Extension One Revising Game (2 Viewers)

[midifile]

was that the catholic trial? (I remember you saying you did the catholic trial)

 

[vds700]

was that the catholic trial? (I remember you saying you did the catholic trial)

yeah it was. Though my school replaced the projectuile question in the 2008 cssa ext1 paper with this one (which i think is from a past cssa paper). Got no idea why

 

[Pwnage101]

I ndont think the question is TOO hard, given enough time, but it does take some serious thinking, and is certainly aimed at the top 5% of the state

 

[melonkitten]

k, thats too hard. now my question.


There are 144 students(62 boys, 82 girls) sitting in a circle.

What is the probability that: 8 friends consisting of 3 boys and 5 girl are NOT sitting together ??

 

[bored of sc]

k, thats too hard. now my question.


There are 144 students(62 boys, 82 girls) sitting in a circle.

What is the probability that: 8 friends consisting of 3 boys and 5 girl are NOT sitting together ??

Probability of 3 boys and 5 girls sitting together:

[8! / (3!.5!) . 135!]/143!

= Ans

Prob of 3 boys and 5 girls not sitting together:

1 - Ans


I think that's wrong but whatever.


My question:

Differentiate y = 5x² - 3 from first principles.

 

[midifile]

dy/dx=lim(h-->0) [f(x+h)-f(x)]/h
=lim(h-->0)[5x² + 10xh + 5h² - 3 - 5x² + 3]/h
=lim(h-->0)[h(10x + 5h)]/h
=lim(h-->0) [10x + 5h]
=10x

Use Newtons method to find a second approximation to the positive root of x - 2sinx = 1. Use x=1.7 as the first approximation

 

[conics2008]

Heres an excellent question frokm my MX1 trial

A particle is projected with velocity v at an angle <META content=Word.Document name=ProgId><META content="Microsoft Word 11" name=Generator><META content="Microsoft Word 11" name=Originator><LINK href="file:///C:%5CUsers%5CAndrew%5CAppData%5CLocal%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml" rel=File-List><STYLE> <!-- /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0cm; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman";} @page Section1 {size:595.3pt 841.9pt; margin:72.0pt 90.0pt 72.0pt 90.0pt; mso-header-margin:35.4pt; mso-footer-margin:35.4pt; mso-paper-source:0;} div.Section1 {page:Section1;} --> </STYLE>θ. Gravity g ms^-2

(i) Show that the range of the projectile is given by R = v^2 sin2<META content=Word.Document name=ProgId><META content="Microsoft Word 11" name=Generator><META content="Microsoft Word 11" name=Originator><LINK href="file:///C:%5CUsers%5CAndrew%5CAppData%5CLocal%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml" rel=File-List><STYLE> <!-- /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0cm; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman";} @page Section1 {size:595.3pt 841.9pt; margin:72.0pt 90.0pt 72.0pt 90.0pt; mso-header-margin:35.4pt; mso-footer-margin:35.4pt; mso-paper-source:0;} div.Section1 {page:Section1;} --> </STYLE>θ/g

(ii)A lawn on horizontal ground is rectangular in shape with lenght 50m and breadth 20m. A garden sprinkler is located at one corner, S of the lawn. It rotates horizontally and delivers water at a speed of 20 m/s at an angle of elevation between 15 degrees and 45 degrees above the horizontal. Taking g = 10, find the area of the garden that ca be watered by the sprinkler, giving your answer in simplest exact form.

hey that was in my trial paper..

you need to use a sector of two circles.

xD

 

[conics2008]

that question wasn't really hard, just need to imagine how it turns out to be.. but seriously not enough time, that was the thing about it.

 

[conics2008]

Use Newtons method to find a second approximation to the positive root of x - 2sinx = 1. Use x=1.7 as the first approximation

x(new)= 1.7 - ( 1.7-2sin(1.7)-1)/(1.7-2cos(1.7))

x(new) = 2.3

keeping in mind to use rad in your cal

new question

When a poly is divided by (x-2) the remainder is 5 and when it is divided by (x+5) the remainder is 10, find the remainder when it is divided by
(x-2)(x+5) ?

 

[Darrow]

I have never come across a question like that
I am at a complete loss, not even the internet is helping me =/

 

[tommykins]

Use Newtons method to find a second approximation to the positive root of x - 2sinx = 1. Use x=1.7 as the first approximation

x(new)= 1.7 - ( 1.7-2sin(1.7)-1)/(1.7-2cos(1.7))

x(new) = 2.3

keeping in mind to use rad in your cal

new question

When a poly is divided by (x-2) the remainder is 5 and when it is divided by (x+5) the remainder is 10, find the remainder when it is divided by
(x-2)(x+5) ?

That's 4unit mate, not 3unit.

 

[kaz1]

That's 4unit mate, not 3unit.

We got a question like that in our last three unit exam. I think you have to use simultaneous equations.

 

[lyounamu]

Correct

Solution is attached

It was a bitch of a question, no-one got full marks for it

That question was a bitch. It was lucky that I pulled 3 marks out of 4.

 

[lolokay]

When a poly is divided by (x-2) the remainder is 5 and when it is divided by (x+5) the remainder is 10, find the remainder when it is divided by
(x-2)(x+5) ?

would the answer be -5/7 x + 45/7 ?

 

[tommykins]

That is correct lolokay.

 

[Pwnage101]

i'm at a loss with that poly Q, although polynomials are my weak point, can u post full solution?

 

[tommykins]

same here, although polynomials are my weak point, can u post full solution?

Sure

When a poly is divided by (x-2) the remainder is 5 and when it is divided by (x+5) the remainder is 10, find the remainder when it is divided by
(x-2)(x+5) ?

When P(x) is divided by (x-2)(x+5), then -

P(x) = (x-2)(x+5).Q(x) + R(x) where R(x) = Ax+B.
Now, P(2) = (2-2)(2+5)...+2a+B = 5
2A+B = 5
P(-5) = (-5-2)(-5+5)..-5A+B = 10

(1) 2A+B = 5
(2) -5A+B = 10
(1)-(2)
7A = -5
A = -5/7

25/7 + B = 10
25 + 7B = 70
7B = 45
B = 45/7

Therefore R(x) remainder is -5x/7 + 45/7

 

[Pwnage101]

cheers

 

[conics2008]

hey yeap thats the correct answer..

you need to state that the remainder was ax+b and that you had to write it in this form

P(x) = (x-2)(x+5)Q(x)+ax+b

then go on to get two equations...

and then find your a and b and then thats your remainder..

its not a 4 UNIT quesiton its just a 3unit quesiton.

 

[vds700]

heres a question

Solve:

5/[(2 - x)(x + 2)] >1