a line that is used to make or mark measurements. The x and y-axis are two important lines of the coordinate plane. The x-axis is a horizontal number line and the y-axis is a vertical number line. These two axes intersect perpendicularly to form the coordinate plane. The x-axis is also called the abscissa and the y-axis is called the ordinate.
Any point on the coordinate plane can be located or represented using these two axes in the form of an ordered pair of the form (x,y). Here, x represents the location of the point with respect to the x-axis and y represents the location of the point with respect to the y-axis.
The origin is where the two axes intersect and is written as (0,0).
## Plotting Points on X and Y Axis
Let us learn how to plot a point on the graph by using the X- and Y-axis.
For example: Let’s try to plot the point B(3,4) on the graph.
Here, the x-coordinate of B is 3. So we will start from the origin and move 3 units to the right on x-axis.
Now, the y-coordinate of B(3,4) is 4, so we will go 4 spaces up from this point.
And thus we have plotted our point B(3,4) on the graph using the axes.
## Representing a Linear Equation on X- and Y-Axis
To understand how to represent a linear equation on the graph using the X- and Y-axis,
let us consider a linear equation, y = x + 1.
Now, let’s build a table to represent the corresponding values of y for different values of x and create their ordered pairs:
The next step is to plot these ordered pairs on the coordinate plane graph.
As a final step, we will join these points to form a straight line and that will be the representation of the equation y = x + 1.
## Solved Questions On X and Y Axis
Question 1: Which of the following points lie on the x-axis?
(0, 1) (4, 0) (7, 7) (−5, 0) (−4, 4) (0, −5) (8, 0) (6, 0)
Answer: Since the coordinates lying on x-axis have their y coordinate zero (0), the following points will lie on x-axis:
(4, 0) (−5, 0) (8, 0) (6, 0)
Question 2: Two different points are to be plotted on a graph. If the given points are (3,2) and (2,3), then plot these two points on the X- and Y-axis. Also, find out the point where the straight line going through these points meets the x-axis.
Answer: For (3,2), as we can see, the x-coordinate point is 3, and the y-coordinate point is 2.
Similarly we can plot the point (2,3).
Now, we can join both points with a straight line when we have plotted both points. After extending the straight line, we see that this line intersects the x-axis at point (5,0).
Question 3: For a linear equation y = 2x + 6, find the point where the straight line meets y-axis on the graph.
Answer: On y-axis, the x-coordinate of the point is 0. Therefore, we can find the intersection point of y-axis and y = 2x + 6 by simply putting the value of x as 0 and finding the value of y. y = 2(0)+6 = 0 + 6 = 6.
So the straight line of the equation y = 2x + 6 meets the y-axis at (0,6).
## Practice Problems On X and Y Axis
1
### What is the x-axis called?
Ordinate
Abscissa
Applicate
None of the above
CorrectIncorrect
Correct answer is: Abscissa
x - axis is also called abscissa.
2
### What is the correct way of representing a point on a graph?
(X-coordinate, X-coordinate)
(Y-coordinate, X-coordinate)
(Y-coordinate, Y-coordinate)
(X-coordinate, Y-coordinate)
CorrectIncorrect
Correct answer is: (X-coordinate, Y-coordinate)
(X-coordinate, Y-coordinate) is the correct way to represent a point.
3
### How is the origin point represented on a graph?
(0,0)
(0,x)
(y,0)
(x,y)
CorrectIncorrect
Correct answer is: (0,0)
(0, 0) is the coordinates of origin on the graph.
4
### A point (0,5) will be lie on the
X-axis
Y-axis
Origin
None of the above
CorrectIncorrect
Correct answer is: Y-axis
As abscissa (x- coordinate) is 0. So, given point lies on the y-axis.
## Frequently Asked Questions On X and Y Axis
The X- and Y-axis are essential for a graphical representation of data. These axes make the coordinate plane. The data is located in coordinates according to their distance from the X- and Y-axis. Graphical representation helps in solving complicated equations.
A coordinate plane is a two-dimensional plane formed by the intersection of two number lines. One of these number lines is a horizontal number line called the x-axis and the other number line is a vertical number line called the y-axis (or ordinate). These two number lines intersect each other perpendicularly and form the coordinate plane.
The two number lines divide the coordinate plane into 4 regions. These regions are called quadrants. The quadrants are denoted by roman numerals and each of these quadrants have their own properties. X and Y have different signs in each quadrant.
• Quadrant I: (x,y)
• Quadrant II: (-x,y),
• Quadrant III: (-x,-y),
• Quadrant IV: (x,-y).
The X-axis gives the horizontal location of a point, and the Y-axis gives a vertical location of a point.
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# Practical Application: Calculating the Standard Deviation
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
Knowing the standard deviation of a set of data is important as it provides a good basis for deciding whether or not a certain data point fits the rest of the data or not.
## Standard Deviation
When a teacher says that her students' test scores all follow a normal distribution, she means that the majority of her test scores fall within one standard deviation. This range of test scores changes for each class as it is based on the test scores for each class.
According to the lesson Calculating the Standard Deviation, the steps to calculate the standard deviation are these:
1. Find the mean or average of the data set.
2. Subtract the mean from each data point.
3. Square each of the differences.
4. Average the squares to find the variance.
5. Take the square root of the variance to find the standard deviation.
One standard deviation is found by adding and subtracting the standard deviation from the mean. Two standard deviations is found by adding and subtracting the standard deviation from the one standard deviation range.
## Questions to Consider
Practice finding the standard deviation with the following three scenarios. Use these questions as an aid when calculating the standard deviation.
• Did you square each of the differences?
After finding the difference of each data point from the average, make sure to square each difference.
• Did you perform an average calculation twice?
The first average calculation is of all the data points and the second average calculation is that of the squares of the differences.
• Are you taking the square root of the variance?
Make sure to get the variance and then take the square root of that to find the standard deviation.
## Test Scores
This first scenario is inside a classroom. Jen, the high school science teacher has just finished grading the chapter test. Find the standard deviation of Jen's science class with these results.
Test Scores
98
97
87
80
99
100
100
89
92
94
Following the steps, the calculation starts with finding the mean.
• (98 + 97 + 87 + 80 + 99 + 100 + 100 + 89 + 92 + 94) / 10 = 93.6
Then, this mean is subtracted from each data point.
98 - 93.6 = 4.4 97 - 93.6 = 3.4 87 - 93.6 = -6.6 80 - 93.6 = -13.6 99 - 93.6 = 5.4 100 - 93.6 = 6.4 100 - 93.6 = 6.4 89 - 93.6 = -4.6 92 - 93.6 = -1.6 94 - 93.6 = 0.4
After this step, then these differences are squared.
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Chapter 6.
Introduction to Probability Sampling Distributions:
From Jagged Complexity to Streamlined Simplicity
"Simplicity, simplicity, simplicity! I say, let
your affairs be as two or three, and not a
hundred or a thousand. ... Simplify, simplify."
—Thoreau, Walden
In the historical development of probability theory, the first step in the streamlining occurred in connection with what are known as binomial probabilities. Although the term is perhaps new to you, the concept it describes is simply an extension of the matters we have just been discussing in Chapter 5. A binomial probability situation is one, such as the mere-chance occurrence of heads when tossing coins, or the mere-chance recovery of patients from a disease, for which you can specify
p the probability that the event or outcome in question will occur in any particular instance E.g., p=.5, the probability that any particular tossed coin will come up as a head; p=.4, the probability that any particular patient will spontaneously recover. q the probability that the event or outcome in question will not occur in any particular instance E.g., q=.5, the probability that any particular tossed coin will not come up as a head; q=.6, the probability that any particular patient will not spontaneously recover. N the number of instances in which the event or outcome has the opportunity to occur
To show how the streamlining came about, we will begin by considering the very simple binomial probability situation in which N is equal to 2, and then work our way up through some more complex binomial situations. If you are tossing 2 coins with elemental probabilities of p=.5 for the outcome of getting a head on any particular one of the coins (labeled below as "H") and q=.5 for the outcome of not getting a head (labeled as "--"), the possible conjunctive outcomes and their associated probabilities are
Coin Sub-pathwayProbability Numberof Heads Main PathwayProbability A B -- -- .5x.5=.25 0 .25 (25%) --H H-- .5x.5=.25.5x.5=.25 1 .50 (50%) H H .5x.5=.25 2 .25 (25%) Totals 1.0 (100%)
Similarly, if you are considering 2 randomly selected patients with elemental probabilities of p=.4 for any particular one of the patients spontaneously recovering (labeled below as "R") and q=.6 for not spontaneously recovering (labeled as "--"), the possible conjunctive outcomes and their associated probabilities are
Patient Sub-pathwayProbability Number ofRecoveries Main PathwayProbability A B -- -- .6x.6=.36 0 .36 (36%) --R R-- .6x.4=.24.4x.6=.24 1 .48 (48%) R R .4x.4=.16 2 .16 (16%) Totals 1.0 (100%)
In Figure 6.1 we display these two sets of probabilities side-by-side in the form of two histograms. In each case, you can think of the total area of the histogram as representing 100% of the total probability that applies to that particular situation. Thus, for both examples there is a 100% chance that the outcome will include either zero or 1 or 2 of the particular events in question—heads or patient recoveries. All that differs is the way in which the 100% total is divided up by the 3 possible outcomes: 25%|50%|25% for the coin-toss example versus 36%|48%|16% for the patient-recovery example.
Figure 6.1. Two Binomial Sampling Distributions for the Case where N=2
These two sets of probabilities are illustrative of a large class of theoretical structures collectively known as probability sampling distributions, so named because they describe the probabilities for the entire range of outcomes that are possible in the particular situations that are under consideration.
Following this chapter is an appendix that will dynamically generate the graphical outlines and numerical details of the binomial sampling distribution for any values of p and q, and for any value of N between 1 and 40, inclusive.
Thus, the sampling distribution for the coin-toss situation (N=2, p=.5, q=.5) specifies that any particular randomly selected sample of 2 tossed coins has a 25% chance of including zero heads, a 50% chance of including exactly 1 head, and a 25% chance of including 2 heads. The sampling distribution for the patient-recovery situation (N=2, p=.4, q=.6) specifies that any particular sample of 2 randomly selected patients who have come down with this disease has a 36% chance of ending up with zero recoveries, a 48% chance of ending up with exactly 1 recovery, and a 16% chance of ending up with 2 recoveries. By extension, these two sampling distributions would also specify that any particular randomly selected sample of 2 tossed coins has a 50%+25%=75% chance of including at least 1 head, and that any particular sample of 2 randomly selected patients has a 48%+16%=64% chance of ending up with at least one recovery.
Another way of interpreting a probability sampling distribution would be to say that it describes the manner in which the outcomes of any large number of randomly selected samples will tend to be distributed. Thus, a large number of randomly selected samples of 2 tossed coins would tend to have 25% of the samples including zero heads, 50% including exactly 1 head, and 25% including 2 heads. A large number of randomly selected 2-patient samples would tend to have 36% of the samples with zero recoveries, 48% with exactly 1 recovery, and 16% with 2 recoveries.
The interpretation of central tendency and variability for a sampling distribution is essentially the same as for any other distribution. The central tendency of the distribution describes the average of all the outcomes, and its variability is the measure of the tendency of individual outcomes to be dispersed away from that average. For the particular case of a binomial probability sampling distribution, these parameters can be easily determined according to the formulas given below. [Note the new symbols used here. When referring to an entire population of potential outcomes, the convention is to use (lower-case Greek letter "mu") for the mean and (lower-case Greek letter "sigma") for the variance and standard deviation.]
mean: = Np
variance: 2 = Npq
standard deviation: = sqrt[Npq]
Thus, for the coin-toss example, with N=2, p=.5, and q=.5, you have
mean: = 2x.5 = 1.0
variance: 2 = 2x.5x.5 = 0.5
standard deviation: = sqrt[0.5] = ±0.71
And for the patient-recovery example, with N=2, p=.4, and q=.6, you have
mean: = 2x.4 = 0.8
variance: 2 = 2x.4x.6 = 0.48
standard deviation: = sqrt[0.48] = ±0.69
Although we are illustrating these various points about probability sampling distributions with the specific examples of coin tosses and patient recoveries, please keep in mind as we proceed that these distributions are abstract and general, in the sense that they apply to any probability situation at all that has certain defining properties. Thus, the sampling distribution shown in the left-hand histogram of Figure 6.1 is not limited to the example of tossing two coins. It applies to any situation at all in which the probability of a certain event is p=.5, the complementary probability of its non-occurrence is q=.5, and there are N=2 opportunities for it to occur. Similarly, the sampling distribution shown in the right-hand histogram of Figure 6.1 applies to any situation at all in which the defining properties are p=.4, q=.6, and N=2.
At any rate, in examining Figure 6.1 you can hardly fail to notice that the two sampling distributions have rather different shapes. Both are unimodal, but whereas the one for the coin-toss situation (p=.5, q=.5) is symmetrical, the one for the patient-recovery example (p=.4, q=.6) has a pronounced positive skew. But now see what happens when the size of the sample is increased from N=2 to N=10. The specific probabilities for the outcomes depicted by the histogram columns in Figure 6.2 have been calculated using the factorial and exponential formula examined in Chapter 5,
P(k out of N) = N!k!(N—k)! x pk x qN-kT
substituting N=10 along with the appropriate values of p, q, and k.
Figure 6.2. Two Binomial Sampling Distributions for the Case where N=10
Please take a moment to compare the outlines of Figure 6.2 with those of Figure 6.1, for the transformation is quite remarkable. As a result of increasing the size of the sample to N=10, both sampling distributions have grown smoother, and the one for the patient-recovery example has in addition grown much more symmetrical. You will also surely recognize the smooth curve that has been superimposed upon the two sampling distributions in Figure 6.2 as the by-now familiar outline of the normal distribution.
In Figure 6.3 we increase the size of the sample to N=20, and there you can see these trends carried even further. The principle illustrated by Figures 6.1, 6.2, and 6.3 is that, as the size of N increases, the shape of a binomial sampling distribution comes closer and closer to the shape of the normal distribution. Eventually it comes so close as to be equivalent to the normal distribution, for all practical purposes—and that is the point where all the jagged complexity that we have been describing suddenly gives way to a smooth, streamlined simplicity. The reason for this remarkable convergence is explained in SideTrip 6.1
Figure 6.3. Two Binomial Sampling Distributions for the Case where N=20
If the elemental probabilities, p and q, are both equal to .5, as in the coin-toss example, a sufficiently close approximation to the normal distribution is reached at the point where N is equal to 10. If p and q are anything other than .5, it is reached at some point beyond N=10. In general, a binomial sampling distribution may be regarded as a sufficiently close approximation to the normal distribution if the products of Np and Nq are both equal to or greater than 5. That is
Np>5 and Nq>5 Criterion for determining that a binomial sampling distribution is a sufficiently close approximation of the normal distribution.
For elemental probabilities of .4 and .6, as in our patient-recovery example, the point of sufficiently close approximation is reached at N=13; for .3 and .7, it is reached at N=17; for .2 and .8, it is reached at N=25; and so on. The simple way of determining these points is to divide 5 by either p or q, whichever is smaller, rounding the result up to the next higher integer if it comes out with a decimal fraction. Thus, for .4 and .6 you have 5/.4=12.5, which rounds up to N=13; for .3 and .7 it is 5/.3=16.67, which rounds up to N=17; and so on.
The graph of the normal distribution that appears in Figure 6.4 will help you see where all this is leading. Once you know that a distribution is normal, or at least a close approximation of the normal, you are then in a position to specify the proportion of the distribution that falls to the left or right of any particular point along the horizontal axis, or alternatively the proportion that falls in-between any two points along the horizontal axis. And specifying the proportion is tantamount to specifying the probability. For the standardized normal distribution, these points along the horizontal axis are marked out in units of z, with each unit of z being equal to 1 unit of standard deviation, .
Figure 6.4. The Unit Normal Distribution
Most of the percentage values that appear in the body of the graph will already be familiar to you. To the right of z=+1 falls 15.87% of the total distribution; to the right of z=+2 falls 2.28% of the total distribution; and so on. And since the normal distribution is precisely symmetrical, the same percentages also fall to the left of corresponding negative values of z. For reasons that will be evident in a moment, Figure 6.4 also shows shaded areas corresponding to the proportions of the normal distribution that fall to the left of z=1.14 and to the right of z=+1.14—in each case, 12.71%. An abbreviated version of the standard table of the normal distribution is shown in Table 6.1. The complete version is given in Appendix A. (Please note that the entries in Table 6.1 are not listed as percentages but as proportions; e.g., .1587 instead of 15.87%.)
Table 6.1. Proportions of the normal distribution falling to the left of negative values of z or to the right of positive values of z. Bold-face entries pertain to examples discussed in the text.
±z 0.00 0.20 0.40 0.60 0.80 1.00 1.14 1.20 1.40 1.60 AreaBeyond ±z .5000 .4702 .3446 .2743 .2119 .1587 .1271 .1151 .0808 .0548 ±z 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00 AreaBeyond ±z .0359 .0287 .0228 .0139 .0082 .0047 .0026 .0013 .0002 .00003
Now turn to Figure 6.5 where you will see this same standardized normal distribution superimposed on the binomial sampling distribution that applies to our patient-recovery example with a sample size of 20. The defining properties of this sampling distribution are N=20, p=.4, q=.6. Hence
mean: = 20x.4 = 8.0
standard deviation: = sqrt[20x.4x.6] = ±2.19
Figure 6.5. Binomial Sampling Distribution for N=20, p=.4, q=.6
You will note that there are two scales along the horizontal axis of Figure 6.5. The first, labeled k, delineates "number of recoveries in 20 patients"; and the second, labeled z, is a direct translation of this binomial k scale into units of standard deviation, with each unit of standard deviation equal to ±2.19 units of k. Any particular one of the discrete values on the k scale—0, 1, 2, ... , 18, 19, 20—can be directly translated into its corresponding value on the z scale by application of the formula
z = (k— )±.5 Note 1. Only the positive value of is used in the denominator of this expression. Note 2. The component ±.5 ('plus or minus .5') in the numerator is a 'correction for continuity', aimed at transforming the zig-zag angularity of the binomial distribution into the smooth curve of the normal distribution. It achieves this effect by reducing the distance between k and the mean by one-half unit. Thus, when k is smaller than the mean you add half a unit (+.5), and when k is larger than the mean you subtract half a unit (—.5).
