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2 years ago

What is the area of the shaded region?

Mathematics

1 answer:

Mekhanik [1.2K]2 years ago

4 0

9514 1404 393

Answer:

434 -49π ≈ 280.1 cm²

Step-by-step explanation:

The shaded area is the difference between the enclosing rectangle area and the circle area.

The rectangle is 14 cm high and 31 cm wide, so has an area of ...

A = WH

A = (31 cm)(14 cm) = 434 cm²

The circle area is given by ...

A = πr²

A = π(7 cm)² = 49π cm²

The shaded area is the difference of these, so is ...

shaded area = rectangle - circle

= (434 - 49π) cm² ≈ 280.1 cm²

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Directions: Calculate the area of a circle using 3.14x the radius

Leokris [45]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

As we know ~

Area of the circle is :

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

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<h3>Problem 1</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 4.4\div 2$

$\qquad \sf \dashrightarrow \:r = 2.2 \: mm$

Now find the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(2.2)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {4.84}^{}$

$\qquad \sf \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 2</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 3.7 \div 2$

$\qquad \sf \dashrightarrow \:r = 1.85 \: \: cm$

Bow, calculate the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (1.85) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 3.4225 {}^{}$

$\qquad \sf \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 3 </h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (8.3) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 68.89$

$\qquad \sf \dashrightarrow \:area \approx216.31 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 4</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 5.8 \div 2$

$\qquad \sf \dashrightarrow \:r = 2.9 \: \: yd$

now, let's calculate area ~

$\qquad \sf \dashrightarrow \:3.14 \times {(2.9)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 8.41$

$\qquad \sf \dashrightarrow \:area \approx26.41 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 5</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 1 \div 2$

$\qquad \sf \dashrightarrow \:r = 0.5 \: \: yd$

Now, let's calculate area ~

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (0.5) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 0.25$

$\qquad \sf \dashrightarrow \:area \approx0.785 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 6</h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(8)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 64$

$\qquad \sf \dashrightarrow \:area = 200.96 \: \: yd {}^{2}$

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8 0

2 years ago

Use the context of a 24-pack of soda to explain why 3/8 is NOT equivalent to 5/12 . Answer in number of cans.

juin [17]

Answer:

In a 24 pack of soda, 1/8 would be 3 cans (24/8 = 3). 1/12 would be 2 cans (24/12=2).

So 3/8 is 9 cans of soda and 5/12 is 10 cans of soda, so they are not equivalent because the number of soda cans is not the same.

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ivann1987 [24]

Answer:

Step-by-step explanation:

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2 years ago

I need help its not 4236

Oduvanchick [21]

Answer: 4236

Step-by-step explanation: I used a calculator. If it's wrong, tell your teacher/instructor.

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2 years ago

PLZZ ANSWER 25 PTS RATED ATAR PROMISE TO GIVE BRAINLIEST PLZZZ ANYONE KNOW THIS???? A 154-lb person burns 420 calories per hour

Novosadov [1.4K]

Answer:

c = 420t . . . . c is calories burned; t is hours riding at 15 mph

Step-by-step explanation:

There is not enough information given to write a function rule relating all the variables to calories burned. If we assume that calories are burned at the constant rate of 420 calories per hour, then total calories will be that rate multiplied by hours:

c = 420·t

where c is total calories burned by the 154-lb person, and t is hours riding at 15 mph.

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In general, rates are related to quantities by ...

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