Answer:
$2.10x + 5= $49.10
x=21
Step-by-step explanation:
the x is used for a variable that changes and in this case the mph changes which lead x being the amount of miles that they had drove which at the end ends up multi. with the $2.10 . as for the 5 it was an initial as in it doesn't change.
for the second part
you subtract 5 from each side so you have the total of 2.10x= 44.10 then u divide 2.10 from each side and that will give you x = 21
Answer:
SAS Similarity
Step-by-step explanation:
given by lines 1, 2, & 3
Answer:
Step-by-step explanation:
b=bench
852=b+(b-98) since the table is less than the bench by 98$
852-b=b-98
950-b=b
950=2b
475=b
Explanation:
Basically, you can do it in many ways. But just, in my opinion, exactly linear algebra was made for such cases.
the optimal way is to do it with Cramer's rule.
First, find the determinant and then find the determinant x, y, v, u.
Afterward, simply divide the determinant of variables by the usual determinant.
eg. 
and etc.
I think that is the best way to solve it without a hustle of myriad of calculations reducing it to row echelon form and solving with Gaussian elimination.
-- 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.
