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3 years ago

What is the additive inverse of 129/7?

Mathematics

2 answers:

andrey2020 [161]3 years ago

8 0

-129/7 is the correct answer

harina [27]3 years ago

6 0

Answer:

-129/7

Step-by-step explanation:

The additive inverse of a number is what you add to a number to create the sum of zero

129/7 + x = 0

It is the opposite of 129/7

-129/7

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FromTheMoon [43]

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3 years ago

Differentiate the function. y = (3x - 1)^5(4-x^4)^5

TiliK225 [7]

Answer:

$\displaystyle y' = -5(3x-1)^4(4 - x^4)^4(15x^4 - 4x^3 - 12)$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Distributive Property

Algebra I

Terms/Coefficients
Factoring

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

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

Step-by-step explanation:

Step 1: Define

Identify

y = (3x - 1)⁵(4 - x⁴)⁵

Step 2: Differentiate

Product Rule: $\displaystyle y' = \frac{d}{dx}[(3x - 1)^5](4 - x^4)^5 + (3x - 1)^5\frac{d}{dx}[(4 - x^4)^5]$
Chain Rule [Basic Power Rule]: $\displaystyle y' =[5(3x - 1)^{5-1} \cdot \frac{d}{dx}[3x - 1]](4 - x^4)^5 + (3x - 1)^5[5(4 - x^4)^{5-1} \cdot \frac{d}{dx}[(4 - x^4)]]$
Simplify: $\displaystyle y' =[5(3x - 1)^4 \cdot \frac{d}{dx}[3x - 1]](4 - x^4)^5 + (3x - 1)^5[5(4 - x^4)^4 \cdot \frac{d}{dx}[(4 - x^4)]]$
Basic Power Rule: $\displaystyle y' =[5(3x - 1)^4 \cdot 3x^{1 - 1}](4 - x^4)^5 + (3x - 1)^5[5(4 - x^4)^4 \cdot -4x^{4-1}]$
Simplify: $\displaystyle y' =[5(3x - 1)^4 \cdot 3](4 - x^4)^5 + (3x - 1)^5[5(4 - x^4)^4 \cdot -4x^3]$
Multiply: $\displaystyle y' = 15(3x - 1)^4(4 - x^4)^5 - 20x^3(3x - 1)^5(4 - x^4)^4$
Factor: $\displaystyle y' = 5(3x-1)^4(4 - x^4)^4\bigg[ 3(4 - x^4) - 4x^3(3x - 1) \bigg]$
[Distributive Property] Distribute 3: $\displaystyle y' = 5(3x-1)^4(4 - x^4)^4\bigg[ 12 - 3x^4 - 4x^3(3x - 1) \bigg]$
[Distributive Property] Distribute -4x³: $\displaystyle y' = 5(3x-1)^4(4 - x^4)^4\bigg[ 12 - 3x^4 - 12x^4 + 4x^3 \bigg]$
[Brackets] Combine like terms: $\displaystyle y' = 5(3x-1)^4(4 - x^4)^4(-15x^4 + 4x^3 + 12)$
Factor: $\displaystyle y' = -5(3x-1)^4(4 - x^4)^4(15x^4 - 4x^3 - 12)$

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

Unit: Derivatives

Book: College Calculus 10e

6 0

3 years ago