From the first equation,
x+5 = 3(y+5)
x = 3y + 15 - 5
Now substitue x in the second equation with (3y +15 - 5).
x-5 = 7(y-5)
(3y+15-5) - 5 = 7(y-5)
3y +5 = 7y - 35
-4y = - 40
y = 10
Since y is 10, and x is (3y +15 - 5),
x = 30 + 15 - 5 = 40
Answer:
x^2 + y^2 + 16x + 6y + 9 = 0
Step-by-step explanation:
Using the formula for equation of a circle
(x - a)^2 + (y + b)^2 = r^2
(a, b) - the center
r - radius of the circle
Inserting the values given in the question
(-8,3) and r = 8
a - -8
b - 3
r - 8
[ x -(-8)]^2 + (y+3)^2 = 8^2
(x + 8)^2 + (y + 3)^2 = 8^2
Solving the brackets
( x + 8)(x + 8) + (y +3)(y+3) = 64
x^2 + 16x + 64 + y^2 + 6y + 9 = 64
Rearranging algebrally,.
x^2 + y^2 + 16x + 6y + 9+64 - 64 = 0
Bringing in 64, thereby changing the + sign to -
Therefore, the equation of the circle =
x^2 + y^2 + 16x + 6y + 9 = 0
Answer:
b the answer is b
Step-by-step explanation:
idk what they mean but i got an 100 and i had this questin so a
We always start from right side in order of operation in any equation
we can set up equation as
step-1:
x divided by three
we can write as
step-2:
six more than x divided by three
so, we get
step-3:
y is six more than x divided by three
so, we get
and this is in y=mx+b form
so, we have
...............Answer
