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Fed [463]
2 years ago
14

Find the ratio of book price to number of pages for each ordered pair. Round your answer to the nearest hundredth. A 3-column ta

ble has 4 rows. Column 1 is labeled x Number of pages with entries 100, 400, 125, 350. Column 2 is labeled y Price of Book (dollar sign) with entries 5. 00, 20. 00, 6. 25, 15. 0. Column 3 is labeled Ration of y colon x with entries 0. 05, a, b, c. A = , b = , c =.
Mathematics
1 answer:
Harlamova29_29 [7]2 years ago
6 0

The ratio of book price to number of pages for the ordered pairs a, b , and c are:

  • <em>A = 0.05</em>
  • <em>B = 0.05</em>
  • <em>C = 0.043</em>

The ratio of book price to number of pages can be evaluated for each ordered pairs as follows;

  • For a = 20.00/400 = 0.05

  • For b = 6.25/125 = 0.05

  • For c = 15.0/350 = 0.043

Therefore, ratio of book price to number of pages for the ordered pairs a, b , and c are:

a = 0.05

b = 0.05

c = 0.043

Learn more about ratio:

brainly.com/question/16981404

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<h3>Correct work shown:</h3>

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y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

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

  • Note 1: A more general solution is y = Ce^x for some constant C.
  • Note 2: It might be tempting to say the general solution is y = e^x+C, but that is not the case because y = e^x+C \to y' = e^x+0 = e^x and we can see that y' = y would only be true for C = 0, so that is why y = e^x+C does not work.
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