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4 years ago

Find the area of the semicircle.

Mathematics

2 answers:

MrMuchimi4 years ago

5 0

Answer:

8 pie

Step-by-step explanation:

area of a circle is pie(radius squared)

radius is 4

4 squared is 16

divide by two because its a semi circle

Kazeer [188]4 years ago

3 0

Answer:

8pi

Step-by-step explanation:

A=pi*r^2

A=pi(4^2)

A=16pi

16pi would be the area if the circle was complete, but since this is a semicircle you just have to divide by two which would give us:

A=8pi

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2(x+7) = -4x + 14<br><br> What is the value of x, will give brainliest to first answer

iVinArrow [24]

Answer:

x = 0

Step-by-step explanation:

$2(x+7)=-4x+14\\x+7=-2x+7\\x=-2x\\3x=0\\x=0$

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3 years ago

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Houses cost £150 000 one year ago they now cost £_____this is a 25% increase

Tanya [424]

187,500
150000x1.25
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3 years ago

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

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3 years ago

don walked 3 3/5 miles on Friday, 3.7 miles on saturday, and 3 5/8 miles on sunday. list the distances from least to greatest

astraxan [27]

Solution:

we are given that

Don walked 3 3/5 miles on Friday

It can be re-written as

$3\frac{3}{5}miles=\frac{18}{5}= 3.6$ miles

3 5/8 miles on sunday.

It can be re-written as

$3\frac{5}{8}miles=\frac{29}{8}= 3.625$ miles

an d He walk 3.7 miles on satarday.

The distances from least to greatest is 3.6, 3.625, 3.7

The distances from least to greatest is $3\frac{3}{5}$ miles, $3\frac{5}{8}$ miles, 3.7 miles.

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3 years ago

A rectangular cooler is 18 inches long, 12 inches wide, and 12 inches deep, what is the volume of the cooler?

MissTica

The volume is 2592 inches

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3 years ago