Math Question

spdcbr definitely doesn't have a brain. It surely cannot be possible. I mean, one cannot be so inane. I refuse to believe anything with a brain could be that inane.
 
Thanks guys! I'm pretty sure I've got it now! I had gotten to the isoceles triangles, and I was going to try subbing in 1 for the radius, I didn't really think of using just x. Thanks again for the help!
 
Thanks guys! I'm pretty sure I've got it now! I had gotten to the isoceles triangles, and I was going to try subbing in 1 for the radius, I didn't really think of using just x. Thanks again for the help!

You don't even need to put the radius as anything. x doesn't represent the radius. You just need to know that the radii are equal to each other.
 
spdcbr definitely doesn't have a brain. It surely cannot be possible. I mean, one cannot be so inane. I refuse to believe anything with a brain could be that inane.

I agree with you. He's practically bragging about how stupid he is by insisting he's right, even though people who both understand the problem and are actually good at math have showed that he was wrong from the start.


By the way, this is a very nice problem. The isosceles triangles are not immediately obvious but they make it work out very nicely with very little algebra.
 
So I'm supposed to take my apology back now? (Since Dene the philosopher and Michael have both agreed upon my original claim?)

By the way I didn't want to reveal all of my solution in the beginning because I wanted to give the rest of you here a shot at the question, but oh well :o
 
For those who haven't already found the solution here it is.
In triangle CAB, CA = CB. Theredore ACB=ABC= x This implies CAB= 180-2x.BAE=2x
In Triangle BAE BA =BE. BAE=BEA =2x.
In triangle BEF BEF=BFE=78. EBF=24.
ABE+BEA+EAB=180=EBC+ABE+EBF
2x+2x=x+24
x=8.
 
Hey, i dont want to make a thread for this but my math teacher gave us some really hard problems yesterday and I need help. There not for a grade but there for extra credit.

1. Every month, a girl gets allowance. Assume last year she had no money, and kept it up to now. Then she spends 1/2 of her money on clothes, then 1/3 of the remaining money on games, and then 1/4 of the remaining money on toys. After she bought all of that, she had $7777 left. Assuming she only gets money by allowance, how much money does she earn every month?

2. If 1/2x +1/2(1/2x + 1/2(1/2x +1/2(1/2x + ... = y,
then x = ?

3. One day, a person went to horse racing area, Instead of counting the number of human and horses, he instead counted 74 heads and 196 legs. Yet he knew the number of humans and horses there. How did he do it, and how many humans and horses are there?

4. If ax*x + bx +c = 0,
then what is x?

5. What number shows up most often when you roll 10 dice?


I seem really confused about this so any help is appreciated.
 
5. Weird. It seems almost like a trick question.
Theoretically, there's a 1/6 chance of any number occurring on all 10 rolls, so...

4. x = (-b + sqr(b^2-4ac))/2a or x = (-b - sqr(b^2-4ac))/2a
It's the quadratic formula.

3. x is the number of humans.
2x + (74-x)x4 = 196
-2x = -100, x = 50
50 humans, 24 horses.

1. First, she spends 1/2 her money so she has 1/2 left. Next, she spends 1/3, so 2/3 left. Then, she spends 1/4, so she had 3/4 left.
(1/2)(1/3)(1/4)x = 7777
She gets $31,108 a month. I wish I had that kind of money...

2. Not sure about this one.
 
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There not for a grade but there for extra credit.

No. That's even worse. Math extra credits are meant to be way too hard for the average student to solve. I've done plenty of them at my school. It actually requires you to sit there yourself and think for a few hours.

Your problems actually look too easy to be extra credit. I would have given that as classwork if I'm the teacher.

1. Very very easy. Definitely regular classwork for pre-Algebra.
2. Looks interesting. Just try it!
3. This is a 1st grade problem, literally.
4. That's extra credit? Look up the formula. Google is your friend.
5. See peterbat's comment below.
 
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