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How do you do these inverses. can you please explain each problem so I know how to do them in the future?? Thanks..

g(x) = (4/3)x - 7/3

h(x) = (18x + 2)/10

f(x) = (2/9)x - 1/9

2007-11-27 10:17:59 · 5 answers · asked by Jaz 'ma' Taz 2 in Science & Mathematics Mathematics

5 answers

so you got y = (4/3)x - 7/3

all you got to do is...flip the x and y! =)

x = (4/3)y - (7/3)

now just solve for y

so add 7/3 to move it over to the other side

x + 7/3 = (4/3)y

now you multiply by reciprocal to get rid of the (4/3)

(3/4)(x+7/3) = (3/4)(4/3)y

so

(3/4)x + (3/4)(7/3) = y < ---when you mult by reciprocal, the fractions cancel out on the right side.

simplify

(3/4)x + (7/4) = y

yay!

2007-11-27 10:22:47 · answer #1 · answered by niiro13 7 · 0 0

I do not know what you mean by inverses???
I can tell you this...g(x) = 4/3x - 7/3 is the same as writing
y = 4/3 x - 7/3
solving for x
3y = 4x -7 and then... 3y + 7 = 4x
4/3 y + 7/3 = x
All that I did was to first get rid of the fraction by multiplying by 3, then added 7 to both sides and divided by 4
The same goes for the rest ... see if you can do it!!!!
Hint.!!!! for the next one....multiply by 10 subtract 2 and divide by 18 ...........

2007-11-27 18:45:26 · answer #2 · answered by ? 3 · 0 0

To do inverse, first it may help to change g(x) to y.

y=(4/3)x - 7/3

Then you switch the x and y.

x=(4/3)y - 7/3

Then you isolate for y (because it always has to be in the y= form)

(4/3)y= x + 7/3
y= 3/4x + 7/4

The rest are done using the same method, hope that helps!

2007-11-27 18:22:22 · answer #3 · answered by Niche 3 · 0 0

g(x) = (4/3)x - 7/3
g(x)/x = (4/3)x/x - 7/3
g = 4/3 - 7/3
g = -3/3 = -1

i think that should give u an idea

2007-11-27 18:40:16 · answer #4 · answered by Anonymous · 0 0

Maybe if you stopped text messaging and talking in class you would know the answers to these instead of getting on the internet and asking people to do your homework!

2007-11-27 18:20:44 · answer #5 · answered by Jaclyn 2 · 0 1

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