Provide an appropraite response.
If y varies inversly as the 4th power of x and y=3 when x=2, what is y when x=0.5?
If Q varies inversly as the sqaure root of R and Q=2 whe R=4, what is Q when R is 25?
1st one:
y = k(1/x^4)
when x=2, y=3
3 = k(1/16) ---> k = 48
equation: y = 48/x^4
when y = .5
.5 = 48/x^4
x^4 = 96
y = 96^(1/4) = 2(6^(1/4) or appr. 3.13
you try the second one, using the same method
Start with
Q = k/√R
is the second one R=625?
To find the value of y when x is 0.5, we can use the inverse variation equation: y = k/x^4. Here's how we can solve it step by step:
Step 1: Plug the given values of y and x into the equation to find the constant of variation, k:
3 = k/2^4
3 = k/16
Step 2: Solve for k by multiplying both sides of the equation by 16:
k = 3 * 16
k = 48
Step 3: Now that we have the value of k, we can find y when x is 0.5:
y = 48/(0.5)^4
y = 48/0.0625
y = 768
Hence, y is equal to 768 when x is 0.5.
For the second question, if Q varies inversely as the square root of R, we can express this relationship as Q = k/sqrt(R). Here's how we can determine the value of Q when R is 25:
Step 1: Substitute the given values of Q and R into the equation to find the constant of variation, k:
2 = k/sqrt(4)
2 = k/2
Step 2: Solve for k by multiplying both sides of the equation by 2:
k = 2 * 2
k = 4
Step 3: Now that we have the value of k, we can find Q when R is 25:
Q = 4/sqrt(25)
Q = 4/5
Q = 0.8
Therefore, Q is equal to 0.8 when R is 25.