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+3146

Lamehe:
Now let's just wait for someone to point out a really stupid mistake I made..

not really a mistake, the reason behind these different results is the interpretation of "-v²". this could be "-(v²)" or "(-v)²", same as the result for "-2^2" could be 4 or -4. web2.0calc interprets your input "-v²" as "(-v)²". I you want the other interpretation you can use parenthesis:

[input]-v²+6v-1/2=0[/input]
[input]-(v²)+6v-1/2=0[/input]

adminMay 25, 2013

whats:
http://web2.0calc.com/questions/posting.phttp://web2.0calc.com/questions/posting.php?mode=post&f=2#hp?mode=post&f=2#

GuestMay 25, 2013

May 23, 2013

+3146

dontknow:
hello, how do I solve this?

1/x - x/5 = 0

[input]1/x - x/5 = 0[/input]

1/x - x/5 = 0 |+x/5
1/x = x/5 | *x
1 = x^2/5 |*5
5 = x^2
x^2 = 5 | sqrt
x= +sqrt(5) and -sqrt(5)

adminMay 23, 2013

May 20, 2013

I don't know if this question is a hoax but here goes... The amount of the drop from 4500 to 2800 is 1700. The percentage of the drop is 1700 / 4500 = 0.377 ad infinitum (to infinity) or 37.77~ percent. So when you subtract 37.77~% from 4800 you get 2800. Therefore 2800 is a 37.77~% decrease from 4500.

GuestMay 20, 2013

Um... 33 and 1/3 percent is 0.3333 ad infinitum (to infinity) percent. This looks like the decimal equivalent to 1/3 except it's a percentage. Therefore 1/3 percent is 0.0033 percent.

GuestMay 20, 2013

All you have to do is press those arrows next to the history bar, of your previous problems. They will then show up at your "Bar" of your math work (The big white space). I hope this helps! Good luck!!

GuestMay 20, 2013

May 17, 2013

WHAT IS THE ANSWER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FRUSTRATED

GuestMay 17, 2013

May 16, 2013

will you plz tell me what are you asking about.....?

johnson33May 16, 2013

+3146

function syntax of pi(1):
[input]2pi(1)^2[/input]

[input]2pi*(1)^2[/input]

[input]2pi*1^2[/input]

adminMay 16, 2013

May 15, 2013

your calculator must be wrong. on my calculator it says 4

GuestMay 15, 2013

(n-10)(n-1)

GuestMay 15, 2013

you must do this order of operations.
6-1x0+2/2 (multiply 1x0 first)
6-0+2/2 (then divide 2/2)
6-0+1 (then add 6+0+1)
7!!! Not 1!!!!

GuestMay 15, 2013

May 13, 2013

GuestMay 13, 2013

$5345.13 if I'm not mistaken.

GuestMay 13, 2013

May 12, 2013

+3146

sin^-1(0.5)
[input]sin^-1(0.5)[/input]

asin(0.5)
[input]asin(0.5)[/input]

adminMay 12, 2013

+42

Look up how many Joules are needed to warm 1 g water 1 degree.
Suppose y J
You will need totally 800 x 88 x y Joules = T Joules
1 Joule = 1 Ws
Divide T trough 2000 and you'll find the number of seconds.

As I remember this it will be about 10 minutes.

ricom39May 12, 2013

+42

sin ( -1) = -0,017452402

ricom39May 12, 2013

GuestMay 12, 2013

May 9, 2013

I guess this sounds easier than it is really.
The thing is, if you want to have exact calculations for things like roots you would have to implement some kind of symbolic calculation.
I don't know any of the source of this application but I'd guess it is all just numerical stuff like you would have in vanilla MATLAB or octave.

To cut it short: there is no solution to this without a great effort.

GuestMay 9, 2013

May 8, 2013

+3146

[input]4*e^x=52[/input]

4*e^x=52
4*e^x/4=52/4
e^x=52/4
e^x=13

assume x>0:
ln(e^x)=ln(13)
x=ln(13)

adminMay 8, 2013

May 7, 2013

+3146

(sin(x)-cos(x))(sin(x)+cos(x))
expand:
sin(x)^2-cos(x)*sin(x)+sin(x)*cos(x)-cos(x)^2
with: cos(x)^2+sin(x)^2=1 is: sin(x)^2=1-cos(x)^2
(1-cos(x)^2)-cos(x)*sin(x)+sin(x)*cos(x)-cos(x)^2
with: -cos(x)*sin(x)+sin(x)*cos(x) = 0
(1-cos(x)^2)-0-cos(x)^2
1-cos(x)^2-cos(x)^2
1-2*cos(x)^2

with: cos(x)^2+sin(x)^2=1
cos(x)^2+sin(x)^2-2*cos(x)^2
1*cos(x)^2+sin(x)^2-2*cos(x)^2
1*cos(x)^2-2*cos(x)^2+sin(x)^2
-cos(x)^2+sin(x)^2
with: cos(2x)=cos(x)^2-sin(x)^2 is: -cos(2x)=-cos(x)^2+sin(x)^2
-cos(x)^2+sin(x)^2
-cos(2x)

adminMay 7, 2013

(6-1)x0+2/2=0+1=1

but

6-1x0+2/2=6-(1x0)+2/2=6-0+1=7

GuestMay 7, 2013

May 6, 2013

K****r:
your mum is dead in a box

trhgfdhg

:lop:

GuestMay 6, 2013

May 3, 2013

why yall talking bout me

GuestMay 3, 2013

c^3-512

GuestMay 3, 2013

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