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4 years ago

The circumference of a circle is 3π in.

Mathematics

1 answer:

11111nata11111 [884]4 years ago

6 0

Answer:

2.25π in²

Step-by-step explanation:

The area can be found from the circumference using the formula ...

A = C²/(4π)

Putting in the given dimension, you have ...

A = (3π)²/(4π) = (9/4)π = 2.25 π . . . . in²

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the pool at the apartment is 30 feet long 20 feet wide and five feet deep it has been filled to 4 feet how many more cubic inche

pickupchik [31]

Answer:

600 cubic feet

Step-by-step explanation:

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3 years ago

The perimeter of a rectangle is 22 inches. Find the dimensions if its length is 3 inches greater than its width.

MaRussiya [10]

Answer:

Length: 7

Width: 4

Step-by-step explanation:

We can create a system of equations for this problem, where $w$ is the width and $l$ is the length.

The perimeter of a rectangle is twice its length added to twice its width.

$2l+2w=22$

The length is 3 more than the width:

$l = w+3$

We can now substitute in $w+3$ as $l$ in the equation $2l+2w=22$ .

$2(w+3) + 2w = 22$

Distribute the first terms:

$2w+6+2w=22$

Combine like terms:

$4w+6=22$

Subtract 6 from both sides:

$4w=16$

Divide both sides by 4:

$w = 4$

Now we know that w = 4. We can now substitute this inside an equation to find $l$ .

$l = w+3\\\\l = 4+3\\\\l=7$

Hope this helped!

7 0

4 years ago

What is the form of the two squares identity?

zvonat [6]

Answer:

Step-by-step explanation:

Consider all options:

A. False, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ab-cd)^2+(ac+bd)^2=a^2b^2-2abcd+c^2d^2+a^2c^2+2abcd+b^2d^2=\\ =a^2b^2+c^2d^2+a^2c^2+b^2d^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2\neq a^2b^2+c^2d^2+a^2c^2+b^2d^2$

B. False, because

$(a^2-b^2)(c^2+d^2)=a^2c^2+a^2d^2-b^2c^2-b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2-b^2c^2-b^2d^2\neq a^2c^2+b^2d^2+a^2d^2+b^2c^2$

C. False, because

$(a^2+b^2)(c^2-d^2)=a^2c^2-a^2d^2+b^2c^2-b^2d^2\\ \\(ac+bd)^2-(ad+bc)^2=a^2c^2+2abcd+b^2d^2-a^2d^2-2abcd-b^2c^2=\\=a^2c^2+b^2d^2-a^2d^2-b^2c^2\\ \\a^2c^2-a^2d^2+b^2c^2-b^2d^2\neq a^2c^2+b^2d^2-a^2d^2-b^2c^2$

D. True, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2= a^2c^2+b^2d^2+a^2d^2+b^2c^2$