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.
Answer:


The measure of the reference angle is 
Step-by-step explanation:
The reference angle of an angle is the magnitude of the smallest angle it makes with the x-axis. In this case, we have
(*Note:
, use
to convert to degrees)
Answer: 100
Step-by-step explanation:
25*4 = 100
Answer:c
Step-by-step explanation:
First off, we factor out the expression:

In the bracket, separate 8 out of the expression.
![\displaystyle \large{y = 2[ ( {x}^{2} - 6x + 8)] }\\ \displaystyle \large{y = 2[ ( {x}^{2} - 6x) + 8]}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7By%20%3D%202%5B%20%28%20%7Bx%7D%5E%7B2%7D%20-%206x%20%2B%208%29%5D%20%7D%5C%5C%20%20%5Cdisplaystyle%20%5Clarge%7By%20%3D%202%5B%20%28%20%7Bx%7D%5E%7B2%7D%20-%206x%29%20%2B%208%5D%7D)
In x^2-6x, find the third term that can make up or convert it to a perfect square form. The third term is 9 because:

So we add +9 in x^2-6x.
![\displaystyle \large{y = 2[ ( {x}^{2} - 6x + 9) + 8]}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7By%20%3D%202%5B%20%28%20%7Bx%7D%5E%7B2%7D%20-%206x%20%2B%209%29%20%20%2B%208%5D%7D)
Convert the expression in the small bracket to perfect square.
![\displaystyle \large{y = 2[ {(x - 3)}^{2} + 8]}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7By%20%3D%202%5B%20%20%7B%28x%20-%203%29%7D%5E%7B2%7D%20%20%20%2B%208%5D%7D)
Since we add +9 in the small bracket, we have to subtract 8 with 9 as well.
![\displaystyle \large{y = 2[ {(x - 3)}^{2} + 8 - 9]} \\ \displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clarge%7By%20%3D%202%5B%20%20%7B%28x%20-%203%29%7D%5E%7B2%7D%20%20%20%2B%208%20-%209%5D%7D%20%5C%5C%20%20%5Cdisplaystyle%20%5Clarge%7By%20%3D%202%5B%20%20%7B%28x%20-%203%29%7D%5E%7B2%7D%20%20%20-%201%5D%7D)
Then we distribute 2 in.
![\displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]} \\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7By%20%3D%202%5B%20%20%7B%28x%20-%203%29%7D%5E%7B2%7D%20%20%20-%201%5D%7D%20%5C%5C%20)
![\displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]} \\ \displaystyle \large{y = [2 \times {(x - 3)}^{2} ]+[ 2 \times ( - 1)] } \\ \displaystyle \large{y = 2 {(x - 3)}^{2} - 2 }](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7By%20%3D%202%5B%20%20%7B%28x%20-%203%29%7D%5E%7B2%7D%20%20%20-%201%5D%7D%20%5C%5C%20%5Cdisplaystyle%20%5Clarge%7By%20%3D%20%5B2%20%5Ctimes%20%20%7B%28x%20-%203%29%7D%5E%7B2%7D%20%5D%2B%5B%202%20%5Ctimes%20%28%20-%201%29%5D%20%7D%20%5C%5C%20%5Cdisplaystyle%20%5Clarge%7By%20%3D%202%20%7B%28x%20-%203%29%7D%5E%7B2%7D%20%20-%202%20%7D)
Remember that negative multiply positive = negative.
Hence the vertex form is y = 2(x-3)^2-2 or first choice.