Isolating to find value.

harem905

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Nov 12, 2013
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I'm working on a lesson in my unit and I came up to a practice quiz where they give you the answer at the end to check your answer and see how you did. I need help being shown to do it to get the answer they give. The question is,

The equation "V =3(t-20)^2" models the relationship between the volume of water (V litres) and time (t minutes). According to this model, when will the volume equal 75 litres?

They gave us the answer, "Answer=15; Explanation=Set V=75 and solve for t."

I guess I just can't properly figure out how to isolate t, to get the answer. Could someone show their work and the process they went through to get 15 as the answer?
 

Subhotosh Khan

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I'm working on a lesson in my unit and I came up to a practice quiz where they give you the answer at the end to check your answer and see how you did. I need help being shown to do it to get the answer they give. The question is,

The equation "V =3(t-20)^2" models the relationship between the volume of water (V litres) and time (t minutes). According to this model, when will the volume equal 75 litres?

They gave us the answer, "Answer=15; Explanation=Set V=75 and solve for t."

I guess I just can't properly figure out how to isolate t, to get the answer. Could someone show their work and the process they went through to get 15 as the answer?
\(\displaystyle \displaystyle{ V \ = \ 3*(t-20)^2}\)

\(\displaystyle \displaystyle{ \frac{V}{3} \ = \ (t-20)^2}\)

\(\displaystyle \displaystyle{ \sqrt{\frac{V}{3}} \ = \ t \ - \ 20}\)

Now continue.....
 

harem905

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Joined
Nov 12, 2013
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\(\displaystyle \displaystyle{ V \ = \ 3*(t-20)^2}\)

\(\displaystyle \displaystyle{ \frac{V}{3} \ = \ (t-20)^2}\)

\(\displaystyle \displaystyle{ \sqrt{\frac{V}{3}} \ = \ t \ - \ 20}\)

Now continue.....
Why do I suck so badly at this? The next step would be to add 20, correct? I keep either ending up with an answer of 25 or an answer of "the square root of 45." What am I doing wrong?
 

Subhotosh Khan

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\(\displaystyle \displaystyle{ V \ = \ 3*(t-20)^2}\)

\(\displaystyle \displaystyle{ \frac{V}{3} \ = \ (t-20)^2}\)

\(\displaystyle \displaystyle{ \sqrt{\frac{V}{3}} \ = \ t \ - \ 20}\)

\(\displaystyle \displaystyle{ \pm 5 \ = \ t \ - \ 20}\)

\(\displaystyle t = 20 \ \pm 5\)

t = 25 or 15
The vessel is emptying out - thus at t=15 it will reach 75 then it will reach V = 0 at t = 20.

After that the vessel will need to fill up and t =25 it will have V = 75 again
 
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