Thus, for k=5 we have
z = (5—8 )+.52.19 = —2.52.19 = —1.14
and for k=11:
z = (11—8 )—.52.19 = +2.52.19 = +1.14
With the calculation of z for a binomial outcome of the general type "k or fewer" or "k or more," you are then in a position to use the normal distribution for assessing the probability associated with that outcome. You simply treat your calculated value of z as though it belonged to the z scale of the normal distribution—and read off the corresponding probability value from the table of the normal distribution. The two particular values of z that we have just calculated are z=1.14 for "5 or fewer" recoveries and z=+1.14 for "11 or more" recoveries. As we saw a moment ago in Figure 6.5 and Table 6.1, the proportions of the normal distribution falling to the left of z=1.14 and to the right of z=+1.14 are in each case equal to .1271. Hence, on the basis of the normal distribution we would judge that with a random sample of 20 patients there is a 12.71% chance that as few as 5 will spontaneously recover, and an equal 12.71% chance that as many as 11 will spontaneously recover. If you work out the exact binomial probabilities for these two outcomes using the factorial and exponential formula given earlier, you will find that they come out to 12.56% for "5 or fewer" and 12.75% for "11 or more." The 12.71% probability values arrived at via the normal distribution do not hit these targets precisely, but for all practical purposes they are close enough. The difference between .1271 and .1256 is only .0015; between .1271 and .1275 it is only .0004.
To appreciate just how useful and labor-saving this streamlining can be, return for a moment to the second experiment of our medical researchers, the one in which there were 430 recoveries out of 1,000 patients. The question is: In a random sample of 1,000 patients, how likely is it that as many as 430 (k>430) would recover by mere chance, if the plant extract had no effectiveness whatsoever? First try to imagine how much it would cost you in time and patience to perform the requisite 571 exact binomial calculations according to the factorial and exponential formula, assuming you had access to a calculator powerful enough to perform huge operations such as 1,000! and .4430x.6570and then reflect on how very much easier it is to reach the same goal on the basis of the normal distribution.
First establish the defining properties of the relevant binomial sampling distribution, making sure that the products of Np and Nq are both equal to or greater than 5:
N=1,000, p=.4, q=.6. Hence
mean: = 1,000x.4 = 400
standard deviation: = sqrt[1,000x.4x.6] = ±15.49
Then perform the simple calculation that translates the details of the binomial situation into the language of the normal distribution:
z = (430—400 )—.515.49 = +29.515.49 = +1.90
And then refer your calculated value of z to the table of the normal distribution. For the moment you need only refer to the abbreviated version shown in Table 6.1, where you will find that the proportion of the normal distribution falling to the right of z=+1.9 is equal to .0287.
The meaning of this proportion is illustrated in Figure 6.6. Of all the possible spontaneous-recovery outcomes that could have occurred within this sample—zero recoveries, 1 recovery, 2 recoveries, and so on—only 2.87% would include as many as 430 recoveries. The mere-chance probability of the observed result is therefore P=.0287, which of course clears the conventional cut-off point for statistical significance (P<.05) by a considerable margin.
Figure 6.6. Location of z=+1.90 within the Unit Normal Distribution
Moving beyond the rather rigid mechanics of the .05 criterion of statistical significance, you can also think of this P=.0287 probability value in terms of confidence. There is only about a 3% likelihood that the observed outcome would have occurred by mere chance coincidence. Hence, you can be about 97% confident that it is the result of something other than mere chance coincidence—presumably some active ingredient of the plant extract. The researchers can now go on to the costly task of isolating and refining that ingredient, with a high degree of confidence that they are on the right track.
As we indicated earlier, the streamlining of binomial probabilities was only the first historical step. In the course of time there came the discovery and detailed analysis of several other families of sampling distributions, each providing a relatively simple streamlined method for answering inferential-statistical questions of the general type: How likely is it that such-and-such might occur by mere chance coincidence? In the remaining chapters of this text we will be making extensive use of some of these families of sampling distributions, in particular, several that are variations on the normal distribution, plus some others that are known as t-distributions, F-distributions, and chi-square distributions. Of all of these, it is the family of chi-square sampling distributions whose logic follows most closely upon the concepts we have been developing in the present chapter, so that is where we will go next, in Chapter 8. First, however, in Chapter 7, another brief interlude on the general concept of statistical significance.
This chapter includes two appendices: Binomial Sampling Distribution Generator, which will produce a graphic and numerical display of the properties of a binomial sampling distribution; and z to P Calculator, which will calculate the probabilities associated with any particular value of z.
End of Chapter 6.
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# 4. The Binomial Theorem
by M. Bourne
A binomial is an algebraic expression containing 2 terms. For example, (x + y) is a binomial.
We sometimes need to expand binomials as follows:
(a + b)0 = 1
(a + b)1 = a + b
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
Clearly, doing this by direct multiplication gets quite tedious and can be rather difficult for larger powers or more complicated expressions.
## Pascal's Triangle
We note that the coefficients (the numbers in front of each term) follow a pattern. [This was noticed long before Pascal, by the Chinese.]
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
You can use this pattern to form the coefficients, rather than multiply everything out as we did above.
## The Binomial Theorem
We use the binomial theorem to help us expand binomials to any given power without direct multiplication. As we have seen, multiplication can be time-consuming or even not possible in some cases.
## Properties of the Binomial Expansion (a + b)n
• There are n + 1 terms.
• The first term is an and the final term is bn.
• Progressing from the first term to the last, the exponent of a decreases by 1 from term to term while the exponent of b increases by 1. In addition, the sum of the exponents of a and b in each term is n.
• If the coefficient of each term is multiplied by the exponent of a in that term, and the product is divided by the number of that term, we obtain the coefficient of the next term.
## General formula for (a + b)n
First, we need the following definition:
Definition: n! represents the product of the first n positive integers i.e.
n! = n(n − 1)(n − 2) ... (3)(2)(1)
We say n! as "n factorial".
### Example 1 - factorial values
Here are some factorial values:
(a) 3! = (3)(2)(1) = 6
(b) 5! = (5)(4)(3)(2)(1) = 120
(c) (4!)/(2!)=((4)(3)(2)(1))/((2)(1))=12
Note: (4!)/(2!) cannot be cancelled down to 2!.
## Factorial Interactive
Instructions: You can use the following interactive to find the factorial of any positive integer up to 30.
For numbers greater than 22!, you'll see output something like this: 2.652528e+32. The "e" stands for exponential (base 10 in this case), and the number has value 2.652528 xx 10^32.
## Binomial Theorem Formula
Based on the binomial properties, the binomial theorem states that the following binomial formula is valid for all positive integer values of n:
(a+b)^n= a^n+na^(n-1)b +(n(n-1))/(2!)a^(n-2)b^2 +(n(n-1)(n-2))/(3!)a^(n-3)b^3 +...+b^n
This can be written more simply as:
(a + b)n = nC0an + nC1an − 1b + nC2an − 2b2 + nC3an − 3b3 + ... + nCnbn
We can use the {::}^nC_r button on our calculator to find these values.
This can also be written nCr.
## Binomial Theorem Interactive
The following interactive lets you expand your own binomial expressions. It shows all the expansions from n=0 up to the power you have chosen.
In the first line of each expansion, you'll see the numbers from Pascal's Triangle written within square brackets, [ ].
The second line of each expansion is the result after tidying up.
Instructions: You can use letters or numbers within the brackets. The maximum power you can use is 6.
Here are the expansions:
### Example 2
Using the binomial theorem, expand (x + 2)6.
### Example 3
Using the binomial theorem, expand (2x + 3)4
### Example 4
Using the binomial theorem, find the first four terms of the expansion (2a-1/x)^11
## Binomial Series
From the binomial formula, if we let a = 1 and b = x, we can also obtain the binomial series which is valid for any real number n if |x| < 1.
(1+x)^n=1+nx+(n(n-1))/(2!)x^2+(n(n-1)(n-2))/(3!)x^3+...
NOTE (1): This is an infinite series, where the binomial theorem deals with a finite expansion.
NOTE (2): We cannot use the {::}^nC_r button for the binomial series. The {::}^nC_r button can only be used with positive integers.
### Example 5
Using the binomial series, find the first four terms of the expansion sqrt(4+x^2).
Here is the graph of what we just did in Example 5. The darker colored curve is
y_1=sqrt(4+x^2)
The lighter colored one is the first 4 terms of the series we found, that is:
y_2=2+x^2/4-x^4/64+x^6/512.
The approximation is quite good between −2 < x < 2, but we would need to take many more terms for a good approximation beyond these bounds.
### Online Algebra Solver
This algebra solver can solve a wide range of math problems. (Please be patient while it loads.)
### Algebra Lessons on DVD
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# How to Determine If an Equation Is a Linear Function Without Graphing?
••• PhotographyFirm/iStock/GettyImages
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A linear function creates a straight line when graphed on a coordinate plane. It is made up of terms separated by a plus or minus sign. To determine if an equation is a linear function without graphing, you will need to check to see if your function has the characteristics of a linear function. Linear functions are first-degree polynomials.
Check that the y, or independent variable, is by itself on one side of the equation. If it is not, rearrange the equation so that it is. For example, given the equation 5y + 6x = 7, move the 6x term to the other side of the equation by subtracting it from both sides. This yields 5y = 7 - 6x. Then divide both side by 5 so you have y = 7/5 - (6/5)x.
Determine whether the equation is a polynomial or not. For an equation to be a polynomial, the power of the independent or "x" variable of each term must be a whole number. The terms can be made up of constants and variables. If the equation is not a polynomial, it is not a linear equation. In the example, y = 7/5 - (6/5)x has one "x" term and its power is 1. Because 1 is a whole number, y = 7/5 - (6/5)x is a polynomial.
Determine whether the equation is a first-degree polynomial. Locate the exponent with the highest degree out of the terms. That exponent is the degree of the polynomial. If it is one, it is a linear equation. Because the highest power of "x" in y = 7/5 - (6/5)x is 1, it is a linear function.
#### Tips
• Make sure no variable is multiplied by another variable in the function. If that's the case, it's not a linear equation.
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# Compiled by : S. Agarwal, Lecturer & Systems Incharge St. Xavier¶s Computer Centre, St. Xavier¶s College Kolkata.
March-2003
Introduction
A number system defines a set of values used to represent quantity. We talk about the number of people attending class, the number of modules taken per student, and also use numbers to represent grades achieved by students in tests. Quantifying values and items in relation to each other is helpful for us to make sense of our environment. We do this at an early age; figuring out if we have more toys to play with, more presents, more lollies and so on. The study of number systems is not just limited to computers. We apply numbers every day, and knowing how numbers work will give us an insight into how a computer manipulates and stores numbers.
The Romans devised a number system which could represent all the numbers from 1 to 1,000,000 using only seven symbols I = 1 V = 5 X = 10 L = 50 C = 100 D = 500 M = 1000 A small bar placed above a symbol indicates the number is multiplied by 1000.
8. 4. Base ten number systems are called decimal number systems. . 0. and 9. 5. 6. In decimal we have ten numerals. 7. 1. 2. 3.The Decimal Number System The primary number system used is a base ten number system.
. And so on. Then we reset the ones place to zero and increment the tens place to get 20. with zeros to the right of the 3. The next number is 10. We continue to count by adding one to the ones place until we run out of digits again at 19. So the number after 299 is 300. That is.Counting in decimal: When we count. We continue in this way until we reach 99. we create a place for the tens digit and put zero in the ones position. Each time we change a digit. all the digits to its right are set to zero. we start at 1 and count up to 9. At that point we create a third digit to hold hundreds and reset the tens and ones place to zero to get 100.
.9000 9 * 1000 9 * 103 + 700 9735 = + 30 is equivalent to 7 * 100 is equivalent to 3 * 10 7 * 102 3 * 101 +5 5*1 5 * 100 So 9735 = (9 * 103) + (7 * 102) + (3 * 101) + (5 * 100).
The Octal Number System
We can use a number system with only eight numerals,
0 through 7. A system with only eight numerals is called octal.
Counting in Octal
In octal, instead of ten digits, we only have eight. So we resort to changing the digits to the left more frequently. Counting in base 8, we get:
0 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 20 21 22 23 24 25 26 27
Since the octal number 10 follows 7, 10 in octal is 8 in decimal. 11 in octal is 9 in decimal. And so on.
octaldecimal 0-0 1-1 2-2 3-3 4-4 5-5 6-6 7-7 octaldecimal 10 - 8 11 - 9 12 - 10 13 - 11 14 - 12 15 - 13 16 - 14 17 - 15 octaldecimal 20 - 16 21 - 17 22 - 18 23 - 19 ... 24 - 20 25 - 21 26 - 22 27 - 23 74 - 60 75 - 61 76 - 62 77 - 63 104 - 68 105 - 69 106 - 70 107 - 71 octaldecimal octaldecimal 70 - 56 71 - 57 72 - 58 73 - 59 octaldecimal 100 - 64 101 - 65 102 - 66 103 - 67 ... octaldecimal
.Octal Digit Positions and Values In base 8. each digit occupies a position worth eight times the position to its right. So 5732 (octal) is 3034 (decimal). So if 5732 is an octal number. instead of ten times as in base 10. it can be read as: 5732 = 5000 (octal) = + 700 (octal) = + 30 (octal) = + 2 (octal) = 5 * 1000 (octal) = 7 * 100 (octal) = 3 * 10 (octal) = 2 * 1 (octal) = 5 * 83 = 7 * 82 = 3 * 81 = 2 * 80 = 5 * 512 (decimal) = 7 * 64 (decimal) = 3 * 8 (decimal) = 2 * 1 (decimal) = 2560 (decimal) 448 (decimal) 24 (decimal) 2 (decimal) Total = 3034 (decimal) We total the decimal values of each octal digit to get the decimal equivalent.
we simply repeat To convert 3034 in decimal to octal. the second remainder will be the second digit from the right. The first remainder will be the last digit in the octal number.To convert a decimal number to dividing by 8 and saving the remainder. octal. all we need to do is write the remainders right-to-left as we divide by 8: Step Divide Equals Remainder (1) (3) (5) (6) 3034 / 8 = 379 / 8 = 47 / 8 = 5/8= 379 47 5 0 2 3 7 5 Digits 2 32 732 5732 . and so on. start by dividing 3034 by 8 to get 379 with a remainder of 2. That is.
then use the letters A for 10. C for 12. . E for 14 and F for 15.The Hexadecimal Number System We can use a number system with sixteen numerals. We can use our usual 0 through 9 for the first ten digits. B for 11. A base 16 number system is call hexadecimal or just hex. D for 13.
we get: 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 20 21 22 23 24 25 26 27 28 29 2A 2B 2C 2D 2E 2F . we count the same same way we did in decimal and octal. we have sixteen. But instead of ten digits or eight digits. Counting in base 16. So we resort to changing the digits to the left less frequently than in decimal or octal. A-F. 0-9.Counting in Hexadecimal In hexadecimal.
246 106 .Since the hex number A follows 9. After F in hex (15 in decimal) we have 10 (16 in decimal). 8-8 18 .19 23 .27 2B .24 28 .32 F0 ..43 FB .258 3-3 13 . B in hex is 11 in decimal.37 F5 .262 7-7 17 .269 F . hex-dec hex-dec hex-dec hex-dec hex-dec hex-dec hex-dec 0-0 10 .15 1F ..18 22 .264 9-9 19 .240 100 .29 2D .39 F7 .30 2E .25 29 .40 F8 .34 F2 .260 5-5 15 .23 27 .255 10F .42 FA .271 .22 26 .265 A .248 108 ..48 FF .244 104 .253 10D .46 FD .241 101 .251 10B .247 107 .252 10C .35 F3 .. And so on.267 C .12 1C .41 F9 .256 1-1 11 .249 109 .259 4-4 14 .243 103 .21 25 .261 6-6 16 .17 21 .10 1A .245 105 .20 24 .14 1E .253 10D .31 2F .263 .33 F1 .13 1D .45 FD .266 B .268 D .269 E .257 2-2 12 .250 10A .28 2C .11 1B .36 F4 .44 FC . A in hex is 10 in decimal.16 20 .242 102 .38 F6 . .26 2A .
So if 3F72 is an hex number.Hex Digit Positions and Values In base 16. it can be read as: 3F72 = 3000 (hex) = + F00 (hex) = + 70 (hex) = + 2 (hex) = 3 * 1000 (hex) = F * 100 (hex) = 7 * 10 (hex) = 2 * 1 (hex) = 3 * 163 = 15 * 162 = 7 * 161 = 2 * 160 = 3 * 4096 (decimal) = 15 * 256 (decimal) = 7 * 16 (decimal) = 2 * 1 (decimal) = 12288 (decimal) 3840 (decimal) 112 (decimal) 2 (decimal) Total = 16242 (decimal) We total the decimal values of each hex digit to get the decimal equivalent. So 3F72 (hex) is 16242 (decimal). each digit occupies a position worth sixteen times the position to its right. . instead of ten times as in base 10 or eight times as in octal.
In the following steps we convert 16242 from decimal to hex: Step (1) (2) (3) (4) Divide 16242 / 16 = 1015 / 16 = 63 / 16 = 3 / 16 = Equals 1015 63 3 0 Remainder 2 = 2 (hex) 7 = 7 (hex) 15 = F (hex) 3 = 3 (hex) Digits 2 72 F72 3F72 So 16242 in decimal is written as 3F72 in hex. The only difference is that we divide by 16 each time since we are working in base 16.Converting Decimal to Hex We can convert a decimal to hex using the same procedure we used to convert decimal to octal. .
