Three consecutive even integers are such that the product of the second integer and third integer is fifteen times the first integer. determine the three integers.

Let's call the three integers

The product of the second and third integers is thus

We know that this quantity equals fifteen times the first integer:

The solutions of this equation are

So, the determinant has to be a perfect square.

This is not the case, because

which is not a perfect square.

So, it is not possible to find three consecutive integers with that property.

answer:

c. the third graph

-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