Since x - 3 < 0 for -2 < x < 3, |x-3| = -(x-3)
Since x + 2 > 0 for -2 < x < 3, |x+2| = x+2
Since x - 5 < 0 for -2 < x < 3, |x-5| = -(x-5).
So y = -(x-3) + (x+2) -(-(x-5)) = 3 - x + x + 2 + x - 5 = x.
y = x, if -2 < x < 3.
Answer:
1 a.) 2t – 2
1 b.) –9z + 4
2 a.) –2v + 3
2 b.) –13k – 9
3 a.) 7w − 2
3 b.) 10s
4 a.) 5
4 b.) –3p + 11
5 a.) –29x + 9
b.) 15v – 60x – 2
a.) 25n + 2
b.) –15n
a.) –71x + 4
b.) –15b
Step-by-step explanation:
- <em>NOTE: I am kind of confused about the way this question is formatted, so I tried to answer everything I could decipher.</em>
1 a.) –2t + 5t – 2 – t → 2t – 2
1 b.) –5z + 4 – 4z → –9z + 4
2 a.) 8v + 3 – 10v → –2v + 3
2 b.) –6 – 6k – 3 – 7k → –13k – 9
3 a.) –6 – 3w + 4 + 10w → 7w − 2
3 b.) 9s + s → 10s
4 a.) 6 + 8b – 1 – 8b → 5
4 b.) 1p + 6 + 5 – 4p → –3p + 11
5 a.) 4x – 3x + 9 – 6x 5 → –29x + 9
b.) 9v + 6v – 2 – 10x 6 → 15v – 60x – 2
a.) n + 2 – (–4)n 6 → 25n + 2
b.) 6n – 3n 7 → –15n
a.) –4x – 4x + 4 – 9x 7 → –71x + 4
b.) –9b + (–6)b → –15b
The answer is never, that is, on a 2 dimensional plane. You can perform an experiment to see why it is the case. On curved surfaces though, two lines can intersect one another more than once. For instance, on the surface of planet Earth, two lines can intersect one another, both at the Earth's North Pole and South Pole.
-2,3 is (x1,y1) and (4,-5) is (x2,y2) so you do y2-y1/x2-x1
y2-y1 is -5-3 which is -8 and x2-x1 is 4-(-2) which is 6 so you get
-8/6= -4/3
(x + y)^2 = (x^2 - 2xy + y^2)
First distribute the ^2 on the left side of the equation to each term inside the parenthesis:
x^2+ 2xy + y^2
Now pick one of the variables to solve for and isolate it:
(solving for x)
x^2 + 2xy + y^2 = x^2 - 2xy + y^2
x^2+ 2xy = x^2 - 2xy
2xy = -2xy
-x = x
x = 0
When you solve for y in the equation it will turn out to be 0 as well
