dansmath > ask dan > questions and answers page
 
 
dan's questions and answers page
I have expanded this section; check out the new questions, or ask your own!
Also you might want to look at some reader comments.
 
Basic Skills Questions (or go to basic lessons)
B1 - Writing a googolplex
B2 - Fraction woes
 
Algebra Questions (or go to algebra lessons)
A1 - I need your help fast
A2 - The 'two trains' problem
A3 - Miminizing a weird function
 
Trig Questions (or go to trig lessons)
T1 - Sine, Cosine, Tangent
 
Calculus Questions (or go to calculus lessons)
C1 - Graphs of functions and derivatives
 
Miscellaneous Questions (or go to 'other' lessons)
M1 - Profit curve for econ?
M2 - Too big for calculator!
M3 - Scarecrow was not a wizard
M4 - Good grapefruit prediction
 
 
Basic Skills Questions
Question B1 - Writing a googolplex [ top of page ]
Dear Dan - Our math professor at U of Louisville posed this question:
If one million zeroes can be written on the front and back of a sheet of
paper, how many sheets of paper are necessary for a googol of zeros?

We have to come up with answer
(which was not a problem - it's 10 to
the 94th power sheets of paper) and we have to be able to explain the
problem and solution to a 12-year old. How do you explain such a
large number to someone - what can you relate it to? - Karen
 
Hi Karen -- Thanks for writing -- interesting question. A googol is 10^100 and a 1 with
a googol zeroes after it is called a 'googolplex.'
 
I thought maybe you could relate the thickness of the paper (a ream of paper, 500 sheets, is
around 2" so each sheet is about .004" thick) and a large distance, say from the earth to the
moon (around 240,000 miles). I'll let you figure out how the number of sheets of paper
from here to the moon compares with 10^94 (still nowhere close?), so talk about how many
trillions of stacks of paper like that you'd need?
 
Another aspect is time: if you wrote 1 number per second (or a printer spit out one sheet
per min), how many centuries would it take if all 6 billion people wrote (or had printers)?
By the way, 1 million seconds = 12 days and 1 billion sec = 31 years, approx!
 
Ask the 12-yr-old how far away tjey've traveled (say New York, 3000 miles), then maybe
relate the earth-moon distance to that (80 times as far). -- Hope this helps!
Meanwhile don't waste paper writing out a googolplex in base 10. ;-}
-- Dan the Math Man -- www.dansmath.com

 


Question B2 - Fraction Woes [ top of page ]
Hello Dave, I am a 7th grader and having a hard time with fractions.
Could you point me in the direction of a helpful site? Thanks, Will
(Coming to you from Colorful Colorado, home of the Awesome Avalanche!!!)
 
Hi Will... (My name's Dan but I'll answer to Dave...)
I was going to point you to my site, www.dansmath.com , but you're right, I don't
have much currently there on fractions. I will give you some pointers but I would
suggest www.google.com -- type in 'Fraction math help' and I swear it will give
you some excellent active choices. Look for some on-line quizzes as well.
Here's one I just found: www.mathleague.com/help/fractions/fractions.htm
You can also go to 'expert sites' such as www.askme.com and look for message
items or ask questions there. I'll be happy to answer specific questions or problems
for you too.
 
As for fractions themselves: here's a quick review: The top is the 'numerator', and the
bottom is the 'denominator'; in 5/7 the 5 is the numerator and the 7 is the denominator.
 
Fractions with the same denominator can easily be added or subtracted: keep the same
denominator and add or subtract the numerators. 5/7 + 3/7 = (5 + 3)/7 = 8/7 or 1 1/7.
 
Fractions can be converted or reduced: 4/6 = 2/3 (divide top & bottom by 2),
or 5/7 = 15/21 (multiply top & bottom by 3).
Fractions with different denoms have to be converted to a common denom (LCM)
2/5 + 1/3 = 6/15 + 5/15 = 11/15 , etc.
Multiplying two fractions: multiply tops and bottoms separately:
2/3 * 5/8 = (2*5)/(3*8) = 10/24 = 5/12.
Dividing two fractions: iinvert the second fraction and then multiply:
(4/9) / (3/7) = (4/9) * (7/3) = 28/27 or 1 1/27.
Then there's the whole connection with fractions and decimals and percents...
 
--Dan the Math Man

 

 
 
Algebra Questions
 
Question A1 - I need your help fast [ top of page ]
Hi Dan, I live in Cleveland, Ohio, and I need your HELP.
Can you teach me algebra? I need to learn, and learn fast,
I am applying for an apprenticeship program and the
requirements are algebra. I really need this opportunity,
so can you please teach me? -- Brian
 
Hi Brian - Algebra is a very important subject ; I call it the 'gateway to the
technical world'. If this apprentice program is asking you to know algebra, it
must mean it's for a job that uses it and depends on it. This means you can't
learn it hastily or all at once. If I wanted to be fluent in speaking Russian, I
wouldn't be able to do it in a few days or weeks.
 
