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27 Dec 2019, 01:58
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A
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y varies directly as the square of x. When x = 4, y = 2. What is the value of y when x = 3/2?
A. 9/32
B. 3/4
C. 16/3
D. 128/9
E. 7/2
A. 9/32
B. 3/4
C. 16/3
D. 128/9
E. 7/2
27 Dec 2019, 03:03
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Its A for me.
Y=kx^2 hence putting values of x and y we get k=1/8. Again putting value of k and x=3/2 we get y= 9/32.
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Y=kx^2 hence putting values of x and y we get k=1/8. Again putting value of k and x=3/2 we get y= 9/32.
Posted from my mobile device
27 Dec 2019, 04:37
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y=cx^2, where c is a constant.
x=4 --> y=2=c*4^2 --> c=1/8
x=3/2 --> y=(1/8)*(3/2)^2 --> x=9/32
Final answer is (A)
Posted from my mobile device
x=4 --> y=2=c*4^2 --> c=1/8
x=3/2 --> y=(1/8)*(3/2)^2 --> x=9/32
Final answer is (A)
Posted from my mobile device
