the altitude of a triangle is increasing at a rate of 1 cm/min while the area in increasing at a rate of 2 cm^2/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100cm^2?

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the altitude of a triangle is increasing at a rate of 1 cm/min while the area in increasing at a rate of 2 cm^2/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100cm^2?

Mathematics
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You got this
What's the formula we use here?
is there a shorcut? i feel like i just have to double something then square it.

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Other answers:

1. Draw diagram, include given info 2. Apply proper formula 3. Take derivative of formula 4. Input values into formula 5. Isolate what you need to find
|dw:1317011985889:dw|
I skipped a step which is to find other necessary values
h': 1 cm/min A': 2 cm^2/min. b'?: h=10 cm, A=100cm^2 A = bh/2 A' = b'h/2 + bh'/2 b' = (A' - bh'/2)(2/h)
given: dA/dt= 1 dAREA/dt=2 A=10 AREA = 100
define b as 2A/h
Amistre, go to bed, lol
im gona try and work it out then ill shoot you guys an answer. give me a sec
b' = (A' - 2Ah'/2h)(2/h) b' = (A' - Ah'/h)(2/h) b' = (2 - 100(1)/10) (2/10) b' = (2 - 10) (1/5) b' = -8/5 maybe?
..... i thought this was bed :)
I believe you
why is b, 2A/h
becasue A = bh/2; Area = base*height/2 ... solveing for base we get; base = 2*Area/height

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