Use base ten blocks to find 182 ÷14 describe the steps you took to find your answer

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

2 years ago

When the net is folded into the rectangular prism shown beside it, which letters will be on the front and back of the rectangula

earnstyle [38]

<h2>Answer:</h2>

<u>The correct option is </u><u>The letter on the front will be N. The letter on the back will be L. </u>

<h2>Step-by-step explanation:</h2>

When we fold the given net, we will get Q,P,M and N on sides. Side M will come to the top, side Q on the right side, side P on the left and side O on the bottom. The side which comes to the front will be N of the observer and similarly the side L will come to the back of the rectangular prism.

4 0

2 years ago

Read 2 more answers

What is the positive solution to this equation? 4x2 + 12x = 135

Bas_tet [7]

Answer:

127/12

Step-by-step explanation:

4 × 2 + 12x = 135

(1. Simplify 4 x 2 to 8.

8 + 12x = 135

(2. Subtract 88 from both sides.

12x= 135 - 8

(3. Simplify 135 - 8 to 127

12x = 127

(4. Divide both sides by 12

x= 127/12

Decimal Form: 10.583333

I think this is the awnser, but don't quote me on that

4 0

2 years ago

Based upon statistical studies, it takes about one-third less time for fans to exit a game as it does for them to enter and find

Mice21 [21]

<span>Given data shows that 1/3 time less time for fans to exit a game as it does for them to enter and find their seats. Ingress for the stadium = 3,000 fans per hour Ingress rate = (1 - 1/3) x Egress rate = 2/3 Egress rate Egress rate = 3/2 ingress rate = 3/2 x 6,000 = 9,000 fans per hour</span>

8 0

2 years ago

Read 2 more answers

A very weak university is on and off probationary status with the accrediting agency. A different procedure is used in July and

tino4ka555 [31]

Answer:

0.3425 = 34.25% probability it will be off probation in February 2020

Step-by-step explanation:

We have these desired outcomes:

Off probation in July 2019, with 0.25 probability, then continuing off in January, with 1 - 0.08 = 0.92 probability.

Still in probation in July 2019, with 1 - 0.25 = 0.75 probability, then coming off in January, with 0.15 probability.

What is the probability it will be off probation in February 2020?

$p = 0.25*0.92 + 0.75*0.15 = 0.3425$

0.3425 = 34.25% probability it will be off probation in February 2020

6 0

2 years ago