Question

ALGEBRA HELP! (ATTACHED)

Answer 1

?

Answer 2

@amistre64

Answer 3

@dan815

Answer 4

i think your set notation is off, but its been awhile ... you also need to fill in your chart

Answer 5

R/{0} i understand, but im not sure its set notation

D = { x \(\in R\) | x\(\ne\) 0}

x in R, such that x not equal to 0

Answer 6

youll have to show some work for the second one, fill in what you can and ask about what you have issues with.

Answer 7

alright thank you

Answer 8

for a I have
x+5=0
x+5-5=0-5
x= -5

for b I have
-1,5
(x-1)(x+5)
(x+5)
(x-1)
1???

Answer 9

a is good

b is your factors are fine, we would expect it to have (x+5) as a factor if it wants us to simplify, and (x-1) is the other.

what this means is, that for all values of x (other than -5) are equaivalent to the graph of x-1

Answer 10

so, C is the graph of x-1, with a hole at x=-5 right?

Answer 11

@amistre64 sorry my computer got messed up yesterday , and right

Answer 12

then you page 2 is worked out, all you really need to finish up is your table on page 1.

show me what youve accomplished and ill rechk it.

Answer 13

idk how to fill out the table this is what im having the most problems with

Answer 14

do you know how to square a number?

Answer 15

since f(x) = 5/x^2

when x=say -6, f(x) = 5/(-6)^2 = 5/36

all you are doing is squareing the denominator

Answer 16

oh ok, ill try it and get back to you

Answer 17

k for the first one I got -5/16 so would I put 16 in the first square or the whole number

Answer 18

@amistre64

Answer 19

site was not playing right yesterday, kept crashing on me.

not -5/16,

5/(-4)^2 = 5/16
5/(4)^2 = 5/16

5/(-3)^2 = 5/9
5/(3)^2 = 5/9

5/(-2)^2 = 5/4
5/(2)^2 = 5/4

etc ...

Answer 20

thanks for helping, appreciate it!!! turned in the assignment today @amistre64