<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>
First find the slope.
-6--2=-4
-2--1=-1
The slope is -4/-1 or 4.
Now plug in one point and the slope into point slope
y-y1=m(x-x1)
y+2=4(x+1)
Answer:
x = 2/3, -2/3
Step-by-step explanation:
Solve the equation for x by finding a
, b
, and c of the quadratic then apply the quadratic formula.
I’m pretty sure the answer is d!!
