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2 years ago

If you are a teenager, then you are younger than 20.

Mathematics

2 answers:

MissTica2 years ago

7 0

A biconditional

brainliest? :)

lana [24]2 years ago

6 0

Answer:

What? I’m 15

Step-by-step explanation:

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Becky wants to fence in a rectangular play area for her dog. If she wants the length of the play area to be 9 feet longer than h

Firdavs [7]

Answer:

The dimension of the play area written algebraically is y²/2 + 9y

Step-by-step explanation:

Let the be x and the width be y

If the length is to be 9 feet longer than half the width,

hence y/2 + 9 = x

hence dimension of the area = (y/2 + 9) × y = y²/2 + 9y

4 0

3 years ago

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What’s the answer to this? Find 1/8 of 88

guapka [62]

Answer:

Step-by-step explanation:

$\frac{1}{8}$ of 88

= $\frac{88}{8}$

= 11

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2 years ago

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Please help ASAP! Will give BRAINLIEST! Please read the question THEN answer correctly! No guessing.

wlad13 [49]

Answer:

Step-by-step explanation:

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3 years ago

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