Given data:
The first set of equations are x+y=4, and x=6.
The second set of equations are 3x-y=12 and y=-6.
The point of intersection of first set of te equations is,
6+y=4
y=-2
The first point is (6, -2).
The point of intersection of second set of te equations is,
3x-(-6)=12
3x+6=12
3x=6
x=2
The second point is (2, -6).
The equation of the line passing through (6, -2) and (2, -6) is,
Thus, the required equation of the line is y=x-8.
Answer:
x=7
Step-by-step explanation:
3(2x - 5) = 9(10 - x)
6x-15=90-9x
6x+9x=90+15
15x=105
x=7
Discriminant = b^2 - 4ac, where a, b and c come from the form of the quadratic equation as ax^2 + bx + c
Discriminant = (4)^2 - 4(1)(5)
= 16 - 20
= -4
-4 < 0, therefor there are no roots
(If the discriminant = 0, then there is one root
If the discriminant > 0, there are two roots, and if it is a perfect square (eg. 4, 9, 16, etc.) then there are two rational roots
If the discriminant < 0, there are no roots)
Sarah
reason
she has the most consistent scores
Answer:
x=5
Step-by-step explanation:
We can write ratios to help us solve
4x 12 +3
---------- = ------------
3x+1 12
Using cross products
4x*12 = (3x+1) * (15)
48x = 45x +15
Subtract 45x from each side
3x =15
Divide by 3
3x/3 =15/3
x=5
