Im so lost how do i solve:
f varies jointly as q2 and h, and f=64 when q=4 and h=2. Find q when f=90 and h=5
f varies jointly as q2 and h, and f=64 when q=4 and h=2. Find q when f=90 and h=5
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Q=3
here is how you do it
f=(k)(q^2)(h)
64=(k)(4^2)(2)
64/32=2
k=2
so then once you got k, you just fill in the problem so it would be...
90=(2)(q^2)(5)
90/10=9
9=q^2
q=3
here is how you do it
f=(k)(q^2)(h)
64=(k)(4^2)(2)
64/32=2
k=2
so then once you got k, you just fill in the problem so it would be...
90=(2)(q^2)(5)
90/10=9
9=q^2
q=3
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f = q2 * 2h
q2 = f/(2h)
=90/10
=9
q = 3
Symbols:
* < Multiplication
/ < Division
You have to figure out the formula first. Ask your teacher if they could help. Trust me, it's not easy.
~Celina
q2 = f/(2h)
=90/10
=9
q = 3
Symbols:
* < Multiplication
/ < Division
You have to figure out the formula first. Ask your teacher if they could help. Trust me, it's not easy.
~Celina
-
Hell...
I have never seen or done problems like this, I would say,
f = 2 * q^2 * h
for
q = 4
h = 2
f = 2 * 16 * 2 = 64
so when,
f = 90
h = 5
q = sqrt( 90/(2 * 5)) = 3
I have never seen or done problems like this, I would say,
f = 2 * q^2 * h
for
q = 4
h = 2
f = 2 * 16 * 2 = 64
so when,
f = 90
h = 5
q = sqrt( 90/(2 * 5)) = 3
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ƒ = K • q² • h
64 = K • 4² • 2
K = 2
ƒ = K • q² • h
ƒ = 2 • q² • h
90 = 2 • q² • 5
q = 3
64 = K • 4² • 2
K = 2
ƒ = K • q² • h
ƒ = 2 • q² • h
90 = 2 • q² • 5
q = 3