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3 years ago

5

The original price of a pair of jeans is $32 the sales price 15% off the original price what is the amount off the original pric

e

1 answer:

ivolga24 [154]3 years ago

3 0

$32+15%=32.15

Hope this helps ^_^

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Use substitution to solve each of the following pairs of simultaneous equations.

Ronch [10]

Answer:

I will go for a part the rest You can do by yourself

Step-by-step explanation:

Multiply the second equation to get same amount of x in both equations

2x=2-3y/2

multiply both equations by 2 to get rid of the fraction in second equation

4x+14y=34

4x=4-3y

subtract equation(2) from equation(1)

14y=30+3y

11y=30

y=30/11, x=-23/22

6 0

3 years ago

Read 2 more answers

Irrational number between 7.7 and 7.9

monitta

That would be 7.8 (7.800) hope this helps! ;)

3 0

3 years ago

Please help I forgot how to do this

timofeeve [1]

Answer:

See answer below.

Step-by-step explanation:

2≥ g/-4

2( -4) ≥ g

-8 ≥ g

graph : Closed circle at -8 and shade to the left of -8

3 0

2 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:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

Distributive Property

<u>Algebra I</u>

Terms/Coefficients
Factoring

<u>Calculus</u>

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:

<u>Step 1: Define</u>

<em>Identify</em>

y = (3x - 1)⁵(4 - x⁴)⁵

<u>Step 2: Differentiate</u>

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

Can someone help me pls?

gladu [14]

Answer:

A

Step-by-step explanation:

4 0

2 years ago