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3 years ago

(30PTS) What are the slopes of the asymptotes of a hyperbola with equation x^2/2^2-y^2/6^2=1

Mathematics

2 answers:

Akimi4 [234]3 years ago

4 0

Answer:

2x2−3y2=6or2x26−3y26=1 or. x23−y22=1or(x−0)2(√3)2−(y−0)2(√2)2=1. This is standard form of the equation of a hyperbola with center.

Step-by-step explanation:

c2 = a2 + b2. Every hyperbola has two asymptotes. A hyperbola with a horizontal transverse axis and center at (h, k) has one asymptote with equation y = k + (x - h) and the other with equation y = k - (x - h).

Crazy boy [7]3 years ago

4 0

Answer:

(3,-3)

Step-by-step explanation:

edge 2020 2021

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sp2606 [1]

<h3>The first time when the ball will reach a height of 20 feet is 0.42 seconds</h3>

Solution:

Given that,

The height of a ball thrown into the air after t seconds have elapsed is:

$h = -16t^2 + 40t + 6$

What is the first time, t, when the ball will reach a height of 20 feet?

Substitute h = 20

$20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0$

Solve by quadractic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=8,\:b=-20,\:c=7$

$t = \frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:8\cdot \:7}}{2\cdot \:8}\\\\t = \frac{20 \pm \sqrt{176}}{16}\\\\t = \frac{20 \pm 4\sqrt{11}}{16}\\\\t = \frac{ 5 \pm \sqrt{11}}{4}\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084$

Rounding off we get,

t = 2.08 , t = 0.42

Thus the first time when the ball will reach a height of 20 feet is 0.42 seconds

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