Preview activity 3.5.1. a spherical balloon is being inflated at a constant rate of 20 cubic inches per second. how fast is the radius of the balloon changing at the instant the balloon's diameter is 12 inches? is the radius changing more rapidly when d =12 or when d =16? why?

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

Because the radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Let be the radius, the diameter, and the volume of the spherical balloon.

We know and we want to find

The volume of a spherical balloon is given by

Taking the derivative with respect of time of both sides gives

We now substitute the values we know and we solve for :

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

When d = 16, r = 8 and is:

The radius is increasing at a rate of approximately 0.025 in/s when the diameter is 16 inches.

Because the radius is changing more rapidly when the diameter is 12 inches.

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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