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2 years ago

The function C gives the cost, in dollars, to shred w pounds of confidential documents of a company.

Mathematics

1 answer:

SSSSS [86.1K]2 years ago

8 0

C’(500)=80 means at 500 pounds of documents, the cost is increasing at a rate of $80 dollars/pound

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Please help solve this with working out :)

Anastaziya [24]

I hope this can help u

3 0

2 years ago

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

5 0

3 years ago

Hank estimated the width of the door to his class room in feet what is a reasonable estimate

timurjin [86]

My reasonable estimate might be about 2 1/2 feet. I know that because taking three rulers up to your bedroom or a normal room and get about 2 or 2 1/2 feet

4 0

3 years ago

Part 1: How's the weather? A. Investigate the weather forecast in your city. Record the high and low temperatures that are forec

victus00 [196]

Using the mean concept, it is found that:

1. The mean change in the forecasted high temperatures is of -0.33ºC.

2. The mean change in the forecasted low temperatures is of 0.5ºC.

-----------------------------

The mean of data-set is given by the <u>sum of all observations divided by the number of observations.</u>

Item 1:

The changes in the high temperature are: 2, 1, -1, -1, -1, -2

Thus:

$M = \frac{2 + 1 - 1 - 1 - 1 - 2}{6} = -0.33$

The mean change in the forecasted high temperatures is of -0.33ºC.

Item 2:

The changes in the low temperature are: 2, -2, 2, 1, 0, 0.

$M = \frac{2 - 2 + 2 + 1 + 0 + 0}{6} = 0.5$

The mean change in the forecasted low temperatures is of 0.5ºC.

A similar problem is given at brainly.com/question/24754716

3 0

1 year ago

Write a decimal that is 1\10 of 3.0

Lapatulllka [165]

3/10=0.3 so it's 0.3.

7 0

3 years ago