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olga2289 [7]
3 years ago
10

I literally have no idea how to do this help...

Mathematics
2 answers:
Anettt [7]3 years ago
4 0

BTW thanks for the crooked picture lol

r(x) = 3x -1

s(x) = 2x+1

(r/s)(x)  means to write r over s  or (3x-1)/(2x+1)

(r/s)(6) means to sub x for 6

(3(6)-1)/(2(6)+1)

18-1 / 12 = 1

17/13

so

its A

hodyreva [135]3 years ago
3 0

r(x) / s(x)

= r/s (x)

= (3x - 1)/ (2x + 1)

so if r/s (6)

= [3(6) - 1] / [(2(6) + 1]

Answer is the first option

3(6) - 1

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

2(6) + 1

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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}
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  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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Unit: Differentiation

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