Cssa mx1 (2 Viewers)

[Hermes1]

i mean because they are roots of the questions. if you substitute back to the original P(x), it should be P(x)=0 right? but apparently it's not.

i just typed them both into my calculator in that equation and they are working for P(x) = 0.
just try it again yourself.

 

[muzeikchun852]

i just typed them both into my calculator in that equation and they are working for P(x) = 0.
just try it again yourself.

i got -6 x 10^(-12) and -1.2 x 10^(-12)

 

[Hermes1]

i got -6 x 10^(-12) and -1.2 x 10^(-12)

i think you should post this question in MX1 forum for drongoski to take a look. bcuz im pretty sure the question is fine but u should get a second opinion.

did u at least leave the answers u got? bcuz testing they equalled zero wasnt a part of the question. so u wont lose any marks.

 

[maths94]

Re: CSSA Maths Ext 1 2011

http://community.boredofstudies.org/showthread.php/262636-Cssa-mx1

+ 1

 

[muzeikchun852]

i think you should post this question in MX1 forum for drongoski to take a look. bcuz im pretty sure the question is fine but u should get a second opinion.

did u at least leave the answers u got? bcuz testing they equalled zero wasnt a part of the question. so u wont lose any marks.

yeah i did.

 

[Dylanamali]

what was the red light answer?

 

[Drongoski]

you get the three roots as:

-4,

and they work.

so the question is fine.

these answers correct.

 

[maths94]

what was the red light answer?

wat did u hav? and i cant remember wat i got

 

[muzeikchun852]

these answers correct.

it's correct even though when you test back to the original equation it doesnt give you zero ?

 

[Dylanamali]

I forgot the specifics of the q, but I'm preety sure p(red light) = 3/10, p(amber) = 1/10, p(green) = 6/10

Therefore p(redlight) = 0.3, q(non redlight) = 0.7

(p+q)^5 =......

Answer: 5C4(0.3)(0.7)^4

 

[Dylanamali]

btw can anyone remember the last 2 questions and the solution, why alpha is the greatest when x^2 = hH or wateva..

 

[Drongoski]

it's correct even though when you test back to the original equation it doesnt give you zero ?

r u sure? I tested and got zero.

 

[Hermes1]

r u sure? I tested and got zero.

i tested and got zero as well.

 

[markrule5]

btw can anyone remember the last 2 questions and the solution, why alpha is the greatest when x^2 = hH or wateva..

for the height i got 4000 and a bit meters
but what u talking about with the alpha shit?

 

[Some Vunt]

wat did ppl get for the time and distance for question one projectile motion ...q7

time was 27 and x was around 33XXm

 

[muzeikchun852]

r u sure? I tested and got zero.

i tested and got zero as well.

got this incredible small number. -6 x 10^(-12) and -1.2 x 10^(-12)

 

[Drongoski]

got this incredible small number. -6 x 10^(-12) and -1.2 x 10^(-12)

show how you got nonzero.

 

[Dylanamali]

for the height i got 4000 and a bit meters
but what u talking about with the alpha shit?

i got 33 seconds i think and 4000+metres

That was 7a) im talking about 7b) the tan one.

 

[muzeikchun852]

show how you got nonzero.

just substitute back to the original equation : i tried wolfram and i got this -7.105427357601×10^-15 but alternative form is 0. maybe it is a root.

 

[Hermes1]