Extension One Revising Game (1 Viewer)

[conics2008]

This is the drill, you answer a question then you ask a question.

ASK ONLY ONE QUESTION AT A TIME, AND WAIT TILL IT HAS BEEN ANSWERED.

Question 1)

Using transformtion


5cos(x)-9sin(x)=2

find all x values such that 0<x<360

 

[tacogym27101990]

This is the drill, you answer a question then you ask a question.

ASK ONLY ONE QUESTION AT A TIME, AND WAIT TILL IT HAS BEEN ANSWERED.

Question 1)

Using transformtion


5cos(x)-9sin(x)=2

find all x values such that 0<x><360

guessing that last bit means 0<x<360 because it was worded really badly/ incorrectly
now
5cos(x)-9sin(x)= rt(106)cos(x+atan(9/5))

so cos(x+atan(9/5))=2/rt106
x+atan(9/5)=acos(2/rt106), 360-acos(2/rt106)
x= </x>acos(2/rt106)-atan(9/5), 360-acos(2/rt106)-atan(9/5)
x=17deg51', 220deg15' (nearest min)

correct me if im wrong, typed the answer straight out onto the computer.

Q2
use the substitution of x=2sin@ to evaluate the integral (S is integral sign)

S rt(4-x^2)dx
fair call tommykins, limits are[0>>1]
nice and easy to start with

EDIT
By the way conics, wheres th trial paper you said you uploaded??

 

[tommykins]

回复: Re: Extension One Revising Game

guessing that last bit means 0<x<360 because it was worded really badly/ incorrectly
now
5cos(x)-9sin(x)= rt(106)cos(x+atan(9/5))

so cos(x+atan(9/5))=2/rt106
x+atan(9/5)=acos(2/rt106), 360-acos(2/rt106)
x= </x>acos(2/rt106)-atan(9/5), 360-acos(2/rt106)-atan(9/5)
x=17deg51', 220deg15' (nearest min)

correct me if im wrong, typed the answer straight out onto the computer.

Q2
use the substitution of x=2sin@ to evaluate the integral (S is integral sign)

S rt(4-x^2)dx

nice and easy to start with

EDIT
By the way conics, wheres th trial paper you said you uploaded??

Shouldn't there be limits considering you're evaluating the integral?

 

[conics2008]

guessing that last bit means 0<X<360 p badly really worded was it because incorrectly<> now
5cos(x)-9sin(x)= rt(106)cos(x+atan(9/5))

so cos(x+atan(9/5))=2/rt106
x+atan(9/5)=acos(2/rt106), 360-acos(2/rt106)
x= </X>acos(2/rt106)-atan(9/5), 360-acos(2/rt106)-atan(9/5)
x=17deg51', 220deg15' (nearest min)

correct me if im wrong, typed the answer straight out onto the computer.

Q2
use the substitution of x=2sin@ to evaluate the integral (S is integral sign)

S rt(4-x^2)dx
fair call tommykins, limits are[0>>1]
nice and easy to start with

EDIT
By the way conics, wheres th trial paper you said you uploaded??

Hey I sent the paper to VDS he has the copy and he can put it in PDF format as I dont know how to do it.

Good luck with the paper.

 

[tacogym27101990]

oi someone do my question 1st!

 

[vds700]

Hey I sent the paper to VDS he has the copy and he can put it in PDF format as I dont know how to do it.

Good luck with the paper.

dont know if i can do that, the only reason the one i uploaded is in PDF is that my scanner software does it automatically

 

[tacogym27101990]

hey conics can you send it to me please??

 

[midifile]

For tacogyms question...

x=2sin@
dx= 2cos@d@
@1 = pi/6
@2 =0
(im leaving out the limits just cos theyre a bitch to keep typing up)

Therefore, S rt(4-x^2)dx = S rt(4-4sin^2@)2cos@d@
= S 4cos^2@d@

But cos2@ = 2cos^2@ - 1
So 2cos^2@ = cos2@ + 1

Therefore S 4cos^2@d@ = 2 S cos2@ + 1 d@
=2 [(sin2@)/2 + @][
=2 (sin(pi/3)/2 +pi/6 – 0 – 0)
=Rt3/2 + pi/3

New question: Use induction to show that 7^n + 2 is divisible by three for all positive integers n>=1

 

[vds700]

New question: Use induction to show that 7^n + 2 is divisible by three for all positive integers n>=1

Step 1, let n = 1

7^1 +2 = 9 which is divisible by 3

Step 2. Assume that

7^k + 2 = 3P

Step 3. Prove that 7^(k + 1) + 2 = 3Q
LHS = 7 . 7^k + 2
=7(3P-2) + 2 (from assumption)
=21P -12
=3(7P -4)
=3Q

therefore true ....

New question: Prove that ⁿ⁺¹C₂ + ⁿ⁺²C₂ is the square of an integer

 

[tacogym27101990]

Therefore S 4cos^2@d@ = 2 S cos2@ + 1 d@
=2 [(sin2@)/2 + @][
=2 (sin(pi/3)/2 +pi/6 – 0 – 0)
=Rt3//2 + pi/3

thats the answer i believe

 

[tacogym27101990]

.

