Multiply the first equation by -2. It turns into 4x+18y=50 and the other one stays -4x-9y=-23. Then you "eliminate" the equations by "adding" them. Set it up like this:
4x+18y=50
+
-4x-9y=-23
4x-4x= 0. 18y-9y= 9y and 50-23= 27
SO: 9y= 27 and y= 3
And then you plug that in to one equation: 4x +18(3)= 50. 4x= 50-(18*3)
4x= -4, so x=-1.
Plug it back in to check!
Hope this helps. Happy solving! :)
Answer:
y = -1/4x^2 - x + 3
Step-by-step explanation:
(-6,0)
(-4,3)
(2,0)
36a - 6b + c = 0
16a -4b + c = 3
4a + 2b + c = 0
c = -4a - 2b
36a - 6b - 4a -2b = 0
16a - 4b -4a - 2b = 3
32a -8b = 0
12a - 6b = 3
4a - b = 0
4a - 2b = 1
b = -1
4a + 1 = 0
4a = -1
a = -1/4
c = 1 + 2 = 3
y = -1/4x^2 - x + 3
Find m∠BOC, if m∠MOP = 110°.
Answer:
m∠BOC= 40 degrees
Step-by-step explanation:
A diagram has been drawn and attached below.
- OM bisects AOB into angles x and x respectively
- ON bisects ∠BOC into angles y and y respectively
- OP bisects ∠COD into angles z and z respectively.
Since ∠AOD is a straight line
x+x+y+y+z+z=180 degrees

We are given that:
m∠MOP = 110°.
From the diagram
∠MOP=x+2y+z
Therefore:
x+2y+z=110°.
Solving simultaneously by subtraction

x+2y+z=110°.
We obtain:
x+z=70°
Since we are required to find ∠BOC
∠BOC=2y
Therefore from x+2y+z=110° (since x+z=70°)
70+2y=110
2y=110-70
2y=40
Therefore:
m∠BOC= 40 degrees
Answer:




Step-by-step explanation:
use circle equation and plug in the centers and radis.