Urgent! Need graphical calculator!

[Wojtek]

Hey all,

I dont know how I did it, but I managed to lose my Ti-83Plus calculator today.

Problem is I have session one's math exam tomorow morning 8:30am and I need a gfx calc!

What I was thinking is to put a gfx calc onto my PDA, but since it has windows ce 1.0, none of the gfx calc aps work.

I know I saw a javascript graphical calculator years ago somewhere on the web, but really cant recall the site :(

Anyone knows where I could get it?
Or even a little plotter that plots a graph based on values. Anything in javascript.

I already found a scientific calc (for cos/sin/tan/blabla) and only need a graphic plotter.

Please help me!

Thanks

Wojtek

[Daniel]

Oooooo....that's $185 down the drain...

Bummer. :(

[DCI]

You mean $185 Canadian?? Because i bought mine for $80 US

[Wojtek]

Originally posted by Daniel
Oooooo....that's $185 down the drain...

Bummer. :(

Still better then 3xx $ for cooking your mobo ;)

Back to the subject!: Anyone has an idea??? Few more hours left! :frown2:

[Gayowulf]

more around 120 cdn. I'd offer you mine, but I may need it in the future.

Try learning math in a manner that doesn't require a calculator :)

You may have to, and quicly, if you cant find a calculator soon.

[Bruce]

One reason I'm glad my school supplied our graphing calculators...

[Wojtek]

Originally posted by Gayowulf
Try learning math in a manner that doesn't require a calculator :)

You may have to, and quicly, if you cant find a calculator soon.
I don't think I can learn by heart cos/sin/tan of 1->360 and everything in 4 hours....

Plus I need the gfx function to double check if a fuction is equal to another function and stuff like that.

[Gayowulf]

Originally posted by Wojtek

I don't think I can learn by heart cos/sin/tan of 1->360 and everything in 4 hours....

Plus I need the gfx function to double check if a fuction is equal to another function and stuff like that.

Just remember the important ones 30 45 60 90 in deg. and rad. if youve only got four hours. even a scientific calculator would be better than nothing, you could just trial and error it.



i can hep you out if you want.

as for checking if functions are the same, just plug in a value for x, and if its the same its all godd.

[Wojtek]

Originally posted by Gayowulf


Just remember the important ones 30 45 60 90 in deg. and rad. if youve only got four hours. even a scientific calculator would be better than nothing, you could just trial and error it.



i can hep you out if you want.

as for checking if functions are the same, just plug in a value for x, and if its the same its all godd.
From what we did I dont think general 30 45 60 90 deg values would be of any help, our math questions tend to be rather 'weird' using some 79 or other not normal numbers....

I'll use the scientific calc of my PDA, but I'm not sure if the teacher will let me use it, you know, I can save some notes in MS Word and read them during the exam :p

Only other thing I have is a crappy 1$ Made In Chine calculator, lol

[YUPAPA]

Find the derivative with respect to the independent variable!

y = sin (3x) SQROOT(cos (3x))

:eek: :biggrin2:

[Coolin]

Buy one the morning before you get to school. I'm not sure what level of Math you're in, but Math 20 in Alberta needs a graphing calculator a lot.

[Wojtek]

Originally posted by YUPAPA
Find the derivative with respect to the independent variable!

y = sin (3x) SQROOT(cos (3x))

:eek: :biggrin2:
What is this used for? :confused:


Coolin, my exam is at 8:30
no store is open yet

[Dusty]

The derivative is the slope of the tangent line at a given point on a graph. There are a few other definitions, but that's what all of them boil down to.


You can't make a graph without using a calculator?


<edit>

Oh, and Yupapa, is this the answer? I'm a bit rusty and not really certain if I even simplified right:

9·cos²(3x)-3 &nbsp; &nbsp; dy
___________ = ___
&nbsp;&nbsp; &nbsp; &nbsp;______
&nbsp;2·√cos(3x)&nbsp;&nbsp; &nbsp; &nbsp;dx

</edit>

[Wojtek]

Yes,

But I like to double check with the calc If I dont have any errors.
Sometimes I can make a little mistake, while the calc dosnt.

