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boyakko [2]
3 years ago
7

Which expression below gives the average rate of change of the function g(x) = -x2 - 4x on the interval 6 ≤ x ≤ 8 ?

Mathematics
2 answers:
Scrat [10]3 years ago
7 0

Answer:

the first selection (see below)

Step-by-step explanation:

The average rate of change (m) on the interval [x1, x2] is given by ...

... m = (g(x2) -g(x1))/(x2 -x1)

For g(x) = -x²-4x and (x1, x2) = (6, 8), the expression is the one attached.

Papessa [141]3 years ago
3 0

Answer:

\frac{[-8^2 - 4(8)]- [-6^2 - 4(6)]}{8-6}

Step-by-step explanation:

average rate of change of the function g(x) = -x^2 - 4x on the interval 6 ≤ x ≤ 8

To find average rate of change we use formula

Average =\frac{g(x_2)-g(x_1)}{x^2-x_1}

use the given interval 6<=-x<=8

x2=8  and x1= 6

we replace the value in the given formula

g(x) = -x^2 - 4x

g(8) = -8^2 - 4(8)

g(6) = -6^2 - 4(6)

x2-x1 is 8-6

So equation becomes

\frac{[-8^2 - 4(8)]- [-6^2 - 4(6)]}{8-6}

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2 years ago
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DochEvi [55]

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therefore

<u>the answer Part a) is</u>

the speed is equal to 3.5\frac{m}{sec}

<u>Part b) </u>Find the velocity

we know that

<u>Velocity </u>is a vector quantity; both magnitude and direction are needed to define it

in this problem we have

the magnitude is equal to the speed

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<u>the answer Part b) is</u>

the velocity is 3.5\frac{m}{sec}\ North\ East\ (NE)

Part c)

we know that

the acceleration is equal to the formula

a=\frac{V2-V1}{t2-t1}

in this problem we have

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t2=15\ sec\\t1=0

substitute in the formula

a=\frac{0-3.5}{15-0}

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therefore

<u>the answer Part c) is</u>

the acceleration is -\frac{7}{30}\frac{m}{sec^{2}}

This is an example of negative acceleration

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