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3 years ago

Calculate the average rate of change over the interval 2

Mathematics

1 answer:

zimovet [89]3 years ago

6 0

Your picture unfortunately cuts off the upper limit of the interval, so let's set it to $b>2$ . Then the average rate of change of a function $y=f(x)$ over the interval $[2,b]$ is

$\dfrac{f(b)-f(2)}{b-2}$

We have

$y=3x+5\implies\dfrac{(3b+5)-(3\cdot2+5)}{b-2}=\dfrac{3b-6}{b-2}=\dfrac{3(b-2)}{b-2}=3$

(where we can cancel the factors of $b-2$ because we assume $b>2$ )

$y=3x^2+1\implies\dfrac{(3b^2+1)-(3\cdot2^2+1)}{b-2}=\dfrac{3b^2-12}{b-2}=\dfrac{3(b-2)(b+2)}{b-2}=3(b+2)$

$y=3^x\implies\dfrac{3^b-3^2}{b-2}=\dfrac{3^b-9}{b-2}$

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Help please will give the brainliest

Mice21 [21]

Answer:

Every angle of Square = 90°

diagonal bisects the angle in half.

=> angle FGI = 45°

=> 7x+3=45

=>7x=42

therefore, x=42/7=6

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3 years ago

Read 2 more answers

C. Find the value of x below:<br> 15.r + 5<br> -2 + 16.

nydimaria [60]

Answer:

Step-by-step explanation:

(15x + 5) and (-2 + 16x) represents the measures of corresponding angles.

Measures of corresponding angles are equal.

Therefore,

- 2 + 16x = 15x + 5

16x - 15x = 5 + 2

x = 7

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3 years ago

Angle a and angle b are both supplementary angles. What must be true about the angles?

melamori03 [73]

Not sure if there're any answer choices of some sort,

but if the angle a and angle b are supplementary, that means when you add their measures together they must equal 180.

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3 years ago

Consider the function f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0

dybincka [34]

I suppose you mean

$f(x)=\cos(x^2)$

Recall that

$\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$

which converges everywhere. Then by substitution,

$\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}$

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

$f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}$

(starting at $n=1$ because the summand is 0 when $n=0$ )

b. Naturally, the differentiated series represents

$f'(x)=-2x\sin(x^2)$

To see this, recalling the series for $\sin x$ , we know

$\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}$

Multiplying by $-2x$ gives

$-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}$

and from here,

$-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}$

$-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)$

c. This series also converges everywhere. By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}$

The limit is 0, so any choice of $x$ satisfies the convergence condition.

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4 years ago

Rashau made a rectangular frame for her latest oil painting. The length is 28 centimeters more than triple the width. The perime

postnew [5]

The length is about 28 more than three times the width. Denoted as (a) width, and as (b) the length. Then move on to lay the system of equations, as done below:

$\begin{cases}2a+2b=136\\b=3a+28\end{cases}$
$\begin{cases}2a+2b=136|:2\\b=3a+28\end{cases}$
$\begin{cases}a+b=68\\b=3a+28\end{cases}$
$\begin{cases}a+3a+28=68\\b=3a+28\end{cases}$
$\begin{cases}4a=40|:4\\b=3a+28\end{cases}$
$\begin{cases}a=10\\b=3*10+28\end{cases}$
$\begin{cases}a=10\\b=58\end{cases}$

The short side has a length of 10 cm, and 58 cm long.

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3 years ago