All categories

3 years ago

Can y’all help me find the width

Mathematics

1 answer:

tatuchka [14]3 years ago

7 0

Answer:

Step-by-step explanation:

To find the width, you have to divide the height and length by the volume

You might be interested in

Corinne earns an hourly wage on top of her weekly gas allowance for her delivery job. Her weekly earnings can be shown using the

sveta [45]

How much money he gets per hour

8 0

4 years ago

Read 2 more answers

How is 600:450 and 4:3 equalivent

romanna [79]

Answer:

when you simplify 600:450 to the smallest it can be, its 4:3. For proof, multiply 4:3 by 150. 4 times 150 is 600 and 3 times 150 is 450.

5 0

3 years ago

Order the numbers from least to greatest -3/4, 0.5, 2/3, -7/3, 1.2

BaLLatris [955]

-7/3, -3/4, 0.5, 2/3, 1.2

Brainliest please

5 0

2 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

Select the expressions that have a value less than -1.

yulyashka [42]

Answer:

B and E

Step-by-step explanation:

A. -7 ÷ (-4) = -7/-4 = 7/4 = 1.75

B. -(3 ÷ 2) = -(1.5) = -1.5

C. -8/5 * (-5/8) = (8/5) * (5/8) = 1

D. -5/-3 = 5/3 ≈ 1.667

E. -9 ÷ 6 = -9/6 = -1.5

5 0

2 years ago