hello :<span>
<span>an equation of the circle Center at the
A(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : a =0 and b = 0 (Center at the origin)
r = OP....p(-8,3)
r² = (OP)²
r² = (-8-0)² +(3-0)² = 64+9=73
an equation of the circle that satisfies the stated conditions.
Center at the origin, passing through P(-8, 3) is : x² +y² = 73</span></span>
2(3x+4y) = 2(24)
6x+8y = 48
Add both the equations
6x + 8y = 48
+ x - 8y = -20
—————————
7x = 28
x = 28/7
x = 4
Plug in this value in the second equation
(4) - 8y = -20
-8y = -24
y = -24/-8
y = 3
Therefore:
x = 4
y = 3
For m it’s 1,1/3and -5/3 you will get different values when plug the different values of n
Answer:
2 solutions:
x=5 and y=7
x=-1 and y=-5
Step-by-step explanation:
y=2x-3
y=x^2-2x-8
y=y so:
2x-3=x^2-2x-8
simplifying to:
x^2-4x-5=0
factoring to:
(x-5)(x+1)=0
x=5 or x=-1
if x=5 then y=2(5)-3=7
if x=-1 then y=2(-1)-3=-5
