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Irina-Kira [14]
3 years ago
13

What is the subset of 2/3??

Mathematics
1 answer:
sineoko [7]3 years ago
3 0

It belongs to the rational numbers.  

It is the ratio of two integers, 2 and 3.

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Find the area of the shaded triangle
emmasim [6.3K]

Answer:

I think its 8. SORRY IF WRONG

Step-by-step explanation:

3 0
3 years ago
Solve the following inequality for t.<br><br> 12.3+t&lt;5.6
luda_lava [24]
The answer in inequality form is t< -6.7
4 0
3 years ago
Read 2 more answers
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
2. Check the boxes for the following sets that are closed under the given
son4ous [18]

The properties of the mathematical sequence allow us to find that the recurrence term is 1 and the operation for each sequence is

   a) Subtraction

   b) Addition

   c) AdditionSum

   d) in this case we have two possibilities

       * If we move to the right the addition

       * If we move to the left the subtraction

The sequence is a set of elements arranged one after another related by some mathematical relationship. The elements of the sequence are called terms.

The sequences shown can be defined by recurrence relations.

Let's analyze each sequence shown, the ellipsis indicates where the sequence advances.

a) ... -7, -6, -5, -4, -3

We can observe that each term has a difference of one unit; if we subtract 1 from the term to the right, we obtain the following term

        -3 -1 = -4

        -4 -1 = -5

        -7 -1 = -8

Therefore the mathematical operation is the subtraction.

b) 0. \sqrt{1}. \sqrt{4}, \sqrt{9}, \sqrt{16}, \sqrt{25}  ...

In this case we can see more clearly the sequence when writing in this way

      0, \sqrt{1^2}. \sqrt{2^2}, \sqrt{3^2 } . \sqrt{4^2} , \sqrt{5^2}

each term is found by adding 1 to the current term,

      \sqrt{(0+1)^2} = \sqrt{1^2} \\\sqrt{(1+1)^2} = \sqrt{2^2}\\\sqrt{(2+1)^2} = \sqrt{3^2}\\\sqrt{(5+1)^2} = \sqrt{6^2}

Therefore the mathematical operation is the addition

c)   ... \frac{-10}{2}. \frac{-8}{2}, \frac{-6}{2}, \frac{-4}{2}. \frac{-2}{2}. ...

      The recurrence term is unity, with the fact that the sequence extends to the right and to the left the operation is

  • To move to the right add 1

           -\frac{-10}{2} + 1 = \frac{-10}{2}  -   \frac{2}{2}  = \frac{-8}{2}\\\frac{-8}{2} + \frac{2}{2} = \frac{-6}{2}

  • To move left subtract 1

         \frac{-2}{2} - 1 = \frac{-4}{2}\\\frac{-4}{2} - \frac{2}{2} = \frac{-6}{2}

         

Using the properties the mathematical sequence we find that the recurrence term is 1 and the operation for each sequence is

   a) Subtraction

   b) Sum

   c) Sum

   d) This case we have two possibilities

  •  If we move to the right the sum
  •  If we move to the left we subtract

Learn more here: brainly.com/question/4626313

5 0
3 years ago
Help me please!!! :(
lana66690 [7]

Answer:

the greatest common factor is 5^2 * 7^3

(I think!!!)

3 0
3 years ago
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