Answer:
Vertical line test is used to identify that the given graph is function or not
In this a vertical line is drawn at any part of the graph
If it cuts only one line then it is a function and it cuts more than one line then it is not a function
Answer:
the answer to part is 112.5 and part b is 4612.5 dont completely trust me though
the absolute value of a number is always positive. Therefore you cannot have an absolute value equal to a negative number.
the answer would be "no solution"
(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 .
