Question

math problem

Answer 1

@AZ

Answer 2

Here you are told that \(2x-1\) applies when x is less than one

If you were to plot the line you'd see something like this

(Note the less than one, it should stop at 1 with an empty circle)

Answer 3

To do the next one follow the same step of graphing it without boundaries then "cutting" the graph off at 1 (Only the point is included here)

Answer 4

@snowflake0531

Answer 5

Well, we can first eliminate two of them, because two of them aren't even separated at -1, rather, they are separated at 1

So, which two can you eliminate

Answer 6

a and c?

Answer 7

actually, I'm not really sure about what letters... since you only showed the graph?

Answer 8

the first graph and the third graph

Answer 9

The first graph is at -1,

Answer 10

Which two are weirded out at x=-1

Answer 11

b and d?

Answer 12

no

Answer 13

It's the third and fourth graph for me, if I"m in the right order

These two can be eliminated:

Answer 14

alright

Answer 15

So, now that we're left with the first two graphs, we have to actually look at the linear equations.

I think that you can do that
When x is smaller than 1, 2x-1
when x is larger than 1, x+1

Graph those lines
and see which one fits

Answer 16

graph 2x-1?

Answer 17

Graph both lines, see which graph fits it

Answer 18

a?

Answer 19

yep, that's what i think too

Answer 20

its not right lol

Answer 21

kek

Answer 22

are you sure you chose this one

Answer 23

yep

Answer 24

well, that's the computer's fault then, it's right xd

Answer 25

@dude
comments xd, help

Answer 26

its alright i have anoher uestion

Answer 27

You eliminated the wrong one :s

\(\color{#0cbb34}{\text{Originally Posted by}}\) dude
Here you are told that \(2x-1\) applies when x is less than one

If you were to plot the line you'd see something like this

(Note the less than one, it should stop at 1 with an empty circle)
\(\color{#0cbb34}{\text{End of Quote}}\)

The one I posted showed one of the equations graphed. Your jump is at x=1

This would mean it *would* have to be C or D