But I do have a lot of lessons on my website, in all levels; arithmetic, prealgebra,
beginning alg. and on up. Check them out at www.dansmath.com -->
math lessons --> Basic Skills. There are also lots of links on my site or go to
www.google.com and type 'algebra help.' It's a great way to find stuff.
 
I also have a great textbook that could be used for self-study, check it out at my
site at 'meet dan' then click textbook. If you want to order the book online, get
the details ready (Bach/Leitner Prealgebra, Houghton Mifflin) and then go to
bookcenter.dvc.edu and order with a credit card, they ship! -- DaMathMan

Question A2 - The Two Trains Problem [ top of page ]
Dear Dan - two trains leave stations 100 miles apart.
"A" train leaves at 12:00 and "B" train leaves at 1:00.
when, and at what mile, will they collide? there will be
no loss of life. > mark
i would appreciate any help. big test on sat am. thank you.
Hi Mark - I mean mark.
If we knew the speed of each train then we could set up distances:
t = hours past 12:00 ; and the trains have gone dA and dB:
dA = (speed of A)*(t hours) ; dB = (speed of B)*(t - 1 hours)
because B starts at 1:00, 1 hr later, so 1 hr less time. (both are d = r t.)
 
Now our equation is dA + dB = 100 miles, but we can't solve for t unless
we have some real numbers for those speeds. If you forgot to include those,
that's where you'd use them; if the speeds weren't given in the problem
then it wasn't a well-phrased question.
 
One thing we do know is if A is going more than 100 mph, the collision
will occur at B's home station, maybe with loss of life! - Dan the Math Man

 


Question A3 - Miminizing a Weird Function [ top of page ]
Dear Dan, How would you find the minimum of
y = square root of (x^2 + 4) + square root of (x^2 - 6 +10).
Do you have to use calculus? - Tara
 
Hi Tara.
You can use calculus, to minimize y = sqrt(x^2 + 4) + sqrt(x^2 - 6x + 10)
by taking the derivative dy/dx and setting it equal to zero. That looked icky,
but do-able. But you'd prefer a calculus-free approach...
 
Or you can type the function into your graphing calculator or some
computer algebra system like Mathematica (see my website), and try
to spot a min. This one looks like x = 2 gives a smallest value of about
y = 4.25 or so. Then you could go back to the formula and see if you
could complete the square to see the min at x = 2:

y - sqrt[x^2 + 4] = sqrt[(x-3)^2 + 1] ; square both sides, simplify,
isolate the square root term, square both sides again, and the result should
have some sort of factor of (x - 2) if my intuition is right.
 
I'll let you have fun with the algebra. Hope this helps! - Dan the Man

 

 
 
Trigonometry Questions
 
Question T1 - Sine, Cosine, Tangent [ top of page ]
Dan, please help!! I have been trying to figure out sine, cosine, and tangent
for about 2 months, and I can't quite get it. I just don't understand how to
get to the answer from: sin/cos/tan(angle)=(side length over side length)

I also do not know how to do sin/cos/tan when you only have one side length.
Please, please, please help!!! ~ Michelle ~
 
Hi Michelle - Have you heard of 'SOHCAHTOA' ?
Draw a right triangle with right angle at the lower right; the angle T at the left;
the Opposite (vertical) and the Adjacent (base) and the Hypotenuse (rising diagonal).
 
It means Sine = Opp / Hyp , Cosine = Adj / Hyp , Tangent = Opp / Adj .
You might like sin(T) = y / r , cos(T) = x / r , tan(T) = y / x .
 
Along with the Pythagorean Theorem x^2 + y^2 = r^2 , you can figure
out the whole story just by knowing any one side, and one other angle.
 
Please visit my (free!) web lessons at www.dansmath.com for more ; go to the
Precalculus area and click Trig. There are pictures and even some animations
there for you to enjoy / learn from. And write back any time with further or more
specific questions.-- Dan the Math Man strikes again
 

Calculus Questions
 
Question C1 - Graphs of Functions and Derivatives [ top of page ]
Dear Dan: Could you explain the principles on graphs with
relation to their derivative? - Shane
 
Hi Shane - Happy to 'explane'; thanks for asking.
 
The derivative of y = f(x) at a point (a, b) is the slope of the tangent line
to the graph at the point (a, b). This means that b = f(a) and that the
graph and the tangent line have the same slope at x = a.
 
