Question

I need help with these pleaaaaseeee

Answer 1

Problem 1

Answer: Choice D

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Step-by-step explanation:

The center is (0,0), so this means h = 0 and k = 0.

The transverse axis runs through the two vertices. Since it's vertical, this means we will go with this general template.

`-((x-h)^2)/(a^2)+((y-k)^2)/(b^2) = 1\\\\`

The fraction with the positive in front will indicate if we have a horizontal or vertical transverse axis (x = horizontal, y = vertical).

So,

`-((x-h)^2)/(a^2)+((y-k)^2)/(b^2) = 1\\\\-((x-0)^2)/(8^2)+((y-0)^2)/(9^2) = 1\\\\-(x^2)/(64)+(y^2)/(81) = 1\\\\(y^2)/(81)-(x^2)/(64) = 1\\\\`

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Problem 2

Answer: 28 Megameters (choice B)

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Step-by-step explanation:

Let's divide both sides by 176,400 to get that right hand side equal to 1.

`225x^2 + 576y^2 = 176,400\\\\(225x^2 + 576y^2)/(176,400) = (176,400)/(176,400)\\\\(225x^2)/(176,400) + (576y^2)/(176,400) = 1\\\\(x^2)/(784) + (y^2)/(306.25) = 1\\\\(x^2)/(28^2) + (y^2)/((17.5)^2) = 1\\\\`

The last equation is in the form
`(x^2)/(a^2) + (y^2)/(b^2) = 1\\\\` which represents an ellipse centered at the origin with semimajor and semiminor axis lengths of 'a' and 'b'. The larger of the 'a' and b values tells us which is the semimajor axis length. In this case, that would be a = 28. This is the distance from the center to the ellipse, and this is the longest such distance to travel. Therefore, you picked the correct answer in 28 Megameters.

This converts to 28000 km which is roughly 17398.393 miles. To get a sense of how big this is, the circumference of the earth is roughly 24901.461 miles around the equator.

Side note: the moon is at the apogee point when it is the farthest away from the planet.

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Problem 3

Answer: Choice D

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Step-by-step explanation:

The general ellipse template is

`((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) = 1\\\\`

where (h,k) is the center and a,b describe half the lengths of the horizontal and vertical axis respectively.

The major axis is vertical, which indicates that b > a. In other words, the denominator for the y term will be larger than the denominator for the x term. Your teacher mentions a = 7 and b = 5, but I think it's better to keep the 'a' always with x so you don't mix things up. Though of course that's just my opinion. Anyways, so because b > a, we'll have b = 7 and a = 5.

Plugging those in, along with the given center, and we'll get this:

`((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) = 1\\\\((x-1)^2)/(5^2) + ((y-(-3))^2)/(7^2) = 1\\\\((x-1)^2)/(25) + ((y+3)^2)/(49) = 1\\\\((y+3)^2)/(49) + ((x-1)^2)/(25) = 1\\\\`

We arrive at choice D as the final answer.