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Yuliya22 [10]
4 years ago
9

For which function is the average rate of change over the interval 1 < x < 5 greater than the average rate of change over

the same interval for the function g(x) = 1.8x2?
Mathematics
1 answer:
-Dominant- [34]4 years ago
4 0

Answer:

The average rate of change of g(x) = 1.8x² with interval 1 < x < 5 is 10.8

Step-by-step explanation:

Given

g(x) = 1.8x²

Interval 1 < x < 5

The average rate of change of the function g (x) on the interval [a,b] is calculated using the following formula:

Average Rate of change = (g(b) - g(a))/(b - a)

Where a and b are values from the interval.

a = lower Interval = 1

b = upper Interval = 5

First, we need to calculate g(b) and g(a)

Given that g(x) = 1.8x²

g(a) = g(5) = 1.8 * 1²

g(a) = 1.8 * 1

g(a) = 1.8

Then we calculate g(a)

g(b) = g(5) = 1.8 * 5²

g(b) = 1.8 * 25

g(b) = 45

We then calculated the average Rate of change by substituting values in= (g(b) - g(a))/(b - a)

Average Rate of Change = (45 - 1.8)/(5 - 1)

Average Rate of Change = 43.2/4

Average Rate of Change = 10.8

Hence, the average rate of change of g(x) = 1.8x² with interval 1 < x < 5 is 10.8

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Suppose that the function g is defined, for all real numbers, as follows.
svetoff [14.1K]

Answer:

<em>a) g(-2) = 0</em>

<em>b) g(-1) =1</em>

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Step-by-step explanation:

Given data

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<u><em>Step( i )</em></u>:-

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put x = -2

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<u><em>Step(ii)</em></u>:-

g(x) = - (x+1)^{2} +1  if -2\leq x\leq 2

Put x = -1

g(-1) = -(-1+1)^{2} +1 = -(0)^{2}+1 = -0+1 =1

<em>g(-1) =1</em>

<u><em>Step(iii)</em></u>:-

g(x) = 2 if x >2

<em>g(4) = 2</em>

<em></em>

7 0
3 years ago
Solve for m 1/C+1/m=1/z
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Answer:  " m = zC / (C − z) " .
___________________________________
Explanation:
_________________________
Given:  1/C + 1/m = 1/z ;  Solve for "m".

Subtract  "1/C" from each side of the equation:
____________________________________
1/C + 1/m − 1/C = 1/z − 1/C  ;

to get:  1/m = 1/z − 1/C ;
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mzC {1/m = 1/z − 1/C} ;

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zC/ (C − z) = m ;   ↔   m = zC/ (C − z) .
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