Answer:
A.
Step-by-step explanation:
If
AND
y = x + 7, then by the transitive property of equality:
We can solve for the values of x by getting everything on one side of the equals sign and then solving for x:
We can factor out the common x to get:
x(x + 1) = 0
which tells us by the Zero Product Property that either
x = 0 and/or x + 1 = 0 and x = -1
We are expecting 2 solutions for x since this is a second degree polynomial. We will sub both -1 and 0 into y = x + 7 to solve for the corresponding values of y
y = 0 + 7 so
y = 7 and the coordinate is (0, 7)
y = -1 + 7 so
y = 6 and the coordinate is (-1, 6)
-6/5,-236/5 would be the minimum
Answer: option B is correct.
Step-by-step explanation:
The equation of a straight line can be represented in the slope-intercept form, y = mx + c
Where c = intercept
Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis represent
change in the value of y = y2 - y1
Change in value of x = x2 -x1
y2 = final value of y
y 1 = initial value of y
x2 = final value of x
x1 = initial value of x
The line passes through (- 5, - 2) and (3, - 1),
y2 = - 1
y1 = - 2
x2 = 3
x1 = - 5
Slope,m = (- 1 - - 2)/(3 - - 5) = 1/8
To determine the intercept, we would substitute x = 3, y = - 1 and
m = 1/8 into y = mx + c. It becomes
- 1 = 1/8 × 3 + c
- 1 = 3/8 + c
c = - 1 - 3/8 = - 11/8
The equation becomes
y = x/8 - 11/8
I:2x – y + z = 7
II:x + 2y – 5z = -1
III:x – y = 6
you can first use III and substitute x or y to eliminate it in I and II (in this case x):
III: x=6+y
-> substitute x in I and II:
I': 2*(6+y)-y+z=7
12+2y-y+z=7
y+z=-5
II':(6+y)+2y-5z=-1
3y+6-5z=-1
3y-5z=-7
then you can subtract II' from 3*I' to eliminate y:
3*I'=3y+3z=-15
3*I'-II':
3y+3z-(3y-5z)=-15-(-7)
8z=-8
z=-1
insert z in II' to calculate y:
3y-5z=-7
3y+5=-7
3y=-12
y=-4
insert y into III to calculate x:
x-(-4)=6
x+4=6
x=2
so the solution is
x=2
y=-4
z=-1
