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Gennadij [26K]
3 years ago
7

1. Identify the vertical asymptotes of

Mathematics
1 answer:
kaheart [24]3 years ago
7 0
1. x1/2=\frac{-3\pm \sqrt{9+40} }{2}
VA(the vertical asymptotes): x=-5,  x=2
2. x1/2=\frac{9\pm \sqrt{81-72} }{2}
VA: x=3, x=6
3.  \lim_{n \to \infty}  \frac{4x}{7}=∞
There is no HA.
4. \lim_{n \to \infty} \frac{7x+1}{2x-9}= \frac{7}{9}
HA : y=7/2
5. \lim_{n \to \infty}  \frac{ x^{2} +5x+3}{4x-1} =∞
There is no HA.
6. m=\lim_{n \to \infty}  \frac{ x^{2} -4x+8}{ x^{2} +2x}
m = 1
b=\lim_{n \to \infty}  \frac{ x^{2} -4x+8- x^{2} -2x}{x+2}
b = -6
OA: y = x - 6
7. Same method:
m = 2
b = 0
OA: y = 2x
8. m = 0
There is no OA.
m = 4 
b=-1
OA: y=4x-1

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Ray : Line NA, Line NB, Line AB

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B)Relative error = 0.0575

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A) Formula for circumference is: C = 2πr

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r is small, so we can write;

ΔC/Δr = 2π

So, Δr = ΔC/2π

We are told that ΔC = 0.5.

Thus; Δr = 0.5/2π = 0.25/π

Now, formula for Volume of a sphere is;

V(r) = (4/3)πr³

Differentiating with respect to r, we have;

dV/dr = 4πr²

Again, r is small, so we can write;

ΔS/Δr = 4πr²

ΔV = 4πr² × Δr

Rewriting, we have;

ΔV = ((2πr)²/π) × Δr

Since C = 2πr, we now have;

ΔV = (C²/π)Δr

ΔV will be maximum when Δr is maximum

Thus, ΔV = (C²/π) × 0.25/π

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Thus;

ΔV = (82²/π) × 0.25/π

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Relative error = 170.32/((4/3)πr³)

Relative error = 170.32/((4/3)C³/8π³)

Relative errror = 170.32/((4/3)82³/8π³)

Relative error = 170.32/2963.744

Relative error = 0.0575

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