5. Which point could not be part of the function that includes (1, 3) (3, 7), (5, 11) (7, 15), (9, 19) ? a) (4, 3) b) (1, 10) c) (8, 17) d) (2, 13)

-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.

step-by-step explanation:

step 1 distribute 2 to every term in the parentheses

step 2 add 10 on both sides of the equal sign

the answer is 8/21

hope this