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madreJ [45]
3 years ago
15

Which of the following equations could be the result of using the comparison method to solve the system shown?

Mathematics
2 answers:
faltersainse [42]3 years ago
7 0

Answer:

Option (2) is correct.

Step-by-step explanation:

 Given : two a system of linear equation as

   x+ y = 5 and

   2x+ y = 7

We have to choose out of given options  that could be the result of using comparison method to solve the given system .

COMPARISON METHOD

In comparison method we find a value for the single variable in term of others in given equations and compare the two equations.

Here, we first find the value of y in both equation and then compare,

Thus,  x+ y = 5 ...........(1)

          2x+ y = 7  .........(2)

From (1) , y = 5- x  ..........(3)

And from (2) , y = 7- 2x ......(4)

Now, comparing (3) and (4) , we have,

5- x = 7- 2x

Thus,  option (2) is correct.

HACTEHA [7]3 years ago
7 0

Answer: 5-x=7-2x

Step-by-step explanation:

Given equations: x+y=5\\\Rightarrow\ y=5-x..........(1)\\2x+y=7\\\Rightarrow\ y=7-2x.........................(2)

Comparing equation (1) from equation (2), we get

5-x=7-2x

  • Comparison method is a method for solving systems of independent equations by starting rewriting both equations with the same variable and then compare.

Hence, 5-x=7-2x is the  result of using the comparison method to solve the system shown.

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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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2 years ago
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