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Agata [3.3K]
2 years ago
15

Mark is baking a pie that has a diameter of 10 inches. he wants to cover the top of the pie with a pie crust . what is the area

of the crust that will cover the top of the pie?
Mathematics
1 answer:
AnnZ [28]2 years ago
8 0

Answer:

79 square inches.

Step-by-step explanation:

Mark is baking a pie that <u>has a diameter of 10 inches</u>. he wants to cover the top of the pie with a pie crust. <u>what is the area</u> of the crust that will cover the top of the pie?

Pie = circle (shape)

What is the area of a circle 10 inches in diameter? ≈ 79 square inches. and r denotes the radius. = 78.5 square inches. ≈ 79 square inches.

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It is 36.75$.

Step-by-step explanation:

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It costs $324 to buy a rug that measures 10’ by 12’. If the cost is proportional to the area, what does it cost to buy a 15’ by
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a bag contains 5 red marbles 4 blue marbles and 6 yellow marbles if 3 marbles are selected in succession what is the probability
Alex

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5/243

Step-by-step explanation:

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3 years ago
Find the value of x by using law of sines.
Pavel [41]

Answer:

x = 7.7

Option B

Step-by-step explanation:

\frac{a}{Sin~A} &=& \frac{b}{Sin~b} \\ \frac{8}{Sin~73^\circ} &=& \frac{x}{Sin~67^\circ} \\ \frac{8}{0.96} &=& \frac{x}{0.92} \\ 0.96x &=& 8 \times 0.92 \\ 96x &=& 8 \times 92 \\ 96x &=& 736 \\ x &=& \frac{736}{96} \\ x&=& \frac{23}{3} \\ &\boxed{x=7.7}&

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3 0
3 years ago
The pinata weighs 11/4 pounds without candy.He adds 8bags that weighs 3/8,3/8,1/2,1/2,1/2,3/4,3,11/4 how much will the pinata we
Natalka [10]

Answer:

Weight = \frac{23}{2}

Step-by-step explanation:

Given

Weight of Pinata = 11/4 lb

Weight of candies = 3/8, 3/8, 1/2 , 1/2, 1/2, 3/4, 3, 11/4

Required

Weight of Pinata when filled with candies

To solve this question, we simply need to add all weights together

First, we need to add up the weight of the candies

Candies =  \frac{3}{8} + \frac{3}{8} + \frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+ \frac{3}{4} + 3 + \frac{11}{4}

Take LCM

Candies = \frac{3 + 3 + 4 +4 + 4 + 6 + 24 + 22}{8}

Candies = \frac{70}{8}

Simplify:

Candies = \frac{35}{4}

Next; Add the weight of the candies to the weight of pinata

Weight = Candies + Pinata

Weight = \frac{35}{4} + \frac{11}{4}

Take LCM

Weight = \frac{35 + 11}{4}

Weight = \frac{46}{4}

Simplify

Weight = \frac{23}{2}

<em>Hence, the weight of the Pinata after it is filled with candies is 23/2 lb</em>

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