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natima [27]
3 years ago
6

The triangle shown on the graph above is rotated 90 degrees clockwise about the original to form triangle P’Q’R which of the fol

lowing are the (x,y) coordinates of the point P’

Mathematics
1 answer:
azamat3 years ago
4 0

Hey there! I'm happy to help!

When rotating a point 90 degrees clockwise about the origin, our original point (x,y) becomes (-y,x), because it is now at a negative y-value.

We see that our point P is at (1,2). We can use this rotation formula to find the coordinates of P' (the new spot where P is)/

(x,y)⇒(-y,x)

(1,2)⇒(-2,1)

Therefore, the coordinates of the point P' are (-2,1).

Have a wonderful day! :D

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Solve the inequality for V.<br><br> -8.3 _&gt;_ V - 1.9
Arada [10]
First keep in mind that the given value is negative and that it is greater than or equal to whatever '<em>v</em>' is.

-8.3  \geq v - 1.9

When solving for a variable in any equation, you do something to both sides in order to keep it equal. Here, <em>v</em> is being subtracted by 1.9; therefore we can add 1.9 to both sides in order to isolate <em />the variable.

-8.3 + 1.9  \geq v - 1.9 + 1.9
-6.4 \geq v

Despite not needing a value for the question, it is worth noting that since this is an inequality, <em>v </em>can be any value from -6.4 to ∞ in order to make it true.
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3 years ago
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
4x = 2x + 6<br> What’s the solution?
Flura [38]
Answer:

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Explanation:

4x = 2x +6

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meriva
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