The midpoint of \overline{\text{AB}} AB
is M(3, -2)M(3,−2). If the coordinates of AA are (8, 4)(8,4), what are the coordinates of BB?

Step-by-step explanation: I am trying to interpret the question: I'll assume:

Line AB has a midpoint, M.

A is (8,4)

B is unknown, but desired

M is (3,-2)

We'll assume a straight line of the form y=mx+b, where m is the slope. m can be calculated by the "Rise" over the "Run" between the points A and M:

Rise = (4-(-2))=6

Run = (8-3)=5

Slope = (Rise/Run) = (6/5)

The line becomes y=(6/5)x+b

Calculate b, the y-intercept (the value of y when x is 0), by entering one of the two known points. I'll choose (8,4).

y=(6/5)x+b

4 = (6/5)(8)+b

b = 4 - (6/5)(8)

b = 4 - (48/5)

b = 4 - (9 3/5) or -5 3/5

The equation becomes:

y=(6/5)x - 5 3/5

Point B will have an x value of -2 [The change in x from A to M was (3-8) = -5. From M to 5 would be another -5 increment, so x goes from 3 to (3-5) or -2.

Find the value of y for x=-2:

y=(6/5)x - 5 3/5

y=(6/5)*(-2) - 5 3/5

y=(-12/5) - (28/5)

y = -(40/5) or - 8

B has coordinates of (-2, -8)

answer: 180

step-by-step explanation:

12-2=10

20-2=18

18x10=180

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct