Answer:
b
Step-by-step explanation:
The area of the shaded region is going to be the area of the rectangle minus the area of the square.
Area of a rectangle is L * W.
A = L * W
A = (x + 10)(2x + 5)
A = x(2x + 5) + 10(2x + 5)
A = 2x^2 + 5x + 20x + 50
A = 2x^2 + 25x + 50 .....this is the area of the rectangle
area of a square is : A = a^2...where a is one side
A = (x + 1)^2
A = (x + 1)(x + 1)
A = x(x + 1) + 1(x + 1)
A = x^2 + x + x + 1
A = x^2 + 2x + 1
now we subtract the area of the square from the area of the rectangle to get the area of the shaded region.
2x^2 + 25x + 50 - (x^2 + 2x + 1) =
2x^2 + 25x + 50 - x^2 - 2x - 1 =
x^2 + 23x + 49 <== the area of the shaded region
Answer:
Yes, as the change in y repeats continuously
40x+8x^2=0 can be solved for x (there are two solutions):
Divide all 3 terms by the greatest common multiple (which is 8x):
40x+8x^2=0
------------- -----
8x 8x
5 + x = 0 produces the root x = - 5.
Setting 8x = 0 and solving for x produces the root x = 0.
Be certain to check these results. substitute x = -5 into 40x+8x^2=0. Is the resulting equation true or false? Next, subs. x=-5 into 40x+8x^2=0. Is the resulting equation true or false?
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