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2 years ago

100 POINTS!!! must be correct!

Mathematics

2 answers:

bixtya [17]2 years ago

7 0

Answer:

$we \: have \: 5x + 6x + y + y + 3y + 4 + 6 + 2 \\ x \: terms = 5x + 6x = 11x \\ y \: terms = y + y + 3y = 5y \\ constant = 4 + 6 + 2 = 12 \\ then \: 11x + 5y + 12 \\ option \: (c) \\ thank \: you$

suter [353]2 years ago

5 0

Answer:

C. 11x + 5y + 12

Step-by-step explanation:

You have to combine all like terms For example:

5x + 6x + y + y + 3y + 4 + 6 + 2

First add 5x and 6x

5x + 6x = 11x

Then you get:

11x + y + y + 3y + 4 + 6 + 2

Next add all of the y's

y + y + 3y = 5y

then you get

5x + 6x + 5y + 4 + 6 + 2

Finally add 4, 6, and 2

4 + 6 + 2 = 12

So your final solution should be 11x + 5y + 12

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3 years ago

Find the derivative.

krek1111 [17]

Answer:

$\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Expanding/Factoring

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
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:

Step 1: Define

Identify

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

Step 2: Differentiate

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