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3 years ago

A circular picture is 8 inches in diameter.

Mathematics

1 answer:

Karolina [17]3 years ago

6 0

Answer:

Part A C. 16π square inches

Part B C. 36π square inches

Step-by-step explanation:

If we know the diameter, we can calculate the radius

d= 2r

8 = 2r

Divide by 2

8/2 =2r/2

4=r

The radius is 4 inches

Area of a circle is

A = pi *r^2

A = pi *4^2

A = 16 pi inches^2

Part B

If the frame is 2 inches wide, we add 2 to the radius

r = 4+2=6

A = pi r^2

A = pi *6^2

A = 36pi inches ^2

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3 years ago

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Answer:

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3 years ago

Read 2 more answers

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andrezito [222]

Answer:

i think its identity

Step-by-step explanation:

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3 years ago

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WORTH 20 POINTS HELP

vfiekz [6]

Below represents the proof that the quadrilateral QRST is a parallelogram

<h3>How to prove that QRST is a parallelogram?</h3>

The coordinates are given as:

Q = (-1,-1)

R = (2,9)

S = (-4,5)

T = (-7,-5)

Calculate the length of each side using:

$d = \sqrt{(x_2 -x_1)^2 + (y_2 -y_1)^2}$

So, we have:

$QR = \sqrt{(-1 - 2)^2 + (-1 -9)^2} = \sqrt{109}$

$RS = \sqrt{(-4 - 2)^2 + (5 - 9)^2} = \sqrt{52}$

$ST = \sqrt{(-7 + 4)^2 + (-5 - 5)^2} = \sqrt{109}$

$TQ = \sqrt{(-7 + 1)^2 + (-5 + 1)^2} = \sqrt{52}$

The above computations show that opposite sides are equal.

Next, we determine the slope of each side using:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$QR = \frac{9 + 1}{2 + 3} = \frac{10}3$

$RS = \frac{5-9}{-4 - 2} = \frac{2}3$

$ST = \frac{5+5}{-4 + 7} = \frac{10}3$

$TQ = \frac{-5+1}{-7 + 1} = \frac{2}3$

The above computations show that opposite sides are parallel, because they have equal slope

Hence, the quadrilateral QRST is a parallelogram