anonymous
  • anonymous
i need to know the answer to this suppose x varies directly as r, and x varies inversely as t. find t when r equals 10 and x equals -5, if t equals 20 when x equals 4 and r equals 2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
So we are given this: x = r/t
amistre64
  • amistre64
or really: x = kr/t where k is any constant, but lets make that constant a "1" for simplicities sake :)
amistre64
  • amistre64
t = r/x then: t = 10/-5

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amistre64
  • amistre64
r = x/t r = 4/20
anonymous
  • anonymous
its not rate over time. i need to find t!
amistre64
  • amistre64
ack!! r = xt lol i forgot how to multiply :)
amistre64
  • amistre64
r = 4(20)
amistre64
  • amistre64
i see i forgot how to read the question as well..... let me start over :)
amistre64
  • amistre64
Find t when r=10 and x= -5, if t=20 when x= 4 and r=2 whats this mean?
anonymous
  • anonymous
its a variation function
amistre64
  • amistre64
are you looking for the Constant? or what?
amistre64
  • amistre64
4 = k(2/20) 40 = k x = 40r/t
anonymous
  • anonymous
just finding the answer for t
anonymous
  • anonymous
?
anonymous
  • anonymous
alright, hold on a minute and I will help, I just need to refresh my memory.
anonymous
  • anonymous
First you need to set up a proportion. Since x varies directly as r and inversely as t, it will look like this:\[x_{1}t _{1} / r _{1} = x _{2}r _{2} / r _{2}\]
anonymous
  • anonymous
You plug in your values, which would make \[(4)(20) / 2 = 5(t)/10\]
anonymous
  • anonymous
or, 80 / 2 = 5t/10.
anonymous
  • anonymous
you then solve the proportion by cross multiplying, like so: \[(80)(10) = 2(5t)\]
anonymous
  • anonymous
or, 800 = 10t ; solving the equation leaves you with t = 80.
anonymous
  • anonymous
hope that helps! ^-^
amistre64
  • amistre64
-5 i think it was :) which makes it -80 right?
anonymous
  • anonymous
ah true, thanks for pointing that out, I always get my signs messed up
amistre64
  • amistre64
t = 400/-5 = -80
amistre64
  • amistre64
i was just wondering if i did it right.... :)

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