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Afina-wow [57]
2 years ago
8

!!!!!! URGENT PLS HELP ME ANSWER !!!!

Mathematics
1 answer:
laila [671]2 years ago
8 0

Answer:

1) A = 547.1136, C = 82.896

2) A = 678.5226, C = 92.316

3) A = 844.5344, C = 102.992

4) A = 1554.49625, C = 139.73

5) A = 1561.4906, C = 140.044

6) A = 994.8776, C = 111.784

Step-by-step explanation:

1)

A = \Pi (\frac{D}{2})^2 = 3.14 * (\frac{26.4}{2})^2 = 3.14 * 13.2^2 = 547.1136

C = \Pi D = 3.14 * 26.4 = 82.896

2)

A = \Pi r^2 = 3.14*(14.7)^2 = 678.5226

C = 2\Pi r = 14.7*2 * 3.14 = 92.316

3)

A = \Pi (\frac{D}{2})^2 = 3.14 * (\frac{32.8}{2})^2 = 3.14 * 16.4^2 = 844.5344

C = \Pi D = 3.14 * 32.8 = 102.992

4)

A = \Pi (\frac{D}{2})^2 =  3.14 * (\frac{44.5}{2})^2 = 3.14 * 22.25^2 = 1554.49625

C = \Pi D = 44.5*3.14 = 139.73

5)

A = \Pi r^2 = 3.14*22.3^2 = 1561.4906

C = 2\Pi r = 2 * 3.14 * 22.3 = 140.044

6)

A = \Pi r^2 = 3.14 * 17.8^2 = 994.8776

C = 2\Pi r = 2 * 3.14 * 17.8 = 111.784

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Answer:

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Answer:

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Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

                           x^2y' +xy = 3

- A general solution to the above ODE is also given as:

                          y = \frac{3Ln(x) + C  }{x}

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

                          y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x)  }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

                          -\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3

- Therefore, the complete solution to the given ODE can be expressed as:

                        y ( x ) = \frac{3Ln(x) + 3 }{x}

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)

- Therefore, the complete solution to the given ODE can be expressed as:

                        y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}

                           

Download docx
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