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3 years ago

Find the area of the shaded region. Round to nearest hundredth.

Mathematics

1 answer:

coldgirl [10]3 years ago

5 0

210/360 = 7/12, meaning the shaded area is 7/12 of the entire circle.

The circle has radius 12 in, so its total area is π (12 in)^2 = 144π in^2.

Then the area of the shaded region is 7/12 of this, or

7/12 * (144 π in^2) = 84π in^2 ≈ 263.89 in^2

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A cookie recipe for 5 dozen cookies calls for 4 cups of flour. How many is flour is needed to make 90 cookies

azamat

6cups of flour will be needed
5*12/4= 15
90/15=6

6 0

4 years ago

Sandra wants to purchase necklaces that cost $4.99. She already purchased 3 bracelets for $8.97. Which of the following function

Volgvan

The answer is A

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6 0

3 years ago

I need help with the last two !!!

Dmitriy789 [7]

Step-by-step explanation:

G(x) = - x^2 + 10x

a + G(a) = a - a^2 + 10a = - a^2 + 11a or 11a - a^2 or a(11 - a)

1/G(a) = 1/(10a - a^2) = 1/a(10 - a)

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2 years ago

What is the volume of a shipping cube with dimensions of 2 and 1/2 feet

Fantom [35]

For this case we have that by definition, the volume of a cube is given by:

$V = l ^ 3$

Where:

l: It's the side of the cube

According to the statement data:

$l = 2 \frac {1} {2} = \frac {2 * 2 + 1} {2} = \frac {5} {2}$

Substituting in the formula we have:

$V = (\frac {5} {2}) ^ 3\\V = \frac {5 ^ 3} {2 ^ 3}\\V = \frac {125} {8}\\V = 15.625$

Thus, the shipping cube volume is $15.625 \ ft ^ 3$

Answer:

$15.625 \ ft ^ 3$

6 0

3 years ago

A pan is heated to 393°F, then removed from the heat and allowed to cool in a kitchen where the room temperature is a constant 6

deff fn [24]

Answer:

The approximate temperature of the pan after it has been away from the heat for 9 minutes is 275.59°F.

Step-by-step explanation:

The formula for D, the difference in temperature between the pan and the room after t minutes is:

$D = 324\cdot e^{-0.05t}$

Compute the approximate difference in temperature between the pan and the room after 9 minutes as follows:

$D = 324\cdot e^{-0.05t}$

$=324\times e^{-0.05\times 9}\\\\=206.59$

Then the approximate temperature of the pan after it has been away from the heat for 9 minutes is:

D = P - R

206.59 = P - 69

P = 206.59 + 69

P = 275.59°F

Thus, the approximate temperature of the pan after it has been away from the heat for 9 minutes is 275.59°F.

6 0

3 years ago