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antoniya [11.8K]
3 years ago
9

Peter is reading a 193-page book. He has read three pages more than one-fourth of the number of pages he hasn’t yet read. How ma

ny pages has he not yet read?
Mathematics
2 answers:
viva [34]3 years ago
5 0
N + n/4 + 3 = 193
5n/4 = 190
n = 152
At 8 pages per day it will take 152 pages/ 8 pages/day = 19 days
FinnZ [79.3K]3 years ago
5 0
He has not read 152 pages.

If y is the number of pages he has not read and x is the number of pages he has read, then:

x+y=193
x=1/4y+3 (the number of pages he has read is three pages more than one fourth the number of pages he hasn't read)

Since we have two equations, solving for one variable will allow us to substitute that value into the other equation. Since x is already isolated in the second equation, we will use that, substituting the value of x from the second equation into the first.

(1/4y+3)+y=193
5/4y+3=193
5/4y=190
y=152
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\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}

General Formulas and Concepts:

<u>Algebra I</u>

Terms/Coefficients

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<u>Calculus</u>

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Basic Power Rule:

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  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]:                                                                           \displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle f(x) = \frac{\sqrt{x}}{e^x}

<u>Step 2: Differentiate</u>

  1. Derivative Rule [Quotient Rule]:                                                                   \displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}
  2. Basic Power Rule:                                                                                         \displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}
  3. Exponential Differentiation:                                                                         \displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}
  4. Simplify:                                                                                                         \displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}
  5. Rewrite:                                                                                                         \displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}
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