Dedicated Mathematician

[EvoN1020v]

Any comments on post #75 please?

 

I looked on the Internet for geometric sequence formula, and it appeared to be the same with my teacher's.

 

I input my answer on the graphic calculator and it gave me the perfect number of 52, 470.

 

Any help?

[Connor]

if you use n rather than n-1 then the population starts not at the original amount, but the amount on the second day, so you should use your teacher's equation

[Connor]

so... any more problems for us?

[CanadaAotS]

I had that stupid cube cut in half problem when I did the fermat's last year. Exact same problem needless to say I didnt get the answer lol

 

I have a problem from another fermat exam I found on the internet (section c).

 

Bill and Jill are hired to paint a line on a road. If Bill works by himself he could paint the line in B hours. If Jill works by herself, she could paint the line in J hours. Bill starts painting the line from one end, and Jill begins painting the line from the other end one hour later. They both work until the line is painted. Which of the following is an expression for the number of hours that Bill works?

 

(A) [math]\frac{B(J+1)}{B+J}[/math] (B) [math]J+1[/math] © [math]\frac{BJ}{B+J}+1[/math] (D) [math]\frac{B+J-1}{2}[/math] (E) [math]\frac{B(J-1)}{B+J}[/math]

[EvoN1020v]

I assume the answer to be A, because of the hint: Jill painted the line after 1 hour.

 

Am I right? I only took a few minutes to think about it, because I actually have to go sleep now.

[CanadaAotS]

Could be, but I dont have the answers

[The Thing]

Err, hokay. Let's try this.

 

So let x be the time during which Bill works.

So x-1 will be the time during which both Bill and Jane work.

 

In 1 hour, bill completes [math]\frac{1}{B}[/math] of the job, and same with Jane, completing [math]\frac{1}{J}[/math].

 

So, multiplying them by the time x and x-1 (for Jane).

 

[math]\frac{x}{B}+\frac{x-1}{J}=1[/math], 1 being the entire job, the entire wall being painted.

 

Err I think all to do now is to simplify and solve for x:

 

Multiply the equation by BJ to get rid of the denominators, ending up with [math]xJ+xB-B=BJ[/math].

 

Then factor out the x, and isolate it: [math]x(J+B)-B=BJ[/math], equals to [math]x=\frac{BJ+B}{J+B}[/math].

 

Factor out the B on the top and we're left with [math]\frac{B(J+1)}{J+B}[/math]

 

I think that's A. Assuming I'm correct, then EvoN1022v is correct as well! But he gets all the credit for answering first. How very unjust.

[EvoN1020v]

I love The Thing's method of answering the question! Very brilliant. I solved the question using my locially mind, and estimate the amount of time Bob will paint the wall. Just a couple of scribbles, and a picture on a looseleaf is just good enough for me.

 

I have a question:

If [math]S_{n}>100[/math], what is the least value for n in the provided sequence?

[90, 45, 22.5,...]

 

The formula to solve this kind of question is: [math]S_{n}=\frac{a(1+r^{n})}{1-r}[/math] where a =first term, r =rate.

 

So input the values:

[math]\frac{90(0.5^{n}-1)}{-0.5}>100[/math].

 

I feel like a total idiot because I can't seem to figure out how to solve for n. I know that you have to use logarithms to solve the n. My teacher gave us bunches of questions except this one, because she said that we don't need to know how to do it. I said yeah right, I want to do it to challenge myself. So any help how to solve for the n?

 

The n should be at least 2.

[Connor]

I'm not sure what you're looking for, the lowest possible value in the sequence? Because it would approach 0.

[EvoN1020v]

Can't you read at all, Connor? I said, "any help how to solve for the n?" in post #83.

 

Just show me the steps how to solve the n. Because I know the n have to be at least 2 for the answer. Ok?

[Connor]

okay, I understand

 

90 + 45 = 135 > 100, but you knew that

 

generalized...

 

 

[math]S_{n}=\frac{a(1+r^{n})}{1-r}[/math]

 

[math]r^n=\frac{S_{n}(1-r)}{a}-1[/math]

 

[math]n=log_r(\frac{S_{n}(1-r)}{a}-1)[/math]

 

I guess just round up. Haven't tried it, too tired

[EvoN1020v]

I did exactly what you did! But apparently one of the number became negative, so the logarithm won't work.

[EvoN1020v]

[math]n=log_r(\frac{S_{n}(1-r)}{a}-1)[/math]

 

I got [math]log_{0.5}-0.444444[/math]

 

So the answer have a negative number, so "math ERROR" appeared on my calculator. Any help please?

[EvoN1020v]

Hello? Are you dead?

[Connor]

huh? oh, sorry, I'm real busy, but I'll take a look at it... eventually...

[EvoN1020v]

 

[math]n=log_r(\frac{S_{n}(1-r)}{a}-1)[/math]

 

 

It should have been [math]n = log_r(1 - \frac{S_{n}(1-r)}{a})[/math].

 

It will yield [math]n = 1.169926444[/math].

 

For those who are in confusion: r = 0.5; a = 90; Sn = 100.

[EvoN1020v]

I apologize for my tardiness, because I have been busy with scholarships lately.

 

Here is the next question:

 

A right triangle in the first quadrant is bounded by lines y = 0, y = x, and y = -x+5. Find its area.

 

I know this is a easy math question, but I'll post harder one next time.

[EvoN1020v]

Anyone?

[Klaynos]

integrate wrt x between 0 and 5....... if you want to do it the hard way....

[The Thing]

Aaarg I died horribly at today's Cayley contest. On one question, it wasn't until now, about 3 hours after the contest, did I remember that 0.56cm does NOT equal 56 millimeters. By far the worst reason to lose a whole chunk of 8 marks.

So for the question above, the three vertices are (0,0), (5,0), and (5/2,5/2)? Then just use those vertices to find the area.

[Connor]

6.25?

[The Thing]

That would be what you get using those three vertices. Assuming that those three vertices ARE correct. And after that absolute fiasco regarding the contest, someone should check whether I have the right vertices.

[Connor]

I didn't see your post, and got the same thing

[EvoN1020v]

6.25 units is the correct answer. Good job Connor. I'll post another math problem next week, or you guys can provide one. I'll be busy this weekend because I have Provinicials wrestling tournament.

 

Note to The Thing: My younger brother did the Cayley mathematician contest too, but Math is not his favorite subject. I just wished him some good luck.

[EvoN1020v]

I am wondering how do you find the maximum/minimum of a quadratic equation? I did some exploring on the Internet, and I found out that you can find the devriative, and from that, you'll have the x-symmetry, and then you'll get the value of the maximum/minimum.

 

Can anybody give me an example, and tell me how to find the devriative for it?

 

Thanks.