Y is partly constent and partly varries as the square of x.if y=40 when x=1 and y=13 when x =2.Find the expression for y in terms of x. Ii)the value of y when x=3
How did you get the 4 in a + 4b = 13
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I. Let's break down the problem and find the expression for y in terms of x.
We know that y is partly constant and partly varies as the square of x. Let's denote the constant part as k.
When x = 1, y = 40. Substituting these values into the equation, we get:
40 = k + (1)^2
Simplifying, we find:
k = 39
So, the expression for y is:
y = 39 + x^2
II. Now, we need to find the value of y when x = 3. Let's substitute the value into the equation we found in the previous step:
y = 39 + (3)^2
Simplifying, we get:
y = 39 + 9
y = 48
Therefore, when x = 3, y equals 48.
To find the expression for y in terms of x, we can assume that y is composed of two parts: a constant part and a variable part.
Let's assume that the constant part of y is represented by "c" and the variable part of y is represented by "kx²" (k times the square of x).
So, the expression for y in terms of x can be written as:
y = c + kx²
Now, we need to determine the values of c and k.
Given that when x = 1, y = 40, we can substitute these values into the expression:
40 = c + k(1)²
40 = c + k
Similarly, when x = 2, y = 13:
13 = c + k(2)²
13 = c + 4k
Now we have two equations with two unknowns (c and k). By solving these equations simultaneously, we can find the values of c and k.
Subtracting the second equation from the first equation, we get:
40 - 13 = c + k - (c + 4k)
27 = -3k
k = -9
Substituting the value of k back into the first equation, we get:
40 = c - 9
c = 49
Now we have the constant and variable parts of y:
y = 49 - 9x²
To find the value of y when x = 3, we substitute x = 3 into the expression:
y = 49 - 9(3)²
y = 49 - 9(9)
y = 49 - 81
y = -32
So, when x = 3, y = -32.
Therefore, the expression for y in terms of x is y = 49 - 9x², and the value of y when x = 3 is -32.
y = a+bx^2
Now, solve
a+b = 40
a+4b = 13
and then you can find y when x=3.