01001001010101110101010001010101101 10101010000000111101010101011111010 .The Binary Number System The base 2 number system. called binary is based on powers of 2 and contains only two digits. 0 and 1.
Counting in Binary With only two numerals. 1 (one) and 0 (zero). Just keep in mind the following: 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 10 1 + 1 + 1 = 11 . counting in binary is pretty simple.
5 110 .30 11111 .6 111 .We we would count in binary as follows: bin-dec 0-0 1-1 10 .2 11 .17 10010 .31 .13 1110 .24 11001 .3 100 .10 1011 .9 1010 .27 11100 .15 bin-dec 10000 .22 10111 .18 10011 .23 bin-dec 11000 .11 1100 .16 10001 .25 11010 .21 10110 .7 bin-dec 1000 .28 11101 .14 1111 .8 1001 .19 10100 .12 1101 .29 11110 .20 10101 .26 11011 .4 101 .
it can be read as: . each digit occupies a position worth two times the position to its right. eight times as in octal.Binary Digit Positions and Values In base 2. instead of ten times as in base 10. or 16 as in hex. So if 1101001 is a binary number.
1101001 = 1000000 (bin) = 1 * 26 = 1 * 64 (decimal) = 64 (decimal) + 100000 (bin) = + 00000 (bin) = 1 * 25 = 1 * 32 (decimal) = 32 (decimal) 0 * 24 = 0 * 16 (decimal) = 1 * 8 (decimal) = 0 * 4 (decimal) = 0 * 2 (decimal) = 1 * 1 (decimal) = 0 (decimal) 8 (decimal) 0 (decimal) 0 (decimal) 1 (decimal) + 1000 (bin) = 1 * 23 = + 000 (bin) = + 00 (bin) = + 1 (bin) = 0 * 22 = 0 * 21 = 1 * 20 = TOTAL = 105 (decimal) We total the decimal values of each binary digit to get the decimal equivalent. So 1101001 (binary) is 105 (decimal). .
. In the following steps we convert 105 from decimal to binary: Step Divide Equals Remainder Digits (1) (2) (3) (4) (5) (6) (7) 105 / 2 = 52 / 2 = 26 / 2 = 13 / 2 = 6/2= 3/2= 1/2= 52 26 13 6 3 1 0 1 0 0 1 0 1 1 1 01 001 1001 01001 101001 1101001 So 105 in decimal is written as 1101001 in binary.Converting Decimal to Binary We can convert a decimal to binary by dividing by 2 each time since we are working in base 2.
and 8 is 23.7 .6 111 .26 .30 11111 . the connection is even more obvious: bin-octal-dec 0-0-0 1-1-1 10 .20 10101 .15 bin-octal-dec 10000 .13 1110 .12 1101 .19 10100 .29 11110 .9 1010 . That is.15 .24 .26 11011 . octal and hex is simple.32 .28 11101 .21 .25 11010 . Octal is base 8.31 .8 1001 . If we line up the binary numbers and octal numbers.27 .4 101 .5 110 .31 .7 bin-octal-dec 1000 . Octal and Binary Converting between binary.2 11 .23 bin-octal-dec 11000 .10 .11 .33 .11 1100 .13 .Converting Between Hex.14 .25 .21 10110 .10 1011 .27 11100 .18 10011 .17 .14 1111 .5 .34 .37 . it takes exactly three binary digits to make one octal digit.16 10001 .3 .2 .16 .12 .22 .6 .36 .17 10010 .35 . Binary is base 2.3 100 .22 10111 .23 .4 .24 11001 .30 .20 .
we get: 10 110 100 111 100 101 101 001 011 2 6 4 7 4 5 5 1 3 That is. If we take the digits in groups of three from right to left and convert. . 10110100111100101101001011 (binary) is 264745513 (octal). Consider the binary number 10110100111100101101001011.What this means is that we can convert from binary to octal simply by taking the binary digits in groups of three and converting.
all we have to do is convert each octal digit to three binary digits.Converting from octal to binary is just as easy.) . Converting 7563021 in octal to binary goes as follows: 7 111 5 101 6 110 3 011 0 000 2 010 1 001 So 7563021 (octal) is 111101110011000010001 (binary. Since each octal digit can be expressed in exactly three binary digits.
For humans. smaller. octal is very useful to programmers. And since it is so easy to convert between binary and octal.Since (almost!) all computers have a binary architecture. and less prone to errors than binary. easier to work with. octal is a favored number system for programmers. octal is more concise. .
By grouping in fours and converting. hex is base 16 and 16 is 24.In the same way. By taking binary digits in groups of four (right to left) we can convert binary to hex. it takes exactly four binary digits to make a hex digit. Consider once more the binary number 10110100111100101101001011. we get: 10 2 1101 D 0011 3 1100 C 1011 B 0100 1011 8 B So 10110100111100101101001011 (binary) is the same number as 2D3CB8B (hex). . That is. and the same number as 264745513 (octal).
. 6F037C2 (hex) is 110111100000011011111000010 (binary).Converting from hex to binary. In this way we can convert 6F037C2: 6 0110 1111 F 0000 0 0011 3 0111 7 1100 C 2 0010 Since we can drop the leading zero. simply write each hex digit as four binary digits.
These days. octal tends to be used by programmers who come from a minicomputer background or who work with unicode. Hex tends to be preferred by programmers with a mainframe background or who work with colors. Many programmers are at home with either. .
Maximum number in a nibble : 1111 = 15 A nibble can be conveniently represented by one hexadecimal digit. .nibble In computers and digital technology. a nibble is four binary digits or half of an eight-bit byte.
or BCD. A decimal digit is represented by four binary digits. . as shown below: The binary combinations 1010 to 1111 are invalid and are not used. is a method of using binary digits to represent the decimal digits 0 through 9.BCD Binary-coded decimal.
4 9 0100 1001 = 01001001BCD (PACKED BCD) Each decimal digit is converted to its binary equivalent. 4910 in binary is 1100012. For example. four bits are used to keep one digit.UNPACKED & PACKED BCD In unpacked BCD. but 4910 in BCD is 01001001BCD. BCD and binary are not the same. one in the lower half and one in the upper half of the byte. so two digits are stored in one byte. . a decimal digit is stored in one byte and in packed BCD.
below each digit write the BCD equivalent of that digit: The BCD equivalent of 26410 is 001001100100BCD . First. For example.BCD Conversion Conversion of decimal to BCD or BCD to decimal is similar to the conversion of hexadecimal to binary and vice versa. let's go through the conversion of 26410 to BCD. then. We'll use the block format that you used in earlier conversions. write out the decimal number to be converted.
simply reverse the process as shown: The DECIMAL equivalent of 100110000011BCD is 98310 .To convert from BCD to decimal.
numbers. . The Extended ASCII Character Set also consists of 128 decimal numbers and ranges from 128 through 255 representing additional special.ASCII Code ASCII. pronounced "ask-key". mathematical. The standard ASCII character set consists of 128 decimal numbers ranging from zero through 127 assigned to letters. and foreign characters. punctuation marks. and the most common special characters. is the common code for microcomputer equipment. graphic.
.The Standard ASCII character set is divided into four groups of 32 characters. The first 32 characters. We call them control characters because they perform various printer/display control operations rather than displaying symbols. form a special set of noncharacters printing characters called the control characters.
There is very little standardization among output devices.Examples of common control characters include: carriage return which positions the cursor to the left side of the current line of characters. line feed which moves the cursor down one line on the output device back space which moves the cursor back one position to the left Unfortunately. different control characters perform different operations on different output devices. you will need to consult its manual. To find out exactly how a control character affects a particular device. .
special characters. and the numeric digits. By subtracting 30h from the ASCII code for any particular digit you can obtain the numeric equivalent of that digit.The second group of 32 ASCII character codes comprise various punctuation symbols. . The most notable characters in this group include the: space character numeric digits 0 through 9 Note that the numeric digits differ from their numeric values only in the high order nibble.
Since there are only 26 different alphabetic characters. group of 32 ASCII character codes are reserved for the lower case alphabetic symbols. and final. The fourth. The ASCII codes for the characters "A" through "Z" lie in the range 41h through 5Ah.The third group of 32 ASCII characters is reserved for the upper case alphabetic characters. . the remaining six codes hold various special symbols. five additional special symbols. and another control character (delete).
The differences between scripts primarily are in their written forms.ISCII Indian Script Code for Information Interchange The ISCII code table is a super-set of all the characters required in the ten Brahmi-based Indian scripts. . where different combination rules get used. An optimal keyboard overlay for these scripts is made possible by the phonetic nature of the alphabet.
. while the Indian script keyboard overlay is designed for the standard English can co-exist with Indian scripts.The 8-bit ISCII code retains the standard ASCII code. so long as 8-bit character codes are allowed. This approach also makes it feasible to use Indian scripts along with existing English computers and software.
provides a logical and intuitive arrangement of vowels and consonants. Alternating between the English and Inscript overlay is achieved through the CAPSLOCK key. The INSCRIPT keyboard. It is based both on the phonetic properties and the relative usage frequencies of the letters. This overlays fits on any existing English keyboard. Not only does this made the keyboard much easier to learn.The common INSCRIPT keyboard overlay allows typing of all the ten Indian scripts. . but also enables a person to type subsequently in all the Indian scripts.
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whereas the last four bits are the called the digit and identify the specific character. customised for different countries. ASCII are normally used instead. EBCDIC takes up eight bits. The first four bits are called the zone and represent the category of the character. . which are divided in two pieces. There are a number of different versions of EBCDIC.EBCDIC Extended Binary Coded Decimal Interchange Code It is an 8 bit character encoding used on IBM mainframes and AS/400s. Outside of such IBM systems. EBCDIC is generally considered an anachronism. It is descended from punched cards and the corresponding six bit Binary Coded Decimal Code that most of IBM's computer peripherals of the late 1950s and early 1960s used.
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--. 0 0 1 1 0 1 0 1 --.--.--.--0 01 01 10 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 --.--.--.--.--11 10 10 01 10 01 01 00 . The right bit of the result is placed under the column of bits. The left bit is called the "carry out of the column".The Binary Addition Algorithm To add two 1-bit (representations of) integers: Count the number of ones in a column and write the result in binary.--.--.
A A B Carry Digit Unit Digit 0 0 0 0 0 0 0 0 0 1 0 1 1 1 1 1 1 1 0 1 B 0 0 1 1 0 0 1 1 C 0 1 0 1 0 1 0 1 CARRY 0 0 0 1 0 1 1 1 UNIT 0 1 1 0 1 0 0 1 1 0 0 1 .
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Signed Binary Integers We will not be using a minus sign (-) to represent negative numbers. This has some difficulties. we will use a method called two's complement notation which avoids the pitfalls of one's complement. There are a few ways to represent negative binary numbers. . among them the fact that zero can be represented in two different ways (for an eight bit number these would be 0000 0000 and 1111 1111).. where the sign of a binary number is changed by simply toggling each bit (0's become 1's and vice-versa). The simplest of these methods is called ones complement. We would like to represent our binary numbers with only two symbols. 0 and 1.
The value of bits in signed and unsigned binary numbers Bit 7 Unsi 7 2 = 128 gned Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0 26 = 64 25 = 32 24= 16 23= 8 22 = 4 21 = 2 20 = 1 Sign -(27) = -128 26= 64 25 = 32 24= 16 23= 8 ed 22 = 4 21= 2 20 = 1 .To represent an n bit signed binary number the leftmost bit. has a special significance. The difference between a signed and an unsigned number is given in the table below for an 8 bit number.
when Bit 7 is set. . However. the 1111 1111 255 -1 number is always negative.Let's look at how this changes the value of some binary numbers Binary Unsigned Signed If Bit 7 is not set (as in the first example) the 0010 0011 35 35 representation of signed and unsigned numbers 1010 0011 163 -93 is the same. For this reason 1000 0000 128 -128 Bit 7 is sometimes called the sign bit.
Since binary uses only 0 & 1. 11012 = -510 (negative) . | . One solution is to reserve a bit (circuit) that does nothing but represent the mathematical sign: . Extra bit. Extra bit. | . 1=negative) . we don't have a third symbol such as a "minus" sign. 1012 = 510 (positive) . 01012 = 510 (positive) . 1=negative) .NEGATIVE BINARY NUMBER Usually we represent a negative decimal number by placing a minus sign directly to the left of the most significant digit. representing sign (0=positive. just as in the example above. . with -5. . representing sign (0=positive. since these circuits can only be on or off (two possible states).
defining our number limits just as before. However. | .There's another method for representing negative numbers called complementation. a value of negative five would be represented as such: Extra bit. it possesses a negative place-weight value. and the leftmost bit representing either zero or negative eight. we assign the leftmost bit to serve a special purpose. For example. . 10112 = 510 (negative) . (1 x -810) + (0 x 410) + (1 x 210) + (1 x 110) = -510 With the right three bits being able to represent a magnitude from zero through seven. place weight = negative eight . With this strategy. rather. the leftmost bit is more than just a sign bit. . just as we did with the sign-magnitude approach. we can successfully represent any integer number from negative seven (10012 = -810 + 710 = -110) to positive seven (01112 = 010 + 710 = 710). this time.
. Note that the negative binary numbers in the right column. zero 0000 positive one 0001 negative one 1111 positive two 0010 negative two 1110 positive three 0011 negative three 1101 positive four 0100 negative four 1100 positive five 0101 negative five 1011 positive six 0110 negative six 1010 positive seven 0111 negative seven 1001 . don't "count" in the same progression as the positive binary numbers in the left column. being the sum of the right three bits' total plus the negative eight of the leftmost bit.Examples «. the right three bits have to be set at the proper value to equal the desired (negative) total when summed with the negative eight place value of the leftmost bit. Rather.
we would first invert all the bits to obtain 0102 (the "one's complement"). changing all 1's to 0's and visa-versa (to arrive at what is called the one's complement) and then add one! For example. simply invert all the bits of that number. two's complement form. to obtain the two's complement of five (1012). To get the two¶s complement.The two¶s complement : The two's complement for any positive binary number will be whatever value is needed to add to the binary number to make that positive value's negative equivalent. . or -510 in three-bit. then add one to obtain 0112.
but performing the complementation process on all four bits. Then. five (01012).Interestingly enough. converting a positive five to a negative five. inverting all bits to obtain the one's complement: 10102. adding one. . First. including the leftmost (sign) bit at the same time as the magnitude bits. We must be sure to include the 0 (positive) sign bit on the original number. Let's try this with the former example. or -510 expressed in four-bit. two's complement form. generating the two's complement of a binary number works the same if you manipulate all the bits. we obtain the final answer: 10112.
Here. and a negative five as 111110112. If our binary numeration field were such that the eighth bit was designated as the negative-weight bit (100000002). A positive five in this system would be represented as 000001012. the two's complement of five (00001012) would be 11110112. we'd have to determine the two's complement based on all seven of the other bits.It is critically important to remember that the place of the negative-weight bit must be already determined before any two's complement conversions can be done. .
if we can leverage the already familiar (and easier) technique of binary addition to subtract. we'll use those negative binary numbers to subtract through addition. "borrowing" as needed from bits to the left). Here. that would be better. right to left. we can represent negative binary numbers by using the "two's complement" method and a negative place-weight bit. However.Subtraction We can subtract one binary number from another by using the standard techniques adapted for decimal numbers (subtraction of each bit pair. . As we just learned.
all we need is three bits plus the negative-weight bit: positive seven = 01112 negative five = 10112 Now.Carry bits . or positive two. Answer = 00102 Since we've already defined our number bit field as three bits plus the negative-weight bit. which is the correct answer. | . .510 Addition equivalent: 710 + (-510) Represent seven and negative five in binary (two's complemented) form.Subtraction: 710 . 0111 . let's add them together: . -----. + 1011 . 1111 <--. . the fifth bit in the answer (1) will be discarded to give us a result of 00102. 10010 . Discard extra bit .
in two's complement form. we need at least five bits. Converting the answer to decimal form by summing all the bits times their respective weight values. let's add them together . let's represent positive eighteen in binary form. Now. which is twenty-five. we're representing negative twenty-five by using the negative-weight (sixth) bit with a value of negative thirty-two. as it should be. . plus positive seven (binary 1112). plus a sixth bit for the negativeweight bit. 111001 Since there were no "extra" bits on the left. -------. The leftmost bit on the answer is a 1. 1810 = 0100102 . which means that the answer is negative. we must first decide how large our binary bit field must be. we get: (1 x -3210) + (1 x 1610) + (1 x 810) + (1 x 110) = -710 . +2510 = 0110012 (showing all six bits) One's complement of 110012 = 1001102 One's complement + 1 = two's complement = 1001112 -2510 = 1001112 Essentially. showing all six bits: . there are no bits to discard.Another example : add -2510 to 1810 If we want to add -2510 to 1810.Carry bits . To represent the largest (absolute value) number in our problem. 11 <--. + 010010 . 100111 . Now.
we have a representation range of 25. .Overflow One problrm with signed binary numbers is that of overflow. or as low as -3210 (1000002). where the answer to an addition or subtraction problem exceeds the magnitude which can be represented with the alloted number of bits. With the last example problem. or sign. bit. and the result either exceeds +3110 or is less than -3210. Remember that the place of the sign bit is fixed from the beginning of the problem. and the left-most (sixth) bit as the negativeweight. our answer will be incorrect. we used five binary bits to represent the magnitude of the number. If we set up an addition problem with two binary numbers. With five bits to represent magnitude. or thirty-two integer steps from 0 to maximum. the sixth bit used for sign. This means that we can represent a number as high as +3110 (0111112).
