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+--- Thread: Discovery Homework Help Kiosk! (/showthread.php?tid=2313)
Discovery Homework Help Kiosk! - Nimble - 02-20-2008
I've got a problem with maths:
I missed a lesson, therefor i didn't get what the vertical lines in a function mean. I've looked in my textbook but it wasn't explained clearly.
For instance: f(x)=|x-6|
Discovery Homework Help Kiosk! - pchwang - 02-20-2008
Nimblewright, the vertical lines indicate absolute value. That means that any number between the lines is it's absolute counting value.
For instance |6| = 6 and |- 6| = 6.
A function with absolute value marks should appear like a "V" on your graph.
Discovery Homework Help Kiosk! - Nimble - 02-20-2008
thank you, it has been a great help
Discovery Homework Help Kiosk! - Dopamino - 02-28-2008
Ok, I've got a problem
How do you do this?
The problem - g(f(x)) where g(x)=x^(3/4) and f(x)=4x^-5
I get the concept for the most part, but my answer keeps on coming out wrong. The correct answer according to my book is 2^(3/2)x^(-15/4)
I need to see detailed work with explanations please. I need to find out what I'm doing wrong.
Thanks:)
Edit - nvm, got it off yahoo answers.
Here's how to do it if anyone is interested:
g[f(x)]
[4 x^(-5)]^(3/4)
0= 4^(3/4) [x^(-5)]^(3/4)]
0= (2?)^(3/4) x^[(-5) . (3/4)]
0= 2^[2 . (3/4)] x^[(-5) . (3/4)]
0= 2^(3/2)] x^(-15/4)
2^(3/2)] x^(-15/4)
Discovery Homework Help Kiosk! - Heartless - 05-04-2008
hello
so right now I need some math help. We're reviewing for final exams and I need help with 2 more problems on this sheet. They're related to logs and exponents. You DO NOT have to tell me the answers, just guide me as to what to do.
Ok, 1st one.
Daniel's Print Shop purchased a new printer for $35,000. Each year it depreciates (loses value) at a rate of 5%. What will its approximate value be at the end of the fourth year?
a) $33,250.00
b) $30, 008.13
c) $28,507.72
d) $27,082.33
This is what I did: (The formula is y=a(b)^x)
y=x
a=35,000
b=.05 -> (5%)
x=4
So I put this together, x=35000(.05)^4, but I got 2187.5
So there's my problem with that one, not sure what to do, help is really appreciated.
2nd one
On January 1, 1999, the price of gasoline was $1.39 per gallon. If the price of gasoline increased by 0.5% per month, what wsa the cost of one gallon of gasoline, to the nearest cent, on january 1 one year later?
-I really don't know in this one, would the ^x be 4, or 5? It would be I guess x=1.39(0.5)^4, right?
Discovery Homework Help Kiosk! - Guest - 05-04-2008
' Wrote:So I put this together, x=35000(.05)^4, but I got 2187.5
On January 1, 1999, the price of gasoline was $1.39 per gallon. If the price of gasoline increased by 0.5% per month, what wsa the cost of one gallon of gasoline, to the nearest cent, on january 1 one year later?
-I really don't know in this one, would the ^x be 4, or 5? It would be I guess x=1.39(0.5)^4, right?
first one: Should be 35000(.95)^4 = 28507.719
second one: 1.39(1.005)^12 = $1.48
Those may not be right, I haven't done stuff like that in a looooong time.
Edit: Just plugged them in and got the answers, they look pretty reasonable to me.
Discovery Homework Help Kiosk! - Heartless - 05-04-2008
yup, you subtract the decimal percent from 100, I had completely forgotten about that. .95 makes the 1st answer equal to b). And for the 2nd one I got $1.46, which looks right. So thanks alot, this really did save me some time.
Discovery Homework Help Kiosk! - Boss - 05-05-2008
Sorry, folks, this isn't math, but I have a problem with my HTML. I'm trying to get the bloody frames working. Code attached.
Code:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.ord/TR/html4/loose.dtd">
<!--CS-125H-30 Page Layout with Frames
Author: Aaron Porter
Date: 5 May 2008
-->
<html lang="en-US">
<body>
<title>
Freelancer/Discovery Datasite
</title>
<frameset rows="15%,*" framespacing="10" scrolling="no">
<frame name="logo" noresize scrolling="no" src="discologo.html" />
</frameset>
<frameset cols="25%,*" framespacing="10" scrolling="auto">
<frame src="linkframe.html" name="links" noresize scrolling="no" />
<frame src="main.html" name="main" noresize scrolling="yes" />
</frameset>
</body>
</html>In theory, at least, this should make a page with three frames, a row along the top with the Disco logo, a column to the left with links, and a main window in the bottom right.
EDIT:
Killed the body tags. Now I have a page with the disco logo, a list of links, and a "page not found". That last is my fault, haven't made it yet;)
Oh, I also found that adding a second "</frameset>" is a bad idea.
EDIT2:
Final code of the page, please be sure to drop by and see it:
Code:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Frameset//EN" "http://www.w3.org/TR/html4/frameset.dtd">
<!--CS-125H-30 Page Layout with Frames
Author: Aaron Porter
Date: 5 May 2008
-->
<! I set the document type and open the HTML>
<html lang="en-US">
<! Set the title>
<title>
Freelancer/Discovery Datasite
</title>
<frameset rows="25%,*" framespacing="10" scrolling="no">
<frame name="logo" noresize scrolling="no" src="discologo.html" />
<frameset cols="10%,*" scrolling="auto">
<frame src="linkframe.html" name="links" noresize scrolling="no" />
<frame src="main.html" name="main" noresize scrolling="yes" />
</frameset>
<! I close the HTML coding and body and save the file>
</html>Discovery Homework Help Kiosk! - Heartless - 05-11-2008
Hi, more math help plez!:)
I think I'm doing better on these practice regents. But I need help with one thing.
3cos2θ+sinθ-1=0
So to me the sinθ doesn't seem like it's enough, would it be just 1?
Discovery Homework Help Kiosk! - Nevermore - 05-11-2008
Oh, you got a hard nut there:P
Anyway, that's my solution. Basic idea: Use a sum formula for cosinus, use the identity sin^2 + cos^2 = 1, substitute sin(x), solve the resulting quadratic equation and resubstitute.
First off, I don't like to type Theta all the time, so I just use x.
That would make:
3cos(2x) + sin(x)-1 = 0 <=>
3cos(x+x) + sin(x) = 1 <=>
3(cos^2(x)-sin^2(x)) + sin(x) = 1 <=> //Used the sum formula
3((1-sin^2(x))-sin^2(x)) + sin(x) = 1 <=> //Identity cos^2 = 1-sin^2(x)
3 - 6sin^2(x) + sin(x) = 1 <=>
sin^2(x) - (1/6)sin(x) = 1/3 <=>
u^2 - (1/6)u = 1/3 with substitution u=sin(x) <=>
(u - (1/12))^2 = 49/144 <=> //quadratic complement
|u - (1/12)| = 7/12 <=>
u1 = -1/2, u2 = 2/3
Resubstitution:
u1 = sin(x1) <=> x1 = arcsin(-1/2) = pi/6
u2 = sin(x2) <=> x2 = arcsin(2/3) = 0,7297276562...
Edit: Ah, I forget the other solutions... So, everything in the form of 2pi*k + x1 or 2pi*k + x2 are solutions, with k being an integer.