VCE Methods Question Thread!

[LiquidPaperz]

Guessing this is further, which I don't do so not sure if my answer is right. I'm not the greatest at geometry. You're better off posting to the further thread.

What was your working?

I've used area of a triangle = 1/2(b)(c)sin(theta) but i'm not sure if you're allowed to use that either.

Area of bench = Outer pentagon - Inner pentagon = 5(1/2(TA)(TB)sin(72)-1/2(TD)(TC)sin(72))

TA=TB=2.55

TD=TC=AT-AD=2.00

Therefore Area=5.95 units^2

yeah correct answer, but why is the shaded thing not a trapezium, and if it were how would you find the area of it treating it as one

[psyxwar]

yeah correct answer, but why is the shaded thing not a trapezium, and if it were how would you find the area of it treating it as one

It is a trapezium, but to find its area you'd need to use the formula 5(1/2*(DC+AB)*h) which is more work because it means you need to find what h, the height of the trapezium is, and what DC is.

[LiquidPaperz]

It is a trapezium, but to find its area you'd need to use the formula 5(1/2*(DC+AB)*h) which is more work because it means you need to find what h, the height of the trapezium is, and what DC is.

its not working

so AB is 3m, AD = 0.55, DC is 2.35 which is found from using triangle TDC, where angle t is 72 degrees and the sides DT and CT are 2m.

then knowing DC which is 2.35 and AB which is 3, i isolated and made two triangles on the sides by doing 3-2.35 = 0.65 and divided this by 2 which got me 0.325. then 0.55 squared - 0.44 squared = 0.1969, square root this got me 0.44 which is now the height

using 5(0.5(2.351127389+4)x0.4436214603) = 5.94m which is 0.01m off.....

why this be?

[psyxwar]

eh probably just a rounding error somewhere if your answer is that close

[LiquidPaperz]

i took all the decimal places which the calculator gave me. maybe is it because using this method gives a very slightly different answer?

cause i know this exists when you use 0.5bh vs herons formula vs 0.5bcsina

[knightrider]

On my casio classpad whenever i type simple operations such as x times x my casio clasppad keeps on saying incorrect arguement anyone know how to fix this

[BLACKCATT]

Just a quick question.

Pr(X=2 | X ≥ 1)
=Pr(X=2) / Pr (X  ≥ 1) Right?

Not, [Pr(X=1) + Pr(X=2)] / Pr(X  ≥ 1)?

So would Pr(X=3 | X ≥ 1) = [Pr(X=2) + Pr(X=3)] / Pr(X ≥ 1) ? 

[silverpixeli]

Just a quick question.

Pr(X=2 | X ≥ 1)
=Pr(X=2) / Pr (X  ≥ 1) Right?

Not, [Pr(X=1) + Pr(X=2)] / Pr(X  ≥ 1)?

So would Pr(X=3 | X ≥ 1) = [Pr(X=2) + Pr(X=3)] / Pr(X ≥ 1) ?

Pr(A | B) = Pr(A intersection B)/Pr(B)

so
Pr(X=2 | X ≥ 1) = Pr(X=2 intersection X ≥ 1)/Pr(X ≥ 1)

but the intersection of X=2 and X ≥ 1 is X=2, (to be in the intersection, X has to be greater than 1 AND equal to 2, which is the same as if it just had to be equal to 2)
so this simplifies to;
Pr(X=2)/Pr(X ≥ 1)

with the same formula,
Pr(X=3 | X ≥ 1) = Pr(X=3 intersection X ≥ 1)/Pr(X ≥ 1)
the intersection if X=3 and X ≥ 1 is X=3 (same logic as above)
so this one simplifies to;
Pr(X=3)/Pr(X ≥ 1)

(not sure where you got [Pr(X=2) + Pr(X=3)] / Pr(X ≥ 1) unless you're trying to add the two things together, then yes)

[bobisnotmyname]

hey guys, just a quick check. is a stationary point of inflection when the gradient is zero at that point like a x^3 graph and a non-stationary point of inflection when the gradient isn't zero like on a sin or cos graph?

[lzxnl]

hey guys, just a quick check. is a stationary point of inflection when the gradient is zero at that point like a x^3 graph and a non-stationary point of inflection when the gradient isn't zero like on a sin or cos graph?

Not needed for methods, but yeah. Stationary point of inflection is like x=0 at y=x^3
Non-stationary point of inflection is like x=0 at y=sin x (the slope is at its max or min)

[knightrider]

On my casio classpad whenever i type simple operations such as x times x my casio clasppad keeps on saying incorrect arguement anyone know how to fix this

[nhmn0301]

On my casio classpad whenever i type simple operations such as x times x my casio clasppad keeps on saying incorrect arguement anyone know how to fix this

What exactly did you type in? Can you give me an example?

[knightrider]

What exactly did you type in? Can you give me an example?

I typed in x times x and it says incorrect arguement

[nhmn0301]

I typed in x times x and it says incorrect arguement

Hmm that's weird. I'm not entirely sure neither but maybe you define your x value as something dodgy from the previous calculation and it doesn't work? Sorry can't help much .

[#J.Procrastinator]

Hi, can someone explain how I would do this... And what if they asked for one solution, or an infinite number of solutions? It comes up in almost every prac exam I've done and I still get it wrong 

Consider the following simultaneous equations,

mx + y= 2
2x + (m-1)y =m

Find the value(s) of m for which there are no solution.