1 11 <--. 100100 The answer (1001002). interpreted with the sixth bit as the -3210 place. 1710 = 100012 1910 = 100112 . is actually equal to -2810. + 010011 . -------. (Showing sign bits) 010001 . .Let's try adding 1710 and 1910 to see how this overflow condition works for excessive positive numbers: .Carry bits . not +3610 as we should get with +1710 and +1910 added together! .
Simply put. we have an overflow error. not +3610 as we should get with +1710 and +1910 added together! Obviously. What went wrong? The answer lies in the restrictions of the six-bit number field within which we're working Since the magnitude of the true and proper sum (3610) exceeds the allowable limit for our designated bit field. so whatever figure we obtain using the strategy of discarding the leftmost "carry" bit will be incorrect. is actually equal to -2810. this is not correct.The answer (1001002). interpreted with the sixth bit as the -3210 place. six places doesn't give enough bits to represent the correct sum. .
1 1111 <--. . -1710 = 1011112 -1910 = 1011012 . Discard extra bit FINAL ANSWER: 0111002 = +2810 The (incorrect) answer is a positive twenty-eight.A similar error will occur if we add two negative numbers together to produce a sum that is too low for our six-bit binary field. Let's try adding -1710 and -1910 together to see how this works (or doesn't work. The fact that the real sum of negative seventeen and negative nineteen was too low to be properly represented with a five bit magnitude field and a sixth sign bit is the root cause of this difficulty. -------. as the case may be!): . . 1011100 .Carry bits . + 101101 . | . (Showing sign bits) 101111 .
. and allowing the use of 6 bits for representing the magnitude: . . except this time using the seventh bit for a sign bit. . .Let's try these two problems again. . . . . . . . 1710 + 1910 1 11 0010001 + 0010011 --------0100100 (-1710) + (-1910) 11 1111 1101111 + 1101101 --------11011100 | Discard extra bit 10111002 = -3610 ANSWERS: 01001002 = +3610 By using bit fields sufficiently large to handle the magnitude of the sums. we arrive at the correct answers.
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# Finding the equation of a circle given two points on the circle
11. Find the equation of the circle which touches $x^{2} + y^{2} - 6x + 2y + 5 = 0$ at $(4, -3)$ and passes through $(0, 7)$.
My textbook has a worked example for obtaining the equation of a circle from three points on the circle. It also talks you through obtaining the equation of a circle if you're given two points on the circle and if it touches an axis (in this case you know a coordinate of the centre will be $\pm r$, where $r$ is your radius.)
I am reasonably well-practised at using these two techniques. I have also been practising finding the length of a tangent from a given point.
However, I have so far been unable to make the "jump" to this question, I suspect there's something I'm not seeing. So I was hoping for a hint that would help me work out how to approach this question.
Lets say the equation of the circle is given by:$$(x-a)^2+(y-b)^2=r^2$$Make use of the fact that the circle passes through $(4,-3)$ and $(0,7)$ to form two equations.
You are also told that it touches the circle:$$x^{2} + y^{2} - 6x + 2y + 5 = 0$$at $(4,-3)$ which mean the tangents of both circles at this point must be equal. This gives you a third equation.
You now have three equations and three unknowns which you should be able to solve.
• So we have $(4 - a)^{2} + (-3 - b)^{2} = r^{2}$, $a^{2} + (7 - b)^{2} = r^{2}$ and, unless I'm much mistaken, we also have the equation of the tangent to the circle at $(4, -3)$, which is $4y - 7x + 40 = 0$. But, I'm sorry, I'm not sure how to go from here to three valid equations pertaining to our circle in $a$, $b$ and $r$. – Au101 Apr 27 '15 at 22:24
• Use implicit differentiation to work out the slope of the tangent line at $(4,-3)$ for both circles. These slopes must be equal - this gives you your third equation. – Mufasa Apr 27 '15 at 22:30
let the circle $$(x-3)^2+ (y+1)^2 = 5$$ has a tangent at $(4, -3)$ with the circle $C.$ the line connecting the centers of the circle has slope $$\frac{-1-(-3)}{3-4} = -2$$.therefore the center of $C$ is at $(3+t, -1-2t)$ for some $t.$
since both $(4,-3), (0,7)$ are on $C,$ equating the radius squared, we have $$(3+t-4)^2 (-1-2t+3)^2 =(3+t-0)^2 +(-1-2t-7)^2$$ that gives $$-8(3+t)+16 + -6(1+2t)+9=14(1+2t)+49\to 48t=-68, t = -17/7$$
$$\text{ center of } C \text{ is }(4/7, 27/7), \text{ radius is } 7.666$$
• Sounds good, but - forgive me - why $(3 + t, -1 - 2t)$. I understand that the centre of $(x - 3)^{2} + (y + 1)^{2} = 5$ (the given circle) is $(3, -1)$, but why do we add $t$ to the $x$-coordinate and take $2t$ from the $y$-coordinate, I don't think I've followed. – Au101 Apr 27 '15 at 23:14
• @Au101, the reason is the the centers and the common point of contact are collinear. please see my edit. – abel Apr 27 '15 at 23:41
I've come back to this question after a while and have found a solution which agrees with that in the textbook. My method is based primarily on the tips given by @Mufasa.
Let the circle which touches $x^{2} + y^{2} − 6x + 2y + 5 = 0$ at $(4,−3)$ and passes through $(0,7)$ be $C_{1}$.
Let the centre of $C_{1}$ be $(p, q)$ and let the radius be $a$.
$(4 - p)^{2} + (-3 - q)^{2} = a^{2}$
$p^{2} + (7 - q)^{2} = a^{2}$
Let the circle $x^{2} + y^{2} − 6x + 2y + 5 = 0$ be $C_{2}$.
The centre of $C_{2}$ is $(3, -1)$.
The gradient of the radius of $C_{2}$ to $(4, -3)$ is $-2$.
$\therefore$ The gradient of the tangent to $C_{2}$ at $(4, -3)$ is $\frac{1}{2}$.
$\therefore$ The gradient of the tangent to $C_{1}$ at $(4, -3)$ is $\frac{1}{2}$.
$\therefore$ The gradient of the radius of $C_{1}$ to $(4, -3)$ is $-2$.
$\therefore \dfrac{q - (-3)}{p - 4} = -2$
$2p + q - 5 = 0 \qquad (1)$
$(4 - p)^{2} + (-3 - q)^{2} = a^{2} \qquad (2)$
$p^{2} + (7 - q)^{2} = a^{2} \qquad (3)$
Equating (2) and (3), we ultimately get:
$p = \dfrac{5}{2}q - 3 \qquad (4)$
Substituting $p$ into (1), we get:
$q = \dfrac{11}{6}$
Substituting $q$ into (4), we get:
$p = \dfrac{19}{12}$
Substituting $p$ and $q$ into (3) we ultimately get:
$a^{2} = \dfrac{4,205}{144}$
This gives us our equation of $C_{1}$:
$$\left(x - \frac{19}{12}\right)^{2} + \left(y - \frac{11}{6}\right)^{2} = \frac{4,205}{144}$$
Expanding and simplifying, we get:
$$6x^{2} + 6y^{2} - 19x - 22y - 140 = 0.$$
|
# Sampling distribution
title
Sampling Distribution
author
Yolande Tra
## Introduction
### Learning Objectives。
1. Define inferential statistics
2. Graph a probability distribution for the mean of a discrete variable
3. Describe a sampling distribution in terms of "all possible outcomes"
4. Describe a sampling distribution in terms of repeated sampling
5. Describe the role of sampling distributions in inferential statistics
6. Define the standard error of the mean
### Define inferential statistics。
Suppose you randomly sampled 10 people from the population of women in Houston, Texas, between the ages of 21 and 35 years and computed the mean height of your sample. You would not expect your sample mean to be equal to the mean of all women in Houston. It might be somewhat lower or it might be somewhat higher, but it would not equal the population mean exactly. Similarly, if you took a second sample of 10 people from the same population, you would not expect the mean of this second sample to equal the mean of the first sample.
Recall that inferential statistics concern generalizing from a sample to a population. A critical part of inferential statistics involves determining how far sample statistics are likely to vary from each other and from the population parameter. (In this example, the sample statistics are the sample means and the population parameter is the population mean.) As the later portions of this chapter show, these determinations are based on sampling distributions.
### Discrete Distributions。
We will illustrate the concept of sampling distributions with a simple example. Figure 1 shows three pool balls, each with a number on it. Two of the balls are selected randomly (with replacement) and the average of their numbers is computed.
Figure 1. The pool balls. All possible outcomes are shown below in Table 1.
```Table 1. All possible outcomes when two balls are sampled with replacement.
Outcome Ball 1 Ball 2 Mean
1 1 1 1.0
2 1 2 1.5
3 1 3 2.0
4 2 1 1.5
5 2 2 2.0
6 2 3 2.5
7 3 1 2.0
8 3 2 2.5
9 3 3 3.0
```
Notice that all the means are either 1.0, 1.5, 2.0, 2.5, or 3.0. The frequencies of these means are shown in Table 2. The relative frequencies are equal to the frequencies divided by nine because there are nine possible outcomes.
```Table 2. Frequencies of means for N = 2.
Mean Frequency Relative Frequency
1.0 1 0.111
1.5 2 0.222
2.0 3 0.333
2.5 2 0.222
3.0 1 0.111
```
### 。
Figure 2 shows a relative frequency distribution of the means based on Table 2.
This distribution is also a probability distribution since the Y-axis is the probability of obtaining a given mean from a sample of two balls in addition to being the relative frequency.
Figure 2. Distribution of means for N = 2. The distribution shown in Figure 2 is called the sampling distribution of the mean. Specifically, it is the sampling distribution of the mean for a sample size of 2 (N = 2).
For this simple example, the distribution of pool balls and the sampling distribution are both discrete distributions. The pool balls have only the values 1, 2, and 3, and a sample mean can have one of only five values shown in Table 2.
### 。
There is an alternative way of conceptualizing a sampling distribution that will be useful for more complex distributions. Imagine that two balls are sampled (with replacement) and the mean of the two balls is computed and recorded. Then this process is repeated for a second sample, a third sample, and eventually thousands of samples. After thousands of samples are taken and the mean computed for each, a relative frequency distribution is drawn. The more samples, the closer the relative frequency distribution will come to the sampling distribution shown in Figure 2. As the number of samples approaches infinity, the relative frequency distribution will approach the sampling distribution. This means that you can conceive of a sampling distribution as being a relative frequency distribution based on a very large number of samples. To be strictly correct, the relative frequency distribution approaches the sampling distribution as the number of samples approaches infinity.
### Continuous Distributions。
In the previous section, the population consisted of three pool balls. Now we will consider sampling distributions when the population distribution is continuous. What if we had a thousand pool balls with numbers ranging from 0.001 to 1.000 in equal steps? (Although this distribution is not really continuous, it is close enough to be considered continuous for practical purposes.) As before, we are interested in the distribution of means we would get if we sampled two balls and computed the mean of these two balls. In the previous example, we started by computing the mean for each of the nine possible outcomes. This would get a bit tedious for this example since there are 1,000,000 possible outcomes (1,000 for the first ball x 1,000 for the second). Therefore, it is more convenient to use our second conceptualization of sampling distributions which conceives of sampling distributions in terms of relative frequency distributions. Specifically, the relative frequency distribution that would occur if samples of two balls were repeatedly taken and the mean of each sample computed.
When we have a truly continuous distribution, it is not only impractical but actually impossible to enumerate all possible outcomes. Moreover, in continuous distributions, the probability of obtaining any single value is zero. Therefore, as discussed in the section "Introduction to Distributions," these values are called probability densities rather than probabilities.
### Sampling Distributions and Inferential Statistics。
As we stated in the beginning of this chapter, sampling distributions are important for inferential statistics. In the examples given so far, a population was specified and the sampling distribution of the mean and the range were determined. In practice, the process proceeds the other way: you collect sample data and from these data you estimate parameters of the sampling distribution. This knowledge of the sampling distribution can be very useful. For example, knowing the degree to which means from different samples would differ from each other and from the population mean would give you a sense of how close your particular sample mean is likely to be to the population mean. Fortunately, this information is directly available from a sampling distribution. The most common measure of how much sample means differ from each other is the standard deviation of the sampling distribution of the mean. This standard deviation is called the standard error of the mean. If all the sample means were very close to the population mean, then the standard error of the mean would be small. On the other hand, if the sample means varied considerably, then the standard error of the mean would be large.
To be specific, assume your sample mean were 125 and you estimated that the standard error of the mean were 5 (using a method shown in a later section). If you had a normal distribution, then it would be likely that your sample mean would be within 10 units of the population mean since most of a normal distribution is within two standard deviations of the mean.
Keep in mind that all statistics have sampling distributions, not just the mean.
In later sections we will be discussing the sampling distribution of the variance, the sampling distribution of the difference between means, and the sampling distribution of Pearson's correlation, among others.
```Questions
```
1 When does the sampling distribution equal the relative frequency distribution?
When the distribution is discrete When the distribution is continuous When there are at least 20 samples When there are infinite samples When it's the sampling distribution of the mean
The more samples, the closer the relative frequency distribution will come to the sampling distribution. The sampling distribution only equals the relative frequency distribution exactly when there is an infinite number of samples.
2 Select all that apply. Which of these statistics has a sampling distribution?
Mean Median Range Standard Deviation Pearson's r
Every statistic has a sampling distribution.
3 What is the standard error of the mean?
The standard deviation of the sampling distribution of the mean The standard deviation of the standard normal distribution The amount the scores in the population vary from the mean The difference between the mean of your first sample and the mean of your second sample
The standard error of the mean is the standard deviation of the sampling distribution of the mean.
## Sampling Distribution of the Mean。
### Learning Objectives
1. State the mean and variance of the sampling distribution of the mean
2. Compute the standard error of the mean
3. State the central limit theorem
The sampling distribution of the mean was defined in the section introducing sampling distributions. This section reviews some important properties of the sampling distribution of the mean introduced in the demonstrations in this chapter.
### Mean
The mean of the sampling distribution of the mean is the mean of the population from which the scores were sampled. Therefore, if a population has a mean μ, then the mean of the sampling distribution of the mean is also μ. The symbol μM is used to refer to the mean of the sampling distribution of the mean. Therefore, the formula for the mean of the sampling distribution of the mean can be written as:
μM = μ
### Variance
The variance of the sampling distribution of the mean is computed as follows:
That is, the variance of the sampling distribution of the mean is the population variance divided by N, the sample size (the number of scores used to compute a mean). Thus, the larger the sample size, the smaller the variance of the sampling distribution of the mean.
The standard error of the mean is the standard deviation of the sampling distribution of the mean. It is therefore the square root of the variance of the sampling distribution of the mean and can be written as:
The standard error is represented by a σ because it is a standard deviation. The subscript (M) indicates that the standard error in question is the standard error of the mean.
### Central Limit Theorem。
The central limit theorem states that: Given a population with a finite mean μ and a finite non-zero variance σ2, the sampling distribution of the mean approaches a normal distribution with a mean of μ and a variance of σ2/N as N, the sample size, increases.
The expressions for the mean and variance of the sampling distribution of the mean are not new or remarkable. What is remarkable is that regardless of the shape of the parent population, the sampling distribution of the mean approaches a normal distribution as N increases.
If you have used the "Central Limit Theorem Demo," you have already seen this for yourself. As a reminder, Figure 1 shows the results of the simulation for N = 2 and N = 10. The parent population was a uniform distribution. You can see that the distribution for N = 2 is far from a normal distribution. Nonetheless, it does show that the scores are denser in the middle than in the tails.
For N = 10 the distribution is quite close to a normal distribution. Notice that the means of the two distributions are the same, but that the spread of the distribution for N = 10 is smaller.
Figure 1. A simulation of a sampling distribution. The parent population is uniform. The blue line under "16" indicates that 16 is the mean. The red line extends from the mean plus and minus one standard deviation.
Figure 2 shows how closely the sampling distribution of the mean approximates a normal distribution even when the parent population is very non-normal. If you look closely you can see that the sampling distributions do have a slight positive skew. The larger the sample size, the closer the sampling distribution of the mean would be to a normal distribution.
Figure 2. A simulation of a sampling distribution. The parent population is very non-normal.
## Sampling Distribution of Difference Between Means。
### Learning Objectives
1. State the mean and variance of the sampling distribution of the difference between means
2. Compute the standard error of the difference between means
3. Compute the probability of a difference between means being above a specified value
Statistical analyses are very often concerned with the difference between means. A typical example is an experiment designed to compare the mean of a control group with the mean of an experimental group. Inferential statistics used in the analysis of this type of experiment depend on the sampling distribution of the difference between means.
The sampling distribution of the difference between means can be thought of as the distribution that would result if we repeated the following three steps over and over again:
1. (1) sample n1 scores from Population 1 and n2 scores from Population 2
2. (2) compute the means of the two samples (M1 and M2)
3. (3) compute the difference between means, M1 - M2. The distribution of the differences between means is the sampling distribution of the difference between means.
As you might expect, the mean of the sampling distribution of the difference between means is:
which says that the mean of the distribution of differences between sample means is equal to the difference between population means. For example, say that the mean test score of all 12-year-olds in a population is 34 and the mean of 10-year-olds is 25. If numerous samples were taken from each age group and the mean difference computed each time, the mean of these numerous differences between sample means would be 34 - 25 = 9.
## Sampling Distribution of Difference Between Means (cont)。
From the variance sum law, we know that:
which says that the variance of the sampling distribution of the difference between means is equal to the variance of the sampling distribution of the mean for Population 1 plus the variance of the sampling distribution of the mean for Population 2. Recall the formula for the variance of the sampling distribution of the mean:
Since we have two populations and two samples sizes, we need to distinguish between the two variances and sample sizes. We do this by using the subscripts 1 and 2. Using this convention, we can write the formula for the variance of the sampling distribution of the difference between means as:
Since the standard error of a sampling distribution is the standard deviation of the sampling distribution, the standard error of the difference between means is:
Just to review the notation, the symbol on the left contains a sigma (σ), which means it is a standard deviation. The subscripts M1 - M2 indicate that it is the standard deviation of the sampling distribution of M1 - M2.
