Not a mastercam question, but....

[Bill]

Could the teacher have possibly wanted to see if kids know that 8 being outside of the () should be calculated (X) to each integer hence 160+8y+8x. It seems to me 7th grade is just starting algebra and the basics.

And 1/3 multiplied by 3 = 1 but 1 divided by 3 =.33333... and .3333~ x 3=.999~

Have a good one!

[gstephens]

I couldn't resist this one...

.99999... = 1

Proof:

let n = .9999...

10n = 9.9999....

10n - n = 9.9999... - .9999....

9n = 9

n = 1

(Also a B.S. in Pure Mathematics who thought about teaching)

 

[Guest CNC Apps Guy 1]

I was wondering when you'd chime in on this gstephens. Nice addition.

I dread it when my son asks me math questions. Usually I have to get on the computer (Excel or Mastercam) and do the equasion, then break it down on paper.

This forum is great. I learn something in here every day.

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James M.

[Andy]

Josh,

First this is not an equation (no = sign)

Therefore the teacher is not looking for a solution for x or y.

I think what is presented implies they want the student to do distributive multiplication to learn how to deal with paranthesis.

Hence: 8(20+y+x) is 8*20 + 8*y + 8*x

or simply: 160+8y+8x

as Bill noted.

Good luck,

Andy

Hey Bill, yes, change your career to math!

------------------

Gordon Engineering

Eureka, Ca.

V8 Lathe and Mill

[s0l0seven]

hay guys how come its not 160xy

8(20+y+x) = 8(20xy) = 160xy

That’s my story and I’m sticking to it

[s0l0seven]

i will study harder

[This message has been edited by s0l0seven (edited 06-07-2001).]

[biss03]

hey seven,

that will work if your allowed to change it from addition to multiplication (i don't see that anywhere in the math rules).

to everyone who responded,

thanks for all your input. I think that just about all of the answers are probably acceptable. unfortunately it wasn't my child who needed help so i don't know much about the type of math class it was. all i know is it was a review question for the final exam. I think maybe the teacher was having thoughts of what they were going to do during there next 3 months of no work(wouldn't that be nice!!) when this problem was created.

anyways thanks again and don't stop posting if you have new ideas!

josh

[Bill]

You are welcome. And to think if I had stayed with it (teaching degree) I would be having a summer off also. DOH! DOH! DOH!

Oh well back to programming.

[Simtech]

Here's another one:

3 Mastercam programmers go to a Mastercam conference and book a room for $30. They pay each $10 for the room. The owner of the place decides to credit them $5. So the bell boy (who didn't finish his math class) decides to pocket $2 and give a $1 back to each programmer.

If each programmer paid $10 and was credited $1, that means they paid $9 x 3 = $27. The bellboy pocketed $2 making it $29. Where's the other dollar????

[altimu]

To easy! The bellboy pocketed $2 so you subtract that from the $27 to get $25 which is what was paid. You subtracted the one dollar for each programmer then tried to add the $2 for the bellboy - Not possible!

[Jody]

No! What really happened was the Gov't took the other dollar. But what i don't get is that they normally take more.

Jody

[Guest CNC Apps Guy 1]

LOL!

.....usually take more......

 

LOL!

------------------

James M.

[Simtech]

Here's yet another one,

Two walls are separated by a distance "x".

To determine this distance, a contractor placed a 10 ft ladder at the bottom of the left wall and let it rest on the right side wall.

With a 12 ft ladder, he placed the ladder at the bottom of the right side wall and let it rest on the left side wall.

He mesured the intersection height "h" of the two ladders from ground level to be 6 ft.

How would you calculate the distance "x" between these two walls?

[BernieT]

Simtech, that's a 'doosy'!!

I'll have to work on that one all weekend!!

BerTau

[marco]

Simtech,

I'm still working on that problem.

[gary dunn]

dern, i forgot how to do that kind of stuff 30 years ago. once we started using computers, that made it even worse. now i can barely add on paper.

[Mike-T]

Is this a trick question ??

oops....had to re-read the question !!

[This message has been edited by Mike-T (edited 06-08-2001).]

[Michael Reynolds]

Hey Simtech, is the answer 93 inches??

Mike R.

[Guest]

i dont think thats physically possible with the dimension you gave.

but if it was you can determine the distance by solving the simultaneous equations from the similar triangles created by the intersection points

[Michael Reynolds]

Hey there, scratch my last message, stupid mistake. This is driving me nuts, though. It better not be a TRICK question......!!!

 

Mike R.

[Ambassador]

If the contractor had a tape measure long enough to measure the height of the intersection of the ladders, he will have enough to measure between the walls.

Ambassador

[BernieT]

Ambassador has the right idea !

And YES it is a trick question !!!

The walls have to be '0' dist apart for those conditions and you could NEVER get a ladder in between the walls!!

(grumble-grumble %#&^%# I just wasted the last 2 hours working on this)

LOL - BerTau

BTW (by the way) Mike, IMHO is 'in my humble opinion'

[This message has been edited by BerTau (edited 06-10-2001).]

[This message has been edited by BerTau (edited 06-10-2001).]

[Michael Reynolds]

2 hours? I've been agonizing over it since it was posted. Called everyone and their brother!!!

Thanks fer the insite on IMHO, bertau, but what's LOL?

The ignorant one,

Mike R.

[BernieT]

LOL - laughing(laughed) out loud

ROFL - rolling on floor laughing

Etcetera - Etcetera, BerTau

[Guest]

i came in on a saturday because that question was bugging me so much!! it wasnt till i drew it up in mastercam that i noticed it wasnt possible. you guys are funny, very funny.

and why would you hire a contractor that didnt have a tape measure???

[This message has been edited by mbowling (edited 06-11-2001).]