6x^2 + 37x - 60 = 6x^2 - 8x + 45x - 60 = 2x(3x - 4) + 15(3x - 4) = (3x - 4)(2x + 15)
Answer:
13.98 in²
Step-by-step explanation:
I don't understand it, either.
Point N is part of a "segment" that above and to the right of chord MO. It is the sum of the areas of 3/4 of the circle and a right triangle with 7-inch sides. The larger segment MO to the upper right of chord MO has an area of about 139.95 in², which <u>is not</u> an answer choice.
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The remaining segment, to the lower left of chord MO does not seem to have anything to do with point N. However, its area is 13.98 in², which <u>is</u> an answer choice. Therefore, we think the question is about this segment, and we wonder why it is called MNO.
The area of a segment is given by the formula ...
A = (1/2)(θ -sin(θ))r² . . . . . . where θ is the central angle in radians.
Here, we have θ = π/2, r = 7 in, so we can compute the area of the smaller segment MO as ...
A = (1/2)(π/2 -sin(π/2))(7 in)² = 24.5(π/2 -1) in² ≈ 13.9845 in²
Rounded to hundredths, this is ...
≈ 13.98 in²
6x + 7(1-x) = -4(x-4)
6x + 7 - 7x = -4x + 16
-x + 7 = -4x + 16
-x + 4x = 16 - 7
3x = 9
x =9/3
x = 3
If its just 5.35 then no. But if its a number like pie then yes.
