Y is partly constant and partly varies as the square if X giveb that Y=16 when X=3 and Y=12 when X=2,
a) Write down an equation connecting X and Y
b) Find Y when X=15
a) Let the constant part be k and the varying part be mx^2. Thus, we have:
Y = k + mx^2
Using the two given data points, we can create a system of equations:
16 = k + 9m
12 = k + 4m
Subtracting the second equation from the first, we get:
4 = 5m
Solving for m, we get:
m = 4/5
Substituting this back into one of the equations, we can solve for k:
16 = k + 9(4/5)
16 = k + 7.2
k = 8.8
Therefore, the equation connecting X and Y is:
Y = 8.8 + (4/5)x^2
b) To find Y when X = 15, we simply plug it into the equation we found:
Y = 8.8 + (4/5)(15^2)
Y = 8.8 + 360
Y = 368.8
Therefore, when X = 15, Y = 368.8.
a) Let's assume that Y is partly constant and partly varies as the square of X. We can express this relationship as:
Y = a + bX^2
where a is the constant part and b is the coefficient of the variable part.
Given that Y = 16 when X = 3, we can substitute these values into the equation:
16 = a + b(3^2)
16 = a + 9b
Similarly, given that Y = 12 when X = 2, we can substitute these values into the equation:
12 = a + b(2^2)
12 = a + 4b
Now we have a system of two equations:
16 = a + 9b
12 = a + 4b
b) To find Y when X = 15, we can substitute X = 15 into the equation we derived earlier:
Y = a + b(15^2)
However, to find the specific values of 'a' and 'b', we need to solve the system of equations.
From the first equation, we can express 'a' in terms of 'b':
a = 16 - 9b
Substituting this expression for 'a' into the second equation:
12 = (16 - 9b) + 4b
12 = 16 - 5b
-4 = -5b
b = 4/5 or 0.8
Now, substituting the value of 'b' back into the expression for 'a':
a = 16 - 9(0.8)
a = 16 - 7.2
a = 8.8
Finally, substituting X = 15 into the equation:
Y = a + b(15^2)
Y = 8.8 + 0.8(225)
Y = 8.8 + 180
Y = 188.8
Therefore, when X = 15, Y is equal to 188.8.