Answer:
D. Malik substituted 60 for y instead of x.
Step-by-step explanation:
(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 .
Answer:
12 12 12 12 12 12 12 12 12 1 2 12 12
Answer:
84%, 86%, 89%
Step-by-step explanation:
The three numbers, expressed as percentages, are ...
- 86%
- (47/50)×100% = 84%
- 0.89×100% = 89%
Smallest to largest, they are 84%, 86%, and 89%. In the original form, they are 47/50, 86%, and 0.89.
Just something that is true or are there choices? If just something true then dialated by 1 is always the same as the original shape.
