4, 8, and 3 must all go into the number. find the LCM(4,8) is 8. the LCM(8,3) is 24. that's your answer, 24
Answer:
C, E
Step-by-step explanation:
Here, we want to change the equation of a circle from general form to standard form. This is done by making the leading coefficients 1, completing the squares, and then rewriting the equation in standard form.
The leading coefficients of the given equation are 2, so we first want to divide by 2. This gives ...
x² +y² -4x -6y +8 = 0
Subtracting 8 puts us in better position to complete the squares.
x² +y² -4x -6y = -8
Now, we can add the squares of half the coefficients of the linear terms.
(x² -4x +4) +(y² -6x +9) = -8 +13 . . . . . . matches C
And we can simplify this to the standard form equation:
(x -2)² +(y -3)² = 5 . . . . . matches E
Just keep adding 1/6 to the given term and that should help you figure out the rule.
Quotient is basically the answer produced when dividing two numbers.
For example, the quotient of 6 and 3 is 2 because when I divide 6 by 3, I get 2.
Eight less than the quotient of x and 3 =
-- 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.
