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3 years ago

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The net for a cube is shown below. All the edges of one cube are 3 units long, and the edges of a second cube are 4 units long.

What is the difference between the surface areas of the two cubes? 6 square units 12 square units 28 square units 42 square units

1 answer:

sveticcg [70]3 years ago

5 0

6 square 6 square units

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Help differentiate this

Arte-miy333 [17]

Answer:

$\displaystyle y' = 20x^3 + 6x^2 + 70x + 9$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Distributive Property

<u>Algebra I</u>

Terms/Coefficients
Expand by FOIL
Functions
Function Notation

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

<em /> $\displaystyle y = (x^3 + 7x - 1)(5x + 2)$ <em />

<em />

<u>Step 2: Differentiate</u>

Product Rule: $\displaystyle y' = \frac{d}{dx}[(x^3 + 7x - 1)](5x + 2) + (x^3 + 7x - 1)\frac{d}{dx}[(5x + 2)]$
Basic Power Rule [Derivative Property - Addition/Subtraction]: $\displaystyle y' = (3x^{3 - 1}+ 7x^{1 - 1} - 0)(5x + 2) + (x^3 + 7x - 1)(5x^{1 - 1} + 0)$
Simplify: $\displaystyle y' = (3x^2+ 7)(5x + 2) + 5(x^3 + 7x - 1)$
Expand: $\displaystyle y' = 15x^3 + 6x^2 + 35x + 14 + 5(x^3 + 7x - 1)$
[Distributive Property] Distribute 5: $\displaystyle y' = 15x^3 + 6x^2 + 35x + 14 + 5x^3 + 35x - 5$
Combine like terms: $\displaystyle y' = 20x^3 + 6x^2 + 70x + 9$

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

Unit: Derivatives

Book: College Calculus 10e

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