Spinout007
Spinout007 HalfDork
4/6/10 7:55 p.m.

My wife is trying to figure out the last few problems in her math homework. Any math geeks here wanna try and help?

Ok....I need help....f f(x)=the sqare root of x+4 and G(x)= 2-x find the simplified equation for H(x)=(f(g(x))

Quoted from her fb page.

Thanks guys

billy3esq
billy3esq SuperDork
4/6/10 8:19 p.m.

If by f(x) you mean sqrt(x+4), then:

H(x) = sqrt(2-x+4) = sqrt(6-x).

OTOH, if by f(x) you mean sqrt(x)+4, then:

H(x) =sqrt(2-x)+4.

In other words, to find f(g(x)), just plug in g(x) for x in f.

Spinout007
Spinout007 HalfDork
4/6/10 8:23 p.m.

not sure, I told her to long in and reply, she's at work, I'm at home, fixing to goto bed. lol 2:30am just comes waaay to early

Maxnscrappers
Maxnscrappers New Reader
4/6/10 8:24 p.m.
billy3esq wrote: If by f(x) you mean sqrt(x+4), then: H(x) = sqrt(2-x+4) = sqrt(6-x). OTOH, if by f(x) you mean sqrt(x)+4, then: H(x) =sqrt(2-x)+4. In other words, to find f(g(x)), just plug in g(x) for x in f.
Maxnscrappers
Maxnscrappers New Reader
4/6/10 8:26 p.m.
billy3esq wrote: If by f(x) you mean sqrt(x+4), then: H(x) = sqrt(2-x+4) = sqrt(6-x). OTOH, if by f(x) you mean sqrt(x)+4, then: H(x) =sqrt(2-x)+4. In other words, to find f(g(x)), just plug in g(x) for x in f.
Maxnscrappers
Maxnscrappers New Reader
4/6/10 8:27 p.m.

THANK YOU I was looking for the sqrt of (X+4) :)

chaparral
chaparral Reader
4/6/10 8:27 p.m.

F(x)=sqrt (x) +4 G(x)=2-x=-x+2 H(x)= F(G(x))

Substitute y = G(x). H(x)=F(G(x))=F(y)

F(y)= sqrt (y) + 4 y=G(x)=-x+2

Substitute back.

F(y)= sqrt (y) + 4

F(G(X))=sqrt (G(x)) + 4 = sqrt (-x+2) +4.

F(G(X)) is imaginary if x >2.

Dr. Hess
Dr. Hess SuperDork
4/6/10 8:28 p.m.

Your inner ing-en-eer is coming out Billy3. You missing honest work?

Spinout007
Spinout007 HalfDork
4/6/10 8:29 p.m.

Thanks again guys, I love this place! It's better than disney!

Maxnscrappers
Maxnscrappers New Reader
4/6/10 8:35 p.m.

In reply to Spinout007:

Now I'm more confused...both would work...hrmm thanks guys...I hate math...give me english anyday:) Thanks I must say the answers here are much more logical than on fb

Marty!
Marty! HalfDork
4/6/10 8:52 p.m.

Spinout007
Spinout007 Dork
4/6/10 8:56 p.m.

WIN! LMAO WOW! I LOVE IT! that dude was a trip,

Maxnscrappers
Maxnscrappers New Reader
4/6/10 9:01 p.m.

We need a like button

John Brown
John Brown SuperDork
4/6/10 9:07 p.m.
Maxnscrappers wrote: We need a like button

LOL, hit the +1

Spinout007
Spinout007 Dork
4/6/10 9:14 p.m.

lmao

billy3esq
billy3esq SuperDork
4/6/10 10:18 p.m.

In reply to Dr. Hess:

Not really, honest work is too hard.

stumpmj
stumpmj Dork
4/6/10 10:51 p.m.

For a second there, I thought it said meth help needed...

GRM does help with everything

VanillaSky
VanillaSky Reader
4/7/10 1:04 a.m.

Miata.

TL;DR

mapper
mapper Reader
4/7/10 11:19 a.m.

Bob Ross R.I.P. I initialy learned to oil paint from his show and books. And I suck at math.

Spinout007
Spinout007 Dork
4/8/10 7:56 p.m.

LOL, my wife says she sucks at painting too.

Tetzuoe
Tetzuoe Reader
4/9/10 6:45 a.m.

speaking of math, check it out: http://www.neatorama.com/2010/04/08/try-this-and-prepare-to-be-amazed/

neatorama said: ake any number with 4 non-repeating digits. Say 1562. Step 1: Arrange the number in ascending and then descending order Step 2: Subtract the smaller number from the bigger number 6521 - 1256 = 5265 Repeat the steps: 6552 - 2556 = 3996 9963 - 3699 = 6264 6642 - 2466 = 4176 7641 - 1467 = 6174 Try any 4-digit number with non-repeating digits, and you'll *always* get 6174. Pretty cool, huh? 6174 is known as Kaprekar's constant. The math operation above, discovered by Indian mathematician D.R. Kaprekar, will reach 6174 after at most 7 steps (if you did more than 7 iterations, check your arithmetics).
Gearheadotaku
Gearheadotaku Dork
4/9/10 7:07 a.m.

solve for XR4Ti x 325xi (MX5) = $2011

Luke
Luke SuperDork
4/9/10 7:15 a.m.
Gearheadotaku wrote: solve for XR4Ti x 325xi (MX5) = $2011

Substitute LS1 and cancel interior, then divide by 2nd-hand race tires.

You'll need to log in to post.

Our Preferred Partners
qCfbyfzY61RDBfGgIy2E3urrHHNnEsqzIf7RTQQeHCNt27aHg6wdW5Fo5dYtQJlL