Ask your own question, for FREE!
Mathematics 9 Online
OpenStudy (anonymous):

Looking for someone to check over my work :)

OpenStudy (anonymous):

Perform the requested operation or operations. f(x) = 4x + 7, g(x) = 3x2 Find (f + g)(x).

OpenStudy (anonymous):

i got 12x3 + 21x

hartnn (hartnn):

you multiplied them ? you had to add them, (f+g)(x)= 4x+7 + 3x^2

hartnn (hartnn):

like if f(x) - x^2, g(x) = e^x (f+g) (x) = x^2+e^x thats it!

OpenStudy (anonymous):

oh wow lol so its this 4x + 7 + 3x2

OpenStudy (anonymous):

lololol im so dumb

OpenStudy (anonymous):

ok i have a couple more do you mind?

hartnn (hartnn):

no i don't mind :)

OpenStudy (anonymous):

ok awesome thank you one sec let me get them

OpenStudy (anonymous):

f(x) = x minus five divided by eight.; g(x) = 8x + 5, find g(f(x)). g(f(x)) = x - five divided by eight.

OpenStudy (anonymous):

ok i got g(f(x)) = x + 10 but im a little skeptical on that one

hartnn (hartnn):

oh, how you got that ? maybe i can find the error in your working ?

OpenStudy (anonymous):

ok yes one sec ill type my work

OpenStudy (anonymous):

so i get to this point \[8(\frac{ x-5 }{ 8 })+5 \] and im not to sure what to do i figured cancel the 8s then foil the -5 to 5 and get 10

OpenStudy (anonymous):

oh wait is it just x?

hartnn (hartnn):

the 8's get cancelled what remains is x-5+5 = x+0 = x just x :)

OpenStudy (anonymous):

AH HA SO i was right :) thanks ok now i have another one similar to these but i don't even know how to start it ill post it one sec

OpenStudy (anonymous):

Find f(x) and g(x) so that the function can be described as y = f(g(x)). y = nine divided by square root of quantity five x plus five. they don't even give me g?

OpenStudy (anonymous):

y= \[\frac{ 9 }{ \sqrt{5x+5} }\]

hartnn (hartnn):

you need to find f and g

OpenStudy (anonymous):

that's what the equation looks like

OpenStudy (anonymous):

how do i even start that?

hartnn (hartnn):

u just need to identify 2 types of function... let me give you an example is 4/x a function ?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

wait how does the vertical line test work again

OpenStudy (anonymous):

wait yes it is

hartnn (hartnn):

correct, its a function and what about sqrt (x+9) ?

OpenStudy (anonymous):

yes

hartnn (hartnn):

good, so in your Question , what 2 function can you identify ?

OpenStudy (anonymous):

5x+15 and sqr 9 over x?

hartnn (hartnn):

how about this ? \(\large f(x)=\dfrac{9}{x},g(x)= \sqrt{5x+15}\) what will be f(g(x)) then ?

OpenStudy (anonymous):

\[\frac{ 9 }{ \sqrt{5x+15} }\]

hartnn (hartnn):

correct! got some idea how to identify the functions ?

OpenStudy (anonymous):

ahh yes so its this one then? f(x) = nine divided by x., g(x) = 5x + 5 but it doesn't have the square rot around the 5x+5

OpenStudy (anonymous):

these are my choices f(x) = nine divided by square root of x., g(x) = 5x + 5 f(x) = square root of quantity five x plus five., g(x) = 9 f(x) = nine divided by x., g(x) = 5x + 5 f(x) = 9, g(x) = square root of quantity x plus five

hartnn (hartnn):

ok, there are many answers possible here. from the given choices, try the 1st one, what would u get f(g(x)) ?

OpenStudy (anonymous):

but in math youre wrong or right xD

hartnn (hartnn):

let me give u example : for 1/e^(x^2-4) the 2 functions could be 1/e^x and x^2-4 or 1/x and e^(x^2-4) both will give you 1/e^(x^2-4) both are correctt :)

OpenStudy (anonymous):

bot i don't get that because only one of those answers can i recvie my points for

hartnn (hartnn):

what will you get f(g(x)) for 1st option ? if u get the required f(g(x)) then its correct one

hartnn (hartnn):

the 3rd one isn't correct...

OpenStudy (anonymous):

ok im not sure what you mean but ill just do what you tell me

OpenStudy (anonymous):

omg lol its a

OpenStudy (anonymous):

it have to look like the y = did correct

hartnn (hartnn):

yeah, find f(g(x)) for a) exactly! it is a :)

OpenStudy (anonymous):

heheh :) ok now there is another section i don't quite get. if you have time and know anything about exact values?

OpenStudy (anonymous):

i can post it as an other question if you would like a medal or if you don't feel it ill just close this and post it :0

hartnn (hartnn):

i am not here for medals lol am just here to help. if i can help, i will :)

OpenStudy (anonymous):

ok awesome ill post it

OpenStudy (anonymous):

Find the exact value of the real number y. y = csc-1(-1)

OpenStudy (anonymous):

π negative pi divided by two pi divided by four 2π

hartnn (hartnn):

you know what csc is ?

OpenStudy (anonymous):

yes the opposite of cos

OpenStudy (anonymous):

\[\cos ^{-1}\]

hartnn (hartnn):

no...

OpenStudy (anonymous):

lol sec

OpenStudy (anonymous):

not cos

OpenStudy (anonymous):

everything i said but sec

OpenStudy (anonymous):

SIN

OpenStudy (anonymous):

wait im lost

OpenStudy (anonymous):

one sec let me figure thois out

hartnn (hartnn):

ok

OpenStudy (anonymous):

is it arccos?

hartnn (hartnn):

\(\csc x = \dfrac{1}{\sin x}\)

OpenStudy (anonymous):

is that an identity?

hartnn (hartnn):

yes \(\csc^{-1}x = y \implies x=\csc y\)

OpenStudy (anonymous):

ok gotcha

hartnn (hartnn):

so u need that value of csc that gives you -1

hartnn (hartnn):

that angle of csc***

hartnn (hartnn):

what is csc pi =... ?

OpenStudy (anonymous):

im not sure :(

hartnn (hartnn):

what is sin pi ?

OpenStudy (anonymous):

.0548

hartnn (hartnn):

sin pi = 0 take pi in radians sin pi = sin 180 degrees = 0

OpenStudy (anonymous):

ahh ok

hartnn (hartnn):

so, csc pi = 1/0 = infinity so pi is not the answer similarly try other choices ?

OpenStudy (anonymous):

so pie over 4?

hartnn (hartnn):

whats sin (pi/4) is it -1 ?

OpenStudy (anonymous):

ahh no so the only one that can work is

OpenStudy (anonymous):

- pie over 2

hartnn (hartnn):

yes.

OpenStudy (anonymous):

you are honestly the best

hartnn (hartnn):

not actually , but thanks :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!