New question: Prove that ⁿ⁺¹C₂ + ⁿ⁺²C₂ is the square of an integer

ⁿ⁺¹C₂ + ⁿ⁺²C₂
= (n+1)!/(2!.(n-1)!) +(n+2)!/(2!(n!))
=(n(n+1)(n-1)!)/2!(n-1)! + ((n+2)(n+1)n)/(2!n!)
=((n+1)n+(n+2)(n+1))/2!
=(n+1)(2n+2)/2!
=(n+1)²

therefore ⁿ⁺¹C₂ + ⁿ⁺²C₂ is the square of n+1

New question:
find the coefficient of x⁵ in the expansion of (x²-2/x)⁷

 

[midifile]

Therefore S 4cos^2@d@ = 2 S cos2@ + 1 d@
=2 [(sin2@)/2 + @][
=2 (sin(pi/3)/2 +pi/6 – 0 – 0)
=Rt3//2 + pi/3

thats the answer i believe

yea... typo (i typed it straight into the computer)

ill change it now

 

[midifile]

New question:
find the coefficient of x⁵ in the expansion of (x²-2/x)⁷

T_k+1=⁷C_k(x²)^7-k(-2/x)^k
Coefficient of x⁵ when 2(7 -k) -k = 5
9=3k
k=3
Therefore coefficient=⁷C₃(1)⁴(-2)³
=-280

Question: A particle moves on the x axis with velocity v given by v = (x + 1)². Initially the particle is at the origin. Find the initial acceleration

 

[hon1hon2hon3]

For the above questions:

Question: A particle moves on the x axis with velocity v given by v = (x + 1)<SUP>2</SUP>. Initially the particle is at the origin. Find the initial acceleration.

v = (x+1)^2
(v^2)/2 = ((x+1)^4 )/2
d(v^2)/2 / dx = 2(x+1)^3

when x = 0 t = 0.

a = 2(x+1)^3 (Sub when x = 0)
a = 2

 

[vds700]

Heres an excellent question frokm my MX1 trial

A particle is projected with velocity v at an angle <meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 11"><meta name="Originator" content="Microsoft Word 11"><link rel="File-List" href="file:///C:%5CUsers%5CAndrew%5CAppData%5CLocal%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <wunctuationKerning/> <w:ValidateAgainstSchemas/> <w:SaveIfXMLInvalid>false</w:SaveIfXMLInvalid> <w:IgnoreMixedContent>false</w:IgnoreMixedContent> <w:AlwaysShowPlaceholderText>false</w:AlwaysShowPlaceholderText> <w:Compatibility> <w:BreakWrappedTables/> <w:SnapToGridInCell/> <w:WrapTextWithPunct/> <w:UseAsianBreakRules/> <wontGrowAutofit/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--><!--[if gte mso 9]><xml> <w:LatentStyles DefLockedState="false" LatentStyleCount="156"> </w:LatentStyles> </xml><![endif]--><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><!--[if gte mso 10]> <style> /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman"; mso-ansi-language:#0400; mso-fareast-language:#0400; mso-bidi-language:#0400;} </style> <![endif]-->θ. Gravity g ms^-2

(i) Show that the range of the projectile is given by R = v^2 sin2<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 11"><meta name="Originator" content="Microsoft Word 11"><link rel="File-List" href="file:///C:%5CUsers%5CAndrew%5CAppData%5CLocal%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <wunctuationKerning/> <w:ValidateAgainstSchemas/> <w:SaveIfXMLInvalid>false</w:SaveIfXMLInvalid> <w:IgnoreMixedContent>false</w:IgnoreMixedContent> <w:AlwaysShowPlaceholderText>false</w:AlwaysShowPlaceholderText> <w:Compatibility> <w:BreakWrappedTables/> <w:SnapToGridInCell/> <w:WrapTextWithPunct/> <w:UseAsianBreakRules/> <wontGrowAutofit/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--><!--[if gte mso 9]><xml> <w:LatentStyles DefLockedState="false" LatentStyleCount="156"> </w:LatentStyles> </xml><![endif]--><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><!--[if gte mso 10]> <style> /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman"; mso-ansi-language:#0400; mso-fareast-language:#0400; mso-bidi-language:#0400;} </style> <![endif]-->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.

 

[Pwnage101]

the fact that it is a rectangular garden is what makes this Q a bitch

 

[Pwnage101]

is the answer: (200root3+100pi/3) m^2??? will post sol. if so

 

[midifile]

the fact that it is a rectangular garden is what makes this Q a bitch

Yea. If both sides were longer than 40m it would be heaps easier.

But i cant seem to work out what to do wth that 20

 

[tommykins]

hahah fuck, that's a cunt of a question.

 

[vds700]

is the answer: (200root3+100pi/3) m^2??? will post sol. if so

Correct

Solution is attached

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