Let "a" and "b" be some number where:
a - b = 24
We want to find where a^2 + b^2 is a minimum. Instead of just logically figuring out that the answer is where a=b=12, I'll just use derivatives.
So we can first substitute for "a" where a = b+24
So we have (b+24)^2 + b^2 = b^2 +48b +576 + b^2
And that equals 2b^2 +48b +576
Then we take the derivative and set it equal to zero:
4b +48 = 0
4(b+12) = 0
b + 12 = 0
b = -12
Thus "a" must equal 12.
So:
a = 12
b = -12
And the sum of those two numbers squared is (12)^2 + (-12)^2 = 144 + 144 = 288.
The smallest sum is 288.
A: Yes because for each x value, there is only one corresponding y value.
B: For the first relation, y is half of x, so f(x)=x/2. If x is 8, y is 4.
Plug in the value of x for the second function. 3(8)-10=24-10=14.
The second relation has a greater value.
C: 80=3x-10,
90=3x
x=30
This is a true statement.
Answer:
y=mx+b is 32x53 divide by 10 and =8.31=b+423.1/74=mx=647.bd-5634=y add them up and is your answer.
Step-by-step explanation:
