<span>1. c, b
2. c, b
1.
To solve this problem, you really need to get a piece of graph paper and draw the triangles. For the triangle RST, you'll see that it's a right triangle with R being the right angle. The long leg of the triangle is close to the origin and the short leg is immediately below R. Then when you draw the triangle R'S'T', you'll see that it's still a right triangle. But now the long leg is parallel to the y axis and S' is now above R' instead of S being to right of R in the original triangle. So it looks like the triangle was rotated 90 degrees counter clockwise which is choice "c". So, draw a new triangle R"S"T" by rotating triangle RST 90 degrees counter clockwise. You'll see that point R" is at (1,-4), S" at (1,-1), and T" at (2,-4). All three of those points are located 2 units below the points R' S' T'. So you need to translate the triangle 2 units higher, which is choice b.
2. This is the exact same question with the exact same choices as #1 above. So the answer is exactly the same.</span>
Answer:
-2<x<4
Step-by-step explanation:
For 
, note that either the numerator or denominator is positive and the other must be negative for this inequality to be satisfied.
This means x+2>0 and x-4<0 or x+2<0 and x-4>0.
Lets look at the first scenario (x+2>0 and x-4<0)
x+2>0 and x-4<0
x>-2 and x<4
This means that -2<x<4.
Let’s look at the other scenario (x+2<0 and x-4>0)
x+2<0 and x-4>0
x<-2 and x>4
This means that x must be <-2 and >4 simualtaneously, which is impossible.
Therefore, this only occurs when -2<x<4.
I hope this helps! :)
Answer:
Aand C I think ......................
Answer:
1 triangle
Step-by-step explanation:
With three forced sides you can only have 1 triangle as to make it different you would have to change a side length which wouldn't be possible. Where as if it was 3 angles then it would be infinitely many.
