if x varies directly as y^2, and x=4 when y=3, then what is the value of x when y=12?

Question

i got 16 but the correct answer according the the book is 64. can someone explain how to solve this with step-by-step work?

i put that the correct answer is 64, not 16 -___- please read what i wrote carefully

the answer is 64 people! thats the value of x! x=64

Answer 1

The statement, "x varies directly as y^2" can be written as an equation, like this:

x = a(y^2)

where "a" is some constant (whose value we don't yet know).

So, we have "x=4 when y=3". Plug those into the above equation, and we get this:

4 = a(3^2)

or:

4 = 9a

or:

a = 4/9

Now that we know "a", we can rewrite the first equation:

x = (4/9)y^2

And that lets us answer, "what is the value of x when y=12?"

x = (4/9)y^2
= (4/9)12^2
= (4/9)(144)
= 64

Answer 2

64 is the correct answer, as a number of people pointed out.

The reason for is that x varies directly with y *SQUARED*.

When y = 3, x = 4.

So, when y = 12, it's 4 times as much as when it was 3.

Therefore, X must by 4² times as much, or 16 times as much.

16 x 4 = 64.

Answer 3

considering the fact that y varies at as quickly as because of the fact the sq. of x: y = kx^2, for some consistent ok. to locate ok, we use the reality that y = 40 8 while x = 4. this provides the value of alright to be: 40 8 = ok(4)^2 ==> ok = 3. So, the equation that relates x and y is: y = 3x^2. for this reason, while x = 5: y = 3(5)^2 = seventy 5, and the respond is (D). i'm hoping this facilitates!

Answer 4

Another way of stating the relationship is the square root (SQRT) of X varies proportionately to Y.

The SQRT of 4 is 2.

Mutliply Y and SQRT(X) by 4:
3 x 4 =12.
2 x 4 = 8 this is the SQRT of X, remember,
8 ^2 = 64. When Y=12, X=64. QED

Answer 5

x=y^2
x=4 when y=3
x varies directly as y^2 in such a way that if x=4 then y=3, the result of which is,

(i) 4=3^2=9

Now what is the value of x when y=12

(ii) x=y^2=12^2=144

We can equate the ratios of (i) and (ii) to keep the variations of x and y^2

x/144=4/9
x=144X4/9=16X4=64
x=64

Answer 6

let x=k*y^2 where k is a constant

also when x=4,y=3
so 4=9k
so k=4/9

so when y=12,
x=k*y^2=(4/9)*144=64

Answer 7

x=16

x=4 when y=3 ==> multiply both by 4
y=3*4 =12 will result in x=4*4 = 16

Answer 8

x=ky^2
when x=4,y=3:
4=9k
k=4/9
x=4/9 y^2
whn y=12
x=4/9*144=4*16=64ANS.
so your answer is not correct.

Answer 9

x1/x2=y1^2/y2^2
4/x2=9/144
x2=144*4/9=64