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HACTEHA [7]
3 years ago
13

Nola hiked down a trail at a steady rate for 10 minutes. Her change in elevation was -200 ft. Then she continued to hike down fo

r another 20 minutes at a different rate. Her change in elevation for this part of the hike was -300 ft. During which portion of did she walk down at a faster rate? Explain your reasoning.
Mathematics
1 answer:
umka21 [38]3 years ago
7 0
Her first rate was faster. The 10 minutes with a change in elevation of -200ft.
In order to solve this you need to have the same minutes. In this problem, I just simply took the 10 minutes and the -200ft and doubled it. Now both rates have 20 minutes. The first rate would have gone -400ft in 20 minutes while the second rate was only -300ft in 20 minutes. I hope this wasn't confusing, and I'm willing to explain more.
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Answer:

A. There is a focus at (0,−10).

Step-by-step explanation:

Assume the hyperbola is like the one below.

The hyperbola is vertical and centred on the y-axis, so its general equation is

\dfrac{y^{2}}{a^{2}} - \dfrac{x^{2}}{b^{2}} = 1

The vertices of your parabola are (0,±8) so a = 8.

The covertices are (±6,0), so b = 6.

Calculate c

\begin{array}{rcl}a^{2} + b^{2} & = & c^{2}\\8^{2} + 6^{2} & = & c^{2}\\64 + 36 & = & c^{2}\\100 & = & c^{2}\\c & = & \mathbf{10}\\\end{array}

A. Foci

The foci are at (0, ±c) = (0, ±10)

TRUE. There is a focus at (0,-10).

B. Foci

The foci are at (0,±10).

False. There is no focus at (0,12)

C. and D. Asymptotes

The equations for the asymptotes are

y = \pm\dfrac{a}{b}x = \pm\dfrac{8}{6}x = \pm\dfrac{4}{3}x

So, y = ±x are not asymptotes.

False.

E. and F. Directrices

The equations of the directrices are  

y = ±a²/c = ±64/10 = ±6.4

y = 6.4 is a directrix.

E is false. x = cannot be a directrix

F is uncertain. Your equation for the directrix is incomplete.

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Answer:

x = 32

y = 16sqrt(5)

z = 8sqrt(5)

Step-by-step explanation:

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x/16 = 2

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y^2 = 16^2 + 32^2

y^2 = 1280

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Dylan purchased a new car in 1993 for $26,300. The value of the car has been depreciating exponentially at a constant rate. If t
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The predicted value of the car in the year 2006 to the nearest dollar would be  $651.

<h3>What is the predicted value of the car?</h3>

The first step is to determine the rate of depreciation

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