x varies partly as y and partly y^2 when y =4,x=52.8 ,and when y=5,x=81 find x when y=6
However, in the interests of passing this class, "varies partly" means
x = ay + by^2
Now plug in your points to find a and b:
4a + 16b = 52.8
5a + 25b = 81
Then use your function to find x(6)
Your use of "partly" is not understood. If your teachers uses this term, I suggest drop the class, you are headed for more made up math that is not understood by many math users.
Good job
Well, let's see. We know that x varies partly as y and partly y^2. So, logically speaking, x might also have a little bit of a crush on y, but is completely smitten by y^2.
When y = 4 and x = 52.8, we can assume that x is partly attracted to 4 and partly head over heels for 4^2. So, let's break out the "relationship math" here:
x = k * (y + y^2) [Where k is the factor of attraction]
Plugging in the values we know:
52.8 = k * (4 + 4^2)
52.8 = k * (4 + 16)
52.8 = k * 20
Solving for k, we find that k is 52.8/20.
Moving on to the next part of the question, when y = 5, we can use the same equation to find x:
x = k * (5 + 5^2)
x = k * (5 + 25)
x = k * 30
Now it's time to find out what x is with y = 6. Let's plug it into the equation one more time, shall we?
x = k * (6 + 6^2)
x = k * (6 + 36)
x = k * 42
So, with y = 6, x would be 42 times the factor of attraction k.
But, since we didn't calculate the value of k, I'm afraid I don't have the exact answer for you. Can we find k, substitute it back into the equation, and get the final answer? That way we would be able to crack this relationship math code together!
More explanation
To find the value of x when y = 6, we need to determine the relationship between x and y. From the given information, we know that x is partly dependent on y and partly dependent on y^2.
Let's start by finding the relationship between x and y when y = 4. We have x = 52.8 when y = 4.
Next, let's find the relationship between x and y^2. We can find this by comparing the two sets of values when y = 4 and y = 5.
When y = 4, x = 52.8, and when y = 5, x = 81.
Comparing these two sets of values, we can see that when y increases by 1 from 4 to 5, x increases from 52.8 to 81.
So, the relationship between x and y^2 is linear, as y increases from 4 to 5, x increases in a consistent manner.
Now, let's use this linear relationship between x and y^2 to find the value of x when y = 6.
Using the values we have, we can calculate the slope of the line:
slope = (change in x) / (change in y^2) = (81 - 52.8) / (5^2 - 4^2) = 28.2 / (25 - 16) = 28.2 / 9 = 3.1333 (rounded)
Now, we have the slope of the line, and we can use it to find the value of x when y = 6:
x = y^2 * slope + (intercept)
Since we only have one data point, we don't know the intercept. However, we can assume that the intercept is the same as when y = 4, which is x = 52.8.
Substituting the values into the equation, we get:
x = 6^2 * 3.1333 + 52.8
= 36 * 3.1333 + 52.8
= 112.7992 + 52.8
= 165.5992
Therefore, when y = 6, x is approximately 165.5992.