(F.g)(x)= (x^2+2x-5)(2x+4)
(F.g)(x)= 2x^3+4x^2+4x^2+8x-10x-20
(F.g)(x)= 2x^3+8x^2-2x-20=> C
Answer: 3
Step-by-step explanation:
Answer:
Your answer is either C or D.
Step-by-step explanation:
The picture shown is a line. Go for c. If not, then definitely D.
3421880 number of the different meals are possible.
<h3>What is the combination?</h3>
The arrangement of the different things or numbers in the number of the ways is called as the combination.
It is given that:-
Number of appetizers is 7, the number of entres is 10, and that there's 4 choices of desserts.
So let's take each course by itself. You can choose 1 of 7 appetizers. So we have n = 7 After that, you chose an entre,
so the number of possible meals to this point is
n = 7 x 10 = 70
Finally, you finish off with a dessert, so the number of meals is:
n = 70 x 4 = 280
Therefore the number of possible meals you can have is 280. Note: If the values of 77, 1010 and 44 aren't errors, but are actually correct, then the number of meals is
n = 77 x 1010 x 44 = 3421880
Therefore 3421880 different meals are possible.
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Answer:
4+x=7 this you make a t chart then you (subtract word bank) 4 because of the (addition word bank) sign to get rid of four and keep the variable rember do it to both sides.
4+x l 7
-4 l -4
x = 3
Step-by-step explanation:
