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2 years ago

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Mathematics

2 answers:

elixir [45]2 years ago

7 0

Answer:

the answer is A cause i just took it

Step-by-step explanation:

Klio2033 [76]2 years ago

6 0

Answer:

The first one.

Step-by-step explanation:

The third and fourth ones can't be right because at when the line takes its first curve (just before .25 hours) it doesn't keep increasing in height, and she walked during that time.

The second one can't be right because when she started out she wasn't immediately between 1.25 and 1.5 miles.

That leaves only the first one.

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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

2 years ago

(-2.655) · (18.44)<br><br> What’s the answer

stepan [7]

Answer:

-48.9582

Step-by-step explanation:

Negative times a positive is negative.

4 0

2 years ago

Read 2 more answers

Find the range of y=3/2cos4x-1

MrRissso [65]

Answer:

Range = [- 2.5, 0.5] = [ - 5/2, 1/2]

Step-by-step explanation:

Smallest value of cos α = - 1,

largest value of cos α = 1.

When cos 4x = - 1, y=3/2cos4x-1 = 3/2*(-1) - 1 = - 5/2 = - 2 1/2 = - 2.5

When cos 4x = 1, y=3/2cos4x-1 = 3/2*1 - 1 = 1/2 = 0.5

Range = [- 2.5, 0.5] = [ - 5/2, 1/2]

5 0

2 years ago

Marissa ate six slices of pizza from the 16-piece large pizza. What percent of the pizza did she eat?

olya-2409 [2.1K]

Answer:

0.96%

Step-by-step explanation:

look on google and search dollar times it helps with problems like this

hope this helps

7 0

2 years ago

The figure below shows the dimensions of a city park in feet. Shannon thinks the area of city park is 73,084 square feet. Do you

Andreyy89

Answer:

We need to solve for the 4th side

4th side base = 75.5 -60.5 = 10 feet

4th side height = 16

4th side LENGTH^2 = 10^2 + 16^2

4th side = sq root (356) = 18.8679622641

Trapezoid Area = [(sum of the bases) / 2 ] * height

Trapezoid Area = [(136)/2] * 16

Trapezoid Area = 1,088 square feet, which is MUCH smaller

than 73,084

Step-by-step explanation:

3 0

2 years ago