[Daniel]

Originally posted by Wojtek
Yes,

But I like to double check with the calc If I dont have any errors.
Sometimes I can make a little mistake, while the calc dosnt.

Grow up and learn math without a damn calculator. :mad:

[Daniel]

Originally posted by Wojtek


Still better then 3xx $ for cooking your mobo ;)



And why would I care? Who lost the $500 bike you got for your birthday? ;)

[Bruce]

Originally posted by Daniel
math without a damn calculator. :mad: Math without a calculator?? Are you mad? I can't even multiply or divide without a calculator let alone plot graphs.

[Daniel]

Originally posted by Bruce
Math without a calculator?? Are you mad? I can't even multiply or divide without a calculator let alone plot graphs.

Long division your friend. :)

[Dusty]

I never used a calculator until my senior year of highschool, and even then we never used it on test or quizes or anything, it was just to speed up classwork.

I never memorized the multiplication tables as such. They attempted to each them in the third grade, but I've never been one to take kindly to memorization without logical backing. I know the more common ones, but that's all. For anything else I'd make a tree and just add.


7x8 = 7 7 7 7 7 7 7 7
\ / \ / \ / \ /
14 14 14 14
\ / \ /
28 28
\ /
56

(I'd've actually started at the 28's, but I wrote out the whole thing for example)

Dividing is done the same way in reverse, though plain ole long division is easier than that.

Roots could be done by differentials easy enough:

____ ____
\/ 51 = ? \/ 49 = 7

1 1
dy = ________ . 2 = ___
____
2 \/ 49 7
____
\/ 51 = 8/7 (approx.)

But we would never go through all that trouble. If we got the square root of 51 for an answer, that's what we'd write-- we wouldn't brother getting its value.

Sketching graphs is the easy bit, it's getting the values that takes time. Find the level places, use the second derivative to see if it's concave or convex, and you're done. We never were too precise about graphs, as long as it looked similar it was okay-- each and every point didn't have to be exactly spot on.


You don't need a calculator, just use your brain and a pencil. Might want some paper, too.

[Bruce]

Originally posted by Dusty
You don't need a calculator, just use your brain and a pencil. Might want some paper, too. But don't you see? Our entire math program they're using at my school is nothing but teachign us how to use a calculator. They don't expect, or even want, us to do it out. The entire program is done using calculators...

Math Connections (http://mathconnections.com) is our entire high school math curriculum. No more, Algebra, Geometry, Calculus or anything... just this retarted book which is supposed to help us relate math to teh "real world" yet manages to teach us absolutely nothing...

[Ben]

Unless my teacher says no calculators, we use one. Alegbra I.

By the way, does anyone understand word problems? Like this:

The Browns and the Williamses were neighbors, and they left for the same resort at the same time on the same day. The Browns got there in 4 hours and the Williamses an hour later. One family travelled at a speed difference of 12mph of the other family. How fast was each family going?

Sorry for the bad grammar, I'm trying to recall it from memory:chinese2:

[Dusty]

Ah, so that's where all those cashiers come from that can't figure out how to add tax when the power's out and the registers don't work. I figured they were just stupid, but I guess they were taught to be helpless.

Imagine if there were a "Connections" program for everything! Why, the moment anything went wrong we'd be reduced to living in caves waiting for lightning to make fire.

Now I just feel sad.

[Dusty]

The Browns and the Williamses were neighbors, and they left for the same resort at the same time on the same day. The Browns got there in 4 hours and the Williamses an hour later. One family travelled at a speed difference of 12mph of the other family. How fast was each family going?They traveled the same distance. The Browns arrived first, so they were the ones going 12 miles faster. The distance the Browns traveled was 4(x+12), where 4 is how long it took in hours and x+12 is how fast they were going in miles per hour. The distance the Williams traveled was 5(x), were 5 is how long it took in hours and x is how fast they were going in miles per hour. Since it was already established both distances were the same, we can do:

4(x+12)=5(x)
4x+48=5x
48=x

Therefore, the Williams were going 48 miles per hour and the Browns were going 48+12, or 60, miles per hour.

