Are my answers correct?
Variable y varies directly with x2, and y = 96 when x = 4. What is the value of y when x = 2? answer= 48
Yes they are all correct.
Variable y varies directly with x2, and y = 196 when x = 7. Enter the constant of variation in the quadratic variation equation. y = 4 x2
y varies directly with x^2 implies y = k * x^2 when x = 4, y = 96 96 = k * (4)^2 = 16k k = 96/16 = 6 y = 6x^2 when x = 2, y = 6 * (2)^2 = 6 * 4 = 24.
for the first one?
yes.
yes the second one is correct
ok, what about this one Variable y varies directly with x2, and y = 48 when x = 4. Which graph represents the quadratic variation?
Yes, the constant of variation is 4 for the second one.
I choose the 4th one
@aum or anyone else..
Not correct. First find the constant of variation. y = k * x^2 when x = 4, y = 48. What is k?
192
How? Show your work.
i just times 4 to 48
y = k * x^2 ---- equation (1) when x = 4, y = 48 put x = 4 and y = 48 in first equation and solve for k.
16k
Earlier in the thread we have solved two other similar problems. Use the exact same method. If you learn to solve this problem just once you can answer all similar problems on your own.
Put both the x and y values into equation (1) and then solve for k.
48 = k * 4^2 48= 16k
correct. Therefore, k = ?
3
correct. So the equation is y = 3x^2. when x = 1, y = ?
I am choosing x = 1 because all graphs have the point x = 1 on them. So what is the y value when x = 1 in the equation y = 3x^2 ?
x2
We started with the equation y = kx^2 we solve for k as k = 3 so y = 3x^2 is the equation. when x = 1, y = 3 * 1^2 = 3 Therefore, when x = 1, y = 3 which means the point (1,3) must be on the graph. Which graph has the point (1,3) ON it?
the 2nd one
The second one says it has the point (1,0.5) on it. Not (1,3)
nvm the third one
Yes. Please close this question.
ok, thx
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