320 is on the Fourth quadrant:)
Answer:
C = (2,2)
Step-by-step explanation:
B = (10 ; 2)
M = (6 ; 2)
C = (x ; y )
|___________|___________|
B (10;2) M (6;2) C ( x; y)
So:
dBM = dMC
√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 6)^2]
(2-2)^2 - (6-10)^2 = (y-2)^2 + (x - 6)^2
0 + (-4)^2 = (y-2)^2 + (x - 6)^2
16 = (y-2)^2 + (x - 6)^2
16 - (x - 6)^2 = (y-2)^2
Also:
2*dBM = dBC
2*√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 10)^2]
4*[(0)^2 + (-4)^2] = (y-2)^2 + (x - 10)^2
4*(16) = (y-2)^2 + (x - 10)^2
64 = (y-2)^2 + (x - 10)^2
64 = 16 - (x - 6)^2 + (x - 10)^2
48 = (x - 10)^2 - (x - 6)^2
48 = x^2 - 20*x + 100 - x^2 + 12*x - 36
48 = - 20*x + 100 + 12*x - 36
8*x = 16
x = 2
Thus:
16 - (x - 6)^2 = (y-2)^2
16 - (2 - 6)^2 = (y-2)^2
16 - (-4)^2 = (y-2)^2
16 - 16 = (y-2)^2
0 = (y-2)^2
0 = y - 2
2 = y
⇒ C = (2,2)
Answer:
Step-by-step explanation:
xy = 42
x+y = - 2 Substitute into the top equation
y = -2 - x Put in for y
x(-2 - x) = 42 Remove the brackets
-2x - x^2 = 42 Subtract 42 from both sides.
-2x - x^2 - 42 = 0 Put in the more normal order.
-x^2 - 2x - 42 = 0 Multiply by -1
x^2 + 2x + 42 = 0
This cannot be factored. It gives complex roots as it is written. I will give you the answer but I kind of doubt the question is correct.
x1 = - 1 + 6.40i
x2 = -1 - 6.40i
Leave a comment if you have a correction.
Answer:
idk
Step-by-step explanation:
we are nt on that yet
