Maths Game (1 Viewer)

[pwoh]

Good question. You can use the pythagoras' theorem because it's just easiest way to approach the question. I really don't see how I can approach this without using that method.

In the question, you are only interested in the value of the tanx. Since the answer you get from the question conforms to the value of the sine (which was initially given), your answer is valid, hence your working out is valid (but not necessarily the case).

Thanks o___o;;


Something a bit easier:
Simplify 3 - (k - p)/(k+p)

 

[lyounamu]

Thanks o___o;;


Something a bit easier:
Simplify 3 - (k - p)/(k+p)

eh, simplify that? It looks simplifid enough...

unless you want to go a bit far:

3 - (k-p)/(k+p)
= ((3k+3p) - (k-p))/(k+p)
= (2k+4p)/(k+p)
= 2 + 2p/(k+p)

 

[pwoh]

eh, simplify that? It looks simplifid enough...

unless you want to go a bit far:

3 - (k-p)/(k+p)
= ((3k+3p) - (k-p))/(k+p)
= (2k+4p)/(k+p)
= 2 + 2p/(k+p)

A '10er was meant to answer that, but never mind

This thread is dying from lack of participation from '10ers...maybe I should start studying properly...

 

[lyounamu]

A '10er was meant to answer that, but never mind

This thread is dying from lack of participation from '10ers...maybe I should start studying properly...

Um, sorry.

 

[madsam]

lets get this party started then,

find (13 + 4i)/7 in the form x + iy

and hence find the imaginary roots of (13 + 4i)/7

 

[Kabuzo01]

lets get this party started then,

find (13 + 4i)/7 in the form x + iy

and hence find the imaginary roots of (13 + 4i)/7

Is that correct pwoh? Damn the year 11 questions =D

 

[lyounamu]

lets get this party started then,

find (13 + 4i)/7 in the form x + iy

and hence find the imaginary roots of (13 + 4i)/7

Are you serious? That's 4 Unit Mathematics. This question would just throw all the yr 10 students out the window...

Well, here we go:

(13+4i)/7 = 13/7 + 4i/7

so 4i/7 is the imaginary part.

 

[Kabuzo01]

this is 2 unit level maths from yr 11. But it can be perceived as a harder yr 10 maths too.

tanA is negative when it goes to the 2nd quadrant (i.e. obtuse angle basically).

So since sine = opp/hyp

adjacent = sqrt(24) (because 7^2 = x^2 + 5^2)

tanA = opp/adj = 5/-sqrt(24) = -5sqrt(24)/24

I guess thats a point for you?

Change of rules...

 

[lyounamu]

I guess thats a point for you?

Point for lyounamu added

lol?

what the hell is the point?

 

[Kabuzo01]

pwoh I need you to tell me which one is correct T.T I'm lost =D Oh yeah an addition to the rules, who gets the answer correct first gets the point and also make sure if you edit your answer no one has posted it already.

By that I mean say Person 1 posted an incorrect answer
Person 2 posts the correct answer
Person 1 sees this and changes the answer to be the same as Person 2.

This can be a mistake if someone accidently edits their post to change wordings or something. PM me if you dont understand.

 

[pwoh]

Which question are you referring to? If it's the 4unit maths one I have no idea lol.

 

[madsam]

Are you serious? That's 4 Unit Mathematics. This question would just throw all the yr 10 students out the window...

Well, here we go:

(13+4i)/7 = 13/7 + 4i/7

so 4i/7 is the imaginary part.

ok, now slightly harder, find the real roots of the equation
sqrt{(13+4i)/7}

 

[pwoh]

ok, now slightly harder, find the real roots of the equation
sqrt{(13+4i)/7}

I thought this was an SC thread >_>;

 

[Kabuzo01]

So did I o.o

 

[shaon0]

wtf is evry1 doing? This is SC not yr 12 MX2.

 

[Kabuzo01]

I've edited the rules and from now all questions must be SC level or general knowledge, all previous unanswered questions will not be counted.

 

[pwoh]

Solve:

(m+5)/3m = 2m

 

[suneeta =)]

Solve:

(m+5)/3m = 2m

m+5 = 6m
5m = 5
m = 1


yay !

 

[pwoh]

m+5 = 6m
5m = 5
m = 1


yay !

Wrong...well the answer if half right but the working is wrong.

You forgot that 3m x 2m = 6m², not 6m.

The answers are 1 and -5/6. Try the working again?

 

[suneeta =)]

roflol.

and this is why i lose marks in exams for stupid mistakes.

thanks for pointing that out :]

m+5 = 6m^2
0=6m^2-m-5
0=(m-1)(m+5/6)