I have some questions that I don't understand and need to be checked please?
Slove the problem.
1. y varies directly as z and y=187 when z=17. Find y when z=15
-This is how I did it: 187=k*17, I divided by 17 on both sides and got k=11, then I plugged it back into the equation for k and got 165.
If y varies inversely as x, find the inverse variation equation for the situation.
4.y=2 when x=6 (I don't understand this question)
Solve the problem.
10. x varies inversely as v, and x=8 when v=4. Find x when v=16
-This is what I got: 8=k/4, I multiplied each side by 4, I got k= 32, I plugged it back into the equation, x=32/16, x=2
12. y varies inversely as y^2, and x=3 when y=8. Find x when y=2
This is what I got: 3=k/(8^2), 3=k/64, I multiplied each side by 64, I got k=192, I plugged back into the equation, x=192/(2^2), x=48
#1,10 and 12 are correct, you probably meant to type "x varies inversely as y^2.." in #12
for #4 all you have to do is say
y = k/x , sub in the x and y values given
so
2=k/6
k = 12
then the equation you are asked for would be
y = 12/x
If A varies directly as B, and A =7 when B =2. Find B when A =21
For the first problem, you correctly used the direct variation formula y = kz, where k is the constant of variation. By plugging in the given values y=187 and z=17, you found the value of k to be 11. To find y when z=15, you can simply substitute these values into the equation: y = 11 * 15 = 165. Therefore, when z=15, y=165.
For the second problem, the statement "y varies inversely as x" means that y and x are related by the equation y = k/x, where k is the constant of variation. In this case, you are given the values y=2 and x=6. To find the inverse variation equation, substitute these values into the equation: 2 = k/6. By multiplying both sides by 6, you get 12 = k. Therefore, the inverse variation equation for this situation is y = 12/x.
Moving on to the third problem, you correctly used the inverse variation formula x = k/v and found k to be 32 by substituting x=8 and v=4 into the equation. To find x when v=16, substitute these values into the equation: x = 32/16 = 2. Hence, when v=16, x=2.
Lastly, for the fourth problem, you are told that "y varies inversely as y^2". This can be written as y = k/y^2, where k is the constant of variation. By substituting x=3 and y=8 into the equation, you obtained 3 = k/(8^2) = k/64. By multiplying both sides by 64, you found k to be 192. To find x when y=2, substitute these values into the equation: x = 192/(2^2) = 48. Therefore, when y=2, x=48.
It seems like you have understood the concept of direct and inverse variation correctly and solved these problems accurately. Well done!