Answer:
Step-by-step explanation:
(-3 ab²)³=(-3)³(a)³(b²)³=-27 a³b^(2×3)=-27 a³b^6
Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
Answer: 3(x-3y)
Step-by-step explanation: first you have to take the 3x and split it up so it can make 3(x-3y) then you have to rewrite it to -9 as 3x3 and then factor out common term 3 which would be your final answer 3(x-3y)
Sorry if this is messy but I hope it helps you
Lol you like maths? Honestly, I don't like it much b r u h