## Sampling Distribution of Difference Between Means (cont)。
Now let's look at an application of this formula. Assume there are two species of green beings on Mars. The mean height of Species 1 is 32 while the mean height of Species 2 is 22. The variances of the two species are 60 and 70, respectively and the heights of both species are normally distributed. You randomly sample 10 members of Species 1 and 14 members of Species 2. What is the probability that the mean of the 10 members of Species 1 will exceed the mean of the 14 members of Species 2 by 5 or more? Without doing any calculations, you probably know that the probability is pretty high since the difference in population means is 10. But what exactly is the probability? First, let's determine the sampling distribution of the difference between means. Using the formulas above, the mean is
The standard error is:
The sampling distribution is shown in Figure 1. Notice that it is normally distributed with a mean of 10 and a standard deviation of 3.317. The area above 5 is shaded blue.
Figure 1. The sampling distribution of the difference between means.
The last step is to determine the area that is shaded blue. Using either a Z table or the normal calculator, the area can be determined to be 0.934. Thus the probability that the mean of the sample from Species 1 will exceed the mean of the sample from Species 2 by 5 or more is 0.934.
As shown below, the formula for the standard error of the difference between means is much simpler if the sample sizes and the population variances are equal. When the variances and samples sizes are the same, there is no need to use the subscripts 1 and 2 to differentiate these terms.
This simplified version of the formula can be used for the following problem: The mean height of 15-year-old boys (in cm) is 175 and the variance is 64. For girls, the mean is 165 and the variance is 64.
If eight boys and eight girls were sampled, what is the probability that the mean height of the sample of girls would be higher than the mean height of the sample of boys? In other words, what is the probability that the mean height of girls minus the mean height of boys is greater than 0?
## Sampling Distribution of Difference Between Means (cont)。
As before, the problem can be solved in terms of the sampling distribution of the difference between means (girls - boys). The mean of the distribution is 165 - 175 = -10. The standard deviation of the distribution is:
A graph of the distribution is shown in Figure 2. It is clear that it is unlikely that the mean height for girls would be higher than the mean height for boys since in the population boys are quite a bit taller. Nonetheless it is not inconceivable that the girls' mean could be higher than the boys' mean.
Figure 2. Sampling distribution of the difference between mean heights.
A difference between means of 0 or higher is a difference of 10/4 = 2.5 standard deviations above the mean of -10. The probability of a score 2.5 or more standard deviations above the mean is 0.0062.
Questions
Population 1 has a mean of 20 and a variance of 100. Population 2 has a mean of 15 and a variance of 64. You sample 20 scores from Pop 1 and 16 scores from Pop 2. What is the mean of the sampling distribution of the difference between means (Pop 1 - Pop 2)?
{{{1}}}
Population 1 has a mean of 20 and a variance of 100. Population 2 has a mean of 15 and a variance of 64. You sample 20 scores from Pop 1 and 16 scores from Pop 2. What is the variance of the sampling distribution of the difference between means (Pop 1 - Pop 2)?
{{{1}}}
The mean height of 15-year-old boys is 175 cm and the variance is 64. For girls, the mean is 165 and the variance is 64. If 8 boys and 8 girls were sampled, what is the probability that the mean height of the sample of boys would be at least 6 cm higher than the mean height of the sample of girls?
{{{1}}}
{{{1}}}
## Sampling Distribution of Pearson's r。
### Learning Objectives
1. State how the shape of the sampling distribution of r deviates from normality
2. Transform r to z'
3. Compute the standard error of z'
4. Calculate the probability of obtaining an r above a specified value
Assume that the correlation between quantitative and verbal SAT scores in a given population is 0.60. In other words, ρ = 0.60. If 12 students were sampled randomly, the sample correlation, r, would not be exactly equal to 0.60. Naturally different samples of 12 students would yield different values of r. The distribution of values of r after repeated samples of 12 students is the sampling distribution of r.
The shape of the sampling distribution of r for the above example is shown in Figure 1. You can see that the sampling distribution is not symmetric: it is negatively skewed. The reason for the skew is that r cannot take on values greater than 1.0 and therefore the distribution cannot extend as far in the positive direction as it can in the negative direction. The greater the value of ρ, the more pronounced the skew.
Figure 1. The sampling distribution of r for N = 12 and ρ = 0.60.
Figure 2 shows the sampling distribution for ρ = 0.90. This distribution has a very short positive tail and a long negative tail.
Figure 2. The sampling distribution of r for N = 12 and ρ = 0.90. Referring back to the SAT example, suppose you wanted to know the probability that in a sample of 12 students, the sample value of r would be 0.75 or higher. You might think that all you would need to know to compute this probability is the mean and standard error of the sampling distribution of r. However, since the sampling distribution is not normal, you would still not be able to solve the problem. Fortunately, the statistician Fisher developed a way to transform r to a variable that is normally distributed with a known standard error.
The variable is called z' and the formula for the transformation is given below.
```z' = 0.5 ln[(1+r)/(1-r)]
```
The details of the formula are not important here since normally you will use either a table or calculator to do the transformation. What is important is that z' is normally distributed and has a standard error of
where N is the number of pairs of scores.
Let's return to the question of determining the probability of getting a sample correlation of 0.75 or above in a sample of 12 from a population with a correlation of 0.60. The first step is to convert both 0.60 and 0.75 to their z' values, which are 0.693 and 0.973, respectively.
The standard error of z' for N = 12 is 0.333. Therefore the question is reduced to the following: given a normal distribution with a mean of 0.693 and a standard deviation of 0.333, what is the probability of obtaining a value of 0.973 or higher? The answer can be found directly from the applet "Calculate Area for a given X" to be 0.20. Alternatively, you could use the formula:
```z = (X - μ)/σ = (0.973 - 0.693)/0.333 = 0.841
```
and use a table to find that the area above 0.841 is 0.20.
Questions
# What is the shape of the sampling distribution of r? {
{{{1}}}
r = 0 r = .6 r = .2 r = -.8
## Sampling Distribution of p。
### Learning Objectives
1. Compute the mean and standard deviation of the sampling distribution of p
2. State the relationship between the sampling distribution of p and the normal distribution
Assume that in an election race between Candidate A and Candidate B, 0.60 of the voters prefer Candidate A. If a random sample of 10 voters were polled, it is unlikely that exactly 60% of them (6) would prefer Candidate A. By chance the proportion in the sample preferring Candidate A could easily be a little lower than 0.60 or a little higher than 0.60. The sampling distribution of p is the distribution that would result if you repeatedly sampled 10 voters and determined the proportion (p) that favored Candidate A.
The sampling distribution of p is a special case of the sampling distribution of the mean. Table 1 shows a hypothetical random sample of 10 voters. Those who prefer Candidate A are given scores of 1 and those who prefer Candidate B are given scores of 0. Note that seven of the voters prefer candidate A so the sample proportion (p) is p = 7/10 = 0.70
As you can see, p is the mean of the 10 preference scores.
```Table 1. Sample of voters.
Voter Preference
1 1
2 0
3 1
4 1
5 1
6 0
7 1
8 0
9 1
10 1
```
The distribution of p is closely related to the binomial distribution. The binomial distribution is the distribution of the total number of successes (favoring Candidate A, for example) whereas the distribution of p is the distribution of the mean number of successes. The mean, of course, is the total divided by the sample size, N.
Therefore, the sampling distribution of p and the binomial distribution differ in that p is the mean of the scores (0.70) and the binomial distribution is dealing with the total number of successes (7).
The binomial distribution has a mean of μ=nπ
Dividing by N to adjust for the fact that the sampling distribution of p is dealing with means instead of totals, we find that the mean of the sampling distribution of p is: μp = π
The standard deviation of the binomial distribution is:
Dividing by N because p is a mean not a total, we find the standard error of p:
Returning to the voter example, π = 0.60 and N = 10. (Don't confuse π = 0.60, the population proportion and p = 0.70, the sample proportion.) Therefore, the mean of the sampling distribution of p is 0.60. The standard error is
The sampling distribution of p is a discrete rather than a continuous distribution. For example, with an N of 10, it is possible to have a p of 0.50 or a p of 0.60 but not a p of 0.55.
The sampling distribution of p is approximately normally distributed if N is fairly large and π is not close to 0 or 1. A rule of thumb is that the approximation is good if both Nπ and N(1 - π) are greater than 10. The sampling distribution for the voter example is shown in Figure 1. Note that even though N(1 - π) is only 4, the approximation is quite good.
Figure 1. The sampling distribution of p. Vertical bars are the probabilities; the smooth curve is the normal approximation.
### Questions。
The binomial distribution is the distribution of the total number of successes whereas the distribution of p is: - the distribution of the mean number of successes - the distribution of the total number of failures - the distribution of the ratio of successes to failures - a distribution with a mean of .5 {
|
# How to find parallelogram diagonal length
In a quadrangle of tops opposite each other the creation of its diagonals is result of connection. There is a general formula connecting lengths of these pieces with another dimensions of a figure. On it, in particular, it is possible to find parallelogram diagonal length.
## Instruction
1. Construct a parallelogram, having chosen if necessary scale so that all known measurements as much as possible corresponded to initial data. Good understanding of statements of the problem and creation of the evident schedule – guarantee of speed of the decision. Remember that are in pairs parallel and equal in this figure of the party.
2. Carry out both diagonals, having connected opposite tops. These pieces have several properties: they are crossed in the middle of the lengths, and any of them divides a figure into two symmetrically identical triangles. Lengths of diagonals of a parallelogram are connected by a formula of the sum of squares: d1² + d2² = 2 • (and² + b²), where and both b – length and width.
3. It is obvious that to know only lengths of the main measurements of a parallelogram insufficiently to calculate at least one diagonal. Let's consider a task in which the parties of a figure are set: and = 5 and b = 9. It is also known that one of diagonals is more another twice.
4. Work out two equations with two unknown: d1 = 2•d2d1² + d2² = 2 • (and² + b²) = 212.
5. Substitute d1 from the first equation in the second: 5 · d2² = 212 → d2 ≈ 6.5; Find length of the first diagonal: d1 = 13.
6. Special cases of a parallelogram are the rectangle, a square and a rhombus. Diagonals of the first two figures represent equal pieces, therefore, the formula can be rewritten in simpler look: 2 · d² = 2 • (and² + b²) → d = √ (and² + b²), where and both b – length and width of a rectangle; 2 · d² = 2•2 • and² → d = √2 • and², where and – the party of a square.
7. Lengths of diagonals of a rhombus are not equal sizes, however its parties are equal. Proceeding from it, the formula can be simplified too: d1² + d2² = 4 • and².
8. These three formulas can be brought also out of separate consideration of triangles into which figures are divided diagonals. They rectangular, so it is possible to apply Pythagorean theorem. Diagonals are hypotenuses, legs – the parties of quadrangles.
Author: «MirrorInfo» Dream Team
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# 5th grade word problems and fractions pd
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• Activity 15.8Slicing SquaresGive students a worksheet with four squares in a row, each approximately 3 cm on a side. Have them shade in the same fraction in each square using vertical dividing line. You can use the context of a garden or farm. For example, slice each square in fourths and shade three-fourths as in Figure 15.20. Next, tell students to slice each square into equal-sized horizontal slices. Each square must be partitioned differently, using from one to eight slices. For each sliced square, they record an equations showing the equivalent fractions. Have them examine their equations and drawings to look for any patterns. You can repeat this with four more squares and different fractions.What product tells how many parts are shaded?What product tells how many parts in the whole?Notice that the same factor is used for both part and whole
• Give students an equation expressing an equivalence between two fraction but with one of the numbers missing and ask them to draw a picture to solve. Here are four different examples:5/3 = _/62/3 = 6/_8/12 = _/39/12 = 3/_The missing number can be either a numerator or a denominator. Furthermore, the missing number can either be larger or smaller that the corresponding part of the equivalent fraction. (All four possibilities are represented in the examples.) The examples shown involve simple whole-number multiples between equivalent fractions. Next, consider pairs such as 6/8 = _/12 or 9/12 = 6/_. In these equivalences, one denominator or numerator is not a whole number multiple of the other.
• ### 5th grade word problems and fractions pd
1. 1. 5th Grade Fractions &Word Problems Laura Chambless RESA Consultantwww.protopage.com/lchambless
2. 2. CCSS and Gaps What are your gaps in curriculum?1. Review CCSS for Fractions2. Think about your resources3. Think about your teaching – Highlight anything your resources covers well in YELLOW. – Highlight any part of the standard you would like more clarification on in BLUE.
3. 3. Learning TargetUse equivalent fractions as a strategy to add and subtract fractions. 5.NF.1, 5.NF.2Apply and extend previous understandings of multiplication and division to multiply and divide fractions. 5.NF.3, 5.NF.4, 5.NF.5, 5.NF.6, 5.NF.7
4. 4. Fraction Word Problem40 students joined the soccer club. 5/8 of the students were boys. How many girls joined the soccer club? Draw a picture and solve it.1. 2 min. working problem on own2. 5 min. sharing with group3. Class discussionFound at: http://www.mathplayground.com/wpdatabase/Fractions1_3.htm
5. 5. Problem Solving with Bar Diagrams1. Understand: Identify what is known and what is unknown. Draw the bar diagram to promote comprehension and demonstrates understanding. (Situation vs. Solution Equation)2. Plan: Decide how you will solve the problem (find the unknown). Analyze the bar diagram to find a solution plan.3. Solve: Execute the plan. Use the bar diagram to solve.4. Evaluate: Assess reasonableness using estimation or substitution. Substitute the solution for the unknown in the bar diagram.
6. 6. Bar DiagramsWatch Introduction Videohttp://www.mhschool.com/math/com mon/pd_video/mathconnects_bardi agram_p1/index.htmlhttp://www.mhschool.com/math/com mon/pd_video/mathconnects_bardi agram_p2/index.html
7. 7. Practice Bar DiagramsTo: Rani earned \$128 mowing lawns and \$73 babysitting. How much money did Rani earn?With: Jin had \$67 in his pocket after he bought a radio controlled car. He went to the store with \$142. How Much did Jin spend on the car?By: There are 9 puffy stickers. There are 3 times as many plain stickers as puffy stickers. How many plain stickers are there?You pick 2 more to do by yourself. Share with partnerDraw Your Way to Problem Solving Success Handout, Robyn Silbey
8. 8. Thinking Blockshttp://www.mathplayground.com/think ingblocks.html Explore the site When done exploring go to my Protopage and look at your grade level math tab.
9. 9. Fractions Stand and ShareMake a list of what you know and any connections you have about the fraction ¼.
10. 10. Build Connections to Whole Numbers 0 1 2 3 4 51+1+1+1+1=5 1/4 1/2 3/4 0 1 ¼ +¼ +¼+¼ =1
11. 11. Fractions Fraction ActivityPaper Strips Fraction Kit: 1, ½, 1/4 , 1/8, 1/16Add to Fraction Kit: 1/3, 1/6, 1/12Add to Fraction Kit: 1/5, 1/10 Compare/Add/Subtract/Multi./Divide with Strips READ and DO: 5.NF.1, 5.NF.2, 5.NF.3, 5.NF.4, 5.NF.5, 5.NF.6, 5.NF .7Smaller Answer Wins (need dice)• Prove with Fraction Strips
12. 12. Lunch
13. 13. Definition of Fractions1. Make a list of what you would like to have in a definition of a fraction2. Partner up and compare lists3. Group discussion
14. 14. Definition of a FractionCreate a working definition of a fraction – Watch Dev-TEAM video 3:5 and 3:6 – Give article: Definitions and Defining in Mathematics and Mathematics Teaching by: Bass and Ball
15. 15. Definition Of Fractions• Identify the whole• Make d equal parts• Write 1/d to show one of the equal parts• If you have d of 1/d, then you have the whole• If you have n of 1/d, then you have n/d• n and d are whole numbers• d does not equal 0Dev-TE@M • School of Education • University of Michigan • (734)408-4461 • dev-team@umich.edu For review only - Please do notcirculate or cite without permission
16. 16. Ordering Fractions Order Fractions 8/6, 2/5, 8/10, 1/12How did you figure out what order they went in?
17. 17. Fractions Prove with Fraction StripsNumber Line: (Benchmarks) 0, ½, 1Equivalent Fractions: Same Name FrameCompare (>/<): same numerator or same denominator
18. 18. Strategies for Comparing Fractions• Dev-TE@M session 9
19. 19. Fraction On A Number LineWriting about Fractions: Draw a number line. Place 3/6 and 7/12 on the number line. Compare the two fractions- why did put them where you did?
20. 20. Conventions Of A Number LineDev-TE@M • School of Education • University of Michigan • (734) 408-4461 •dev-team@umich.edu For review only - Please do not circulate or cite withoutpermission
21. 21. Model of Number Line Talk• Watch Dev-TE@M Session 4:5 (video)
22. 22. Talking Through A Number Line1. Understand the problem.2. Think about which representation you are going to use.3. Describe your thinking process while constructing the number line.4. Sum up the solution that proved your answer.Model Example: 3/10 & 6/8
23. 23. Fraction On A Number LineUsing a number line, compare 5/6 and 3/8 and tell which one is greater . Have a partner listen to you as you construct the fractions and find the answer.
24. 24. FractionsWhat conceptual understanding do students need? 1. Begin with simple contextual tasks. 2. Connect the meaning of fraction computation with whole number computation. 3. Let estimation and informal methods play a big role in the development of strategies. 4. Explore each of the operations using models.Van De Walle Book: Number Sense and Fraction Algorithms Pg. 310
25. 25. Equivalence with Fraction Strips• Fraction Strips ½+¼= ¾ + 1/3 =
26. 26. Methods for Generating andExplaining Equivalent Fractions• Dev-TE@M session 9
27. 27. Add/Subtract Fractions with Unlike Denominators Developing Equivalent Fractions• Slicing Squares Van de Walle book: pg. 304-305 3 x = 3 x 4 = 4 3 x 3 x = = 4 4
28. 28. Add/Subtract Fractions with Unlike Denominators Developing Equivalent Fractions• Missing-Number Equivalencies Van de Walle book: pg. 304-305 5 2 6 = = 3 6 3
29. 29. Fraction Multiplication StrategiesTOOLKIT for Multiplication of Fractions1. Skim over TOOLKIT2. Read assigned page (2 min)3. 30 second report: What are the important part of your page?4. Questions from audience
30. 30. FractionsMultiply a fraction by a whole number• Work as a group• Use Fraction strips to show answers 4 x 1/3 ¼ x 12• What connection can you make to multiplication? What other representations can you use? Can you use a number line?
31. 31. Multiple a Fraction by a Whole Number 4 x 1/3 (4 groups of 1/3) = 4/3 = 1 1/3I want 4 ribbons each at 1/3 of a yard. How much ribbon will I need to purchase? 1/3 2/3 3/3 4/3 ¼ x 12 (1/4 of 12) = 3I have 12 cookies and want each of my friends to have ¼ of them. How many cookies will each friend get?
32. 32. Scaling (resizing)• 5.NF.5 – Read learning targets and discuss – Prove greater/less than given number statements with last slide. – Making equivalent fractions
33. 33. Multiply Fraction by FractionAIMS• Fair Squares and Cross ProductsMMPI• Worksheet 1: Show different representations 2/3 of ¾ ¾ of 2/3
34. 34. Multiply Fractions and Mixed NumbersMMPI• Area Model Rectangular Multiplication PPThttp://www.michiganmathematics.org/
35. 35. Fraction as Division (a/b = a ÷ b)• I can explain that fractions (a/b) can be represented as a division of the numerator by the denominator (a ÷ b) can be represented by the fraction a/b.• I can solve word problems involving the division of whole numbers and interpret the quotient- which could be a whole number, mixed number, or fraction – in the context of the problem.• I can explain or illustrate my solution strategy using visual fraction models or equations that represent the problem.
36. 36. Divide Fraction by Whole Number½÷6=6÷¼=4 ÷ 2 = (how to connect division of whole numbers with fractions)
37. 37. Divide Fraction by Whole Number½ ÷ 6 = If I have ½ cup of sugar and divide it among 6 people, how much sugar does each person have? 1/121 2 3 4 5 6 7 8 9 10 11 126 ÷ ¼ = If I have 6 candy bars and divide each one into fourths, how many pieces will I have? 24
38. 38. MOPLS http://mi.learnport.org Search: MOPLS Math (navigate by using top tabs)Look at Concepts Tab– Introduction– Math Behind the Math– Misconceptions– Tasks & Strategies
39. 39. Fractions OnlineCheck out some sites on my 5th grade math Protopage
40. 40. Learning TargetUse equivalent fractions as a strategy to add and subtract fractions. 5.NF.1, 5.NF.2Apply and extend previous understandings of multiplication and division to multiply and divide fractions. 5.NF.3, 5.NF.4, 5.NF.5, 5.NF.6, 5.NF.7
41. 41. Closer ActivityList something you learn about story problems and fractions today.
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# SMART Packet #1. Measures of Central Tendency: Mean, Median, Mode. Student: Teacher: Standards
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1 Unit 5: Probability & Statistics SMART Packet #1 Measures of Central Tendency: Mean, Median, Mode Student: Teacher: Standards A.S.4 A.A.6 A.S.16 Compare and contrast the appropriateness of different measures of central tendency for a given data set Analyze and solve verbal problems whose solution requires solving a linear equation in one variable or linear inequality in one variable Recognize how linear transformations of one-variable data affect the data s mean, median, mode, and range
2 Measures of Central Tendency Measure of Central Tendency MEAN MEDIAN MODE Definition The average. The middle value of the data, when it is in numerical order. The number that appears the most. Steps to Find the Measure of Central Tendency 1. Add up all of the numbers in the data set. 2. Divide the total found in step 1 by the amount of numbers in the set. 1. Put the numbers in order from low to high. The number in the middle is the median. 2. If there are two numbers in the middle, add them together and divide by Find the number that appears the most. 2. If no number appears the most, write no mode. Example 1: Find the mean, median and mode for the following data: 5, 15, 10, 16, 8, 10, 20, 12 Mean: sum of all numbers in the data amount of numbers in the data 96 = = 12 8 The mean is 12. Median: and 12 are both in the middle = 11 2 The median is 11. Mode: 10 appears more than any other number. The mode is 10.
3 Example 2: The prices of seven race cars sold last week are listed in the table below. (a) What is the mean value of these race cars, in dollars? mean = = = sum of all numbers in the data amount of numbers in the data 126, (140,000) + 180, (400,000) + 819, ,205,000 7 = 315,000 The mean value of these races cars is 315,000. (b) What is the median value of these race cars, in dollars? 126, , , , , , , ,000 is the median (c) State which of these measures of central tendency best represents the value of the seven race cars. Justify your answer. The median best represents the value of the seven race cars because the data set includes an outlier, a number that is much larger than the rest data in the set.
4 PRACTICE, PART I 1. Sara's test scores in science were 70, 78, 80, 92, 84, 76, and 80. Determine the mean, the median, and the mode of Sara's test scores. Mean: Median: Mode: 2. The weekly salaries of six employees at McDonalds are \$140, \$220, \$90, \$180, \$140, \$200. Find the mean, median, and mode for these six salaries. Mean: Median: Mode:
5 3. The ages of five children in a family are 3, 3, 5, 8, and 18. Which statement is true for this group of data? (1) mode > mean (3) median = mode HINT: Find the mean, (2) mean > median (4) median > mean median, and mode. Then compare your results. 4. Which statement is true about the data set 3, 4, 5, 6, 7, 7, 10? (1) mean = mode (3) mean = median (2) mean > mode (4) mean < median 5. Seth bought a used car that had been driven 20,000 miles. After he owned the car for 2 years, the total mileage of the car was 49,400. Find the average number of miles he drove each month during those 2 years.
6 6. The values of 11 houses on Washington St. are shown in the table below. (a) Find the mean value of these houses in dollars. (b) Find the median value of these houses in dollars. (c) State which measure of central tendency, the mean or the median, best represents the values of these 11 houses. Justify your answer
7 Example 3: On his first 5 biology tests, Bob received the following scores: 72, 86, 92, 63, and 77. What test score must Bob earn on his sixth test so that his average (mean score) for all six tests will be 80? Show how you arrived at your answer. Step 1: Set up an equation to represent the situation using all 6 test scores. Let x represent the unknown test score x = 80 6 Step 2: Solve for x x = x = 80 6 Combine like terms x = 480 Multiply both sides of the equation by 6. x = 90 Bob needs to earn a 90 on his sixth test. Subtract 390 from both sides. PRACTICE, PART II 7. TOP Electronics is a small business with five employees. The mean (average) weekly salary for the five employees is \$360. If the weekly salaries of four of the employees are \$340, \$340, \$345, and \$425, what is the salary of the fifth employee? =
8 8. The students in Woodland High School's meteorology class measured the noon temperature every schoolday for a week. Their readings for the first 4 days were Monday, 56 ; Tuesday, 72 ; Wednesday, 67 ; and Thursday, 61. If the mean (average) temperature for the 5 days was exactly 63, what was the temperature on Friday? 9. In his first three years coaching baseball at High Ridge High School, Coach Batty's team won 7 games the first year, 16 games the second year, and 4 games the third year. How many games does the team need to win in the fourth year so that the coach's average will be 10 wins per year? (1) 13 (3) 3 (2) 10 (4) 9
9 10. The exact average of a set of six test scores is 92. Five of the scores are 90, 98, 96, 94, and 85. What is the other test score? (1) 92 (3) 89 (2) 91 (4) For five algebra examinations, Maria has an average of 88. What must she score on the sixth test to bring her average up to exactly 90? (1) 98 (3) 94 (2) 92 (4) If 6 and x have the same mean (average) as 2, 4, and 24, what is the value of x? (1) 5 (3) 10 (2) 36 (4) 14
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# Systems of Linear Equations in Two Variables
## Solving for two variables using elimination.
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Systems of Two Equations and Two Unknowns
The cost of two cell phone plans can be written as a system of equations based on the number of minutes used and the base monthly rate. As a consumer, it would be useful to know when the two plans cost the same and when is one plan cheaper.
Plan A costs $40 per month plus$0.10 for each minute of talk time.
Plan B costs $25 per month plus$0.50 for each minute of talk time.
Plan B has a lower starting cost, but since it costs more per minute, it may not be the right plan for someone who likes to spend a lot of time on the phone. When do the two plans cost the same amount?
### Solving Systems of Equations with Two Unknowns
There are many ways to solve a system that you have learned in the past including substitution and graphical intersection. Here you will focus on solving using elimination because the knowledge and skills used will transfer directly into using matrices.
When solving a system, the first thing to do is to count the number of variables that are missing and the number of equations. The number of variables needs to be the same or fewer than the number of equations. Two equations and two variables can be solved, but one equation with two variables cannot.
Get into the habit of always writing systems in standard form: \begin{align*}Ax+By=C\end{align*}. This will help variables line up, avoid +/- errors and lay the groundwork for using matrices. Once two equations with two variables are in standard form, decide which variable you want to eliminate, scale each equation as necessary by multiplying through by constants and then add the equations together. This procedure should reduce both the number of equations and the number of variables leaving one equation and one variable. Solve and substitute to determine the value of the second variable.
Here is a system of two equations and two variables in standard form\begin{align*}5x+12y=72\end{align*} and \begin{align*}3x-2y=18\end{align*}. Notice that there is an \begin{align*}x\end{align*} column and a \begin{align*}y\end{align*} column on the left hand side and a constant column on the right hand side when you rewrite the equations as shown. Also notice that if you add the system as written no variable will be eliminated.
Equation 1: \begin{align*}5x+12y=72\end{align*}
Equation 2: \begin{align*}3x-2y=18\end{align*}
Strategically choose to eliminate \begin{align*}y\end{align*} by scaling the second equation by 6 so that the coefficient of \begin{align*}y\end{align*} will match at 12 and -12.
\begin{align*}5x+12y &= 72\\ 18x-12y &= 108\end{align*}
\begin{align*}23x &= 180\\ x &= \frac{180}{23}\end{align*}
The value for \begin{align*}x\end{align*} could be substituted into either of the original equations and the result could be solved for \begin{align*}y\end{align*}; however, since the value is a fraction it will be easier to repeat the elimination process in order to solve for \begin{align*}x\end{align*}. This time you will take the first two equations and eliminate \begin{align*}x\end{align*} by making the coefficients of \begin{align*}x\end{align*} to be 15 and -15. Scale the first equation by a factor of 3 and scale the second equation by a factor of -5.
Equation 1: \begin{align*}15x+36y=216\end{align*}
Equation 2: \begin{align*}-15x+10y=-90\end{align*}
\begin{align*}0x+46y &= 126\\ y = \frac{126}{46} &=\frac{63}{23}\end{align*}
The point \begin{align*}\left(\frac{180}{23}, \frac{63}{23} \right)\end{align*} is where these two lines intersect.
### Examples
#### Example 1
Plan A costs $40 per month plus$0.10 for each minute of talk time.
Plan B costs $25 per month plus$0.50 for each minute of talk time.
If you want to find out when the two plans cost the same, you can represent each plan with an equation and solve the system of equations. Let \begin{align*}y\end{align*} represent cost and \begin{align*}x\end{align*} represent number of minutes.
\begin{align*}y &= 0.10x + 40\\ y &= 0.50x+25\end{align*}
First you put these equations in standard form.
\begin{align*}x-10y &= -400\\ x-2y &= -50\end{align*}
Then you scale the second equation by -1 and add the equations together and solve for \begin{align*}y\end{align*}.
\begin{align*}-8y &= -350\\ y &= 43.75\end{align*}
To solve for \begin{align*}x\end{align*}, you can scale the second equation by -5, add the equations together and solve for \begin{align*}x\end{align*}.
\begin{align*}-4x &= -150\\ x &= 37.5\end{align*}
The equivalent costs of plan A and plan B will occur at 37.5 minutes of talk time with a cost of \$43.75.
#### Example 2
Solve the following system of equations:
\begin{align*}6x-7y &= 8\\ 15x-14y &= 21\end{align*}
Scaling the first equation by -2 will allow the \begin{align*}y\end{align*} term to be eliminated when the equations are summed.
\begin{align*}-12x+14y &= -16\\ 15x-14y &= 21\end{align*}
The sum is:
\begin{align*}3x &= 5\\ x &= \frac{5}{3}\end{align*}
You can substitute \begin{align*}x\end{align*} into the first equation to solve for \begin{align*}y\end{align*}.
\begin{align*}6 \cdot \frac{5}{3}-7y &= 8\\ 10-7y &= 8\\ -7y &= -2\\ y &= \frac{2}{7}\end{align*}
The point \begin{align*}\left(\frac{5}{3}, \frac{2}{7}\right)\end{align*} is where these two lines intersect.
#### Example 3
Solve the following system using elimination:
\begin{align*}5x-y &= 22\\ -2x+7y &= 19\end{align*}
Start by scaling the first equation by 7 and notice that the \begin{align*}y\end{align*} coefficient will immediately be eliminated when the equations are summed.
\begin{align*}35x-7y &= 154\\ -2x+7y &= 19\end{align*}
Add, solve for \begin{align*}x=\frac{173}{33}\end{align*}. Instead of substituting, practice eliminating \begin{align*}x\end{align*} by scaling the first equation by 2 and the second equation by 5.
\begin{align*}10x-2y &= 44\\ -10x+35y &= 95\end{align*}
Add, solve for \begin{align*}y\end{align*}
Final Answer: \begin{align*}\left(\frac{173}{33}, \frac{139}{33}\right)\end{align*}
#### Example 4
Solve the following system of equations:
\begin{align*}5 \cdot \frac{1}{x}+2 \cdot \frac{1}{y} &= 11\\ \frac{1}{x}+\frac{1}{y} &= 4 \end{align*}
The strategy of elimination still applies. You can eliminate the \begin{align*}\frac{1}{y}\end{align*} term if the second equation is scaled by a factor of -2.
\begin{align*}5 \cdot \frac{1}{x}+2 \cdot \frac{1}{y} &= 11\\ -2 \cdot \frac{1}{x}-2 \cdot \frac{1}{y} &= -8\end{align*}
Add the equations together and solve for \begin{align*}x\end{align*}.
\begin{align*}-3 \cdot \frac{1}{x}+0 \cdot \frac{1}{y} &= 3\\ -3 \cdot \frac{1}{x} &= 3\\ \frac{1}{x} &= -1\\ x &= -1 \end{align*}
Substitute into the second equation and solve for \begin{align*}y\end{align*}.
\begin{align*}\frac{1}{-1}+\frac{1}{y} &= 4\\ -1+\frac{1}{y} &= 4\\ \frac{1}{y} &= 5\\ y &= \frac{1}{5}\end{align*}
The point \begin{align*}\left(-1, \frac{1}{5}\right)\end{align*} is the point of intersection between these two curves.
#### Example 5
Solve the following system using elimination:
\begin{align*}11 \cdot \frac{1}{x}-5 \cdot \frac{1}{y} &= -38\\ 9 \cdot \frac{1}{x}+2 \cdot \frac{1}{y} &=-25\end{align*}
To eliminate \begin{align*}\frac{1}{y}\end{align*}, scale the first equation by 2 and the second equation by 5.
To eliminate \begin{align*}\frac{1}{x}\end{align*}, scale the first equation by -9 and the second equation by 11.
Final Answer: \begin{align*}\left(-\frac{1}{3}, 1 \right)\end{align*}
### Review
Solve each system of equations using the elimination method.
1. \begin{align*}x+y=-4; -x+2y=13\end{align*}
2. \begin{align*}\frac{3}{2}x-\frac{1}{2}y=\frac{1}{2}; -4x+2y=4\end{align*}
3. \begin{align*}6x+15y=1; 2x-y=19\end{align*}
4. \begin{align*}x-\frac{2y}{3}=\frac{-2}{3}; 5x-2y=10\end{align*}
5. \begin{align*}-9x-24y=-243; \frac{1}{2}x+y=\frac{21}{2}\end{align*}
6. \begin{align*}5x+\frac{28}{3}y=\frac{176}{3}; y+x=10\end{align*}
7. \begin{align*}2x-3y=50; 7x+8y=-10\end{align*}
8. \begin{align*}2x+3y=1; 2y=-3x+14\end{align*}
9. \begin{align*}2x+\frac{3}{5}y=3;\frac{3}{2}x-y=-5\end{align*}
10. \begin{align*}5x=9-2y;3y=2x-3\end{align*}
11. How do you know if a system of equations has no solution?
12. If a system of equations has no solution, what does this imply about the relationship of the curves on the graph?
13. Give an example of a system of two equations with two unknowns with an infinite number of solutions. Explain how you know the system has an infinite number of solutions.
14. Solve:
\begin{align*}12 \cdot \frac{1}{x}-18 \cdot \frac{1}{y} &= 4\\ 8 \cdot \frac{1}{x}+9 \cdot \frac{1}{y} &= 5\end{align*}
15. Solve:
\begin{align*}14 \cdot \frac{1}{x}-5 \cdot \frac{1}{y} &= -3\\ 7 \cdot \frac{1}{x}+2 \cdot \frac{1}{y} &= 3\end{align*}
To see the Review answers, open this PDF file and look for section 8.1.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
elimination The elimination method for solving a system of two equations involves combining the two equations in order to produce one equation in one variable.
Periodic Decimal A periodic decimal is a decimal number that has a pattern of digits that repeat. The decimal number 0.146532532532..., is a periodic decimal.
scale To scale an equation means to multiply every term in the equation by the same constant.
