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2 years ago

Find the approximate area of a circle that has a diameter of 11 inches. Round your answer to the nearest hundredth

Mathematics

1 answer:

OLga [1]2 years ago

4 0

Answer:

95.03in.^2

Step-by-step explanation:

Use the formula, A= pi r^2 and D=2r

A=1/4pi d^2=1/4 times pi time 11 squared

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Is this all the information provided or are you missing something?

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3 years ago

“An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and wor

Artist 52 [7]

Answer:

Part 1) The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Part 2) The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

Step-by-step explanation:

Step 1

Find the radius of each sphere

we know that

The circumference of a circle is equal to

$C=2\pi r$

Find the radius of the sphere with a 5.50-meter circumference

For $C=5.50\ m$

assume

$\pi =3.14$

substitute and solve for r

$5.50=2(3.14)r$

$r=5.50/[2(3.14)]=0.88\ m$

Find the radius of the sphere with a 7.85-meter circumference

For $C=7.85\ m$

assume

$\pi =3.14$

substitute and solve for r

$7.85=2(3.14)r$

$r=7.85/[2(3.14)]=1.25\ m$

step 2

Find the surface area of each sphere

The surface area of sphere is equal to

$SA=4\pi r^{2}$

Find the surface area of sphere with a 5.50-meter circumference

For $r=0.88\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(0.88)^{2}$

$SA=9.73\ m^{2}$

Find the surface area of sphere with a 7.85-meter circumference

For $r=1.25\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(1.25)^{2}$

$SA=19.63\ m^{2}$

step 3

Find the cost of finishing each sphere

we know that

To find out the cost , multiply the surface area by $92 per square meter

Find the cost of sphere with a 5.50-meter circumference

$9.73*(92)=\$895.16$

therefore

The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Find the cost of sphere with a 7.85-meter circumference

$19.63*(92)=\$1,805.96$

therefore

The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

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3 years ago

What type of coordinate system are polar equations graphed in?

dybincka [34]

The polar coordinate system.

6 0

3 years ago

Let f(x)=x^3 and g(x)= √x, and h(x)=x/3. Find each of the following: f(g(h(6)))

rewona [7]

Answer:

go from inside out.

h(x)=x/3

$(\sqrt{2})^{3}$ => h(6) = 6/3 = 2

g(x) = $\sqrt{x}$

=> g(h(6)) = g(2) = $\sqrt{2}$

f(x) = $x^{3}$

=> f(g(h(6))) = f( $\sqrt{2}$ ) = $(\sqrt{2} )^{3}$

6 0

3 years ago

Kiwi and Corn got into an argument over if 1.75 was a reasonable estimate for 6. Do you think 1.75 was a reasonable estimate for

snow_tiger [21]

Answer:

No, it is not.

Step-by-step explanation:

comparing the two given values, 1.75 and 6, estimating 1.75 for 6 is not reasonable. This is due to the fact that converting 1.75 to the nearest whole number gives 2 which is far away from 6. Since,

6 - 2 = 4

So, estimating 1.75 for 6 would involve a large value of error. Which make it unreasonable. It would have been more reasonable to estimate 1.75 for 2.

4 0

4 years ago