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Nonamiya [84]
3 years ago
14

Will mark brainliest and give 20 points! In what ways can vertical, horizontal, and oblique asymptotes be identified? Please use

your own example to identify.

Mathematics
1 answer:
Reil [10]3 years ago
3 0
Vertical asymptote:
When you have a rational expression in which the denominator is zero, you have a vertical asymptote. So to find vertical asymptotes, just set the denominator of your rational expression equal to zero, and then, solve for x:
\frac{x-1}{x-3}
Set the denominator equal to zero:
x-3=0
Solve for x:
x=3 is the vertical asymptote of our rational expression.

Horizontal asymptote:
Here we have two scenarios.
1) Is the degree of the denominator is higher than the degree of the numerator, you will have a horizontal asymptote at y=0:
y= \frac{x-1}{x^{2}+3}
Since the degree of the denominator is higher of the degree of the numerator, our rational expression will have an asymptote at y=0
2) If the degree of both denominator and numerator is the same, the rational expression will have an horizontal asymptote at the ratio of the leading coefficients:
\frac{3x^{2}+5}{2x^2-3x+1}
Leading coefficients: 3 and 2
Ratio of leading coefficients:
\frac{3}{2}. Our rational expression will have an horizontal asymptote at y= \frac{3}{2}

Oblique asymptote:
If the degree of the numerator is higher than the degree of the numerator, you will have an oblique asymptote. To find it, we are going to perform long division; the quotient (without the remainder) will be the equation of the oblique asymptote line:
\frac{x^2+5x+2}{x+1}
The quotient of the long division is x-1 with a remainder of 2; therefore, the equation of the oblique asymptote line will be:
y=x+4

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The radius of a right circular cylinder is increasing at the rate of 7 in./sec, while the height is decreasing at the rate of 6
Arlecino [84]

Answer:

\approx \bold{6544\ in^3/sec}

Step-by-step explanation:

Given:

Rate of change of radius of cylinder:

\dfrac{dr}{dt} = +7\ in/sec

(This is increasing rate so positive)

Rate of change of height of cylinder:

\dfrac{dh}{dt} = -6\ in/sec

(This is decreasing rate so negative)

To find:

Rate of change of volume when r = 20 inches and h = 16 inches.

Solution:

First of all, let us have a look at the formula for Volume:

V = \pi r^2h

Differentiating it w.r.to 't':

\dfrac{dV}{dt} = \dfrac{d}{dt}(\pi r^2h)

Let us have a look at the formula:

1.\ \dfrac{d}{dx} (C.f(x)) = C\dfrac{d(f(x))}{dx} \ \ \ (\text{C is a constant})\\2.\ \dfrac{d}{dx} (f(x).g(x)) = f(x)\dfrac{d}{dx} (g(x))+g(x)\dfrac{d}{dx} (f(x))

3.\ \dfrac{dx^n}{dx} = nx^{n-1}

Applying the two formula for the above differentiation:

\Rightarrow \dfrac{dV}{dt} = \pi\dfrac{d}{dt}( r^2h)\\\Rightarrow \dfrac{dV}{dt} = \pi h\dfrac{d }{dt}( r^2)+\pi r^2\dfrac{dh }{dt}\\\Rightarrow \dfrac{dV}{dt} = \pi h\times 2r \dfrac{dr }{dt}+\pi r^2\dfrac{dh }{dt}

Now, putting the values:

\Rightarrow \dfrac{dV}{dt} = \pi \times 16\times 2\times 20 \times 7+\pi\times 20^2\times (-6)\\\Rightarrow \dfrac{dV}{dt} = 22 \times 16\times 2\times 20 +3.14\times 400\times (-6)\\\Rightarrow \dfrac{dV}{dt} = 14080 -7536\\\Rightarrow \dfrac{dV}{dt} \approx \bold{6544\ in^3/sec}

So, the answer is: \approx \bold{6544\ in^3/sec}

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Step-by-step explanation:hope this helps :)

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what is that I think that is a fraction?

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