Question 3 of 3 Will wanted to solve the problem
468
÷
12
using partial quotients.

Which of the methods could be used to find the quotient? Select all that apply.

A.
Subtract (
10
×
12
) from 468 to get 348. Subtract (
3
×
12
) from 348 to get 312. Subtract (
9
×
12
) from 292.

B.
Subtract (
30
×
12
) from 468 to get 108. Subtract (
9
×
12
) from 108.

C.
Subtract (
20
×
12
) from 468 to get 228. Subtract (
10
×
12
) from 228 to get 108. Subtract (
3
×
12
) from 108 to get 72. Subtract (
6
×
12
) from 72.

D.
Subtract (
30
×
12
) from 468 to get 108. Subtract (
4
×
12
) from 108 to get 60. Subtract (
4
×
12
) from 60.

phosphate, calcium ion, chloride ion, neon atom

step-by-step explanation:

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

a) 0,0; 1,1; and 2,2

step-by-step explanation:

these have the exact same coordinates, thus they are the correct function.