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3 years ago

Why would you calculate for volume to determine if a pizza would fill you up. yes

Mathematics

1 answer:

Novay_Z [31]3 years ago

8 0

Yes I would why not and yea so I would

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Please help:)!!! thanks sm

castortr0y [4]

The answer is 12 because I just did the math

6 0

4 years ago

PLEASE HELP !!! The height of a rectangle is twice the width and the area is 32. Find the dimensions.

Helen [10]

Answer:

The rectangle has a width of 4 and a height of 8

Step-by-step explanation:

Let the height of the rectangle be H and the width be W.

We know the height of the rectangle is twice the width, so:

H = 2W

The area of a rectangle, A, is given by A = W * H, so in this case:

32 = W * 2W

32 = 2W²

W² = 16

W = 4

Knowing that the width is 4, the height must be 8. This gives us an area of 32.

8 0

4 years ago

Read 2 more answers

Triangle ABC is similar to triangle DEF.

valina [46]

Step-by-step explanation:

angle B = angle E

angle C = angle F

AC= DF

BC = EF

7 0

3 years ago

Find the area and perimeter

pashok25 [27]

Answer:

Part 1) The perimeter is $P=172\pi\ ft$

Part 2) The area is $A=1,204\pi\ ft^2$

Step-by-step explanation:

Part 1) Find the perimeter

we know that

The perimeter is equal to the circumference of the exterior circle plus the circumference of the interior circle

$P=2\pi (50)+2\pi (36)$

$P=2\pi (86)\\P=172\pi\ ft$

Part 2) Find the area

we know that

The area of the figure is equal to the area of the larger circle minus the area of the smaller circle

$A=\pi (50)^2-\pi (36)^2$

$A=\pi (1,204)$

$A=1,204\pi\ ft^2$

8 0

3 years ago

In parallelogram DEFG, DH equals X +3, HF equals 3Y, GH equals 2X -5 and HE equals 5Y plus to find the values of X and Y

daser333 [38]

The values of X and Y are 30 and 11 respectively

<h3>How to determine the values of X and Y?</h3>

The figure that represents the complete question is added as an attachment

The given parameters are:

DH = X +3

HF = 3Y

GH = 2X -5

HE = 5Y

From the attached parallelogram, we have:

DH = HF

GH = HE

Substitute the known values in the above equation

X + 3 = 3Y

2X - 5 = 5Y

Make X the subject in X + 3 = 3Y

X = 3Y - 3

Substitute X = 3Y - 3 in 2X - 5 = 5Y

2(3Y - 3) - 5 = 5Y

Expand

6Y - 6 - 5 = 5Y

Evaluate the like terms

Y = 11

Substitute Y = 11 in X = 3Y - 3

X = 3*11 - 3

Evaluate

X = 30

Hence, the values of X and Y are 30 and 11 respectively