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Anvisha [2.4K]
3 years ago
12

Shaded areas please help guys ❤❤❤​

Mathematics
2 answers:
pentagon [3]3 years ago
3 0

Answer:

77cm

Step-by-step explanation:

Area of a rectangle=L×B

So

7×11=77cm

Dmitrij [34]3 years ago
3 0

Answer:

64.44 cm²

Step-by-step explanation:

A = rectangle - circle

(7 x 11) - (4π) = 64.44 cm²

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8 0
3 years ago
7 1/2 lb =_oz<br> please help me
KonstantinChe [14]
1 lb is equal to 16 oz, therefore 16 * 7 is 112oz then add 1/2 lb which is 8oz so the final answer is 120 oz
3 0
3 years ago
PLEASE HELP!!!<br> Solve the inequality: <br> 5 (2x-3) - 6x &gt; -2 (x - 6) +3
jeyben [28]

Answer:

x>5

Step-by-step explanation:

Step 1: Simplify both sides of the inequality.

4x−15>−2x+15

Step 2: Add 2x to both sides.

4x−15+2x>−2x+15+2x

6x−15>15

Step 3: Add 15 to both sides.

6x−15+15>15+15

6x>30

Step 4: Divide both sides by 6.

3 0
3 years ago
Read 2 more answers
A rectangle is twice as long as it is wide. if its length is increase by 4 cm and its width is decreased by 3 cm, the new rectan
strojnjashka [21]
Now the width is w.
It's twice as long as wide, so now the length is 2w.

If the length is increased by 4 cm, the length will be 2w + 4.
The width is decreased by 3 cm, so the width will be w - 3.

The are of the new rectangle is 100 cm^2.

area = length * width

area = (2w + 4)(w - 3)

The area of the new rectangle is 100, so we get

(2w + 4)((w - 3) = 100

2w^2 - 6w + 4w - 12 = 100

2w^2 - 2w - 112 = 0

w^2 - w - 56 = 0

(w - 8)(w + 7) = 0

w - 8 = 0   or   w + 7 = 0

w = 8   or   w = -7

A width cannot be negative, so discard w = -7.

w = 8

The width is 8 cm.
The length is twice the width, so the length is 16 cm.
5 0
3 years ago
An octagonal swimming pool has a base area of 22 square feet The pool is 3 feet deep How many cubic feet of water can the pool h
LenKa [72]

The pool can hold 65.84 ft³ of water

<u>Explanation:</u>

Given:

Shape of pool = octagonal

Base area of the pool = 22 ft²

Depth of the pool = 3 feet

Volume, V = ?

We know:

Area of octagon = 2 ( 1 + √2) a²

22 ft² = 2 ( 1 + √2 ) a²

\frac{11}{1+\sqrt{2} } = a^2

a² = \frac{11}{2.42}

a² = 4.55

a = 2.132 ft

Side length of the octagon is 2.132 ft

We know:

Volume of octagon = 2(1+\sqrt{2} ) X (a)^2 X h

V = 2(1+\sqrt{2})X (2.132)^2 X 3\\ \\V = 2 ( 2.414) X 4.5454 X 3\\\\V = 65.84 ft^3

Therefore, the pool can hold 65.84 ft³ of water

8 0
3 years ago
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