All Answers 358145 Answers

Hi Dragonlance,

I do have an account on here --  I’m GingerAle.  The new forum kicks me off as soon as I log in.  I don’t know why.   Maybe it doesn’t like ginger ale. 

I have read about CPhill and his quest for the Roman Zero. I’m sure someday he will find it.

I’ve read about his fine art of coconut chopping too 

 Your life seems very interesting –you seem very interesting. 

Your children’s church story reminds me of funny stories from when I was in Sunday school and church. I think if I went to church with you I would never forget you.   I would remember you with a lifetime of laughter. I will anyway. 

I'm looking forward to your next story posting.

Take care, Lance.

If I get my account working again, I’ll send you a message. 

10+9=19 which is correct

I'm not quite sure what you're looking for, but try this one.

For the given hyperbola, (though the result is true for any hyperbola, prove the general result if you want), show that if the tangent at any point P cuts the asymptotes at Q1 and Q2, then P is the mid-point of the line joining Q1 to Q2.

@@ What is Happening?  [Wrap4]   Wed 30/12/15   Sydney,  Australia Time 11:00 pm   ♪ ♫

Hi all our dedicated Christmas forumers.   

Well we have had some fabulous answers from CPhill, Solveit, Alan, tiferon2 and geno3141.  Thanks guys  

Technical Issues:

How can you post a picture when the picture upload on the forum is not working properly ?

Thanks very much Alan for these instructions.

http://web2.0calc.com/questions/ven-diagram-need-help-for-questions#r3

Interest Posts: 

If you ask or answer an interesting question, you can private message the address to me (with copy and paste) and I will include it.  Of course only members are able to do this.  I quite likely will not see it if you do not show me.  

1)  Happy birthday Gino   

Yes and that may have been a good answer too guest.    

The slopes of the red and blue triangles are different.  When the two sets are plotted together the "hypotenuses" have a gap between them (see below).  I'll leave it to others to confirm that the area of the gap is one square unit.

You are right Alan

I meant Standard Form NOT standard notation.

All these terms were tossed around when I was a kid and it was very confusing.   

Thanks Guest, it is displaying now and I looked at what you presented.

Yes I like that, I'd like some more to them to play with though. ://

Thanks guest, I would be very interested in what you have to say but your post will not display properly.

It did display for about 2 seconds but then it disappeard again.   

Try: http://onlinegraphingcalculators.net/texas-instruments/ 

how do you calculate (6.3)*(-0.8)*(0.7)

(6.3)*(-0.8)*(0.7) = -3.528

Well if you have a prism and the front fac is fixed but the legth of it can change. 

Like different lengths of a standard plumming pipe for instance.

Then  if the length doubles the weight will double also etc.

I wonder if that helps you?

Hi Melody.

You probably have this sorted by now, but if you haven't here's the usual approach.

Begin with a change of variables, let

\(\displaystyle X=x+2\qquad Y=y-5\)

so

\(\displaystyle \frac{X^{2}}{121}-\frac{Y^{2}}{81}=1\) ,

which in the limit (sort of) as \(X\) and \(Y\) go to infinity becomes

\(\displaystyle Y = \pm \frac{9}{11}X\) .

Those are the asymptotes relative to the \(\displaystyle X\) and  \(Y\) axes.

Now switch back to the original co-ordinate system,

\(\displaystyle y-5 = \pm \frac{9}{11}(x+2)\) .

Tha Auxiliary Equation is

\(\displaystyle m^{2} + 1 = 0\) ,

which has roots

\(\displaystyle \pm \imath\) ,

which means that the general solution is

\(\displaystyle y = Ce^{\imath x}+De^{-\imath x}\)

which is more conveniently written as

\(\displaystyle y = A\cos x+B \sin x\) .

A comment on this statement of Melody's:  "This is in scientific notation (or standard notation if you are from the United Kingdom)"

An internet search reveals some ambiguity in the use of the term "standard notation".  

Some sites use it as an alternative to "scientific notation", as noted by Melody.

