Extension One Revising Game (2 Viewers)

[midifile]

new question

Show that for all positive integers n:
x[(1+x)^n-1 + (1+x)^n-2 + ... + (1+x)² + (1+x) + 1] = (1+x)ⁿ -1

 

[loller]

new question

Show that for all positive integers n:
x[(1+x)^n-1 + (1+x)^n-2 + ... + (1+x)² + (1+x) + 1] = (1+x)ⁿ -1

no.

new question:

how many boys can fit into a cup full of water if p = q+3

 

[lolokay]

new question

Show that for all positive integers n:
x[(1+x)^n-1 + (1+x)^n-2 + ... + (1+x)² + (1+x) + 1] = (1+x)ⁿ -1

nCr = (n-1)C(r-1) + (n-1)Cr
= (n-1)C(r-1) + (n-2)C(r-1) + (n-2)Cr
= ...
= Sum[1 < a < n-r+1] (n-a)C(r-1)

in the bracketed section, [(1+x)^n-1 +...], the coefficient of any given (r-1)th power of x is Sum[1 < a < n-r+1] (n-a)C(r-1) = nCr
but this term is then multiplied by x, which adds 1 to the power. So the coefficient of the rth power of x is nCr. The powers of x are from 1 to n

but on the RHS the coefficient of the rth power of x is also nCr. (1+x)ⁿ contains powers of x from 0 to n, but nC0 x⁰ = 1, so this term is cancelled out.

So, the RHS and LHS both have the same powers of x, and for any given power the coefficient is the same on each side, therefore they are equivalent.

 

[lolokay]

no.

new question:

how many boys can fit into a cup full of water if p = q+3

i get 78.32 to 2 dp but I think my working has mistakes in it can you please confirm ??? ??? ???

 

[vds700]

nCr = (n-1)C(r-1) + (n-1)Cr
= (n-1)C(r-1) + (n-2)C(r-1) + (n-2)Cr
= ...
= Sum[1 < a < n-r+1] (n-a)C(r-1)

in the bracketed section, [(1+x)^n-1 +...], the coefficient of any given (r-1)th power of x is Sum[1 < a < n-r+1] (n-a)C(r-1) = nCr
but this term is then multiplied by x, which adds 1 to the power. So the coefficient of the rth power of x is nCr. The powers of x are from 1 to n

but on the RHS the coefficient of the rth power of x is also nCr. (1+x)ⁿ contains powers of x from 0 to n, but nC0 x⁰ = 1, so this term is cancelled out.

So, the RHS and LHS both have the same powers of x, and for any given power the coefficient is the same on each side, therefore they are equivalent.

omg you're a freak! Well done

 

[midifile]

nCr = (n-1)C(r-1) + (n-1)Cr
= (n-1)C(r-1) + (n-2)C(r-1) + (n-2)Cr
= ...
= Sum[1 < a < n-r+1] (n-a)C(r-1)

in the bracketed section, [(1+x)^n-1 +...], the coefficient of any given (r-1)th power of x is Sum[1 < a < n-r+1] (n-a)C(r-1) = nCr
but this term is then multiplied by x, which adds 1 to the power. So the coefficient of the rth power of x is nCr. The powers of x are from 1 to n

but on the RHS the coefficient of the rth power of x is also nCr. (1+x)ⁿ contains powers of x from 0 to n, but nC0 x⁰ = 1, so this term is cancelled out.

So, the RHS and LHS both have the same powers of x, and for any given power the coefficient is the same on each side, therefore they are equivalent.

I never thought of doing it that way...

you can also do it using the sum of a GP formula

 

[lolokay]

you can also do it using the sum of a GP formula

haha I didn't even notice that. that would've made it a lot simpler.

 

[bored of sc]

Simplify
(6ⁿ + 3ⁿ) / (2ⁿ⁺¹ + 2)

 

[lolokay]

Simplify
(6ⁿ + 3ⁿ) / (2ⁿ⁺¹ + 2)

Spoiler

(6ⁿ + 3ⁿ) / (2ⁿ⁺¹ + 2)
= (2ⁿ3ⁿ + 3ⁿ)/(2(2ⁿ + 1))
= 3ⁿ(2ⁿ + 1)/2(2ⁿ+1)
= 3ⁿ/2

 

[M@ster P]

Just got this question from some book, don't even understand it.

If A is acute, and cosA = 1 - x, where x is so small that x^2 is negligible compared with unity, prove that cos2A is aprroximately equal to 1 - 4x, and that cos3A is approximately equal to 1 - bx, where b is a constant. Find the value of b.

 

[vds700]

Just got this question from some book, don't even understand it.

