Question

Can somebody please help me! I will fan and medal!

Answer 1

First off, do you know how to do this partially/somewhat/or not at all?

Answer 2

Not at all. But I'm willing to learn!

Answer 3

@wmj259

Answer 4

a) 4+v<5 and 5>4+v

Answer 5

What about B and C @dyna-marie99

Answer 6

solve
4+v<5
-4 -4
v<1

Answer 7

@Dyna-Marie99 That is the same equation written in a different way. They want two different inequalities.

\[\large |x-a|<b \implies -b<x-a<b \implies -b+a<x<b+a\]
From there you can split it and say\[\huge -b+a<x\]and\[\huge x<b+a\]

Answer 8

Wait wait, okay so for the B column I write : 4+v<5? and solve it?

Answer 9

@Ms-Brains Do you understand the difference between Dyna-Marie99's answer and mine for a)? There's a huge difference.

Answer 10

Um..kind of. I'm not sure which one is correct though..@CShrix

Answer 11

Which one should I use I mean?

Answer 12

4+v<5 and 5>4+v are both the same exactly inequalities, except rearranged. Do you see that? In both scenarios, 5 is greater than 4+v. This isn't what the questing is asking for. It wants two separate inequalities that describe the interval.

The first thing we have to do is get rid of the absolute value bar. In order to do this, we make the following procedure:\[\huge |x-a|<b \implies -b<x-a<b \]All we did was take the negative of b and bring it on the left side of the inequality and copy the sign if the inequality (< sign).

Answer 13

Do you understand so far?

Answer 14

Yeah, a little. :)
I wrote 4 + v < 5 and 5>4+v as the A part. @CShrix

Let's move on to B

Answer 15

That's wrong.

Answer 16

What am supposed to write then..?

Answer 17

I was trying to tell you, but you wanted to move on to B. Reread the post above about the inequality and then tell me when you're ready to continue.

Answer 18

Like this??? |4 - v| < 5 ==> -4 < v - 5 < 4 @CShrix

Answer 19

Close, but not quite!
\[\huge |4-v|<5 \implies -5 < 4-v < 5\]

Answer 20

Okay, I get it! Thanks! :)

Now can we try B?

Answer 21

From there, we need to look at both sides of the inequality and break it up wherever there is an inequality.

So the first part we break up is the left side!
\(\huge -5<4-v\)

Now let's look at the right side:
\(\huge 4-v<5\)

These are the two inequalities we're looking for, for a).

Answer 22

for b).

If we solve the left side for v, then we get the following procedure:
\(\huge -5<4-v\)
\(\huge v<9\)

If we solve for v, then we get the following:
\(\huge 4-v<5\)
\(\huge -1<v\)

The solution set is the combination of these two intervals, AKA
\[\huge -1<v<9\]

Answer 23

So in the b column i write -1 < V <9

Answer 24

@CShrix

Answer 25

Yes. Now graph c and I will check your work.

Answer 26

like this...?

Answer 27

So close! Since the interval is open (less than or greater than, but no equal to), then the circles are open. Either than that, it's good!

Answer 28

Can you please show me..? :3

Answer 29

The circles are open, indicating an open interval. If it included "greater than or equal to" or "less than or equal to" then they would be closed because it includes those specific values. Since we're saying it's greater than -1, then it cannot actually include -1, but every value greater than -1. We place the open circle at -1 indicate that it doesn't include -1, but every value greater than it. Why, you might ask? Try graphing -0.99999999999999999999999999999999999999999 (it goes infinitely). It's nearly impossible! X)

Answer 30

like this? cx

Answer 31

Yes!

Answer 32

Awesome! can you help with one more question?