I'm confused about the answer to this question one-half times twelve times the quantity thirty-six plus fifteen

-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

step-by-step explanation:

; fx

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: