Preliminary mathematics marathon (1 Viewer)

[nikkifc]

Solve

for .

....

 

[Drongoski]

Wtf. I found the last one easier than the first

Please show your solution.

 

[hscishard]

omgd, i never thought of parametrics that way!

Lol what? Parametrics is all about using a third variable.

 

[hscishard]

show your solution.

The third one?

 

[Drongoski]

The third one?

yes.

 

[roadrage75]

....

you can create a triangle with sides: t, root(1-t^2) and hypotenuse 1 and then solve for t

 

[ohexploitable]

Lol what? Parametrics is all about using a third variable.

but... usually you'd have to find the equation of the line using the given point and the focus then solve simultaneously with the parabola

with pq=-1 you can just sub a=3 and the given point into P(2ap,ap²) and Q(2aq,aq²)

 

[hscishard]

yes.

Can't remember what I did now, and I don't think what I did was right. I somehow jumped to 2x+y=2y+x

 

[Drongoski]

Can't remember what I did now, and I don't think what I did was right. I somehow jumped to 2x+y=2y+x

Don't worry. I'm still trying to figure out how to do it myself.

 

[hscishard]

Don't worry. I'm still trying to figure out how to do it myself.

Please post solution when you solve it.

I think this one involves contructing lines and definitely congruency

 

[lordinance]

Find the number of code words involving 4 distinct letters chosen from the letters in the word ''forest'' if no code word contains both the letters R and S.

thanks

edit: the given answer is 216 but i don't know how to get it

 

[MATHSCAPE]

Find the number of code words involving 4 distinct letters chosen from the letters in the word ''forest'' if no code word contains both the letters R and S.

thanks

i'll take a guess. 336?

 

[cutemouse]

Find the number of code words involving 4 distinct letters chosen from the letters in the word ''forest'' if no code word contains both the letters R and S.

thanks

edit: the given answer is 216 but i don't know how to get it

Well technically 'r' isn't the same as 'R' and 's' isn't the same as 'S'...

Anyway, ignoring char cases...

No of ways = (6*5*4*3) - (4 * 3 * 4C2 * 2!) = 216.

New question:

In a certain cricket club there are 15 players available for selection, including 2 Smith brothers, 3 Brown brothers and 10 others. In how many ways may eleven players be selected for a game, if no more than 1 Smith and 2 Browns may be chosen?

 

[k02033]

In a certain cricket club there are 15 players available for selection, including 2 Smith brothers, 3 Brown brothers and 10 others. In how many ways may eleven players be selected for a game, if no more than 1 Smith and 2 Browns may be chosen?

= all possible team selection - [team with 2 smiths or team with 3 browns]
= C(15,11) - [C(13,9)+C(12,8)-C(10,6)]

C(10,6) is team with 2 smiths and 3 browns. We take it away from C(13,9)+C(12,8) because C(13,9)+C(12,8) itself has counted that team twice, so we make the subtraction to get rid of extra count of the same team.

 

[hscishard]

An initial salary of $50,000 increases by $3,000. In which year will the salary be twice the original salary?

 

[nikkifc]

An initial salary of $50,000 increases by $3,000. In which year will the salary be twice the original salary?

Series and Sequences (and its applications) is a HSC topic

 

[hscishard]

It's an easy question. It's just that the answer in the cambridge book is 18th year. Shouldn't it be 17th?

 

[nikkifc]

An initial salary of $50,000 increases by $3,000. In which year will the salary be twice the original salary?

Series is: 50000+53000+56000+...

A. P. a=50000, d =3000, Tn=?

If Tn = 100000

then 100000 = 50000 + (n-1)3000

3000(n-1)= 50000
n-1=16.66666...
n= 17.6666

ie. His salary will be twice the original salary after 17.666... years.... ie. in the 18th year.

New question: (wasn't answered before)

Solve for .

 

[hscishard]

Rofl. Bloody n-1

 

[hscishard]

Only 36 deg 52' 306deg 52'.