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07 May 2019, 00:00
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D
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Difficulty:
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Question Stats:
73% (01:29) correct
27%
(01:54)
wrong
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If x = 3t – 1 and y = 12t^2, what is y in terms of x?
A. (x + 1)^2
B. 4(x + 1)^2
C. 3(x + 1)^2/4
D. 4(x + 1)^2/3
E. (x - 1)^2
A. (x + 1)^2
B. 4(x + 1)^2
C. 3(x + 1)^2/4
D. 4(x + 1)^2/3
E. (x - 1)^2
07 May 2019, 00:27
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Bunuel
We have \(t=\frac{x+1}{3}\)
So, \(y=12*t^2=12*\frac{(x + 1)^2}{3^2}=\frac{4(x + 1)^2}{3}\)
Ans. (D)
07 May 2019, 00:54
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I started expressing t in terms of x and y
1) x=3t-1; t=(x+1)/3
2) y = \(12t^2\); \(t^2\)=\(\frac{y}{12}\) oops, I cant get rid of the power, because +-\(\frac{y}{12}\) is possible
So I went back to the first expression and expressed it as \(t^2\); \(t^2\)=\((x+1)^2\)/9
Now I have an equation:
\(\frac{y}{12}\)=\((x+1)^2\)/9
y=4\((x + 1)^2\)/3
IMO
Ans: D
1) x=3t-1; t=(x+1)/3
2) y = \(12t^2\); \(t^2\)=\(\frac{y}{12}\) oops, I cant get rid of the power, because +-\(\frac{y}{12}\) is possible
So I went back to the first expression and expressed it as \(t^2\); \(t^2\)=\((x+1)^2\)/9
Now I have an equation:
\(\frac{y}{12}\)=\((x+1)^2\)/9
y=4\((x + 1)^2\)/3
IMO
Ans: D
