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netineya [11]
3 years ago
15

Select the correct answer.

Mathematics
1 answer:
alex41 [277]3 years ago
4 0

Answer:

Option B

Step-by-step explanation:

Average rate of change of a function 'f' in the interval x = a and x = b is given by,

Average rate of change of change = \frac{f(b)-f(a)}{b-a}

Given function is,

g(x) = \frac{5}{x-1}+2

We have to calculate the rate of change in the interval [-4, 3],

Average rate of change = \frac{g(3)-g(-4)}{3-(-4)}

g(3) = \frac{5}{3-1}+1

       = 2.5 + 1

       = 3.5

g(-4) = \frac{5}{-4-1}+1

       = -1 + 1

       = 0

Average rate of change = \frac{3.5-0}{3+4}

                                        = \frac{3.5}{7}

                                        = \frac{1}{2}

Therefore, Option B will be the answer.

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38+25+25=88

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The difference between value of y at x=11 is 385, the correct option is third.

Step-by-step explanation:

The difference between value of y at x=11 is 385, the correct option is third.

The equation of a line which passing through two points is given below,

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Put x=11

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Therefore third option is correct.

3 0
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Shipping rates for Company A and Company B are shown in the tables below. Which company has shipping rates that you can represen
meriva

Company B; the ratios of cost to weight are equivalent.

Step-by-step explanation:

Step 1:

In the equation, y=kx k is the constant of proportionality.

If the values are in accordance with y=kx, the values of k will be constant for all the values.

So we determine the values of k for both the companies and see which has a constant k.

If y=kx, k = \frac{y}{x}. In these tables, y is the total cost and x is the weight in lbs.

Step 2:

For company A,

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when y=10.85, x=2, k = \frac{10.85}{2} = 5.425,

when y=11.15, x=3, k = \frac{11.15}{3} = 3.71666.

For company B,

when y=2.75, x=1, k = \frac2.75}{1} = 2.75,

when y=5.50, x=2, k = \frac{5.50}{2} = 2.75,

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So company B has a constant value of k=2.75.

3 0
2 years ago
What is 1.5 Devided by 1
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Answer:

1.5

Step-by-step explanation:

Anything divided by 1 is itself

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