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umka21 [38]
2 years ago
14

HELPPPPPPPPPP PLEASE ​

Mathematics
1 answer:
Vlad1618 [11]2 years ago
3 0

Answer:

a) 27. b) 8/3

This is your answer ☺️☺️☺️

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<><><><>Simple true and false math<><><><>
skad [1K]

Answer:

true

Step-by-step explanation:

just my opinion but for me it's true

7 0
2 years ago
Explain why the square root of a number is defined to be equal to that number to the 1/2 power<br>​
denis23 [38]

Answer:

Because square root is essentially the same thing as the power to a half

sqrt(x) is the same as (x)^1/2

4 0
2 years ago
A new shop is being built in town and the design shows that it will be a rectangular prism with 8 meters along the front 12 mete
Kisachek [45]

Volume of space inside shop is 288 m³

<u>Given that;</u>

Dimensions of rectangle = 8 meter , 12 meter , 3 meter

<u>Find:</u>

Volume of space inside shop

<u>Computation:</u>

Volume of rectangle = [l][b][h]

Volume of rectangle = Volume of space inside shop

So,

Volume of space inside shop = [8][12][3]

Volume of space inside shop = [96][3]

Volume of space inside shop = 288 m³

Learn more;

brainly.com/question/15019502?referrer=searchResults

8 0
2 years ago
What radius of a circle is required to inscribe a regular hexagon with an area of 166.28 in2 and an apothem of 6.928 in?
Lunna [17]
If you want to inscribe a polygon inside a circle, you have a formula that doesn't have to use the apothem. The formula is:

A = (nr²/2)sin(360/n)

Since the polygon is a hexagon, it has 6 sides. Thus, n = 6. Knowing the area, we can determine the radius of the circle, r.

166.28 = (6r²/2)sin(360/6)
r = 8 inches

Thus, the radius of the circle is 8 inches.
7 0
3 years ago
Read 2 more answers
What’s the correct volume ?
nikitadnepr [17]

Answer:

V = 12π

Step-by-step explanation:

Given radius of cone, r = 3 and y = 5

by Pythagorean theorem (or by recognizing that this is a standard 3-4-5 right triangle), we can determine that height of cone, x = 4 in

volume of cone is given by :

V= (1/3) πr²x

= (1/3) x π x 3² x 4

V = 12π

7 0
3 years ago
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