Acone has a radius of 5 cm and a slant height of 10 cm. the area of the base of the cone is ? cm2. the lateral area of the cone is ? cm2. the surface area of the cone is ? cm2.

Radius: 5 cm

Slant Height: 10 cm

I don't know if you are supposed to round for pi or not, so I will do it both ways for each part.

The base of a cone is a circle, so the formula for the base area of a cone and the area of a circle are the same: A = πr², r = radius and π (or 3.14) = pi.

A = πr²

A = π5²

A = 25π

A ≈ 78.54 cm² (rounded to the nearest hundredths place)

A = (3.14)(5²)

A = (3.14)(25)

A = 78.5 cm²

The formula for finding the lateral area of a cone is A = πrl, r = radius, l = slant height, and π (or 3.14) = pi.

A = πrl

A = π(5)(10)

A = 50π

A ≈ 157.08 cm² (rounded to the nearest hundredths place)

A = πrl

A = (3.14)(5)(10)

A = 157 cm²

The formula for finding the surface area of a cone is A = πrl + πr², r = radius, l = slant height, and π (or 3.14) = pi.

A = πrl + πr²

A = π(5)(10) + π(5)²

A = 50π + 25π

A = 75π

A ≈ 235.62 cm² (rounded to the nearest hundredths place)

A = πrl + πr²

A = (3.14)(5)(10) + (3.14)(5)²

A = 157 + 78.5

A = 235.5 cm²

For each part, just pick the answer that goes with how you were told to solve the problem (rounding for pi or not). And if the answer itself is supposed to have the pi symbol in it, then just use the second-to-last step in the first section for each part. For example, in Part 1, it shows

A = 25π

A ≈ 78.54 cm² (rounded to the nearest hundredths place).

These two answers are equal to each other. The second is just the extended version of the first.

I hope this helps!

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).