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Mathematics 8 Online
OpenStudy (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

OpenStudy (amistre64):

So we are given this: x = r/t

OpenStudy (amistre64):

or really: x = kr/t where k is any constant, but lets make that constant a "1" for simplicities sake :)

OpenStudy (amistre64):

t = r/x then: t = 10/-5

OpenStudy (amistre64):

r = x/t r = 4/20

OpenStudy (anonymous):

its not rate over time. i need to find t!

OpenStudy (amistre64):

ack!! r = xt lol i forgot how to multiply :)

OpenStudy (amistre64):

r = 4(20)

OpenStudy (amistre64):

i see i forgot how to read the question as well..... let me start over :)

OpenStudy (amistre64):

Find t when r=10 and x= -5, if t=20 when x= 4 and r=2 whats this mean?

OpenStudy (anonymous):

its a variation function

OpenStudy (amistre64):

are you looking for the Constant? or what?

OpenStudy (amistre64):

4 = k(2/20) 40 = k x = 40r/t

OpenStudy (anonymous):

just finding the answer for t

OpenStudy (anonymous):

?

OpenStudy (anonymous):

alright, hold on a minute and I will help, I just need to refresh my memory.

OpenStudy (anonymous):

First you need to set up a proportion. Since x varies directly as r and inversely as t, it will look like this:\[x_{1}t _{1} / r _{1} = x _{2}r _{2} / r _{2}\]

OpenStudy (anonymous):

You plug in your values, which would make \[(4)(20) / 2 = 5(t)/10\]

OpenStudy (anonymous):

or, 80 / 2 = 5t/10.

OpenStudy (anonymous):

you then solve the proportion by cross multiplying, like so: \[(80)(10) = 2(5t)\]

OpenStudy (anonymous):

or, 800 = 10t ; solving the equation leaves you with t = 80.

OpenStudy (anonymous):

hope that helps! ^-^

OpenStudy (amistre64):

-5 i think it was :) which makes it -80 right?

OpenStudy (anonymous):

ah true, thanks for pointing that out, I always get my signs messed up

OpenStudy (amistre64):

t = 400/-5 = -80

OpenStudy (amistre64):

i was just wondering if i did it right.... :)

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