Replace ∗ with a monomial to obtain an identity: 1. (∗− 2m)^2 = 100–40m+4m^2 2. (3a+2.5b)^2=9a^2+6.25b^2 + ∗

1. ∗ = 10

2. ∗ = 15ab

Step-by-step explanation:

The identity: (A-B)²=A²-2AB+B² or (A+B)²= A²+2AB+B²

1. In the first problem, A = ∗, which means that ∗² is going to be equal to 100.

∗ = √100= 10

2. In the second problem we already have A and B, and we have both A² (9a²) and B² (6.25b²) and we are missing the "2AB" part in the middle.

∗ = 2AB = 2×3a×2.5b = 15ab

to plot points   for x just start at the origin (middle) and go to the left however many times it say(if the number is a positive) if the number is a negative then go right however many times it says if it is 0 go nowhere

now keep you pencil there and go up however many times it says on y(if it is positive) if it is negative then go down however many times it says. if it says 0 go nowhere.

you have just plotted a point! now when you finish plotting all of them connect them with a ruler

plot each point as (-3, 7), (0,4), (3,1), (8,-4)   in the parenthesis on the left of the comma is x and the right is y.

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

step-by-step explanation: