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1 year ago

The art club had an election to select a president. 15 out of the 75 members of the art club

Mathematics

1 answer:

Karolina [17]1 year ago

7 0

Answer:

15/75=20%

Step-by-step explanation:

My method is to do 100 divided by 75, which is 1 1/3, then we so 15x1 1/3 which comes out to be 20%

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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

Write a rule describing the translation below. (x,y) -> (x+<br> y +

zepelin [54]

Answer: 4 units to the right, 3 units up

Hope this helped! :)

7 0

2 years ago

I will make you the brainleist!!!! Please explain and no guessing. Thank you!! Ps due in 10

aniked [119]

Is this even a question? its 66/100 you always use the first examining over 100.... makes sense

5 0

2 years ago

Read 2 more answers

[5x+10y+15z=60<br> [2x+4y+6z=60<br> [x+2y+3z=60

Alla [95]

9514 1404 393

Answer:

no solution

Step-by-step explanation:

The equations are inconsistent. The set of equations reduces to ...

x + 2y + 3z = 12

x + 2y + 3z = 30

x + 2y + 3z = 60

No values of x, y, and z can satisfy all three equations. There is no solution.

4 0

2 years ago

Use the information to find the value cot 0 =

hjlf

The tangent function is given by:

$\tan (\theta)=\frac{opposite}{_{\text{ }}adjacent}=\frac{5}{12}$

Since the angle is in the 3rd quadrant the opposite side and the adjacent side are negative. So, the cotangent function is given by:

$\cot (\theta)=\frac{1}{\tan (\theta)}=\frac{adjacent}{_{\text{ }}opposite}=\frac{-12}{-5}=\frac{12}{5}$

Answer:

12/5

8 0

1 year ago