Math game!

[Kieee]

Firstly You guys think this will be boring as its math but simply just give a math question to the person below you ill start what 16x13

 

[Cubes]

Rubik's Cube?

 

[RC_4777]

208

Find all zeros of x^3+3x^2-x+4

 

[Cubes]

I can't seem to find any rational roots.
Edit: Grouping, factoring, and synthetic division reveal nothing.

Here's my problem:
If f(x)=sin(x) and f(a)=1/10, find the exact value of the following:
f(a)+f(a+2π)+f(a+4π)
Do not use a calculator.

 

[RC_4777]

I can't seem to find any rational roots.
Edit: Grouping, factoring, and synthetic division reveal nothing.

Here's my problem:
If f(x)=sin(x) and f(a)=1/10, find the exact value of the following:
f(a)+f(a+2π)+f(a+4π)
Do not use a calculator.

You're probably right, I just used a random equation XD

 

[Cubes]

You're probably right, I just used a random equation XD

Meeen
But do you have any idea about mine? (Calculus/Trig for the win)

 

[ProHighJacker]

I can't seem to find any rational roots.
Edit: Grouping, factoring, and synthetic division reveal nothing.

Here's my problem:
If f(x)=sin(x) and f(a)=1/10, find the exact value of the following:
f(a)+f(a+2π)+f(a+4π)
Do not use a calculator.

preferably english thanks cubes

 

[Lively]

3/10 is the answer for Cubes question.

(x+1)^4

 

[TheCornStealer]

WAY TOO COMPLICATED FOR MY IM 13 FOR GOODNESS SAKES!!!

5+5?

 

[jonnysurvives]

3/10 is the answer for Cubes question.

(x+1)^4

x^4+4x^3+6x^2+4x+1. Cheeky binomial expansion.

 

[Cubes]

x^4+4x^3+6x^2+4x+1. Cheeky binomial expansion.

Niiice. Pascal's Triangle would be the most preferred method.

Here's a fun one:
The function f(x)=1-(π/π/2)sec(π/4)(3x-4π)
Is this function Even or Odd?
Also, to what does this function pay respect to?

 

[jonnysurvives]

Niiice. Pascal's Triangle would be the most preferred method.

Here's a fun one:
The function f(x)=1-(π/π/2)sec(π/4)(3x-4π)
Is this function Even or Odd?
Also, to what does this function pay respect to?

I used Pascal's Triangle as my method, but the process where you expand an unknown and a constant / two unknowns to the power of whatever is called a binomial (two number) expansion. Or at least that's how it's taught in England. I get the impression that there are some differences in terminology, judging by the fact that RC called what I would call 'roots' 'zeroes'.

RC_4777 For your question I used the cubic formula and a calculator and got -3.59, 0.295±1.01i where i= sqrt(-1). Safe to say it wasn't a very fun question.