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3 years ago

Find the volume of a rectangular holding tank whose dimensions are 5p, 2t, and 4w. (V = lwh)

Mathematics

1 answer:

Sever21 [200]3 years ago

6 0

5p*2t*4w=10pt*4w=40ptw
hope this helps!

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Which expressions are equivalent to the expression 7(5 + m) - 8?

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Answer:

27 + 7m

Step-by-step explanation:

$7(5 + m) - 8$

open the bracket:

$= 35 + 7m - 8 \\ = 27 + 7m$

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2 years ago

Find the sum 70.6+74.02+121.985=

vova2212 [387]

266.605 is the answer.

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2 years ago

Coefficient of the term 6xy\7

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3 years ago

Explain how the graph of a quadratic function relates to the solutions of the related quadratic equation.

OleMash [197]

Answer:

The zeros of a quadratic function are the solutions of the related quadratic equation. If the graph of a quadratic function crosses the x-axis, there will be two real number solutions. If the graph of a quadratic function just touches the x-axis, there will be one unique real number, or double root, solution. If the graph of a quadratic function does not cross the x-axis, there will be no real number solution.

Step-by-step explanation:

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

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

7 0

2 years ago