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Answer:

Show work in this space.

Using these points, draw a curve.
Show graph here.

2006-10-15 02:28:31 · 6 answers · asked by othe1220 1 in Education & Reference Homework Help

6 answers

y=1/x-2

replace the x with their values
y=1/-2 -2
y=-0.5 -2
y=-2.5 or y= -2 1/2
x= -2 and y= -2 1/2 or (-2,-2 1/2)

y=1/-1 -2
y=-1-2
y=-3
x=-1 and y=-3 or (-1,-3)

y=1/0 - 2
y=0-2
y=-2
x=0 and y=-2 or (0,-2)

y=1/1 -2
y=1-2
y=1
x=1 and y=1 or (1,1)

y=1/2-2
y=0.5-2
y=-1 1/2
x=2 and y= -1 1/2 or (2,-1 1/2)

y=1/3-2
y= i dunno so you solve this one


sry my answer might be wrong but im sure the procces is to replace x with the value and then you will get the value of y

good luck

.

2006-10-15 03:04:55 · answer #1 · answered by ? ? ? bloody_angel ??? 3 · 0 0

Assuming that y = 1/x - 2; we get the values of y as follows:

y = 1/- 2 - 2 = -1/4

y = 1/ -1 -2 = -1/3

y = 1/0 - 2 = -1/2

y = 1/1 - 2 = -1/2

y = 1/2 - 2 = infinity

y = 1

Extremely sorry, but I cannot make a graph here.

2006-10-15 02:47:24 · answer #2 · answered by Aazib A 1 · 0 0

First: replace each form with the x-variable with the intention to discover their corresponding y-variables... a. y = a million/(-2-2) y = a million/- 4 y = -a million/4 b. y = a million/(-a million-2) y = a million/-3 y = -a million/3 c. y = a million/x-2 y = a million/(a million-2) y = a million/-a million y = -a million d. y = a million/x-2 y = a million/(2-2) y = a million/0 undefined e. y = a million/x-2 y = a million/(3-2) y = a million/a million y = a million

2016-11-23 12:48:35 · answer #3 · answered by Anonymous · 0 0

Just plug and chug baby, plug and chug


x=-2

y= 1/-2-2

y=-1/4

do the same for the other points...plug them for x

them just graph the ordred pairs and draw the curve through them

(-2,-1/4) and so on

got it, get it, good!

2006-10-15 02:31:35 · answer #4 · answered by Anonymous · 1 0

y=1/x-2
y=1/x-2/1
cross multiply
y=-2x-1
i hope you get it!!

2006-10-15 02:42:51 · answer #5 · answered by katenz_25 1 · 0 0

Why can't you do it?

2006-10-15 02:30:12 · answer #6 · answered by Anonymous · 0 0

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