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alexandr402 [8]
2 years ago
9

I) Domain, ii) x-intercept, iii) y-intercept, iv) vertical asymptote, and v) horizontal asymptote of

Mathematics
1 answer:
Montano1993 [528]2 years ago
7 0

Answer:

i)

Step-by-step explanation:

i) you can't divide by 0 and when you put x=5 into the bottom part 4x-20 = 4(5)-20 = 20-20 = 0 so x cannot be 5. So the domain is x > 5 and x < 5 OR in other words, when x is not 5.

ii) substitute y=0 into the equation and solve the quadratic on top

iii) substitute x=0 into the equation and write down the value

horizontal asymptote at x=5 because we found out that x cannot be 5 in part i)

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Identify the area of segment MNO to the nearest hundredth. HELP PLEASE!! I don't understand it!
katrin2010 [14]

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.

__

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

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