y is partily a constant and partly varies as the square of x, if y=40 when x=1 and y =13 when x=2 find the value of y, when y=13
When x=13, y=0.
y = mx^2+b
m+b = 40
4m+b = 13
solve for m and b, then
fix your typo to make it a real question.
From the given information, we are told that y is partly a constant and partly varies as the square of x. This can be written as:
y = k + ax^2
where k is the constant part and a is the coefficient of the variable term.
We are also given two sets of values: when x = 1, y = 40 and when x = 2, y = 13. Using these values, we can form two equations to solve for the values of k and a.
When x = 1:
40 = k + a(1)^2
40 = k + a
When x = 2:
13 = k + a(2)^2
13 = k + 4a
We now have a system of two equations:
40 = k + a
13 = k + 4a
To solve this system, we can subtract the first equation from the second equation:
13 - 40 = (k + 4a) - (k + a)
-27 = 4a - a
-27 = 3a
Divide both sides of the equation by 3:
-27/3 = a
-9 = a
Now substitute the value of a back into the first equation to solve for k:
40 = k + (-9)
40 = k - 9
49 = k
So, the value of k is 49 and the value of a is -9.
To find the value of y when y = 13, we substitute this value into the equation:
13 = 49 + (-9)x^2
Simplifying the equation:
13 = 49 - 9x^2
Rearranging the equation:
9x^2 = 49 - 13
9x^2 = 36
Divide both sides of the equation by 9:
x^2 = 4
Taking the square root of both sides:
x = ±√4
x = ±2
So, when y = 13, the value of x can be either 2 or -2.
To find the value of y when y = 13, we can use the information given that y is partly a constant and partly varies as the square of x.
Let's first write down the general equation for y in terms of x:
y = constant + (x^2)
From the given information, we know that y = 40 when x = 1 and y = 13 when x = 2.
Plugging these values into the equation, we get two equations:
Equation 1: 40 = constant + (1^2)
Equation 2: 13 = constant + (2^2)
Now we can solve these equations to find the value of the constant and, subsequently, the value of y when y = 13.
Subtracting (1^2) from both sides of Equation 1 gives us:
40 - (1^2) = constant
40 - 1 = constant
constant = 39
Now we substitute the value of the constant into Equation 2:
13 = 39 + (2^2)
13 = 39 + 4
13 = 43
It seems that there may be an error in the information provided. From the given equations, we obtain 13 ≠ 43, which means either the values of x or y have been stated incorrectly. Please double-check the given information and provide the correct values, if possible.