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The altitude of a triangle is increasing at a rate of 2 centimeters/minute while the area of the triangle is increasing at a rate of 3.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 10 centimeters and the area is 95 square centimeters?

Mathematics
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i really think you should practice related rates, working on your other problem right now
hey
hey will do this one later, i gotta eat

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Other answers:

carra?
yes
want me to solve it
ok
Area of triangle = 1/2 base * altitude
area is changing with respect to time, base is changing with respect to time and altitude is changing with respect to time . so taking derivative we have
product rule , which is complicated... ok lets look for a relationship given
ok product rule it is
dA/dt = 1/2 b * da/dt + 1/2 db/dt * a , where a is altitude
b is base, A is area
da/dt = 2 "The altitude of a triangle is increasing at a rate of 2 centimeters/minute " dA/dt = 3.5 "while the area of the triangle is increasing at a rate of 3.5 square centimeters/minute." At what rate is the base of the triangle changing when the altitude is 10 centimeters and the area is 95 square centimeters? what is db/dt , when a = 10 , and A= 95
dA/dt = 1/2 b * da/dt + 1/2 db/dt * a we have to go back and solve for b here to find db/dt
A = 1/2 * b*a, A = 95, a = 10, so b = 19 using algebra.
A = 1/2 * b*a, A = 95, a = 10, so b = 19 using algebra.
plug back into dA/dt = 1/2 b * da/dt + 1/2 db/dt * a 3.5 = 1/2 * 19 *2 + 1/2 * db/dt * 10 , and solve for db/dt
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