We need to find the center and the radius of
The general circle equation is the following
where
(h,k) is the center and
r is the radius
1. rearrange the equation
2. Add 25 on both sides
3. Factor
Now we have an equation that is very similar to the circle equation, so let's compare them
Center -> (h,k) = (5,-11)
radius -> r = 5
Use y = mx + b to determine the equation of the graph, then decided the inequality symbol.
m = +5/+6 = b = 5
y = x + 5
6y = 5x + 30
-30 = 5x - 6y ⇒ 5x - 6y = -30
Now let's look at the inequality symbol: The shaded area is below so it will be a less than symbol and it is a solid line so the equal sign is included.
Answer: 5x - 6y ≤ -30
Answer:
System A has 4 real solutions.
System B has 0 real solutions.
System C has 2 real solutions
Step-by-step explanation:
System A:
x^2 + y^2 = 17 eq(1)
y = -1/2x eq(2)
Putting value of y in eq(1)
x^2 +(-1/2x)^2 = 17
x^2 + 1/4x^2 = 17
5x^2/4 -17 =0
Using quadratic formula:
a = 5/4, b =0 and c = -17
Finding value of y:
y = -1/2x
System A has 4 real solutions.
System B
y = x^2 -7x + 10 eq(1)
y = -6x + 5 eq(2)
Putting value of y of eq(2) in eq(1)
-6x + 5 = x^2 -7x + 10
=> x^2 -7x +6x +10 -5 = 0
x^2 -x +5 = 0
Using quadratic formula:
a= 1, b =-1 and c =5
Finding value of y:
y = -6x + 5
y = -6(\frac{1\pm\sqrt{19}i}{2})+5
Since terms containing i are complex numbers, so System B has no real solutions.
System B has 0 real solutions.
System C
y = -2x^2 + 9 eq(1)
8x - y = -17 eq(2)
Putting value of y in eq(2)
8x - (-2x^2+9) = -17
8x +2x^2-9 +17 = 0
2x^2 + 8x + 8 = 0
2x^2 +4x + 4x + 8 = 0
2x (x+2) +4 (x+2) = 0
(x+2)(2x+4) =0
x+2 = 0 and 2x + 4 =0
x = -2 and 2x = -4
x =-2 and x = -2
So, x = -2
Now, finding value of y:
8x - y = -17
8(-2) - y = -17
-16 -y = -17
-y = -17 + 16
-y = -1
y = 1
So, x= -2 and y = 1
System C has 2 real solutions