1.) Suppose that y varies directly as x. If y is 5 when x is 30, the constant of variation is 150. True Or False
2.) If w varies directly as z and w = 9 when z = -6, find w when z = 3.
3.) If y varies directly as x and y = 2 when x = 4, find x when y = 4.
false
w=kz k=w/z
w=(9/-6)*3=-9/2
1.) To determine if the statement is true or false, we can use the formula for direct variation, which is y = kx, where k is the constant of variation.
Given that y is 5 when x is 30 and the constant of variation is 150, we can substitute these values into the formula:
5 = 150 * 30
To solve for k, we divide both sides of the equation by 30:
k = 5 / 30
Simplifying, we get:
k = 1/6
So, the constant of variation is 1/6, not 150. Thus, the statement is false.
2.) We are given that w varies directly as z and w is 9 when z is -6. Let's use the formula for direct variation, w = kz.
Substituting the given values into the formula, we have:
9 = k * (-6)
To solve for k, we divide both sides by -6:
k = 9 / (-6)
k = -3/2
Now we can use the value of k to find w when z is 3:
w = (-3/2) * 3
w = -9/2
w = -4.5
Therefore, when z is 3, w is -4.5.
3.) Just like the previous problems, we can use the formula for direct variation, y = kx.
Given that y is 2 when x is 4, we can substitute these values into the formula:
2 = k * 4
To solve for k, we divide both sides by 4:
k = 2 / 4
Simplifying, we have:
k = 1/2
Now we can use the value of k to find x when y is 4:
4 = (1/2) * x
To solve for x, we multiply both sides by 2:
8 = x
Therefore, when y is 4, x is 8.
1.) To determine if the given statement is true or false, we need to calculate the constant of variation.
The direct variation equation is y = kx, where k represents the constant of variation.
Given that y is 5 when x is 30, we can substitute these values into the equation:
5 = k * 30
To solve for k, divide both sides of the equation by 30:
5/30 = k
1/6 = k
Therefore, the constant of variation is 1/6, not 150.
The answer is False.
2.) To find the value of w when z is 3, we need to use the direct variation equation w = kz.
Given that w is 9 when z is -6, we can substitute these values into the equation:
9 = k * (-6)
To solve for k, divide both sides of the equation by -6:
9 / -6 = k
-3/2 = k
Now that we have the value of k, we can substitute it into the equation to find w when z is 3:
w = (-3/2) * 3
Multiply -3/2 by 3:
w = -9/2
Therefore, when z is 3, w is -9/2.
3.) To find the value of x when y is 4, we need to use the direct variation equation y = kx.
Given that y is 2 when x is 4, we can substitute these values into the equation:
2 = k * 4
To solve for k, divide both sides of the equation by 4:
2/4 = k
1/2 = k
Now that we have the value of k, we can substitute it into the equation to find x when y is 4:
4 = (1/2) * x
Multiply 1/2 by x:
4 = x/2
To solve for x, multiply both sides of the equation by 2:
8 = x
Therefore, when y is 4, x is 8.