wildareal's questions thread

[luken93]

Whoops sorry my mistake, I always use the oppostie one for some reason.

Just have cos^2(x) + sin^2(x) = 1
Divide them by sin^2 to get cot^2 + 1 = cosec^2
Divide by cos^2 to get 1 + tan = sec

Yeh I know, I have a habit of writing tan(x) for cos(x)/sin(x) instead of cot(x)

oooohhh haha, fair enough.

[wildareal]

Here's a vectors one:

Prove, using vectors that the midpoint of a hyptenuse of a right-angled triangle is equidistant from all all three vertices.

[b^3]

I hope you can read this.

Since u.u=|u|^2

EDIT: Flipped image the right way up.

[wildareal]

Using the appropriate expansion show that sin(arcos(3/5))+artan(-3/4)=7/25

[wildareal]

^^Thanks so much!

[TrueTears]

Using the appropriate expansion show that sin(arcos(3/5))+artan(-3/4)=7/25

eh i assume you mean sin[arcos(3/5)+artan(-3/4)]=7/25

in that case, just let x = cos^{-1}(3/5) and y = tan^{-1}(-3/4) then expand sin(x+y)

or you can simplify it first to sin[arcos(3/5)-artan(3/4)]=7/25

then let y = tan^{-1}(3/4) and expand sin(x-y).

[wildareal]

If F'(x)=f(x) then an antiderivative of 3f(3-2x) is?

[wildareal]

Hello Wildareal:
Pretend f(X) is just any other function:
F'(3-2x)=f(3-2x)*2

Therefore:
-3F'(3-2x)/2=3f(3-2x)

[b^3]

Is it just me or is the person who answered got the same login "wildareal" showing asking and answering?

[Greatness]

Is it just me or is the person who answered got the same login "wildareal" showing asking and answering?

Yeah i was just gonna post something o.O Could be a glitch? Or he just answered his own question by addressing himself? lol

[b^3]

Is it just me or is the person who answered got the same login "wildareal" showing asking and answering?

Yeah i was just gonna post something o.O Could be a glitch? Or he just answered his own question by addressing himself? lol

Yeh i was about to post too. Thought I'd over done it working and was seeing things or I might have finally gone "crazy".

[AskQuestions]

He normally does this.

Two chemicals, A and B, are put together in a solution where they react to form a compound,
X. The rate of increase of the mass, x kg, of X is proportional to the product of the masses of
unreacted A and B present at time t minutes. It takes 1 kg of A and 3 kg of B to form 4 kg of
X. Initially 2 kg of A and 3 kg of B are put together in solution. One kg of X forms in one
minute.
a Set up the appropriate differential equation expressing dx/dt as a function of x.

Question 2 Chapter review 9 Extended response.

[AskQuestions]

Anyone?

[TrueTears]

Is it just me or is the person who answered got the same login "wildareal" showing asking and answering?

wildareal has a habit of answering his own questions

[taiga]

I think he means to put up the answer in case anyone else is keen.

Or he's crazy