2. | y+4| <1 (1 point)
-5<y<-3
-3<y<5
-4<y<1
1<y<4
3. |2r| -5 =7. (1 point)
T=1 or -1
T=-6 or 6
T= 10 or -10
T= 12 or -12
4. |A| -3/4 = -5/8
1/8 or -1/8
7/8 or -7/8
1 3/8 or -1 3/8
No solution
#2
| y+4| <1
Use your definition:
y+4 < 1 AND -y-4 < 1
y < -3 AND -y < 5
y < -3 AND y > -5
or as they wrote it, -5 < y < -3
I don't see your choices, so how can we check?
clearly not 2D. Just plug in y=2
|2+4| = |6| = 6 Bzzzt
#3 is also wrong. You forgot to divide by 2.
|2r| = 12, not |r| = 12
#4 is correct
The last one is correct, the first two are wrong.
Show me your steps so I can tell where you are going wrong.
Just number 2 I'm not even understanding where to start
#2 is the main one i need help with
Why are you ignoring Reiny's advice?
I said i don't even know where to start with it
#2. Remember that
|x| = x if x >= 0
|x| = -x if x < 0
So, you have
|y+4| < 1
so, either y+4 >= 0, meaning y >= -4
or y+4 < 0, meaning y < -4
If y >= -4
y+4 < 1
y < -3
If y < -4
-(y+4) < 1
-y-4 < 1
-5 < y
But also, we started with y < -4
So the solution is -5 < y < -3
Or, think of what |y+4| means. If y = -4, then y+4=0. So, |y+4| is the distance from y to -4. If that distance is less than 1, then you see the solution. Plot the value -4 on the number line. y must be less than 1 unit away from -4.
My answers
2.D
3.also D
4.A