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AlladinOne [14]
2 years ago
14

Need the answers to 38-41

Mathematics
1 answer:
Volgvan2 years ago
6 0
38. 8•4 =32cm^2
39. 8+8+5+5=26 cm^2
40. Pi•r*2 so = 28.27in^3
41. 15/2=7.5 so pi•7.5^2 so area is 176.71m^3
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The design for the palladium window shown includes a semicircular shape at the top. The bottom is formed by squares of equal siz
9966 [12]

Answer:

(a)\ Area = 3765.32

(b)\ Area = 4773

Step-by-step explanation:

Given

A_1 = 169in^2 --- area of each square

Shade = 4in

See attachment for window

Solving (a): Area of the window

First, we calculate the dimension of each square

Let the length be L;

So:

L^2 = A_1

L^2 = 169

L = \sqrt{169

L=13

The length of two squares make up the radius of the semicircle.

So:

r = 2 * L

r = 2*13

r = 26

The window is made up of a larger square and a semi-circle

Next, calculate the area of the larger square.

16 small squares made up the larger square.

So, the area is:

A_2 = 16 * 169

A_2 = 2704

The area of the semicircle is:

A_3 = \frac{\pi r^2}{2}

A_3 = \frac{3.14 * 26^2}{2}

A_3 = 1061.32

So, the area of the window is:

Area = A_2 + A_3

Area = 2704 + 1061.32

Area = 3765.32

Solving (b): Area of the shade

The shade extends 4 inches beyond the window.

This means that;

The bottom length is now; Initial length + 8

And the height is: Initial height + 4

In (a), the length of each square is calculated as: 13in

4 squares make up the length and the height.

So, the new dimension is:

Length = 4 * 13 + 8

Length = 60

Height = 4*13 + 4

Height = 56

The area is:

A_1 = 60 * 56 = 3360

The radius of the semicircle becomes initial radius + 4

r = 26 + 4 = 30

The area is:

A_2 = \frac{3.14 * 30^2}{2} = 1413

The area of the shade is:

Area = A_1 + A_2

Area = 3360 + 1413

Area = 4773

7 0
3 years ago
Please help me out please
ad-work [718]

this is the answer, I guess.

8 0
3 years ago
A school wishes to form three sides of a rectngular playground using 480 meters of fencing. The playground borders the school bu
nirvana33 [79]

Answer

school building, so the fourth side does not need Fencing. As shown below, one of the sides has length J.‘ (in meters}. Side along school building E (a) Find a function that gives the area A (I) of the playground {in square meters) in

terms or'x. 2 24(15): 320; - 2.x (b) What side length I gives the maximum area that the playground can have? Side length x : [1] meters (c) What is the maximum area that the playground can have? Maximum area: I: square meters

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Find the values for m and n that would make the following equation true.
dimaraw [331]

You would solve it in two parts.

4*n = 44     Divide by 4

4n/4 = 44/4

n = 11

=============

z^m * z^2 = z^9

When the bases are the same (on the left) the powers on the left add

m + 2

The powers on the left side of the equation are the same as the powers on the right when the bases are the same.

m + 2 = 9                 Subtract 2 from both sides.

m + 2 - 2 = 9 - 2

m = 7

=================

Answers

m = 7

n = 11

4 0
3 years ago
What is 5382619 rounded to the nearest 100
Vlad1618 [11]
It would be 5,832,600
6 0
3 years ago
Read 2 more answers
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