(a).
The product of two binomials is sometimes called FOIL.
It stands for ...
the product of the First terms (3j x 3j)
plus
the product of the Outside terms (3j x 5)
plus
the product of the Inside terms (-5 x 3j)
plus
the product of the Last terms (-5 x 5)
FOIL works for multiplying ANY two binomials (quantities with 2 terms).
Here's another tool that you can use for this particular problem (a).
It'll also be helpful when you get to part-c .
Notice that the terms are the same in both quantities ... 3j and 5 .
The only difference is they're added in the first one, and subtracted
in the other one.
Whenever you have
(the sum of two things) x (the difference of the same things)
the product is going to be
(the first thing)² minus (the second thing)² .
So in (a), that'll be (3j)² - (5)² = 9j² - 25 .
You could find the product with FOIL, or with this easier tool.
______________________________
(b).
This is the square of a binomial ... multiplying it by itself. So it's
another product of 2 binomials, that both happen to be the same:
(4h + 5) x (4h + 5) .
You can do the product with FOIL, or use another little tool:
The square of a binomial (4h + 5)² is ...
the square of the first term (4h)²
plus
the square of the last term (5)²
plus
double the product of the terms 2 · (4h · 5)
________________________________
(c).
Use the tool I gave you in part-a . . . twice .
The product of the first 2 binomials is (g² - 4) .
The product of the last 2 binomials is also (g² - 4) .
Now you can multiply these with FOIL,
or use the squaring tool I gave you in part-b .
the parabola has maximum at 9, meaning is a vertical parabola and it opens downwards.
it has a symmetry at x = -5, namely its vertex's x-coordinate is -5.
check the picture below.
so then, we can pretty much tell its vertex is at (-5 , 9), and we also know it passes through (-7, 1)
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \leftarrow \textit{using this one}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-5\\ k=9 \end{cases}\implies y=a[x-(-5)]^2+9\implies y=a(x+5)^2+9](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cleftarrow%20%5Ctextit%7Busing%20this%20one%7D%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D-5%5C%5C%20k%3D9%20%5Cend%7Bcases%7D%5Cimplies%20y%3Da%5Bx-%28-5%29%5D%5E2%2B9%5Cimplies%20y%3Da%28x%2B5%29%5E2%2B9)
![\bf \textit{we also know that } \begin{cases} x=-7\\ y=1 \end{cases}\implies 1=a(-7+5)^2+9 \\\\\\ -8=a(-2)^2\implies -8=4a\implies \cfrac{-8}{4}=a\implies -2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=-2(x+5)^2+9~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bwe%20also%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20x%3D-7%5C%5C%20y%3D1%20%5Cend%7Bcases%7D%5Cimplies%201%3Da%28-7%2B5%29%5E2%2B9%20%5C%5C%5C%5C%5C%5C%20-8%3Da%28-2%29%5E2%5Cimplies%20-8%3D4a%5Cimplies%20%5Ccfrac%7B-8%7D%7B4%7D%3Da%5Cimplies%20-2%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20y%3D-2%28x%2B5%29%5E2%2B9~%5Chfill)
Answer:
y2-y1/x2-x1
Step-by-step explanation:
Answer:
(9sqrt13)/13
Step-by-step explanation:
