SZhang

It was correct! Thank you so much! The explanation cleared it up. 

So 17 then?  

I am so sorry for being this late, but I recently had to solve this problem too. 

If you look at the equation \((a^mx-a^n)(a^py-a^2)=a^4b^4\), you may notice that you can factor the previous equation to be in this form. 

First, you distribute \(a^4\)into the parentheses to get \(a^4b^4 - a^4\). If you look at the other equation, you may see that their answer is very similar to \(a^4b^4 - a^4\). So, you first add \(a^4\) to both sides to get \((a^3x-a^2)(a^4y-a^2)=a^4b^4\). Factoring the left side gives \((a^3x-a^2)(a^4y-a^2)=a^4b^4\). So, \((m,n,p)=(3,2,4)\), which means \(mnp=3\cdot2\cdot4=\boxed{24}\).

You did it! It was right! Good job! 

Let \(x_1\) and \(x_2\) be the roots of the equation \(8x^2+12x-14 \). We want to find \(x_1^2+x_2^2\). Note that \(x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2\). We know that \(x_1+x_2\), the sum of the roots, is equal to \(\frac{-b}{a}\), which for this equation is \(\frac{-12}{8}=\frac{-3}{2}\). Likewise, we know that \(x_1x_2\), the product of the roots, is equal to \(\frac{c}{a}\), which for this equation is \(\frac{-14}{8}=\frac{-7}{4}\). Thus, \(x_1^2+x_2^2=\left(\frac{-3}{2}\right)^2-2\left(\frac{-7}{4}\right)=\frac{9}{4}+\frac{14}{4}=\boxed{\frac{23}{4}}\).

Nevermind. It was -10x. 

Thank you so much! I am really grateful to all of you!

A point lies above \(y=2x+7\) if its y -coordinate is greater than 2 times its x-coordinate plus 7. Checking the given points, we find that (6,20), (12,3), and (20,50) satisfy this condition. The sum of the x-coordinates of these points is \(6+12+20=\boxed{38}\). 

Since the problem only asks for the difference in the \(y \)-coordinates, we can ignore the \(x\)-coordinates.
They originally agreed to meet at the midpoint of \((5,-11)\) and \((-7,13)\), so the -coordinate of the planned location is \( \frac{-11+13}{2}=1\).
The correct meeting location should be at the midpoint of \((5,-11)\) and \((-5,5)\), so the -coordinate should be at \(\frac{-11+5}{2}=-3\). The positive difference is \(1-(-3)=\boxed{4}\).
Alternatively, notice that an 8-unit change in the y-coordinate of Barbara's location resulted in a 4-unit change in the midpoint since the 8 gets divided by 2. \(\frac{-11+5}{2}=\frac{-11+13}{2}-\frac{8}{2}=1-\boxed{4}\).