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2 years ago

3. The area of a rectangle is 24a² b5. The length of the rectangle is 6ab. What is the width of the

Mathematics

1 answer:

adelina 88 [10]2 years ago

5 0

Answer:

4ab^4

Step-by-step explanation:

(24a^2b^5) / (6ab) = 4ab^4

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Look At picture! Below!

eimsori [14]

Answer:

The answer is: the second one

Step-by-step explanation:

(2x + 2)/y = 4w + 2

pass y to the right side 2x + 2 = y(4w + 2)

expand 2x +2 = 4wy + 2y

pass 2 to the right 2x = 4wy + 2y -2

pass 2 to the right x = (4wy + 2y - 2)/2

simplify x = 2wy + y - 1 or = 2yw + y -1

8 0

3 years ago

Solution set for 6a^2-a-5=0

VMariaS [17]

(step 1) (6a+5)(a-1)=0

(step 2) 6a+5=0 or a-1=0

answer is
a= -5/6 or a= 1
i hope this helps! ;)

6 0

3 years ago

Read 2 more answers

Identify the vertex y= —(x — 6)^2 + 2

alisha [4.7K]

(6,2)

Mark brainliest please

Hope this helps you

8 0

3 years ago

Mr.Martinez rents a car for one day.The charge is $36 plus $0.18 per mile.Mr.Martinez has a budget $60.How many miles can he dri

kondor19780726 [428]

1... the wording is a little confusing.
Is the charge $0.18 per mile or is it actually $36.18 per mile?

8 0

3 years ago

Find the derivative of

kirill [66]

Answer:

$\displaystyle y'(1, \frac{3}{2}) = -3$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \frac{3}{2x^2}$

$\displaystyle \text{Point} \ (1, \frac{3}{2})$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y = \frac{3}{2}x^{-2}$
Basic Power Rule: $\displaystyle y' = -2 \cdot \frac{3}{2}x^{-2 - 1}$
Simplify: $\displaystyle y' = -3x^{-3}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{-3}{x^3}$

Step 3: Solve

Substitute in coordinate [Derivative]: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1^3}$
Evaluate exponents: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1}$
Divide: $\displaystyle y'(1, \frac{3}{2}) = -3$

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

Unit: Derivatives

Book: College Calculus 10e

6 0

2 years ago