Special Topics: Due Monday, Please Help!

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Kamunt

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Feb 11, 2006
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OK, these are the ones that I'm having trouble with...which is all of them. :roll: I tried searching for these questions already on this forum, and something came up on Google, but the post didn't exist when I clicked on the result. :(

1) Mathman goes to Math R Us to buy a textbook and a graphing calculator. He notices that the sum of the list prices of the two items is $132.50. The store manager, Ray D. Cal, will give a 25% discount on the textbook, but no discount on the calculator. After the discount is figured and 7% sales tax is added to the purchases, to Mathman's amazement, the bill comes to exactly the same value as the original sum of the list prices ($132.50). What is the list price of the calculator? Give your answer to the nearest cent.

2) A complex number has the form a + bi. If this complex number is multiplied but its conjugate the result is 221. If the original complex number a + bi is squared, the result is 171 - 140i. What is the value of |a| + |b| ? (I should be able to figure this one out considering that we're on imaginary numbers in Pre-Calculus right now, but I can't. :( )

3) The cost of a gallon of gasoline is due three things (contrary to popular belief, gas prices are not random [yes, this was actually in the question]):
  • The cost to manufacture (drilling, transporting, refining).[/*:m:2kxkjts9]
  • State and local taxes.[/*:m:2kxkjts9]
  • Profits (manufacturers and retailers).[/*:m:2kxkjts9]
Last week, the cost of a gallon of unleaded gasoline at Stuckey's, whose motto is, "Eat at Stuckey's and get gas," was $1.12 (lol). If the governor announces a 15% increase in the gasoline taxes the price would change to $1.19 per gallon. If the profits are to be increased by 15%, the cost per gallon would change to $1.13 per gallon. What is the cost needed to manufacture a gallon of gasoline? Round your answer to the nearest cent.

4) If ( n + [1/n] )^2 = 5 , then find n^3 + 1/(n^3) .

I would graciously any and all help that could be provided, so help me, please!
 
For #2.
\(\displaystyle \L
\begin{array}{l}
z \cdot \overline z = |z|^2 \\
a^2 + b^2 = 221 \\
a^2 - b^2 + 2abi = 171 - 140i\quad \Rightarrow \quad a^2 - b^2 = 171\mbox{ and }2ab = - 140. \\
\end{array}\)

Solve for a and b.
 
#4.

Is that \(\displaystyle (\frac{n+1}{n})^{2}=5\)?.

If it is:

\(\displaystyle \frac{n+1}{n}=\sqrt{5}\)

\(\displaystyle 1+\frac{1}{n}=\sqrt{5}\)

\(\displaystyle \frac{1}{n}=\sqrt{5}-1\)

\(\displaystyle n=\frac{1}{\sqrt{5}-1}\)

If you want to tidy it up you can rationalize the denominator.

Now that you have n, enter into your other equation to find out what it is.
 
galactus: No, I'm afraid. It looks like this:
mathproblem.jpg
The first n is only over 1.
 
Find the value of n and sub into the other equation.

Here's a hint: the Golden Ratio.
 
1) The calculator costs c
The text costs t
c+t=132.50
1.07(c+.75t)=132.50
Two equations in two unknowns.

3) the three variables are m, t and p
m+t+p=1.12
m+1.15t+p=1.19
m+t+1.15p=1.13
Three equations in three unknowns which give last week's amounts.
 
galactus said:
Find the value of n and sub into the other equation.

Here's a hint: the Golden Ratio.
Click to expand...

Sigh, I really feel stupid by saying this...but I don't know how to do that. :cry:

Thanks to everyone for helping me out, though. :D You guys have all been a huge help!
 
Well, Kamunt, fret not. Let's step through it.

\(\displaystyle \L\(n+\frac{1}{n})^{2}=5\)

\(\displaystyle \L\(n+\frac{1}{n})=\sqrt{5}\)

Multiply through by the LCD, which is n

\(\displaystyle \L{n^{2}-{\sqrt{5}}n+1}=0\)

Now, you have a quadratic to solve:

Can you do that?.
 
