The altitude (i. e., height) of a triangle is increasing at a rate of 2 cm/minute while the area of the triangle is increasing at a rate of 1.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 12 centimeters and the area is 99 square centimeters

A = 1/2 B H       area = 1/2 base X height

B = 2 A / H = 2 * 99 / 12 = 16.5 cm

dA / d t = 1/2 * (B dH / dt + H dB / dt)

dB / dt = (2 dA / dt - B dH / dt) / H

dB / dt = (2 * 1.5 cm^2 / min - 16.5 cm * 2 cm / min) / 12 cm

dB / dt = (3 - 33) / 12 cm/min = -2.5 cm/min

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

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