Answer:
3, 27 and 9,9
Step-by-step explanation:
3*27 = 81
9*9=81
Technically you could say 1 and 81 but those two answers fit the most.
-- He must have at least one of each color in the case, so the first 3 of the 5 marbles in the case are blue-green-black.
Now the rest of the collection consists of
4 blue
4 green
2 black
and there's space for 2 more marbles in the case.
So the question really asks: "In how many ways can 2 marbles
be selected from 4 blue ones, 4 green ones, and 2 black ones ?"
-- Well, there are 10 marbles all together.
So the first one chosen can be any one of the 10,
and for each of those,
the second one can be any one of the remaining 9 .
Total number of ways to pick 2 out of the 10 = (10 x 9) = 90 ways.
-- BUT ... there are not nearly that many different combinations
to wind up with in the case.
The first of the two picks can be any one of the 3 colors,
and for each of those,
the second pick can also be any one of the 3 colors.
So there are actually only 9 distinguishable ways (ways that
you can tell apart) to pick the last two marbles.
Answer:
f(-2) =3
Step-by-step explanation:
f(x) = 9+3x
Let x=-2
f(-2) = 9+3(-2)
= 9-6
= 3
X / 6=3.5 would be the answer
4. k less than 3.5, so a number k is being taken away from 3.5:
The answer is 3.5-k
5. the sum of g and h, reduced by 11, so you need a set of parentheses to write this one: (g+h)-11
6. 5 less than y, plus g, so do a combination of the strategies used in the previous two problems and you will get: (y-5)+g
7. 5 less than the sum of y and g, you are taking away 5 from the sum of two numbers, y and g, so: (y+g)-5
Hope this helps :-)
