X+y+5=0
y=-x-5
If a solution exists y=y so we can say
x^2-9x+10=-x-5 add x+5 to both sides
x^2-8x+15=0 now factor
x^2-3x-5x+15=0
x(x-3)-5(x-3)
(x-5)(x-3) so x=3 and 5, using y=-x-5
y(3)=-8 and y(5)=-10
So the two solutions are:
(3,-8) and (5,-10)
Step-by-step explanation:
15t + 30 = 20t | -15t
30 = 5t
t = 30/5 = 6
20 + 5t = -6t + 86 | + 6t
20 + 11t = 86 | -20
11t = 66
t = 6
18t + 20 = 24t + 50 | -18t
20 = 6t + 50 | -50
-30 = 6t
t = -30/6 = -5
18t - 2 = 10t + 54 | -10t
8t - 2 = 54 | +2
8t = 56
t = 56/8 = 7
7t - 5 = 15t + 91 | -7t
-5 = 8t + 91 | -91
-96 = 8t
t = -96/8 = -12
-4(-5) = -4*-5 = 20
6 - (-4) = 10
Quotient = 20 /10 = 2 answer
So remember that in a quadratic equation there is a type of quadratic equation of the form ax^2+bx+c=0 so to find what values of a, b, and c you need for the quadratic formula, you need to look at your equation and label a, b, and c. In this equation a=6, b=5 and c=-4. So for the quadratic formula you would use a=6, b=5 and C=-4. I hope this helps!!!!!
From the following figure:
Because the center angle measure 45 degrees, we must rotate 3 times 45 degrees to get point D.
In other words, we must rotate

that is 135 degrees clockwise.