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creativ13 [48]
3 years ago
9

A number is called squarefree if it is not divisible by the square of a positive integer greater than one. find the number of sq

uare free positive integers less than 100.
Mathematics
1 answer:
german3 years ago
3 0
Count the number of positive integers less than 100 that do not contain any perfect square factors greater than 1.

Possible perfect squares are the squares of integers 2-9.
In fact, only squares of primes need be considered, since for example, 6^2=36 actually contains factors 2^2 and 3^2.
Tabulate the number (in [ ])of integers containing factors of 
2^2=4: 4,8,12,16,...96 [24]
3^2=9: 9,18,....99  [11] 
5^2=25: 25,50,75  [3]
7^2=49:  49,98 [2]

So the total number of integers from 1 to 99
N=24+11+3+2=40
=>
Number of positive square-free integers below 100 = 99-40 = 59
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Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +
Veronika [31]

Answer:

Verified

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

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

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}

                           

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6 0
3 years ago
Solve A plumber gave an estimate for the renovation of a kitchen. His hourly pay is S28 per hour and the plumbing parts will cos
Vika [28.1K]
He plans to work 14 hours.

Here's the equation:

490-98= 392
392 divided by 28= 14 hours
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Trying to figure out how many 15pt into cups?
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