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3 years ago

12

WILL GIVE BRAINLIEST AND 10 POINTS!!!!!!!!! PLZ HELP!!!!! Lianne drew a sphere with a radius of 7 in. She calculated its volume

and wrote it below.
StartFraction question mark pi Over 3 EndFraction inches cubed

What value should be multiplied by Pi in the numerator to correctly show the volume of the sphere?

2 answers:

zzz [600]3 years ago

8 0

Answer:

= 1372

i took the test

GarryVolchara [31]3 years ago

3 0

Answer 1372.

4x7x7x7=1372:v=4/3 pi r^3

Step-by-step explanation:

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-12x^3 + 28x^2

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a square garden has a diagonal of 12 m. What is the perimeter of the garden? Express in simplest radical form

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Answer:

The perimeter of the garden, in meters, is $24\sqrt{2}$

Step-by-step explanation:

Diagonal of a square:

The diagonal of a square is found applying the Pythagorean Theorem.

The diagonal of the square is the hypothenuse, while we have two sides.

Diagonal of 12m:

This means that $d = 12$ , side s. So

$s^2 + s^2 = 12^2$

$2s^2 = 144$

$s^2 = \frac{144}{2}$

$s^2 = 72$

$s = \sqrt{72}$

Factoring 72:

Factoring 72 into prime factors, we have that:

72|2

36|2

18|2

9|3

3|3

1

So

$72 = 2^{3}*3^{2}$

So, in simplest radical form:

$s = \sqrt{72} = \sqrt{2^{3}*3^{2}} = \sqrt{2^3}*\sqrt{3^2} = 2\sqrt{2}*3 = 6\sqrt{2}$

Perimeter of the garden:

The perimeter of a square with side of s units is given by:

$P = 4s$

In this question, since $s = 6\sqrt{2}$

$P = 4s = 4*6\sqrt{2} = 24\sqrt{2}$

The perimeter of the garden, in meters, is $24\sqrt{2}$

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3 years ago

Find the similarity ratio and the ratio of perimeters for two regular octagons with areas of 18in2 and 50in2

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Answer:

3. 3 : 5; 3: 5

Step-by-step explanation:

First, let's find the leght of the sides of each octagon.

$A= 18in^{2}$

The area of an octagon is defined by

$A=2(1+\sqrt{2})l^{2}$

Replacing the area

$18=2(1+\sqrt{2})l^{2}\\\frac{18}{2(1+\sqrt{2})} =l^{2}\\l=\sqrt{\frac{18}{3.4} } \approx 2.3 \ in$

Therefore, the side of the first octagon is 1.6 inches long.

Its perimeter is: $P=8(2.3in)=18.4in$

$A=50 in^{2}$

$50=2(1+\sqrt{2})l^{2}\\\frac{50}{2(1+\sqrt{2})} =l^{2}\\l=\sqrt{\frac{50}{3.4} } \approx 3.8 \ in$

Therefore, the side of the second octagon is 3.8 inches long.

Its perimeter is $P=8(3.8in)=30.4in$ .

Now, let's divide to find each ratio:

$\frac{3.8}{2.3} \approx 1.65$ (the ratio between sides).

$\frac{30.4}{18.4} \approx 1.65$ (the ratio between perimeters).

Therefore, the closest ratio is 3. 3 : 5; 3: 5

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