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
Answer:
u kind of forgot the attachment bestie
Answer:
X=2
Step-by-step explanation:
Answer:
Step-by-step explanation:
(a) The function ...

can be evaluated for x=-2√2 to get ...

The point (-2√2, 1) is on the graph of f(x).
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(b) Likewise, we can evaluate for x=2:

The point on the graph is (2, 0.8).
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(c) From part (a), we know that f(-2√2) = 1. Since the function is even, this means that f(2√2) = 1. The graph is a maximum at those points, so there are no other values for which f(x) = 1.
The points (±2√2, 1) are on the graph.
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(d) There are no values of x for which f(x) is undefined. The domain is all real numbers.
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(e) The only x-intercept is at the origin, (0, 0). The x-axis is a horizontal asymptote.
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(f) The only y-intercept is at the origin, (0, 0).