Standard Form The standard form of a line is $Ax + By = C$, where $A, B,$ and $C$ are real numbers.
system of equations A system of equations is a set of two or more equations.
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Multiplying by 6 to 9
How to Multiply by 6 to 9
Multiplication Mini-Review
Multiplication is a shortcut for repeated addition.
When we multiply...
5 × 4
...we add 5 to itself, 4 times:
5 × 4 = 5 + 5 + 5 + 5 = 20
...or add 4 to itself, 5 times:
5 × 4 = 4 + 4 + 4 + 4 + 4 = 20
In the last lesson, you learned how to multiply numbers by 0, 1, 2, 3, 4 and 5.
Now, let's learn how to multiply numbers by 6, 7, 8 and 9.
Multiplying by 6
6 × 4 = ?
To multiply by 6, you can either skip count by 6, or use this trick:
If you know 5 times a number, just add one more of that number to figure out 6 times that number.
So to figure out 6 × 3, you just figure out 5 × 3 and add 3 more.
6 × 3 = 5 × 3 + 3 = 15 + 3 = 18
Why does that work? Take a look:
6 = 5 + 1
We can break multiplying by 6 down into two easier multiplication problems:
Multiplying by 6 = Multiplying by 5 + Multiplying by 1
Can you multiply these numbers?
6 × 2 = ?
Let's try multiplying by 5 times 2 first, and adding one more 2.
6 × 2 = 12
Great work.
Multiplying by 7
7 × 3 = ?
You can either skip count by 7 three times, or you can use a trick.
We know that:
7 = 5 + 2
So:
Multiplying by 7 = Multiplying by 5 + Multiplying by 2
Tip: It's like you can multiply by 5 and skip count up two more times.
Let's see if it works.
7 × 3 =
5 × 3 + 2 × 3 =
15 +
2 × 3 =
15 + 6 =
21
It works.
Let's try one more.
7 × 5 = ?
7 × 5 = 35
Awesome.
Multiplying by 9
9 × 6 = ?
You can skip count by 9, but that can be a little hard.
Luckily, there's a trick.
We know that:
9 = 10 - 1
So we can break it down into two steps:
Multiplying by 9 = Multiplying by 10 - Multiplying by 1
Let's try this trick to solve our example.
9 × 6 = 60 - 6 = 54
Nice work! 👍
Can you now solve this using the trick?
9 × 4 = ?
9 × 4 = 40 - 4 = 36
Very good. 👏
Multiplying by 8
8 × 4 = ?
You can either skip count by 8 four times,
Or you can use a trick. 😃
We know that:
8 = 10 - 2
So,
Multiplying by 8 = Multiplying by 10 - Multiplying by 2
Let's try the trick.
8 × 4 = 40 - 8 = 32
Great job! 👏
Can you solve this on your own?
8 × 7 = ?
You can split it down into two easier multiplication problems and a subtraction, like this:
8 × 7 =
(
10 × 7) - (2 × 7) =
70 - 14 =
56
Way to go.
6 to 9 Times Tables
You just learned a bunch of tricks, but your goal is actually to know all of these by heart, because you'll use them your whole life!
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# Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
To do:
We have to prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Solution:
Let two circles with centres $A$ and $A'$ intersect each other at $B$ and $B'$ respectively.
In $\triangle BAA’$ and $\triangle B'AA’$
$AB = AB'$ (Radii of circle with centre $A$)
$A’B = A’B'$ (Radii of circle with centre $A'$)
$AA’ = AA’$ (Common side)
Therefore, by $SSS$ congruency,
$\triangle BAA’ \cong \triangle B'AA’$
This implies,
$\angle ABA' = \angle AB'A’$
From above,
The line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Hence proved.
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# TI-AIE: Learning through talking: variables and constants
## What this unit is about
Variables and constants are the basic concepts used in mathematical modelling and formulas. Understanding the role of variables and constants allows students to become skilled in algebraic manipulation, which is important in reasoning mathematically and in order to do well in mathematics examinations.
In this unit you will think about the roles of variables and constants in the mathematics curriculum and how understanding this helps students to give meaning to mathematical statements and algebraic expressions.
## What you can learn in this unit
• Some ideas to encourage your students to learn through talking, and to express themselves using mathematical vocabulary and phraseology.
This unit links to the teaching requirements of the NCF (2005) and NCFTE (2009) outlined in Resource 1.
## 1 Variables and constants in school mathematics
Understanding the role of variables and constants is essential for developing mathematical reasoning and understanding. It is required to manipulate algebraic expressions and also enables students ‘ to express mathematical relations in different ways, and know more about them’ (Watson et al., 2013, p. 15).
Technically, there are ‘dependent’ variables, ‘independent’ variables and constants. The unknown, x, is conventionally used to denote the independent variable, and is conventionally plotted along the horizontal axis when drawing a graph.
For example, in the expression:
y = x + 4
where x and y are integers:
• x is the independent variable and can stand for any value in the set for which the expression is defined. In this example this means it can be any integer.
• y is described as a dependent variable. It is dependent because its value will depend on the value of x. It is a variable because, like x, it can stand for any value in the set for which the expression is defined. In this example this means it can be any integer.
• 4 is the constant, that is, a fixed quantity, whatever the values of the independent or dependent variables.
Research suggests that one of the main issues that students encounter in learning about variables and constants, and manipulating algebra in general, is not understanding the relationships between quantities and variables in algebraic expressions. This unit aims to develop this understanding by giving meaning to variables and constants by making students think and talk about connections between numbers and algebraic expressions.
## 2 Learning about variables and constants through talking
A very efficient way for students to develop understanding and give meaning to any mathematical concept is through talking:
Children need to learn how to … use mathematical language to create, control and express their own mathematical meanings as well as to interpret the mathematical language of others.
(Pimm, 1995, p. 179)
Students who do not learn how to ‘talk mathematics’ lose out on many things; in particular, as Pimm says above, they do not have the resources to available to create, control and express their own mathematical ideas.
Encouraging students to talk about mathematics and helping them to develop appropriate vocabulary and phraseology to do so is an important part of learning. Thinking and communicating are so intimately entwined (Sfard, 2010) that it is not possible to know where one stops and the other starts. If you want your students to think about, understand and therefore effectively learn mathematics, they will also need to learn to communicate their mathematical ideas.
Students will also be able to talk about what they are thinking with one another. The act of forming thoughts in order to communicate with another can enable misunderstandings to be corrected. Thoughts that have to be formed into something that can be communicated, have been shown to be much more susceptible to recall by students (Lee, 2006); in other words, they are more likely to have learned those ideas.
The first activity asks students to think about identifying quantities in their experiences from their real life. It uses a picture to trigger their imagination. The activity suggests giving a short time limit for students to come up with their ideas to give a sense of urgency, competitiveness and excitement. This also means they will have little time to worry about doing algebra.
Before attempting to use the activities in this unit with your students, it would be a good idea to complete all, or at least part, of the activities yourself. It would be even better if you could try them out with a colleague as that will help you when you reflect on the experience. Trying them for yourself will mean you get insights into a learner’s experiences which can, in turn, influence your teaching and your experiences as a teacher. When you are ready, use the activities with your students and, once again, reflect and make notes on how well the activity went and the learning that happened. This will help you to develop a more student-focused teaching environment.
### Activity 1: Pictures are worth a thousand words
#### Preparation
This activity is done best in pairs or small groups. Make sure the students in the group are seated so that they can hear each other well. If you feel the students need some more time at any point, give them some bonus or extra time. You may want to look at the key resource ‘Using pair work’.
#### The activity
How many of you have travelled in an autorickshaw? Figure 1 shows Mr Murti travelling in an autorickshaw. In your groups, think of as many (measurable) quantities as you can that are associated with an autorickshaw ride. The group who writes the greatest number of such quantities in four minutes will be the winner. Your time starts … now.
Figure 1 Autorickshaw driver and passenger. (Source: Muhammad Mahdi Karim)
Then, ask for their ideas to share with the rest of the class. This could be organised as follows:
• At the end of the time limit, ask them to put their pencils down.
• Give them ten seconds to count the number of things they wrote.
• Now choose the group which wrote the minimum and the group which wrote the maximum number of quantities.
• Ask two students of the group that wrote the least number of quantities and two students of the group that wrote the most to come and write the quantities they thought of on the blackboard simultaneously – this will save time. Ask the students to stay at the blackboard.
• Ask each group in the class to share any other quantities they came up with that are different to those already on the blackboard. Ask one of the students at the blackboard to write down each of the new suggestions. Having so many scribes at the blackboard at once means that this can be done quickly.
• Soon, you will have a lot of quantities related to an autorickshaw ride on the blackboard. Some of these may include:
• number of passengers
• total fare for a journey
• time taken for a journey
• number of red lights at traffic signals during the journey
• distance of a journey
• number of wheels on an autorickshaw
• number of bolts on each wheel of an autorickshaw
• registration number of the autorickshaw
• speed of the autorickshaw
• cost of the autorickshaw
• mileage of the autorickshaw.
If you can, leave this list on the blackboard and ask your students to copy it. They will need it for Activity 2.
Video: Using pair work
### Case Study 1: Mrs Bhatia reflects on using Activity 1
This is the account of a teacher who tried Activity 1 with his elementary students.
I myself sometimes get confused about variables and constants, and how they relate to each other, so on the one hand I thought it would be good for my own subject knowledge development to do these activities in the class, on the other hand that made me feel a bit fearful. What would happen if I got confused in the class, in front of my students?
So I did Activities 1, 2 and 3 first on my own, and then with a colleague during a lunch break. The difference between doing them on my own, and with my colleague, is that we could talk together, and help each other when something was not clear in our understanding. And also, together we came up with so many funny examples meaning we laughed a lot doing algebra! Having a go at it first made me feel very well prepared to do these activities in class.
To introduce Activity 1, I showed the students the picture of Mr Murti, and I asked them to tell the class a little bit about their own journeys in an autorickshaw. That actually helped getting some extra variables to think about, such as: how far it was to Auntie Anju’s house, how far to the park, the different number of passengers on these journeys, the time it took and how this varied whether the roads were clear or a bit flooded from the monsoon rains. Only then did I ask them to do the activity.
I asked them to work in groups of four. Telling them they had only four minutes really spurred them into immediate action and there was a sense of competitiveness and eagerness. However, I also think that by giving them only four minutes I gave the students who are a bit shy an excuse to not really contribute much. Perhaps next time I will give them a bit longer and the extra instruction of making sure that everyone in the group has contributed at least two ideas.
To feedback to the class, I used the approach that is suggested in the activity. The big advantage of it was that we had loads of examples on the blackboard in record time while still valuing the contributions of the whole class. I thought that was a good idea – I have about 90 students in the class and so I often avoid ‘sharing with the whole class’ because it takes so much time.
### Reflecting on your teaching practice
When you do such an exercise with your class, reflect afterwards on what went well and what went less well. Consider the questions that led to the students being interested and being able to progress, and those you needed to clarify. Such reflection always helps with finding a ‘script’ that helps you engage the students to find mathematics interesting and enjoyable. If they do not understand and cannot do something, they are less likely to become involved. Use this reflective exercise every time you undertake the activities, noting as Mrs Bhatia did some quite small things that made a difference.
Pause for thought Good questions to trigger reflection are: How did it go with your class? What responses from students were unexpected? Why?Did you feel you had to intervene at any point? Did you modify the task in any way? If so, what was your reasoning for this?
## 3 Identifying variables and constants
In the previous activity the students used their experiences from real life and linked these to the mathematical concept of quantities. The second activity now moves the students on to make the distinction between variables and constants based on the quantities they identified in Activity 1.
To help students in developing their understanding of the difference between these concepts it is important that they have the opportunity to talk about it. It is very important to expect the students to use the mathematical words themselves and to create an environment where they have to do so. This will enable your students to recognise, use and communicate with one another about algebraic expressions, variables and constants. It is an important step in learning to ‘speak like a mathematician’ and to understand mathematics rather than just remembering it.
Asking the students ‘What will be the same?’ or ‘What will change?’ as in the next activity can help to trigger discussion and to identify variables and constants.
### Activity 2: What will stay the same, what will change?
This activity is best done in small groups or pairs. It is important that all students have a chance to say what they are thinking and to practise their mathematical vocabulary.
• Ask the students to look at the list of quantities from Activity 1 that they wrote in their books and listed on the blackboard. Then, in their groups or pairs, ask them to identify, discuss and make a note about which of these are:
• quantities whose values will change, called ‘variables’
• quantities whose values do not change, that will stay the same, called ‘constants’.
• As a whole class, discuss with the students their findings and reasons for categorising the quantities as variables or constants.
• Now ask your students to work in small groups, picking three of the variables and discussing what would make the variables change. Ask them to record this in their own way, but they must use the words and phrases ‘variable’, ‘constant’, ‘change’, ‘stay the same’.
• Pick two or three of the variables that were most looked at in the previous step and invite the students to come and write down how they had recorded what would make the variables change. Discuss these records with the class, and discuss how the students think their records could be made clearer.
### Case Study 2: Mrs Mehta reflects on using Activity 2
For the first part of the activity I asked the students to work in pairs or threes so they could all see each other’s writing.
I told them the question and I asked the students to draw a box around the quantities that would change. While they were doing this I walked around the classroom, but without interfering. I overheard a lot of discussions about whether a quantity would always change, or always stay the same, no matter what. At first their explanations in trying to convince each other were not always fluent, but I do think they got better at it as they tried again and again. By the time we had the whole-class discussion about the categorisation most of the students were able to express themselves pretty clearly. Those who sounded a bit muddled I asked to have another go at explaining and that helped in most cases.
We still had the list on the blackboard, and at first I also put boxes around the variables (at the students’ request) but that just looked messy. So I rewrote the list in two columns – one labelled ‘Variables – quantities whose values will change’ and the other ‘Constants – quantities whose values will stay the same’. I thought writing the mathematical terms first would help them in learning these.
They worked in groups of four to six students on the third part of the activity. I gave them a big sheet to write their records and told them we would put these on the wall at the end of the lesson (which we did). I think doing this made them think with greater precision about what they were writing. While they were working on step 3 I walked around the classroom and decided on two variables we would look at in greater detail for the last part of the activity – one was the number of wheels, and the other a rather complicated one of the number of rupees (or the price) a passenger would need to pay as their share when travelling in a group to different places.
By asking the students to come and copy their record of this on the blackboard we ended up with spider diagrams, whole sentences, algebraic expressions, and a mixture of these, all on the blackboard. What I liked about it was how it showed all these connections and different representations, and that we were able to discuss the similarities and differences in these representations.
Pause for thought Mrs Mehta’s lesson involved quite a lot of writing and recording on the backboard, by the students themselves as well as the teacher. What do you think are the advantages and potential disadvantages of this approach?Now think about how the activity went with your students, and reflect on the following questions:What responses from students were unexpected? Why?What questions did you use to probe your students’ understanding? Did you modify the task in any way like Mrs Mehta did? If so, what was your reasoning for this?
## 4 Moving on to write formal algebraic statements and expressions
Professional mathematicians develop models to predict and describe dynamics and changes in what is happening. In doing this they make it possible to foresee what might be needed when changes happen, which is very important in all planning. This mathematical modelling relies on deciding what the variables and constants are, which ones are connected and how they are connected. This has been considered in Activities 1 and 2. The next step is to decide how these variables and constants influence and relate to each other and to record this ‘model’ in a mathematical way by making mathematically expressed statements.
The next activity will develop your thinking about how to make simple versions of such mathematical models, and build on the learning from Activities 1 and 2. These tasks work particularly well for students working in pairs or small groups because this allows more ideas to be generated and students can offer mutual support when they are stuck.
### Activity 3: Variables and constants in algebraic expressions
[Note for the teacher: this task can be simplified by using whole numbers only.]
#### Part 1: Mr Murti calculates his autorickshaw fare
• Remind the students of Mr Murti and his travels around the city in an autorickshaw. To make sure he does not get overcharged, Mr Murti likes to calculate how much he has to pay himself.
• The autorickshaw driver charges Rs. 25 for a journey that is up to 2 km. After that the fare is Rs. 0.80 for every extra 0.1 km.
Ask the students to do the following:
• Calculate the fare that Mr Murti pays for travelling 3.6 km, 6.7 km, 12.3 km, 25.9 km, 31 km, 1,000 km, 1 crore km, and finally ‘x’ km.
• Write down the way that you worked out each answer. Did you need to change your method to find the cost for x km? Check your answers with your classmates.
• Now complete this statement in algebra or in words:
If Mr. Murti travels x km, he will have to pay a fare of ___________.
Note that the whole sentence above, ‘If Mr Murti …’, is called a statement. What the students have to fill in at the end, in this case something similar to 25 + 8(x – 2), where x > 2, is called ‘the algebraic expression’. If students cannot write it yet in algebraic notation they should write it in their own words in a sentence.
#### Part 2: Make up your own statements
Ask the students to use the list of variables and constants they made in Activity 2 to construct their own statements in words or with algebraic expressions using variables and constants for concepts like cost or time.
Your students will probably not all be at the same stage in their understanding of how to construct their own statements using variables and constants. This activity should provide you with an excellent opportunity to monitor their performance and provide them with constructive feedback. You may wish to have a look at the key resource ‘Monitoring and giving feedback’ to help you prepare for this aspect of the activity.
Video: Monitoring and giving feedback
### Case Study 3: Mrs Aparajeeta reflects on using Activity 3
I asked the students to work on part 1 of Activity 3 in pairs or in groups of three because I thought that might help them to get more ideas and to get un-stuck if they did not know how to do it.
While they were working on this, I walked around observing how they went about working out the fare Mr Murti would have to pay. I noticed that they used different methods. I thought it would be nice to share those with the whole class so the students could see that there were several ways of solving a question. So after about five minutes I stopped the class and asked two students that I knew had used different approaches to come to the blackboard and write them down. I then asked who had done it differently and asked them to explain how they had done their calculations.
