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Mamont248 [21]
3 years ago
8

The symmetry of a hyperbola with a center at (h, k) only occurs at y = k. True or False

Mathematics
1 answer:
Vikki [24]3 years ago
6 0
A hyperbola with a center at (0, 0) can be defined  as x²/a² − y²/b² = ±1.<span> 
</span>The statement "<span>The symmetry of a hyperbola with a center at (h, k) only occurs at y = k" </span>is false, because a hyperbola have many different orientations.
It doesn't have to be symmetric about the lines y = k or x = h.
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Step-by-step explanation:

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this is a quadratic equation, and its points of f(t) = 0 can be calculated. there are usual 2 such solutions for a quadratic equation.

so, we need to bring it to a form that ends in "= 0".

-16t² + 64t - 60 = 0

that is also the same as

-4t² + 16t - 15 = 0

the formula for the solutions of such a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

a = -4

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so,

t = (-16 ± sqrt(256 - 4×-4×-15))/(2×-4) =

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t1 = (-16 + 4)/-8 = -12/-8 = 3/2

t2 = (-16 - 4)/-8 = -20/-8 = 5/2

so, it depends on what the unit of time t is. let's assume seconds.

therefore, the rope will reach the top of the building while going up after 3/2 = 1 1/2 = 1.5 seconds.

and then, as the shot rope makes a curve and comes back down again (that is why we have 2 solutions : the rope will reach the height of 120ft during its flight path twice : once while still going up, and once when coming back down again) it will be at the height of the top of the building after 5/2 = 2 1/2 = 2.5 seconds.

so, it depends if the people there can grab it at the first chance, and if the path of the rope would allow them to grab it also again on its way back down. this we don't know.

I therefore guess, that your teacher is aiming for the first solution.

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