Others use it to distinguish it from "expanded notation":  e.g.  365 = standard notation.   3*100 + 6*10 + 5 = expanded notation.

Some sites use "standard form" as the equivalent of "scientific notation".

I'm from the UK and, when I was at school, only ever used the term "scientific notation" (mind you, that was in the days when we did mathematics by chipping blocks of stone with sharpened pebbles!).  Numbers that weren't in scientific notation weren't given a name at all!

Names aren't important (unless some idiotic examination boards use them as questions); concepts are.

I would've said with your fingers. :)

0.2 x (35/15) ^1.8

enter it just like this into the input box

0.2*(35/15)^1.8     [enter]   you may need to press = on the actual calc 

4*2-((-1)to the 3rd -3) to the third = ?

I am never sure whether a power of a third means 'to the power of 3' or to the 'power of 1/3'

After all, if you occupy the third spot in a queue you are number 3 in the queue.

\(4*2-((-1)^3-3) ^3 \\ 8-(-1-3) ^3 \\ 8-(-4) ^3 \\ 8-\;-64 \\ 72 \)

OR

\(4*2-((-1)^{1/3} -3)^{1/3}\\ =4*2-(-1 -3)^{1/3}\\ =4*2-(-4)^{1/3}\\ =8-(-\sqrt[3]{4})\\ =8\;+\;\sqrt[3]{4}\\ \mbox{This is the only REAL solution but there are other complex solutions as well} \)

Thank you very much, Guest.

Tammy drives the distance of 48 km at an average of 72 kph, therefore it takes her:

48/72=2/3 of an hour, or 2/3 X 60=40 mintes.

On her return trip, it takes her: 40 - 10=30 minutes or 1/2 an hour. Therefore, her average speed on the return trip, will average: 48 / (1/2)=48 X 2=96 kph.

Solve ( d^2 y(x))/( dx^2)+y(x) = 0:
Assume a solution will be proportional to e^(lambda x) for some constant lambda.
Substitute y(x)  =  e^(lambda x) into the differential equation:
( d^2 )/( dx^2)(e^(lambda x))+e^(lambda x)  =  0
Substitute ( d^2 )/( dx^2)(e^(lambda x))  =  lambda^2 e^(lambda x):
lambda^2 e^(lambda x)+e^(lambda x)  =  0
Factor out e^(lambda x):
(lambda^2+1) e^(lambda x)  =  0
Since e^(lambda x) !=0 for any finite lambda, the zeros must come from the polynomial:
lambda^2+1  =  0
Solve for lambda:
lambda = i or lambda = -i
The roots lambda  =  ± i give y_1(x) = c_1 e^(i x), y_2(x) = c_2 e^(-i x) as solutions, where c_1 and c_2 are arbitrary constants.
The general solution is the sum of the above solutions:
y(x)  =  y_1(x)+y_2(x)  =  c_1 e^(i x)+c_2/e^(i x)
Apply Euler's identity e^(alpha+i beta) = e^alpha cos(beta)+i e^alpha sin(beta):
y(x)  =  c_1 (cos(x)+i sin(x))+c_2 (cos(x)-i sin(x))
Regroup terms:
y(x)  =  (c_1+c_2) cos(x)+i (c_1-c_2) sin(x)
Redefine c_1+c_2 as c_1 and i (c_1-c_2) as c_2, since these are arbitrary constants:
Answer: | y(x)  =  c_1 cos(x)+c_2 sin(x)

1/72×48=2/3hrs=40min

Home

48km in 1/2 hour

96km/hr

My explanation is a bit scant cause I'm on a phone. Ask if you don't understand ;)

I started by thinking it had sin or cos in it and I narrowed it to y=asin(ix)

Where a is any real number.

I think that is right be maybe there are lots of answers.

Here is another one

Y = ke^(ix+b)

Where k and b are any

real numbers.

Maybe they could be complex numbers too ? Not sure....  

What is X+X+6=2X + 6=2(X + 3)

Learn to use the calculator here. Enter 47 first, then press ! to left and then the = key TWICE. It will give you the answer in scientific notation. Try it.