Y= ax+b
when x=3, y=11 and when x= -2, y=11
3a+b=11
-2a+b=11 | * -1
3a+b=11
2a-b=-11
------------
5a=0, a=0:5, a=0
⇒ b=11
y=11 for every x
So ( 1,11) or (2,11) are also points on the graph
Answer:
The answer in the procedure
Step-by-step explanation:
we have
2x-3y=-1 ----> equation A
3x+3y=26 --> equation B
Solve the system by elimination
Adds equation A and equation B
2x-3y=-1
3x+3y=26
----------------
2x+3x=-1+26
5x=25
x=25/5
x=5
Find the value of y
substitute the value of x in equation A or equation B and solve for y
2(5)-3y=-1
10-3y=-1
3y=10+1
y=11/3
<span>i Ace</span>hello :
let : A(-6,6) B(6,-2)
the center is w((-6+6)/2 , (6-2)/2)....(midel <span>[ AB]
w(0 ,2)
the ridus is : r = AB/2
AB = </span>√(-6-6)²+(6+2)² = √(144+64) =<span>√208/2
</span>an equation in <span>standard form equation of the circle :
</span>(x+0)²+(y-2)² = (√208/2)²=208/4 = 52
