Mathematics, 28.07.2019 12:00 bullockarwen

Let f(x)=−1/2(x+2)^2+5 . what is the average rate of change for the quadratic function from x=−3 to x = 1? enter your answer in the box. show your work

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Answer from: Rusven

Average rate of change= [ f(1)-f(-3) ] / [1-(-3) ]
Average rate of change= [ f(1)-f(-3) ] / (1+3)
Average rate of change= [ f(1)-f(-3) ] / (4)

x=1→f(1)=-1/2(1+2)^2+5
f(1)=-1/2(3)^2+5
f(1)=-1/2(9)+5
f(1)=-(1*9)/2+5
f(1)=-9/2+5
f(1)=(-9+2*5)/2
f(1)=(-9+10)/2
f(1)=1/2

x=-3→f(-3)=-1/2(-3+2)^2+5
f(-3)=-1/2(-1)^2+5
f(-3)=-1/2(1)+5
f(-3)=-(1*1)/2+5
f(-3)=-1/2+5
f(-3)=(-1+2*5)/2
f(-3)=(-1+10)/2
f(-3)=9/2

Average rate of change= [ f(1)-f(-3) ] / (4)
Average rate of change= [ 1/2-9/2 ] / (4)
Average rate of change= [ (1-9) /2 ] / (4)
Average rate of change= [ (-8) /2 ] / (4)
Average rate of change= (-4) / (4)
Average rate of change= -1

The average rate of change for the quadratic function from x=−3 to x = 1 is equal to -1

Answer from: bward04

Ans: Average rate of change of function = -1

Explanation:
The average rate of change of function f over the interval $a \leq x \leq b$ is given as:

=> Average rate of change of function = $\frac{f(b) - f(a)}{b-a}$ --- (A)

Since,
a = -3
b = 1

Given function = $f(x) = - \frac{(x+2)^2}{2} +5$

Therefore,
$f(b) = - \frac{(b+2)^2}{2} +5$
Since b = 1; therefore,
$f(b) = f(1) = - \frac{(3)^2}{2} +5 = \frac{1}{2}$

$f(a) = - \frac{(a+2)^2}{2} +5$
Since a = -3; therefore,
$f(a) = f(-3) = - \frac{(-1)^2}{2} +5 = \frac{9}{2}$

Plug-in the values of f(a), f(b), a, and b in equation (A):

(A) => Average rate of change of function = $\frac{ \frac{1}{2} - \frac{9}{2} }{1-(-3)}$

=> Ans: Average rate of change of function = -1

-i

Answer from: Quest

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this

Answer from: Quest

answer: the answer is (b) amie should have multiplied

$54~\textup{m}^3~\textup{by}~\dfrac{4}{3}\pi.$

step-by-step explanation: given a sphere and a cylinder with same radius and height.

let, 'r' be the radius of the cylinder and sphere and 'h' be the height of the cylinder. according to the question,

r = h.

now, the volume of the sphere will be

$v_s=\pi r^2h=\pi r^3,$

and the volume of the cylinder will be

$v_c=\dfrac{4}{3}\pi r^3=54~\textup{m}^3.$

therefore, we have

$v_c=\dfrac{4}{3}v_s=\dfrac{4}{3}\times 54=72~\textup{m}^3.$

thus, ammie should have multiplied the volume of the sphere $54~\textup{m}^3~\textup{by}~\dfrac{4}{3}\pi.$

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Let f(x)=−1/2(x+2)^2+5 . what is the average rate of change for the quadratic function from x=−3 to...