Which two points satisfy y = -x^2 + 2x + 4 and x + y = 4?

Two points that satisfy the system propose are P1  = (3, 1) and P2 = (0, 4).

Step-by-step explanation:

Despite this equation system could be solve in different ways, as many things in Math do, we can take advantage of the negative symbol of equation   y = -x^{2}+ 2x + 4. Yes, instead of solving this using the quadratic formula, and doing so many other steps, and with the high probability of making a mistake observing that , we can solve this system using the well known logical formula that if x = b and y = b, then x = y.

This equation system could be solve in the following way:

First, we have these equations:

 ( 1 )

y = -x^{2}+ 2x + 4   ( 2 )

Equation ( 2 ) can be rewritten as:

-y = x^{2} - 2x - 4;

And Equation ( 1 ) can be rewritten as:

x + y = 4; y = 4 - x, but also  -y = x - 4;

So,

-y = x^{2} - 2x - 4, and

-y = x - 4;  

That is, the logical formula previously described above.

If -y = x^{2} - 2x - 4, and

  -y = x - 4;  then,

x^{2} - 2x - 4 = x - 4 .

x^{2} - 2x - 4 -( x - 4 ) = 0 .

x^{2} - 2x - 4 -x + 4 ) = 0 .

x^{2} - 2x -x + 4 - 4  = 0, where ( 4 - 4 = 0 ), that is

x^{2} - 3x = 0;

x^{2} - 3x = 0, where x is a common factor, so

x * (x - 3)   = 0.

For this product to be true,

x  = 0  or

x - 3  = 0  ⇒    x  =  3 .

Having these values for x (two values, x = 0 and x = 3), we can find the other values for y, that is:

We know that  x + y = 4 (Equation 1), then

For x = 3;

3 + y = 4 , and

y = 4 - 3 = 1;  y = 1 .

So,  P1 = ( 3, 1 );

For x = 0;

0 + y = 4 or y = 4 .

So, P2 = ( 0, 4 ).

Using equations ( 1 ) and ( 2 ):

For P1 = ( 3, 1 ),

3 + 1 = 4 ( 1 )

1 = -(3)^{2} + 2*(3) + 4  ( 2 ) or

1 = -(9) + 6 + 4 = -9 + 10 or

1 = 1 .

For P2 = ( 0, 4 ),

0 + 4 = 4 ( 1 )

4 = -(0)^{2} + 2*(0) + 4 ( 2 ) or

4 = 0 + 0 + 4 = 4 or

4 = 4 .

12m

step-by-step explanation:

square root means have of a number so have of 3 is 1.5 then multiply it by 8 and you get your answer!

answer: b f g .

step-by-step explanation: