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LenKa [72]
3 years ago
9

How many extraneous solutions does the equation below have? 9 over n squared +1 =n+3 over 4? 0 1 2 3

Mathematics
2 answers:
Annette [7]3 years ago
7 0

Answer:

Option 1 - There is no extraneous solution i.e. 0.      

Step-by-step explanation:

Given : Expression \frac{9}{n^2+1}=\frac{n+3}{4}

To find : How many extraneous solutions does the equation have ?

Solution :

First we solve to expression to determine the extraneous solution,

\frac{9}{n^2+1}=\frac{n+3}{4}

Cross multiply,

9\times 4=(n+3)(n^2+1)

36=n^3+n+3n^2+3

n^3+3n^2+n-33=0

An extraneous solution is defined as a solution, such as that to an equation, that emerges from the process of solving the problem but is not a valid solution to the problem.

The equation form is a cubic function so it has 3 solutions.

Therefore, There is no extraneous solution.

LenKa [72]3 years ago
4 0
The answer is 0. While the answer for n is approximately 2.384503, there are no EXTRANEOUS solutions. Again, the answer is a.) 0
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36

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The circumference of a sphere was measured to be 80 cm with a possible error of 0.5 cm. A) Use differentials to estimate the max
siniylev [52]

Answer:

A) The maximum error in the calculated surface area: 25cm^2

Relative error: 0.013

B) The maximum error in the calculated volume: 162cm^2

Relative error: 0.019

Step-by-step explanation:

A) The formula for the surface area is:

A=4\pi r^2

The measured value is the circumference which is equal to:

C=2\pi r

then the radius is:

r=\frac{C}{2\pi}

Substituting in the formula of the surface:

A=4\pi(\frac{C}{2\pi})^2\\A=4\pi(\frac{C^2}{4\pi^2})\\A=\frac{C^2}{\pi}

Using the formula to calculate the error:

dy=f'(x)dx

Where x is the variable measured and y is a function of x(y=f(x)).

dA=f'(C)dC\\dA=\frac{2C^{(2-1)}}{\pi}dC\\dA=\frac{2C}{\pi}dC

We have C=80cm and dC=0.5cm

dA=\frac{2C}{\pi}dC\\dA=\frac{2(80)}{\pi}(0.5)\\dA=\frac{160}{\pi}(0.5)\\dA=50.9296(0.5)\\dA=25.4648\approx25cm^2

The relative error is the maximum error divide by the total area. The total area is: A=\frac{C^2}{\pi}=\frac{(80)^2}{\pi}=\frac{6400}{\pi}=2037.1833cm^2

\frac{dA}{A}=\frac{25.4648}{2037.1833} =0.0125\approx0.013

B) The formula for the volume is:

V=\frac{4}{3} \pi r^3

Using r=\frac{C}{2\pi}

V=\frac{4}{3} \pi r^3\\V=\frac{4}{3} \pi (\frac{C}{2\pi})^3\\V=\frac{4}{3} \pi (\frac{C^3}{8\pi^3})\\V=\frac{1}{3}(\frac{C^3}{2\pi^2})\\V=\frac{C^3}{6\pi^2}

The maximum error is:

dV=\frac{3C^{3-1}}{6\pi^2}dC\\dV=\frac{C^{2}}{2\pi^2}dC\\dV=\frac{(80)^{2}}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=(324.2278)(0.5)\\dV=162.1139\approx162cm^2

The calculated volume is:

V=\frac{C^3}{6\pi^2}\\V=\frac{(80)^3}{6\pi^2}\\V=\frac{512000}{6\pi^2}\\V=8646.0743

The relative error is:

\frac{dV}{V}=\frac{162.1139}{8646.0743}=0.0188\approx0.019

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