If A is acute, and cosA = 1 - x, where x is so small that x^2 is negligible compared with unity, prove that cos2A is aprroximately equal to 1 - 4x, and that cos3A is approximately equal to 1 - bx, where b is a constant. Find the value of b.

cos2A = 2cos^2 A -1
=2(1-x)^2 - 1
=2(1 - 2x + x^2) - 1
=2 - 4x + 2x^2 - 1
= 1 - 4x as 2x^2 is negligible

cos3A = cos(2A + A)
=cos2AcosA - sin2AsinA
=(1-4x)(1-x) - 2sinAcosA . sinA
=1-5x + 4x^2 -2sin^2 A (1-x)
=1-5x -2(1-cos^2 A)(1-x)
=1-5x -2[1 - (1-2x + x^2)](1-x)
=1-5x - 2(1 - 1 + 2x - x^2)(1-x)
=1-5x - 4x(1-x)
=1-5x -4x + 4x^2
=1 - 9x, hence b = 9

may have made a mistake in the 2nd part

New question:
The rise and fall of the tides can be approximated to SHM. On October 18, the depth of water in a tidal lagoon at low tide is 2m at 11.00 am. At the followiung high tide at 5.20 pm, the depth is 6m. Calculate between what times a yaught could safely cross the lagoon if a minimum depth of 3.5m is needed.

 

[M@ster P]

wow the question makes sense now, also if cosA = 1 - x, can you draw a triangle and use pythagoras theorem to find sinA

 

[conics2008]

New question:
The rise and fall of the tides can be approximated to SHM. On October 18, the depth of water in a tidal lagoon at low tide is 2m at 11.00 am. At the followiung high tide at 5.20 pm, the depth is 6m. Calculate between what times a yaught could safely cross the lagoon if a minimum depth of 3.5m is needed..

Yeah I hate these questions, they are soo annoying.. Can someone help me out in these types of question, I dont even know where to begin. Its a 4unit question.

 

[conics2008]

wow the question makes sense now, also if cosA = 1 - x, can you draw a triangle and use pythagoras theorem to find sinA

yes dont forget the 1

 

[vds700]

New question:
The rise and fall of the tides can be approximated to SHM. On October 18, the depth of water in a tidal lagoon at low tide is 2m at 11.00 am. At the followiung high tide at 5.20 pm, the depth is 6m. Calculate between what times a yaught could safely cross the lagoon if a minimum depth of 3.5m is needed..

Yeah I hate these questions, they are soo annoying.. Can someone help me out in these types of question, I dont even know where to begin. Its a 4unit question.

No this is a 3 unit level question, got it out of the Reddam 2004 3 unit trial.

Hint, start witht the basic equation for SHM, x = Acos(nt + a) work out the period to find n, work out the amplitude etc.

It helps to draw a diagram.

 

[lyounamu]

New question:
The rise and fall of the tides can be approximated to SHM. On October 18, the depth of water in a tidal lagoon at low tide is 2m at 11.00 am. At the followiung high tide at 5.20 pm, the depth is 6m. Calculate between what times a yaught could safely cross the lagoon if a minimum depth of 3.5m is needed..

Yeah I hate these questions, they are soo annoying.. Can someone help me out in these types of question, I dont even know where to begin. Its a 4unit question.

I will actually draw a diagram where 11.00 am is at the origin. Origin can be (4m, 11.00 am) = (0,0) or basically anything that suits you.

So if you follow the origin being (4m, 11.00 am), the graph will look like as if it's a negative cos graph. The graph will start at (2m, 11.00 am) = (-2, 0).

And the high tide is at (6m, 5.20 pm) = (2, 6 1/3)

Drawing that, you can also determine the graph being

y = -2cos(6pi/38 x)
Then the sub in y=-0.5 in there

so -0.5 = -2cos(6pi/38 x)
x = 2.657272719th... & 10.00939395.... since 11.00 am
There you go.

 

[vds700]

I will actually draw a diagram where 11.00 am is at the origin. Origin can be (4m, 11.00 am) = (0,0) or basically anything that suits you.

So if you follow the origin being (4m, 11.00 am), the graph will look like as if it's a negative cos graph. The graph will start at (2m, 11.00 am) = (-2, 0).

And the high tide is at (6m, 5.20 pm) = (2, 6 1/3)

Drawing that, you can also determine the graph being

y = -2cos(6pi/38 x)
Then the sub in y=-0.5 in there

so -0.5 = -2cos(6pi/38 x)
x = 2.657272719th... & 10.00939395.... since 11.00 am
There you go.

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

 

[Dota55]

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.... )

 

[bored of sc]

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

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

3) Find the remainder when f(x) is divided by g(x), given f(x) = x²-3x+6 and g(x) = x-2.

4) Find the exact value of cos75^o.

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

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

8) Show that sin(x+y)/cos(x-y) = (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

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.

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.

 

[kaz1]

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

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

3) Find the remainder when f(x) is divided by g(x), given f(x) = x²-3x+6 and g(x) = x-2.

4) Find the exact value of cos75^o.

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

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

8) Show that sin(x+y)/cos(x-y) = (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

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.

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.

3) 4