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Mathematics 14 Online
OpenStudy (zackattack350):

Need help with this problem, Please explain. Screenshot below!

OpenStudy (zackattack350):

OpenStudy (zackattack350):

i tried to do it myself but I don't understand what i'm supposed to do. I need help with an explanation for it please!

OpenStudy (zackattack350):

@satellite73 hey if you could help with this i'd be really greatful! I have a test tomorrow on these materials and i'm trying to learn this and other questions so i can do it!

OpenStudy (anonymous):

sure from the table,what is \(f\) when \(x=2\)?

OpenStudy (zackattack350):

7?

OpenStudy (mathmale):

A table? Are you (zack) going to post it?

OpenStudy (anonymous):

no not 7? but 7!!

OpenStudy (anonymous):

and what is \(g\) when \(x=2\)?

OpenStudy (zackattack350):

its the screenshot below my original question @mathmale G would be -2!

OpenStudy (zackattack350):

so f=7 g=-2

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

to to get \((f+g)(2)\) add those two numbers

OpenStudy (anonymous):

because what \((f+g)(2)\) means is \(f(2)+g(2)\)

OpenStudy (zackattack350):

((7)(2))+((-2)(2))=(14)(4)

OpenStudy (zackattack350):

56?

OpenStudy (anonymous):

oh no i see your confusion

OpenStudy (anonymous):

it is this \[f(2)+g(2)\] do not multiply anything

OpenStudy (zackattack350):

(7)(2)+(-2)(2)?

OpenStudy (anonymous):

it is much easier than you are making it you have \[f(2)=7\\ g(2)=-2\]so \[f(2)+g(2)=7+(-2)\]that is all `

OpenStudy (zackattack350):

OH!

OpenStudy (anonymous):

easy right?

OpenStudy (zackattack350):

so therefore the answer would be... 5?

OpenStudy (anonymous):

yes it would

OpenStudy (anonymous):

wanna try \((f-g)(4)\)?

OpenStudy (zackattack350):

Thank you! So will i do that for all the other equations i have to evaluate using this??

OpenStudy (zackattack350):

Sure!

OpenStudy (anonymous):

yes that is what you have to do for B, evaluate the functions, then subtract

OpenStudy (zackattack350):

so that would be... (11-6)(4)

OpenStudy (anonymous):

stop multplying !!

OpenStudy (anonymous):

it is not "times" it is "of" f of four minus g of four

OpenStudy (zackattack350):

f(4)+g(4)=11-6?

OpenStudy (anonymous):

bingo

OpenStudy (zackattack350):

so then the answer would be 5

OpenStudy (anonymous):

the less you do, the better it is

OpenStudy (anonymous):

yes

OpenStudy (zackattack350):

so i gotta stop trying to multiply and read it as of

OpenStudy (anonymous):

yup wanna try the next one?

OpenStudy (zackattack350):

yes! ill try and get this one right!

OpenStudy (zackattack350):

(fg)(-2)=((0)(5))-2

OpenStudy (zackattack350):

do we multiply this one?

OpenStudy (anonymous):

yes, but you do not subtract two just \(0\times 5\)

OpenStudy (zackattack350):

so why dont we subtract the 2 like we did the other ones? as the answer is 0.

OpenStudy (anonymous):

we did not subtract the two in the other ones here is a quick example if \(f(5)=7,g(5)=2\) hten \[(f+g)(5)=7+2\\ (f-g)(5)=7-2\\ (fg)(5)=7\times 2\\ \frac{f}{g}(5)=\frac{7}{2}\]

OpenStudy (zackattack350):

oh the -2 in the first one is the actual number! ok that one confused me! lemme try the last one and see if i get it right.

OpenStudy (anonymous):

ok

OpenStudy (zackattack350):

(f/g)(0) f=3, g=0

OpenStudy (zackattack350):

therefore

OpenStudy (anonymous):

...

OpenStudy (zackattack350):

(f/g)(0)=(3/0)=0?

OpenStudy (anonymous):

the \(\frac{3}{0}\) is right the \(0\) is wrong

OpenStudy (zackattack350):

3/0=..... nothing?

OpenStudy (anonymous):

not a number, you cannot divide by zero

OpenStudy (astrophysics):

3/0 = blackhole

OpenStudy (anonymous):

i tried it once, had a headache for a week

OpenStudy (zackattack350):

so how would i input that as an answer?

OpenStudy (anonymous):

cant help you there

OpenStudy (zackattack350):

got it!

OpenStudy (zackattack350):

Thanks for the help!!!

OpenStudy (anonymous):

yw

OpenStudy (zackattack350):

i have another type of question ill make another post! i have a few types... im trying to learn so i know what im doing.

OpenStudy (anonymous):

kk

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