Math HELP!

[Excell]

DSOK....how old are you?

[DSOK]

DSOK....how old are you?

Old enough.

Diapadion, my brother who introduced me to The Way, is probably older than most of you, and I'm while he was the youngest in our family, I was the eldest.

[Glenn47]

Old enough....43!

He's 43. Don't let him try to tell you otherwise. Unless he says he's older, in which case, wonder why someone that age still plays rpgs. Not that there's anything wrong with it. It's just not done much.

Glenn's vote: 43

Glenn's flesh wounds: 6

[Non-Sequitur]

I am very sorry, but my professors were not able to do any better than I did. Good luck with it anyway.

[DSOK]

I am very sorry, but my professors were not able to do any better than I did. Good luck with it anyway.

It was a minor point anyway, one which I had already pretty much given up on.

[EmperorJeramyu]

Ok, new one:

((7.899*10^-45)^53*144*22.9898)/((25689)^4*0.00155*(30.48)^7)

And no, the answer isn't 0. No calculators that I have access to can solve this one. We've been told to use our "carbon calculators", but I'm too damn lazy.

[Fitz Tayo]

Tis a tricky one alright.

[TeppicTheBloodLyn]

42, the number is, 42

[Telephalsion]

actually the answer is 0, got to love the TI-82

[EmperorJeramyu]

Try typing in just (7.899*10^-45)^53 on your calc. It'll give you 0, even though that's clearly not the answer. It goes higher than your calculator can display.

[HolsteinCow]

(5!(66.866^2048)pi((990^-6843)x+55346679^-3)x)°100•£33.333333333333333333333333333333333333333333334

Figure that one out.

[Non-Sequitur]

7.7478155406944174 X 10^-2360

I haven't really checked my work yet so I may have made a mistake. I will check it as soon as I can.

[fwacho]

being a C+ math student pretty much my entire life (but still going up to precal as was never one to let a little absolte confusion stop me) I've found that trial and eror are my best friend... simply plug in the numbers and see what happens. (This got me past the SAT and let me clep college algabra) I really am a much better word smith.

after reading all the posts I couldn't tell if anyone actually posted and answer. (It looked like DSOK might have but i wasn't sure)
Teh question stated it needed a"combination of the two coins" so now my philosphy degree kicks in and I ask does that mean..
1) the solution must consider at least 1 "11" and at least 1 "7"
example I could say that "70" can be gotten to by multipling 7x10 and is therfore not apossible answer
2) a solution need not ... wow Ican't even phrase it...
example 70 can not be arrived at using at least one "7" and at least one "11"
3) does it involve apples and oranges because there is sale going on at the supermarket and that would help with visual representation.

This hurts my head so bad and would probably take me about 20 minutes to trial and eror it but I'll stop now.

[Glenn47]

Why in God's name are we still posting here? I think he's turned in the paper by now...

[Iorek]

"A government issues only two kinds of coins: one of 7 units, and one 11 units. Thus, certain prices cannot be paid exactly, for example: 15 units. What is the highest price that CANNOT be paid with any combination of the two coins? HINT: You may use as many of the coins as needed."

A friend and I have been laboring at this for a while, but we're stupid, so we need help. Perhaps a resident genious (Read: DSOK), could help us out, and provide some reasoning behind the answer. Thanks.

I'm currently in Honors Geometry.

First of all, that's a pretty lousy hint, since anyone who reads the problem should know that already.

This problem's very simple, you don't have to use higher level mathematical terms to explain and solve it like DSOK. You simply have to write out every single combination of 7 and 11. It may sound like it would take a long time, but it seems like Jeramyu spent a very long time to arrive at his bushit (yes, bushit) solution anyway.

Honestly guys, if it "hurts your head" and you have to read it "five times just to kind of get it", please don't post in this thread, as you're likely a C+ regular student or a failing honors student. No offense.