The direct variation equation that relates x and y can be written as y = kx, where k is the constant of variation. To find the value of k, we can use the given information that y = 5 when x = 2.
Substituting these values into the equation, we get:
5 = k(2)
Simplifying the equation, we have:
10 = 2k
Dividing both sides by 2, we find:
k = 5
So, the direct variation equation that relates x and y is y = 5x.
To find the value of y when x = 16, we can plug in this value into the equation:
y = 5(16)
y = 80
Therefore, when x = 16, y has a value of 80.
Suppose y varies directly with x. Write a direct variation equation that relates x and y. Then find the value of y when x=16.
y=5 when x=2
Write a direct variation equation that relates x and y.
y =
(Simplify your answer.)
9 answers
wrong
Apologies for the mistake. Let's correct it.
If y varies directly with x, we can write the direct variation equation as:
y = kx
We need to find the value of k. We are given that y = 5 when x = 2. Substituting these values into the equation, we have:
5 = k(2)
To solve for k, we divide both sides by 2:
k = 5/2
Therefore, the direct variation equation relating x and y is:
y = (5/2)x
Now, we need to find the value of y when x = 16. We can substitute this value into the equation:
y = (5/2)(16) = 40
Therefore, when x = 16, y has a value of 40.
If y varies directly with x, we can write the direct variation equation as:
y = kx
We need to find the value of k. We are given that y = 5 when x = 2. Substituting these values into the equation, we have:
5 = k(2)
To solve for k, we divide both sides by 2:
k = 5/2
Therefore, the direct variation equation relating x and y is:
y = (5/2)x
Now, we need to find the value of y when x = 16. We can substitute this value into the equation:
y = (5/2)(16) = 40
Therefore, when x = 16, y has a value of 40.
incorrect!
Apologies for the incorrect response. Let's correct it.
If y varies directly with x, we can write the direct variation equation as:
y = kx
We are given that y = 5 when x = 2. Substituting these values into the equation, we can solve for the constant of variation k:
5 = k(2)
To solve for k, we divide both sides by 2:
k = 5/2
Therefore, the direct variation equation relating x and y is:
y = (5/2)x
Now, to find the value of y when x = 16, we can substitute this value into the equation:
y = (5/2)(16) = 40
Therefore, when x = 16, y has a value of 40.
If y varies directly with x, we can write the direct variation equation as:
y = kx
We are given that y = 5 when x = 2. Substituting these values into the equation, we can solve for the constant of variation k:
5 = k(2)
To solve for k, we divide both sides by 2:
k = 5/2
Therefore, the direct variation equation relating x and y is:
y = (5/2)x
Now, to find the value of y when x = 16, we can substitute this value into the equation:
y = (5/2)(16) = 40
Therefore, when x = 16, y has a value of 40.
Suppose y varies directly with x. Write a direct variation equation that relates x and y. Then find the value of y when x=16.
y=9 when x=8
Write a direct variation equation that relates x and y.
y
(Simplify your answer.)
y=9 when x=8
Write a direct variation equation that relates x and y.
y
(Simplify your answer.)
The direct variation equation relating x and y can be written as y = kx, where k is the constant of variation.
To find the value of k, we can use the given information that y = 9 when x = 8:
9 = k(8)
Dividing both sides by 8, we find:
k = 9/8
So, the direct variation equation that relates x and y is y = (9/8)x.
To find the value of y when x = 16, we can substitute this value into the equation:
y = (9/8)(16)
y = 18
Therefore, when x = 16, y has a value of 18.
To find the value of k, we can use the given information that y = 9 when x = 8:
9 = k(8)
Dividing both sides by 8, we find:
k = 9/8
So, the direct variation equation that relates x and y is y = (9/8)x.
To find the value of y when x = 16, we can substitute this value into the equation:
y = (9/8)(16)
y = 18
Therefore, when x = 16, y has a value of 18.
Suppose y varies directly with x. Write a direct variation equation that relates x and y. Then graph the equation.
y=4.8 when x=1.2
The equation of variation is y = (simplify your answer)
y=4.8 when x=1.2
The equation of variation is y = (simplify your answer)
Since y varies directly with x, we can write the direct variation equation as y = kx, where k is the constant of variation.
We are given that y = 4.8 when x = 1.2. Substituting these values into the equation, we can solve for the constant of variation k:
4.8 = k(1.2)
To solve for k, we divide both sides by 1.2:
k = 4.8/1.2
k = 4
Therefore, the direct variation equation that relates x and y is:
y = 4x
To graph this equation, we can create a table of values:
| x | y |
|-------|-------|
| 0 | 0 |
| 1 | 4 |
| 2 | 8 |
| 3 | 12 |
Plotting these points on a graph, we can see a straight line passing through the origin (0,0) with a slope of 4.
We are given that y = 4.8 when x = 1.2. Substituting these values into the equation, we can solve for the constant of variation k:
4.8 = k(1.2)
To solve for k, we divide both sides by 1.2:
k = 4.8/1.2
k = 4
Therefore, the direct variation equation that relates x and y is:
y = 4x
To graph this equation, we can create a table of values:
| x | y |
|-------|-------|
| 0 | 0 |
| 1 | 4 |
| 2 | 8 |
| 3 | 12 |
Plotting these points on a graph, we can see a straight line passing through the origin (0,0) with a slope of 4.
1 answer