Acircular pond is to be surrounded by a gravel path. use the diagram to find the square feet of gravel needed if we know the pond has a radius of 200 ft and the radius of the whole space is 250 ft? a. 86,270 ft2 b. 57,234 ft2 c. 70,686 ft2 d. 65,523 ft2

C. 70,686 ft²

There are a couple of ways to get there.

Fairly straightforward is subtracting the pond area from the overall area of path and pond. The area of a circle is given by ...

... A = πr²

so that approach looks like ...

... A = π·250² -π·200² = 196,349.5 - 127,663.7 ≈ 70,685.8 ≈ 70,686 . . . ft²

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You can factor π from the above expression to make the arithmetic a little easier:

... A = π(250² -200²) = π(62500 -40000) = 22500π ≈ 70,686 . . . ft²

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You can further factor the difference of squares to give:

... A = π(250² -200²) = π(250 -200)(250 +200) = π·50·450 ≈ 70,686 . . . ft²

You might notice that this latter formulation is the product of the length of the centerline of the path and its width.

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.