:eek: It seems so simple now. Sometimes I think I'm too spacey for my own good. Yes, I can handle quadratic equations. Thanks SO much for the help, everyone! :D I wasn't quite sure how patient you all would be, so thanks for being so patient. *bookmarks this site*
 
OK, I'm sorry to bump this topic, and it's a bit late now, but I realized that I never finished doing problem number three on my Special Topics...mainly because I still couldn't quite figure out how to do it. :( A system of two equations works fine for me, but a system of three equations with three variables means trouble for me, apparently.

Gene said:
3) the three variables are m, t and p
m+t+p=1.12
m+1.15t+p=1.19
m+t+1.15p=1.13
Three equations in three unknowns which give last week's amounts.
Click to expand...

Could someone please help me work the rest or most of this problem through? I know that it's too late to fix my mistakes on the S.T. to get me full credit, but I'd just like to know so that if I have the chance to fix it when it's handed back next week, I'll be able to do so. Again, thanks SO much everyone for your help.
 
Just subtract the first from the other two.
Or if you prefer
Solve the first for m.
Substitute that for the m in the other two.
The m's are gone and you are down to two in two unknowns.
 
Ah...of course. :oops: Thanks, sometimes I just need a good kick in the head to get me back on track. Thanks a lot!
 
Hello, Kamunt!

There is a back-door approach to #4.
\(\displaystyle \;\;\)Use it to impress/surprise/terrify your teacher . . .

4) If \(\displaystyle \,\left(n\,+\,\frac{1}{n}\right)^2\:=\:5\), then find \(\displaystyle n^3\,+\,\frac{1}{n^3}\)
Click to expand...
We have: \(\displaystyle \,\left(n\,+\,\frac{1}{n}\right)^2\;=\;5\)

Take square roots: \(\displaystyle \,n\,+\,\frac{1}{n}\;=\;\sqrt{5}\;\) [1]

Cube both sides: \(\displaystyle \,\left(n\,+\,\frac{1}{n}\right)^3\;=\;\left(\sqrt{5})^3\;=\;5\sqrt{5}\)


Expand the left side: \(\displaystyle \:n^3\,+\,3\cdot n^2\cdot\frac{1}{n}\,+\,3\cdot n\cdot\frac{1}{n^2}\,+\,\frac{1}{n^3}\;=\;5\sqrt{5}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;n^3\,+\,3n\,+\,\frac{3}{n}\,+\,\frac{1}{n^3}\;=\;5\sqrt{5}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;n^3\,+\,\frac{1}{n^3}\,+\,3n\,+\,\frac{3}{n}\;=\;5\sqrt{5}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;n^3\,+\,\frac{1}{n^3}\,+\,3\left(\underbrace{n\,+\,\frac{1}{n}}\right)\;=\;5\sqrt{5}\)
. . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle \downarrow\)
Substsitute [1]: \(\displaystyle \:n^3\,+\,\frac{1}{n^3}\,+\,3\sqrt{5}\;=\;5\sqrt{5}\)


Therefore: \(\displaystyle \:n^3\,+\,\frac{1}{n^3}\;=\;2\sqrt{5}\)
 
Oro?!

:eek: Wow, that's really cool. I never would have thought of that. *takes notes on soroban's method in his notebook*

I gotta know real quick, though: how do you guys all put your work into those image files? Like, is there a program you all use that automatically turns what you typed into a handy little GIF or something?
 
Re: Oro?!

"Kamunt"how do you guys all put your work into those image files? Like, is there a program you all use that automatically turns what you typed into a handy little GIF or something?
Click to expand...
It is called TeX or LaTeX.
At the top of the page is a tab “Forum Help”.
On that tab are suggestions for TeX.
If you are a Windows user TeXaide is a free download.
 
Sorry, but I couldn't find the tab you spoke of. If you meant FAQ, it wasn't there, either.
 
Kamunt said:
Sorry, but I couldn't find the tab you spoke of. If you meant FAQ, it wasn't there, either.
Click to expand...

The tab is labeled "Forum Help" and it is at the very top of the screen (Above the cute graphic that says Free Math Help).

I just wish there was some easy (and free!) way for Mac users to take advantage of LaTex......
 
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