But the graph of the derivative is different; it records the slope of the graph
at each point, so if the function is increasing, the derivative is positive, and
if the graph of f(x) is decr, then f '(x) is neg.
 
I have a picture on my site of a graph, its deriv, and second deriv. The link is:
http://home.earthlink.net/matica.html ; go to the 'options for graphics' section.
You've inspired me to make a movie of a moving tangent line along a curve
y = f(x), and a second curve recording the slope f '(x) as it goes along.
 
Hope this helps a little; try graphing a few pairs {function , derivative}
on your graphing calculator, like {y = x^2 , y = 2x} to see the relation. -Dan

 

 
 
Miscellaneous Questions

Question M1 - "Profit Curve" [ top of page ]

Dear Dan, Can you tell me how to start and understand how to solve
(simple) economicproblems using curves and diagrams? If the volume
goes up, I realize the price goes down, but how do I state these
circumstances on a graph? --Failed Economics
 
Dear Econ -- I'll answer you with an example.
 
Let's call the volume or quantity 'q' and the unit price 'p', where p is a function of q.
p = f(q) in function notation.
The revenue R is how much money you take in; R = p * q = (price per item) * (number of items).
 
Fixed Cost: For example if I sell q of my Dan's MathClinic CD's, and the price is $50 each,
then p = 50 and the revenue is R = pq = 50q.
This makes a simple revenue curve (in blue in fig.1).
If my costs C(q) are 300 + 10q (see red curve in fig.1; $300 for the CD writer and $10 each in costs per CD),
then my profit P(q) = R(q) - C(q) = revenue minus cost (see dotted line).
Here P(q) = 50q - (300 + 10q) = 40q - 300. Thus if 40q - 300 > 0 , I make a profit.
Solving, 40q > 300 ; q > 300/40 = 7.5 ; I'd need to sell at least 8 CD's to make a profit.
There's no maximum profit here; the more I sell, the more I make.
 
Variable Cost: Like you say, Econ, normally the price goes down as q goes up.
Let's say I lower my price by $2 each (for all the CD's), each 10 CD's I sell.
This means if I sell 10, they're $48 each, sell 20, they're$46 each...
Now p = f(q) = 50 - (2/10)q = 50 - q/5. Then R = pq = (50 - q/5)*q = 50q - (q^2)/5.
This graph is shown in fig.2. This revenue curve has a max at q = 125, p = $25,
R = ($25)(125) = $3125 = max revenue.
If we work in the cost C(q) = 300 + 10q, the profit will be
P(q) = (50q - (q^2)/5) - (300 + 10q) = - (q^2)/5 + 40q - 300.
The max profit is where the revenue curve and the cost curve are farthest apart,
which is at q = 100 ; P(100) = R(100) - C(100) = 3000 - 1300 = $1700 max profit.
 
Another thing is that the slopes of the curves are equal at the max profit point;
that slope is called the "marginal"; the marginal revenue varies, while the
marginal cost is always $10. This is usually called MR = MC (curves are parallel).
 
What price maximizes revenue? If q = 125 then p = 50 - q/5 = $25;
while with a $10 unit cost, if q = 100, then max profit is p = 50 - 20 = $30. -- Dan
 

Question M2 - Too Big for my Calculator! [ top of page ]
Dear Dan, wrote an AOL member, Help me on this:
"Evaluate 514^623 to 5 significant digits. Use scientific notation.
Show work." I tried this but my calculator overflowed.
 
Dear Digits,
 
After you clean up your soggy calculator, you can use logarithms (logs) to solve your dilemma.
The "inverse function" principle says that 10^(log(n)) = n , where log(n) = log base 10 of n.
For example log(100) = 2 because 10^2 = 100 , so 10^(log(100)) = 10^2 = 100.
 
This means that your base 514 = 10^(log(514)).
Thus 514^623 = (10^(log(514))^623 = 10 ^ (log(514) * 623) ,
by the exponent law (a^m)^n = a ^ (mn).
 
Therefore since log(514) = 2.710963119 (plenty good for 5 sig figs), we get
514^623 = 10 ^ (2.710963119 * 623)
= 10 ^ 1688.930023
= 10^0.930023 * 10^1688
= 8.5118337 * 10^1688
= 8.5118 * 10^1688 , rounded to 5 sig figs.
 
This means 514^623 = 851,183,...,..., ,,, ,544. (It has 1689 digits.)
I worked out the last 3 digits for you as a bonus, keeping just the last
three digits as I raised (514^7)^89 = (...504)^89 etc.
 
