To solve using completing square method we proceed as follows:
x^2-10x+8=0
x^2-10x=-8
but
c=(b/2)^2
c=(10/2)^2=25
thus we can add this in our expression to get
x^2-10x+25=8+25
factorizing the LHS we get:
(x-5)(x-5)=33
(x-5)^2=33
getting the square roots of both sides we have:
x-5=+/-√33
x=5+/-√33
Y-y1=m(x-x1)
y-5=10(x-1)
Y-5=10x-10
Y=10x-5
The answer to your question is
No
If RS is the hypotenuse of the triangle RST and point T is in Quadrant 3, then point T must be the intersection of the lines: x = - 4 and y = - 5.
Therefore, the coordinates of point T are ( x, y ) = ( - 4, - 5 )
Answer:
T ( - 4, - 5 )
Answer: You can multiply the top equation by -1 to eliminate the x variable.
And the solution is (2,4/3) in case you need it.
Step-by-step explanation:
2x + 3y = 8
2x + 6y = 12
If you multiply the upper equation or down equation by one, you will be able to eliminate the x variable.
-1( 2x + 3y) = -1(8) New equation: -2x -3y = -8.
Add the new equation you got by multiplying the top equation by -1 to the bottom equation.
Add them: -2x -3y = -8
2x + 6y = 12
3y = 4
y = 4/3
You can now input the value for y into the one of the equations and solve for x.
-2x - 3(4/3) = -8
-2x -4 = -8
+4 +4
-2x = -4
x = 2
