Question

Math help Answers just need correction

Answer 1

@ehuman @UsukiDoll @Zale101

Answer 2

@Compassionate @wio @divu.mkr

Answer 3

Is this for a test?

Answer 4

no its a pre-test I have to go over before I take the test

Answer 5

can you help me go over my answers and correct me if im wrong?

Answer 6

@Compassionate can you help me?

Answer 7

@shamil98

Answer 8

getting there I just got here

Answer 9

How did you approach #3?

Answer 10

4 b
6 c
7 a

Answer 11

for remaining u did good

Answer 12

yeah, sorry, I meant #6, not #3

for 7, to find the discontinuities, find the values of x where the denominator = 0

for 4,

\[\frac{x+2}{2} -\frac{x+1}{3x}\]Multiply terms by \(3x/3x\) and \(2/2\) respectively to get common denominator, then combine, being careful about applying the \(-\) in front of the second term! might rewrite as \[\frac{x+2}{x} - \frac{x}{3x} - \frac{1}{3x}\]to make it clearer.

Answer 13

except that should be
\[\frac{x+2}{2}-\frac{x}{3x}-\frac{1}{3x}\]

Answer 14

@whpalmer4 so for 7 is it a?

Answer 15

still working on 5 but I got the rest. you did good on the first one, but the rest...you need more practice

Answer 16

hey tahnks alot :)

Answer 17

okay

Answer 18

number 5 was asking for LCD right?

Answer 19

yea

Answer 20

I think..maybe it's just me...but the LCD should be

Answer 21

I haven't finished it through though...it may involve some cancellation I'm not sure... but I have collected all the terms in the denominator...

Answer 22

okay i have 2 more cna u help me i have answers

Answer 23

well, that problem is poorly written. the discontinuity is at x = -4. that isn't one of your answer choices :-)

Answer 24

ohh k

Answer 25

is 5 d?

Answer 26

@whpalmer4 I saw x doesn't equal to -4 as one of the choices for #7

Answer 27

Yes, but the discontinuity is only at x = -4!

Answer 28

yeah I know that.

Answer 29

5 is wt heck is it? :S

Answer 30

that's not an answer choice. the question asks for the discontinuties.