Happy mathing! -- Dan the Man -- %;-}
 
P.S. In case you want the whole number, I put it into Mathematica:
 
514^623 =
8511833779452615656486074261588685321885848269727007926473787021726299150613454731118227788880566249
9579341888420948898108481304117665427347156460108301497273089253879842812864182743351005163472035436
9091698147158225289509378576257427198682776595751872026868258927180938200163803994427647392715183966
8438066177414163107908914764766832312856288614285766846216688067422913987245591809525756132033707727
3261381164233095129638631892480829669588620858837489833863435391375254264716570918878595804922844974
2310418481180889249912455348674673062108084846174355138107409273786376749378544392942724724802378363
6951658353324022475138456232761933645669704603048983352602514049705723158039218408255041542765534406
4868815917217198525065278490345695277562166592816967071940449517006873597453967132291299704130560450
5723279949674682963922922526606392473508756303281655716042245944851970688413823936680306848990052843
4172747125372607254400181641278574025661557833730883587376180848061056695516518084197482071334661602
9788779118737691122722049733546225166341998238481750889164581840259239162814525821884767794193810740
1944924038500661882235340706891647456413345490150061465554646972653249409508235179697455196265802208
1857783642375526377369578939516800481364204375582244936236988756677536432530501400853958311462717242
5921231278348064084321169911247088080501465108150050001777239037687892455271721936588067472562372013
9226797016031858987695906723992518615071128448123951828909483023185215617848926913001735240233431571
8047094403631130175011120909154066915853750235015414384029541658961970066661248857253769694559849148
49683683205294629570227145647934661584612656822106000776999057985958189406303177926508544.

 

 

 
Question M3 - Scarecrow was not a Wizard! [ top of page ]
Adam, a student of mine, wrote: Hi Dan,
I don't know if you've seen the movie The Wizard of OZ, but at the end of
the movie when the Scarecrow receives the info that he's already had a brain,
he says this, "The sum of the square roots of any two sides of an isosceles
triangle is equal to the square root of the remaining side. Oh joy, rapture!
I've got a brain!" But this is only true when you have the two sides and are
finding the hypotenuse. If you have the hypotenuse and are finding one of
the other legs then you subtract, so..if the writer of The Wizard of OZ knew
anything about math he woulda said,"The sum of the square roots of the two
legs not including the hypttenuse an isosceles triangle is equal to the square
root of the remaining side. Oh joy, rapture! I've got a brain!" I guess he
really didn't get a brain after all. Hehe... Adam
 
Hey, Adam. Thanks for the reference. Yes I 've seen that scene many times, and it always
bugged me, not because of any subtraction you might need to do, but because of the incorrect
use of the word "isosceles," which isn't "right." But you're right, he shouldn't have said "any
two sides." But he also should have said "square" of the sides, not "square root. -- Dan
 
P.S. I sent this into the "Internet Movie Database Bloopers page."
Ok, what about an "isosceles right triangle"? Are there any? Are there any with all three sides
integers? (Pythag. triple) Anything close, within 1 degree? Think about these, ok?
Also, you can look up the real Pythagorean Theorem on my Trigonometry page.

 

 

Question M4 - Good Grapefruit Prediction [ top of page ]
Dear Dan - Have a question, if you can help me.
It is estimated that 75% of the grapefruit crop is good, the other
25% have rotten centers which cannot be detected unless the
grapefruit are cut open. The grapefruit are sold in sacks of l0.
If you buy one sack, what is the probability that your sack will
contain at least 9 good grapefruit? What is the average number
of good grapefruit per sack? - Thanks; Wolfe
 
Hi Wolfe - Thanks for writing.
This is a question about 'binomial probability', meaning there's a repeated experiment
with probability p, like n repeated coin flips with p = 0.5. Here the experiment is
'choosing a grapefruit', and the probability of success is p = 0.75, with q = 1 - p = 0.25.
 
The average number of 'good' grapefruits in a bag of n = 10 would be m = np = 10(0.75)
= 7.5 good ones per bag (75%). There'll never be 7.5 good fruits but that's the average in
the long run.
 
The number of ways of '9 fruits being good' is 10: any of the 1st thru 10th g-fruit could
be the bad one. And the prob of each case happening is p^9 * q^1 because there are 9
good fruits and 1 bad. So we have P9 = (prob of 9 good) = (10) * (0.75^9) * (0.25^1);
P9 = 10 * 0.07509 * 0.25 = 0.1877, or about an 18.8% chance.
 
The chance that all 10 are good is P10 = 1 * (0.75^10) * (0.25^0) = 0.0563. This must
be added to the previous number: 0.1877 + 0.0563 = 0.2440 , or about 24.4%.
 
There are also tables that can give you this value directly. look for 'binomial probability
tables' in the back of your book. Hope this helps! -- Dan the Stat Man
 

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