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3 years ago

What is the equation of a hyperbola with a = 9 and c = 12? Assume that the transverse axis is horizontal

1 answer:

4 0

The general equation of a hyperbola with a horizontal transverse axis is defined as:
x²/a² - y²/b² = 1

Solving for b², we use the formula: a² + b² = c²
b² = 12² - 9² = 63

Equation of our hyperbola will be:
x²/81 - y²/63 = 1

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shutvik [7]

Given:

The limit problem is:

$\lim_{x\to 3}(x^2+8x-2)$

To find:

The limit of the function by using direct substitution.

Solution:

We have,

$\lim_{x\to 3}(x^2+8x-2)$

Applying limit, we get

$\lim_{x\to 3}(x^2+8x-2)=(3)^2+8(3)-2$

$\lim_{x\to 3}(x^2+8x-2)=9+24-2$

$\lim_{x\to 3}(x^2+8x-2)=33-2$

$\lim_{x\to 3}(x^2+8x-2)=31$

Therefore, the correct option is D.

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Talja [164]

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Step-by-step explanation:

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Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

3 years ago