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3 years ago

FREE BARINLIEST

Mathematics

2 answers:

DerKrebs [107]3 years ago

3 0

The answer would have to be 30%

IceJOKER [234]3 years ago

3 0

The answer would be 30 percent.

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Determine whether the statement is true or false. If it is true, explain why.

Over [174]

True.

Since $y^4 \geq 0$ , $y' = -1 - y^4 \ \textless \ 0$ and the solutions are decreasing functions.

8 0

3 years ago

The logo for a new company is a circle divided into 24 equally sized sectors. Six of the sectors are white, and the others are d

AveGali [126]

Answer:

r = 2.5

Step-by-step explanation:

the area of the whole circle = $\frac{13}{50}$ *π * 24

A = πr² = (13π*24)/50

r = √((13*24)/50) = 2.5 inches

8 0

3 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

Three pieces of artwork were chosen from each grade of Eastlake Elementary School to be displayed at the school art show. If the

kkurt [141]

9×3

=27 pieces of artwork to be displayed.

7 0

2 years ago

Read 2 more answers

Find the value of x and y. WILL MARK BRAINLIEST

Eduardwww [97]

Answer:

x=7

y = 4

Step-by-step explanation:

Since it is a parallelogram

2x-2 =12

and 2y =8

2x-2=12

Add 2 to each side

2x-2+2 =12+2

2x = 14

Divide by 2

2x/2 =14/2

x = 7

2y =8

Divide by 2

2y/2 =8/2

y =4

6 0

3 years ago

Read 2 more answers