Extension One Revising Game (2 Viewers)

[vds700]

1) Find the point which divides the line joining (4,8) and (12,6) externally in the ratio 4:1.

.

Is it right with external division, of m:n, u make it m:-n and put it in the formula? Sorry its been a while since i learnt this.

x =( -4 + 48)/3
=44/3

y = (-8 + 24)/3
=16/3

 

[Dr. Zoidberg]

This last one will be really tough cause a diagram came with it (which I can't upload).
11) P(2p,p²) and Q(2q,q²) are two points that move on the parabola x² = 4y such that OQ [O is the origin (0,0)] is parallel to the tangent to the parabola at P. M is the midpoint of PQ. The equation of the locus M and describe this locus geometrically. It is suggested that you try to draw a diagram.

I got this for my yearly lol.

Midpoint=[q+p, (q^2 +p^2)/2 ]

x = q+p
y= (q^2 + p^2)/2

gradient of line OQ = q/2

gradient of tangent P = p

since the two lines gradients are equal,

q/2 = p

therefore, q=2p

Sub this into x and y above

x = 2p + p = 3p ---> p=x/3
y= (4p^2 +p^2) /2 = (5p^2)/ 2

sub p into y

y = [5(x/3)^2 ]/2

2y = (5x^2)/9

18y=5x^2

therefore, x^2 = (18/5)y

therefore, the locus is a parabola

Not sure if im right though

 

[S1M0]

correct

 

[bored of sc]

Is it right with external division, of m:n, u make it m:-n and put it in the formula? Sorry its been a while since i learnt this.

x =( -4 + 48)/3
=44/3

y = (-8 + 24)/3
=16/3

Yeah. Correct.

 

[tommykins]

Is it right with external division, of m:n, u make it m:-n and put it in the formula? Sorry its been a while since i learnt this.

x =( -4 + 48)/3
=44/3

y = (-8 + 24)/3
=16/3

correct.

 

[bored of sc]

3) 4

Correct. Damn. In the test I got this right but I forgot that you could just sub in x = 2 into the polynomial. Instead I did a long division.

 

[bored of sc]

Question 2 and 4-10 need doing. See above post.

 

[tommykins]

2) Solve for x: 4/(x+6) > 1.

4(x+6) > (x+6)²
0 > (x+6)²-4(x+6)
0 > (x+6)[x+6-4]
0 > (x+6)(x+2)

-6 < x < -2

4) Find the exact value of cos75^o.

cos75 = cos(45+30) = cos45.cos30 - sin45.sin30 = sqrt3/2sqrt2 - 1/2sqrt2 = sqrt3-1/2sqrt2

5) Find the acute angle between the lines y = 3x+6 and 3x-2y = 8 to the nearest minute.

m1 = 3, m2 = 3/2
tan@ = |(3-3/2)/1+9/2| = 3/11
@ = 15*15'18.43'' = 15*15' degrees.

6) Show that [1/(1+tanx)] + [1/(1-tanx)] = tan2x/tanx.

let t = tanx for simplicity.

LHS
= 1/1+t + 1/1-t
= 1-t + 1+t/(1-t²)
= 2/(1-t²)
= 2/(1-t²)*t/t
=2t/(1-t²)*1/t
= tan2x/tanx

7) If X, Y, Z are the roots of the cubic equation 4x³+2x²-x-2 = 0, find the value of
(i) X + Y + Z
(ii) XY + YZ + XZ
(iii) X² + Y² + Z².

i) -1/2
ii) -1/4
iii) 3/4

8) Show that sin(x+y)/cos(x-y) = (tanx + tany) / (1 + tanx.tany).

LHS -
= sinx.cosy + siny.cosx/cosx.cosy+sinx.siny -> divide through by cosx.cosy
= tanx+tany/1+tanx.tany

9) In how many ways can the following letters be arranged if all the P's are to be separated:
M T P P T A

2*4! = 48

10) Sophia invites 8 guests to celebrate achieving 80% in her recent math's text. Everyone is randomly seated at a round table. Find the number of seating arrangements possible if a particular couple, Zac and Joanne sit together.

2! of Zac and joanne.
7 groups.
2!*6! = 1440[/quote]

 

[lyounamu]

Good work but u didn't answer the question, between what times can the yaught safely cross?

I did.

 

[lyounamu]

If you let 0 be the origin when you draw the graph, would the amplitude (a) be 6 - 2 = 4?

(I'm not going to go very well.... )

Amp = (6-2)/2 = 2

Tommy, I don't think your answer for Q9 is correct. The total number of perm is 6!/2!2! = 180

And the total number of Ps together is basically = 5!/2!2! = 30

So the total number of Ps separaed is = 150.

 

[vds700]

I did.

I meant, u need to say the yaught can cross safely between 1.40 pm and 9.00 pm

 

[lyounamu]

I meant, u need to say the yaught can cross safely between 1.40 pm and 9.00 pm

Ah, right. Sorry. I couldn't be bothered to specify...stupid of me.

 

[Trebla]

a) Prove that from k = 1 to n:
Σ(k³ - (k - 1)³) = n³

b) Hence show that from k = 1 to n:
Σk² = n(n+1)(2n+1)/6

 

[lolokay]

a) Prove that from k = 1 to n:
Σ(k³ - (k - 1)³) = n³

b) Hence show that from k = 1 to n:
Σk² = n(n+1)(2n+1)/6

if n=1, the sum = 1 - 0 = 1 = n²
assuming true for n
... + (n+1)³ - n³ = n³ + (n+1)³ - n³
= (n+1)³
true for 1, if true for n then true for n+1 so true for all n

Σ k³ - (k-1)³
= Σ 3k² - 3k + 1 = n³

Σk² = n³/3 + Σk - Σ1/3
= 2n³/6 + 3n(n+1)/6 - 2n/6
= (2n³ + 3n² + 3n - 2n)/6
= n(2n² + 3n + 1)/6
n(n+1)(2n+1)/6

 

[Trebla]

Prove a) without induction.

 

[lolokay]

n³ - (n-1)³ + (n-1)³ - ... -1³ + 1³ - 0³ = n³

is that fine?

 

[tommykins]

Tommy, I don't think your answer for Q9 is correct. The total number of perm is 6!/2!2! = 180

And the total number of Ps together is basically = 5!/2!2! = 30

So the total number of Ps separaed is = 150.

Probably, looking back at it I cbf figuring it out.

</3 perm+comb's

 

[Pratham]

hey guys, this doesnt really have to do with any of the questions, but its a doubt i have so i was wondering if someone can clarify (sorry if im breaking the rules a bit).

Lets say that in a simple harmonic motion we are given a set of initial conditions and we can say something like; in general x= Acos/sin(nt[+/-]alpha) +B. Now lets say the questions says find apha, and we do so by plugging in the intial conditions. Now in most 3u cases we take alpha as the positive value and forget about the various quadrants and where what is positive/negative (A,S,T,C), but sometimes in 4u (and 3u) we have to determine whether alpha is positive or negative....now heres the question (finally!)...the way i do it is by using the intial input values/conditions e.g. at time t=o x(dot)>0 etc. but from what i understand there is a easier way using a unit circle. Firstly, how does this method work and secondly what are the principles behind it...is it kinda like evaluating shm to circular motion???? PLEASE HELP

 

[cutemouse]

How about,

Solve for x:
|x+2| = -5

(If this takes you more than 0.1 of a second to figure out then you're an official novice )

 

[clintmyster]

How about,

Solve for x:
|x+2| = -5

(If this takes you more than 0.1 of a second to figure out then you're an official novice )

no solution?