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2 years ago

The two figures have a scale of 3:1. What is the length of the corresponding side in the second figure?

Mathematics

2 answers:

disa [49]2 years ago

4 0

Answer:

The answer is A) 5

zhuklara [117]2 years ago

3 0

I will say it will be c

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Keisha pumped 6 gallons of water into her pool each minute for 48 minutes what was the total change in the amount of water in th

Eduardwww [97]

The total change in the amount of water in the pool is 288 gallons.

<h3>What is Volume?</h3>

Volume is a scalar quantity expressing the amount of three-dimensional space enclosed by a closed surface.

Here, Rate of water pumped into her pool = 6 gallons / minute.

Total Duration = 48 minutes

Total amount of water = Rate X Time

= 6 X 48

= 288 gallons

Thus, the total change in the amount of water in the pool is 288 gallons.

Learn more about Volume from:

brainly.com/question/1578538

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1 year ago

Hello I need help Find the missing side lengths, leave your answers as radicals in simplest form

Jet001 [13]

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$y = \frac{1}{2} \times \frac{8 \sqrt{3} }{3} \\$

$y = \frac{4 \sqrt{3} }{3} \\$

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$x = \frac{ \sqrt{3} }{2} \times \frac{8 \sqrt{3} }{3} \\$

$x = \frac{8 \times { \sqrt{3} }^{2} }{2 \times 3} \\$

$x = \frac{8 \times 3}{2 \times 3} \\$

$x = \frac{8}{2} \\$

$x = 4$

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4 0

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