To check: 60x4=240, 48x5=240, 240=240, correct.

[Ben]

Thanks, Dusty! :)

(No, that wasn't a homework question, I just didn't know how to do word problems like that)

[conkermaniac]

Originally posted by Wojtek



Only other thing I have is a crappy 1$ Made In Chine calculator, lol

They make graphing calcs in China? :eek:

[conkermaniac]

Originally posted by Ben
Unless my teacher says no calculators, we use one. Alegbra I.

By the way, does anyone understand word problems? Like this:

The Browns and the Williamses were neighbors, and they left for the same resort at the same time on the same day. The Browns got there in 4 hours and the Williamses an hour later. One family travelled at a speed difference of 12mph of the other family. How fast was each family going?

Sorry for the bad grammar, I'm trying to recall it from memory:chinese2:

You're taking Algebra I? I remember doing these type of problems...For some strange reason, I actually like doing word problems. If I'm not mistaken, they were travelling at 48 and 60 mph.

Dusty's explanation was good, btw. Here's how I did it:

Williamses: d = rt
Browns: d = rt

Plug in the values for rate and time...we have no information for the distance traveled, except that the two families traveled the same distance. Let's pretend that the Williamses traveled faster. The problem specifically states that the faster family traveled 12 mph faster than the Browns. So therefore, the Williamses traveled at (r + 12) and the Browns traveled at (r)

Williamses: d = (r+12)t
Browns: d = rt

For the time...we know that the family traveling faster must have arrived first. From the problem, we know the faster family arrived in 4 hours. The slower family arrived one hour later - or 5 hours.

Williamses: d = (r+12)4
Browns: d = (r)5

Now, we know that the distances are the same and we can set the two equations equal to each other.

4(r+12) = 5r

Then, solve. :)

I love explaining Algebra to people! :biggrin2:

[trenzterra]

Until secondary school, we cannot use a calculator.

[conkermaniac]

Originally posted by trenzterra
Until secondary school, we cannot use a calculator.

There's really no need for a calculator until Algebra, and even then, you don't really need a calculator. But when it comes to solving determinants, sin, cosine, and stuff, you'll need a calculator.

[Wojtek]

Originally posted by conkermaniac
I love explaining Algebra to people! :biggrin2:

So help me with this:

Solve this one: 7x+2x²=15

Our teacher showed us this method:

7x+2x²=15
7x+2x²-15=0

then replace the middle term (2x²) with two terms so you can do the double simplification or whatever its called in english.

Basically you have to find 2 numbers that added to each other give 2x² and if you multiply them it gives you -105x (7x*-15) and I can't find those numbers...


This is the only question from today's exam that I didnt do...



Oh and the exam was not that bad, I did it with a scientific calc on my Pda, and No, I didnt have my class notes saved on it :biggrin2:

[Coolin]

It's called "Completing the Square" here.

2x²+7x-15=0
2(x²+7)-15=0
2(x²+7x/2-3.0625)-15+3.0625=0
2(x²+7x/4)²-11.9375=0

Then again, I might be wrong, it's been a while.

[Dusty]

You could factor:


7x+2x²-15=0
2x²+7x-15=0

(x+5)(2x-3)=0

x+5=0 2x-3=0
x=-5 2x=3
x=3/2

answer: -5 and 3/2



Or you could use the quadratic formula:


___________
-b ± \/ b² - 4·a·c
__________________

2·a

7x+2x²-15=0
2x²+7x-15=0

_______________
-7 ± \/ 7² - 4·2·(-15)
______________________ =

2·2

_________
-7 ± \/ 49 + 120
_________________ =

4

_____
-7 ± \/ 169
____________ =

4


-7 + 13 3
________ = ___

4 2

-7 - 13
________ = -5

4

answer: 3/2 and -5

I don't know what your teacher was trying to tell you to do.

[Daniel]

Originally posted by Dusty


I don't know what your teacher was trying to tell you to do.

I can tell you already, in Quebec, the teachers don't teach you squat.