jinny1's Methods Questions Thread :D

[jinny1]

Just seeking clarification on something. A function is a graph that passes the vertical line test right? However an inverse function only exists when we have a 1:1 graph?? Because in an inverse sine graph, we restrict the domain to make it 1:1 when we can also make it 1:many which wud still technically be a function.

Cheers

[Greatness]

Function: vertical and horizontal line test cross once
Yep inverse functions only exists if its 1-1.

[jinny1]

Function: vertical and horizontal line test cross once
Yep inverse functions only exists if its 1-1.

Thanks! I always thought it was justvertical line test?? Because isnt a quadractic a function??

[REBORN]

Quadratic is a function but not a 1:1 function as it fails the horizontal line test

[cranberry]

alright, this question ffrom vcaa 2008...(i know its VCAA so im probably wrong, but...)

Her "8th attempt" is only dependent on her last attempt (her 7th attempt), isnt it??

Shouldnt there only be two possible orders - successful give that she missed prior x successful give that she was successful prior?

If you read the question, it clearly says her 8th attempt, which would have the same chances of success as the 3rd attempt? if not the probabilty of succes would change each attempt so the matrices would be different each attempt, no?

just rambling anyway. 

[xZero]

alright, this question ffrom vcaa 2008...(i know its VCAA so im probably wrong, but...)

Her "8th attempt" is only dependent on her last attempt (her 7th attempt), isnt it??

Shouldnt there only be two possible orders - successful give that she missed prior x successful give that she was successful prior?

If you read the question, it clearly says her 8th attempt, which would have the same chances of success as the 3rd attempt? if not the probabilty of succes would change each attempt so the matrices would be different each attempt, no?

just rambling anyway. 

its a normal markov chain, you're over thinking it

[jinny1]

Hey guys i've been wondering how to integrate an equation with given parameters. Lets say i want to integrate y'=x^2 nd the questions already says y(0)=0 , then how to i make the calculator calculate tthe Constant as well in the process of integrating??

[Vincezor]

Hey guys i've been wondering how to integrate an equation with given parameters. Lets say i want to integrate y'=x^2 and the questions already says y(0)=0 , then how to i make the calculator calculate the Constant as well in the process of integrating??

I'm using the classpad, and for that we use 'dsolve' (I've never used it in methods come to think of it!) but you use it heaps in spesh!

Interactive -> Advanced -> dSolve -> Type in equation -> Inde var = x, Depe var = y -> Check box 'include condition' and type in x=0, y=0

You should get



Hope that helped!

[jinny1]

thanks i have the ti-nspire but it works almost identically...

[jinny1]

nvm

[b^3]

nvm

You should have this:
desolve(y'=x^2 and y(0)=1,x,y)
y=x^3/3 +1

EDIT: I probably should have added inital conditions into the calculator guide.

[jinny1]

nvm

You should have this:
desolve(y'=x^2 and y(0)=1,x,y)
y=x^3/3 +1

EDIT: I probably should have added inital conditions into the calculator guide.

Thanks buddy

just one last one! what would be the calculator syntax to solve this question??

[b^3]

nvm

You should have this:
desolve(y'=x^2 and y(0)=1,x,y)
y=x^3/3 +1

EDIT: I probably should have added inital conditions into the calculator guide.

Thanks buddy

just one last one! what would be the calculator syntax to solve this question??

(Image removed from quote.)

nvm

You should have this:
desolve(y'=x^2 and y(0)=1,x,y)
y=x^3/3 +1

EDIT: I probably should have added inital conditions into the calculator guide.

Thanks buddy

just one last one! what would be the calculator syntax to solve this question??

(Image removed from quote.)

I'm not sure if you can do it in one step on the calc, you have to intergrate the stuff in front of the i,j and k, then find c and sub it back in. once you have displacement with the +c, you can start putting it in the calc and doing it the quick way (i.e. define things, solve for c, plug in t e.t.c)

[jinny1]

i tried desolve( y'= [...i, ....j] and y(0)= [0, 2],x,y)

but didin't work. i dont think u can desolve vector components