I noticed that not all students were listening, so I then asked all the students, in their pairs or groups, to find justification for the methods others had used, and then we shared this again with the whole class. This led them to discover and discuss misunderstandings. For example, Seema and her partner had found multiples of 0.8 km and the costs for each of those multiples. They had then used the one closest to what was asked in the question. She had, however, forgotten that the first two kilometres was a fixed price. Jai’s group examined her method and pointed out that they had forgotten this fixed price. So what then happened was, not only could they see the different ways of getting to a solution, but they could also see what they had missed.
Most of the students were able to complete the statement in words, and about a third of the class attempted to then complete the statement with an algebraic expression. The algebraic expression was seldom correct however. I asked a couple of students to come and write their algebraic expression, together with their statements, in words on the blackboard. We then discussed how they were linked, and whether we could improve on the mathematical notation.
Part 2 of the activity helped them in experimenting with this more, which was useful. I think for the student to be able to talk about algebraic notation and how it relates to a statement in words was really helpful and I realised I had actually never given them a chance to do that before.
Pause for thought Did you feel you had to intervene at any point? Were all your students engaged in the activity?What points did you feel you had to reinforce? Did you modify the task in any way like Mrs Aparajeeta did? If so, what was your reasoning for this?
## 5 Putting algebra into cricket
The National Curriculum Framework (NCF, 2005) lists one of its principles regarding the approach to knowledge in the curriculum as:
Connecting with the local and the contextualised in order to ‘situate’ knowledge and realising its ‘relevance’ and ‘meaningfulness’; to reaffirm one’s experiences outside school; to draw one’s learning from observing, interacting with, classifying, categorising, questioning, reasoning and arguing in relation to these experiences.
(NCF, 2005, p. 33)
Activity 4 aims to address this. The activity can be used as a consolidation exercise, where students can use their learning from the earlier activities in a different context, one they are likely to be familiar with – cricket.
Video: Using local resources
### Activity 4: Runs per over
This activity allows students to become aware of variable quantities in a cricket game.
Figure 2 Students playing cricket.
#### Preparation
For this activity the students could go outside and play the game for real. Alternatively, it could be played inside the classroom by throwing a dice for the number of runs that each player gets for each ball. (Let the number five on the dice be zero runs, since you only rarely score five runs from one ball in cricket!) If the dice falls off the desk, they are out!
#### The activity
Say to the students:
• Let’s play a 5–5 cricket match. For this, we are going to create groups of two teams of five students. Each team is to include both girls and boys. For each team, one student will be designated the score keeper (not always the girls!). Each team gets to bowl five overs.
• After six balls have been bowled, add up the score for that over.
• The two scorers jointly fill in Table 1 by recording the number of runs scored in each over: Table 1 Scoring card.
Over Team 1 Team 2
1
2
3
4
5
Totals
After the match, ask the class to discuss the following questions. In larger classes it works well to ask the students to first discuss these questions in small groups and then share with the whole class.
1. Did each team score the same number of runs in each over? Why?
2. What is the maximum number of runs that could be scored per over? Why? (Note for the teacher: You can score one, two or three runs if you really run between the wickets, four for reaching the boundary having hit the field first, and six for hitting the ball over the boundary without hitting the field first – so the maximum number of runs is six sixes. This is a case of a variable having limited values that it can take.)
3. For each team, is there a visible trend in the number of runs scored in each over? Is the trend the same for both teams? If not, why do you think the trend is different?
4. If this was a six-over match, what could have been the runs scored by each team? Would the result of the game be different or the same if each team got six overs?
5. Which of the following quantities are variables? In other words, what may have varied during the match?
• number of wickets taken by each bowler
• number of overs bowled by each team
• number of boundaries scored by different batsman
• weight of the ball used in the match.
6. What other quantities may have varied during the match? What quantities are constants (which may have remained unchanged during the match)?
Similar to Part 2 in Activity 3, ask the students to use their list of variables and constants to construct their own statements with algebraic expressions.
Video: Involving all
### Case Study 4: Mr Kapur reflects on using Activity 4
Cricket being a game very close to all their hearts, the students could really get into the discussion and contribute a lot to the activity. The day before, I asked the students to bring their cricket bats and balls to the school the next day if that was possible.
I have both girls and boys in the class and I thought the girls might feel a bit left out, or the boys might think this would be all about them! So I introduced the activity by saying I had read in the newspaper about the National Indian women’s cricket team and how well they were doing. I mentioned some names such as Mithali Raj, team captain from Rajasthan, and Jhulan ‘Babul’ Goswami, from Bengal, who, like Mithali Raj also has won the government’s Arjuna Award for excellence in the game and had also been team captain in the past.
I also made sure we discussed some rules of the games that they needed to know to complete the activity, so there would be no issues about that. We then went outside and played the games. We had mixed gender teams, we had girls against boys, we had boys against boys and girls against girls – a real mix.
In preparation for the whole-class discussion I decided to make some changes and work on Questions 1 to 3 first, and only then move on the other questions. I thought otherwise the discussion about variables and constants might get lost and it was important that did not happen as that is about the mathematics I wanted them to learn. So I wrote the Questions 1 to 3 on the blackboard and asked the students to discuss these in their teams of five. We then discussed this as a whole class.
I then wrote the Questions 4 and 5 on the blackboard, asked them to discuss these in their teams again and to make a list in their books with ‘Variables and constants in the cricket game’ as the title. I gave them five minutes to do this. We then had a whole-class discussion. I asked them to use the words ‘variable’ and ‘constant’ all the time so they could get used to the vocabulary.
By the end of the lesson I think I can safely say that most of them were confident of what varying quantities were and what was meant by constants. They also could see that even when denoting the varying quantity by a letter, it was a number. Some of the students managed to write their statements in algebraic notation, others could describe the statements in words. I was happy with that differentiation – they had all learned, and moved on their learning from where they were.
At the end of the lesson I told them to now go and discuss ‘variables and constants in the cricket game’ at home and with their friends. I do not know whether they did this, but they seemed to like the idea!
Pause for thought What questions did you use to probe your students’ understanding? Did you feel you had to intervene at any point? What points did you feel you had to reinforce? Which students might need further reinforcement? Did you modify the task in any way like Mr Kapur did? If so, what was your reasoning for this?
## 6 Summary
This unit has used ideas about teaching variables and constants to focus on two important issues.
The first was the need to help students to be able to talk about their mathematical ideas and algebraic notation. The students need to be able to express their ideas if they are to fully understand them and use them beyond the immediate context of the classroom. Thinking requires language, and students that are helped to use specific mathematical vocabulary and phraseology themselves are at an advantage when asked to think.
The second big idea in this unit was the importance of students understanding the role of constants and variables to allow them to give meaning to mathematical statements and algebraic expressions.
Pause for thought Identify three ideas that you have used in this unit that would work when teaching other topics. Make a note of two topics you have to teach soon where those ideas can be used with some small adjustments.
## Resources
### Resource 1: NCF/NCFTE teaching requirements
This unit links to the following teaching requirements of the NCF (2005) and NCFTE (2009) and will help you to meet those requirements:
• View students as active participants in their own learning and not as mere recipients of knowledge; how to encourage their capacity to construct knowledge; how to shift learning away from rote methods.
• Let students learn important mathematics and see mathematics is more than formulas and mechanical procedures.
• Reaffirm one’s experiences outside school: to draw one’s learning from observing, interacting with, classifying, categorising, questioning, reasoning and arguing in relation to these experiences.
## References
Bruner, J. (1986) Actual Minds, Possible Worlds. Cambridge, MA: Harvard University Press.
Lee, C. (2006) Language for Learning Mathematics: Assessment for Learning in Practice. Buckingham: Open University Press.
National Council for Teacher Education (2009) National Curriculum Framework for Teacher Education (online). New Delhi: NCTE. Available from: http://www.ncte-india.org/ publicnotice/ NCFTE_2010.pdf (accessed 12 April 2014).
National Council of Educational Research and Training (2005) National Curriculum Framework (NCF). New Delhi: NCERT.
Pimm, D. (1995) Symbols and Meanings in School Mathematics. London: Routledge.
Sfard, A. (2010) Thinking as Communicating. Cambridge, UK: Cambridge University Press.
Watson, A., Jones, K. and Pratt, D. (2013) Key Ideas in Teaching Mathematics. Oxford: Oxford University Press.
Zack, V. and Graves, B. (2001). ‘Making mathematical meaning through dialogue: “Once you think of it, the Z minus three seems pretty weird”’, Educational Studies in Mathematics, vol. 46, pp. 229–71.
## Acknowledgements
Except for third party materials and otherwise stated below, this content is made available under a Creative Commons Attribution-ShareAlike licence (http://creativecommons.org/ licenses/ by-sa/ 3.0/). The material acknowledged below is Proprietary and used under licence for this project, and not subject to the Creative Commons Licence. This means that this material may only be used unadapted within the TESS-India project and not in any subsequent OER versions. This includes the use of the TESS-India, OU and UKAID logos.
Grateful acknowledgement is made to the following sources for permission to reproduce the material in this unit:
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# Lesson 11Equations of All Kinds of Lines
Let’s write equations for vertical and horizontal lines.
### Learning Targets:
• I can write equations of lines that have a positive or a negative slope.
• I can write equations of vertical and horizontal lines.
## 11.1Which One Doesn’t Belong: Pairs of Lines
Which one doesn’t belong?
## 11.2All the Same
1. Plot at least 10 points whose -coordinate is -4. What do you notice about them?
1. Which equation makes the most sense to represent all of the points with -coordinate -4? Explain how you know.
2. Plot at least 10 points whose -coordinate is 3. What do you notice about them?
3. Which equation makes the most sense to represent all of the points with -coordinate 3? Explain how you know.
4. Graph the equation .
5. Graph the equation .
### Are you ready for more?
1. Draw the rectangle with vertices , , , .
2. For each of the four sides of the rectangle, write an equation for a line containing the side.
3. A rectangle has sides on the graphs of , , , . Find the coordinates of each vertex.
## 11.3Same Perimeter
1. There are many possible rectangles whose perimeter is 50 units. Complete the table with lengths, , and widths, , of at least 10 such rectangles.
2. The graph shows one rectangle whose perimeter is 50 units, and has its lower left vertex at the origin and two sides on the axes. On the same graph, draw more rectangles with perimeter 50 units using the values from your table. Make sure that each rectangle has a lower left vertex at the origin and two sides on the axes.
3. Each rectangle has a vertex that lies in the first quadrant. These vertices lie on a line. Draw in this line, and write an equation for it.
4. What is the slope of this line? How does the slope describe how the width changes as the length changes (or vice versa)?
## Lesson 11 Summary
Horizontal lines in the coordinate plane represent situations where the value doesn’t change at all while the value changes. For example, the horizontal line that goes through the point can be described in words as “for all points on the line, the value is always 13.” An equation that says the same thing is .
Vertical lines represent situations where the value doesn’t change at all while the value changes. The equation describes a vertical line through the point .
## Lesson 11 Practice Problems
1. Suppose you wanted to graph the equation .
1. Describe the steps you would take to draw the graph.
2. How would you check that the graph you drew is correct?
2. Draw the following lines and then write an equation for each.
1. Slope is 0, -intercept is 5
2. Slope is 2, -intercept is
3. Slope is , -intercept is 1
4. Slope is , -intercept is
3. Write an equation for each line.
4. A publisher wants to figure out how thick their new book will be. The book has a front cover and a back cover, each of which have a thickness of of an inch. They have a choice of which type of paper to print the book on.
1. Bond paper has a thickness of inch per one hundred pages. Write an equation for the width of the book, , if it has hundred pages, printed on bond paper.
2. Ledger paper has a thickness of inch per one hundred pages. Write an equation for the width of the book, , if it has hundred pages, printed on ledger paper.
3. If they instead chose front and back covers of thickness of an inch, how would this change the equations in the previous two parts?
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Home » Worksheets » Graph & Charts » Graphing Proportional Relationships and Identifying Slope of the Line 8th Grade Math Worksheets
Graphing Proportional Relationships and Identifying Slope of the Line 8th Grade Math Worksheets
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Definition:
The concept of proportional relationship and slope of the line are very essential in many aspects especially in economic aspect. In these worksheet, you will further understand these concepts.
Summary:
Proportional relationship follows the equation y = mx, whereas variable y varies directly as x and m is the constant of proportionality. This means that as x decreases, y decreases and vice-versa and that the ratio between x and y always stays the same.
Step1: Given x and y values
Step 2: Plot points of x and y
When the line does not passed through the origin it means that it is a non-proportional relationship.
From the equation y = mx + b:
• It is a proportional relationship when b = 0 between x and y, so y = mx
• It is a non-proportional relationship when b ≠ 0 between x and y.
Step 3 : Connect the dots
As you can see, the points formed a straight line that passes through the origin indicating a proportional relationship.
The constant of proportionality of line (m) is the same as the slope. It can be solved by using the formula y = mx + b. When b = 0 the equation will be y = mx, therefore you can rearrange the formula to m = y / x where y is rise and x is run.
Step 4 : Solve change in y axis or the rise.
Step 5 : Solve change in x axis or the run.
Graphing Proportional Relationships and Identifying Slope of the Line Worksheets
This is a fantastic bundle which includes everything you need to know about Graphing Proportional Relationships and Identifying Slope of the Line across 15+ in-depth pages. These are ready-to-use Common core aligned Grade 8 Math worksheets.
Each ready to use worksheet collection includes 10 activities and an answer guide. Not teaching common core standards? Don’t worry! All our worksheets are completely editable so can be tailored for your curriculum and target audience.
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# Polynomials and Factoring Review By: Ms. Williams.
## Presentation on theme: "Polynomials and Factoring Review By: Ms. Williams."— Presentation transcript:
Polynomials and Factoring Review By: Ms. Williams
Classifying Polynomials 1. Find the degree of the polynomial and name it based on its degree and number of terms. 2x + 5
Answer Degree: 1 Name: Linear Binomial
Adding and Subtracting Polynomials 1. Simplify the expression. (a 3 – 2a 2 ) + (3a 3 – 4a 2 )
Answer 4a 3 – 6a 2
Multiplying a Monomial and Trinomial 1. Simplify the expression. 3x 3 (x 2 + 5x – 8)
Answer 3x 5 + 15x 4 – 24x 3
Finding the GCF 1. Find the GCF of the polynomial. 18x 6 + 12x 3
Answer 6x 3
Factoring out the GCF 1. Factor out the GCF of the polynomial. 9x 4 + 6x 3 – 18x 2
Answer 3x 2 (3x 2 + 2x – 6)
Multiplying Binomials 1. Simplify the product. (x – 10)(x + 1)
Answer x 2 – 9x – 10
Multiplying a Binomial and Trinomial 1. Simplify the product. (x – 4)(x 2 + 2x – 9)
Answer x 3 – 2x 2 – 17x + 36
The Square of a Binomial 1. Simplify the product. (x – 12) 2
Answer x 2 – 24x + 144
The Difference of Squares 1. Simplify the product. (2x + 5)(2x – 5)
Answer 4x 2 – 25
Classifying Polynomials 1. Find the degree of the polynomial and name it based on its degree and number of terms. 3x 3 + 2x 2 – 1
Answer Degree: 3 Name: Cubic Trinomial
Adding or Subtracting Polynomials 1. Simplify the expression. (–4k 4 + 3k 2 + 14) – (4k 4 + 7k 2 – 6)
Answer –8k 4 – 4k 2 + 20
Multiplying a Monomial and Trinomial 1. Simplify the expression. (–6y 4 )(–5y 2 + 11y – 8)
Answer 30y 6 – 66y 5 + 48y 4
Finding the GCF 1. Find the GCF of the polynomial. 21x 2 y 5 + 35x 6 y
Answer 7x 2 y
Factoring out the GCF 1. Factor out the GCF of the polynomial. 27x 3 – 9x 2 + 18
Answer 9(3x 3 – x 2 + 2)
Multiplying Binomials 1. Simplify the product. (2x 2 + 7x)(9 – x)
Answer –2x 3 + 11x 2 + 63x
Multiplying a Binomial and Trinomial 1. Simplify the product. (3x 2 – 4x + 2)(2x + 3)
Answer 6x 3 + x 2 – 8x + 6
The Square of a Binomial 1. Simplify the product. (5x + 4) 2
Answer 25x 2 + 40x + 16
The Difference of Square 1. Simplify the product. (9x 2 + 10)(9x 2 – 10)
Answer 81x 4 – 100
Classifying Polynomials 1. Find the degree of the polynomial and name it based on its degree and number of terms. 45
Answer Degree: 0 Name: Constant Monomial
Adding and Subtracting Polynomials 1. Simplify the expression. (–x 4 + 13x 5 + 6x 3 ) + (6x 3 + 5x 5 – 7x 4 )
Answer 18x 5 – 8x 4 + 12x 3
Multiplying a Monomial and Trinomial 1. Simplify the product. 8m 7 (5m 3 – 7m 2 – 9)
Answer 40m 10 – 56m 9 – 72m 7
Finding the GCF 1. Find the GCF of the polynomial. –12x 4 – 60x 2 – 36x
Answer 12x OR –12x
Factoring out the GCF 1. Factor out the GCF of the polynomial. –12x 4 – 60x 2 – 36x
Answer 12x(–x 3 – 5x – 3) OR –12x(x 3 + 5x + 3)
Multiplying Binomials 1. Simplify the product. (x – 15)(x + 4)
Answer x 2 – 11x – 60
Multiplying a Binomial and Trinomial 1. Simplify the product. (x 2 – 6 – 2x)(6x + 1)
Answer 6x 3 – 11x 2 – 38x – 6
The Square of a Binomial 1. Simplify the product. (4x – 3y) 2
Answer 16x 2 – 24xy + 9y 2
The Difference of Squares 1. Simplify the product. (13a + 5b)(13a – 5b)
Answer 169a 2 – 25b 2
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