The altitude (i. e., height) of a triangle is increasing at a rate of 2.5 cm/minute while the area of the triangle is increasing at a rate of 3.5 square cm/minute. at what rate is the base of the triangle changing when the altitude is 10.5 centimeters and the area is 81 square centimeters? the base is changing at

For this, we need to know the length of the base at the time of interest. It will be

... A = (1/2)bh

... b = (2A)/h = 2(81 cm²)/(10.5 cm) = 108/7 cm

Differentiate the formula for area and plug in the given numbers.

... A = (1/2)bh

... A' = (1/2)(b'h +bh')

... 3.5 cm²/min = (1/2)(b'·(10.5 cm) + (108/7 cm)·(2.5 cm/min))

... 7 cm²/min = 10.5b' cm + 38 4/7 cm²/min . . . . simplify a bit

... -31 3/7 cm²/min = 10.5b' cm . . . . . . . . . . . . . . . subtract 38 4/7 cm²/min

... (-220/7 cm²/min)/(10.5 cm) = b' ≈ -3.0068 cm/min

The base is changing at about -3 cm/min.

answer:   (s+1)(s+4)

step-by-step explanation:

factor s2+5s+4

s2+5s+4

=(s+1)(s+4)

your answer is 125y5