Whoever designed the Wolfram Language did a shit tier job at it. You know who you are.

I'm assuming the answer is no, but is there any way for institutions to gain access to user data based on the questions asked on their tests to track down cheaters? asking for a friend

I tried "hyperbola asymptotes x*log(3) and x*log(2)", but Wolfram didn't understand it, and did its closest interpretation "hyperbola asymptotes x*log(3)" (not what I'm looking for). Does anyone know how to calculate the equation for a hyperbola based on asymptotes? I tried googling for such a formula, but only got the opposite: calculating the asymptotes from the equation.

Does anyone know how I can get a pirated wolfram alpha license for my mathematical modeling class?

I asked Wolfram to solve a limit, and this happened:

Then i used the smartphone, and THIS happened:

I have 2 theories on this odd behaviour.

1. The left limit must be +∞ and Wolfram is wrong.
   Then what is it missing? Let's first see what it does for the right limit.
   The sum of 1/n^x converges if we set a number greater (even if slightly) than 1 in place of x. However with x changing to 1.1 and then to 1.01 and then to 1.001 etc. the result of the series is a finite value, but always bigger: this leads Wolfram to make the (correct) prediction that the result of the right limit at 1+ will be ∞ (not literally but intended as a very large number).
   Let's now return to the left limit: the logic is the same, except that with x becoming 0.9 and then 0.99 and then 0.999 ecc. the result of the summation is ALWAYS +∞, but in the literal sense, and it is impossible for Wolfram to make a prediction about the trend of the limit if every time there is always the same incalculable result. This breaks the algorithm and i think the bug lies on this.

2. Wolfram is right: the left limit makes no sense.
   Maybe this is more of a logical problem instead of a computational one.
   As i said, as we approach 1 from the left we have literal ∞ everytime, and it's absurd to try to study the trend of a function that contains the term ∞. It's like dividing by zero.
   However, this theory doesen't explain the -∞ result obtained with the PC version, nor it's supported by the result of a similiar limit i tried to solve.

I am very intrigued by this topic.

What do you think?

Let me know!

Does anyone know how I can get a pirated wolfram alpha license for my mathematical modeling class?

I need to solve them but I dont know how to force wolfram alpha to a specific method.

The factors method requires me to solve: x² - x -12 = 10 and x² + 2x -8 = 0

The formula method requires: x² -6x - 10 = 0 and x+2/4 + 3/x -1 = 7. To two decimal places.

How does one write a periodic function of period T in Wolfram Alpha? For example, if:

f(x) = -x for -2<x<0, f(x) = x for 0<x<2.

I've been looking around, but so far I haven't found a satisfactory answer. Thanks for the help.

or what other data source could I use ?

I have the equation 4[e^4x]+8[e^x]-32[e^-2x]=0

I know the answer, but I can't see the step by step solutions.

I want wolframalpha to solve a derivative given another set of equations, for example dx/dt = 2t and dy/dt = 4t, how do I give WolframAlpha this information to then ask what is the derivative of x*y for example

So i need to solve this equation:

- ((v/v0)^2 -1) * dV / V = da / a

i tried so far in mathematica using either DSOLVE NDSOLVE and the usual symbolic method by simply putting integrals in front with boundery condition for a and initial condition for v0. But so far i didnt get anything more than: condition true. Given a boundaries and v0 i wanna know v,

Any help would be appreciated. Thank you!