Yeah, just check it the solutions afterwards
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3 Unit Revising Marathon HSC '10 (1 Viewer)
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Shadowdude
Cult of Personality
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Okay fine jm...
^{2}}\\ =\sqrt{1+49}\\ =\sqrt{50}=5\sqrt{2}\\ \\ tan\alpha =\frac{b}{a}=7\\ \alpha=81.86989765\\ \\ sinx-7cosx=5\sqrt{2}sin(x-\alpha)=-5\\ 5\sqrt{2}sin(x-\alpha)=-5\\ \sqrt{2}sin(x-\alpha)=-1\\ sin(x-\alpha)=-\frac{1}{\sqrt{2}}\\ x-\alpha=-\frac{\pi}{4},\:\frac{5\pi}{4}\\ x-81.86989765=-45,\: 215\\ x=36.86989765,\: 296.8698977\\ \\ x=36.87,\: 296.87\; (rounding\,to\,2\,decimal\,places))
Not sure how you'll make radians out of those though.
Not sure how you'll make radians out of those though.
Well that's how my teacher taught me in Year 11.
ninetypercent
ninety ninety ninety
Question: (easy one) prove nCk+nC(k-1)=(n+1)Ck
LHS =nCk+nC(k-1)
= n!/(n-k)!k! + n!(n-k+1)!(k-1)!
= n!/(n-k)!k(k-1)! + n!/(n-k+1)(n-k)!(k-1)!
= [n!(n-k+1) + n!(k)]/(n-k+1)(n-k)!(k-1)!k
=( n!n + n!)/(n-k+1)(n-k)!(k-1)!k
= [n!(n+1)]/[(n-k+1)(n-k)!(k-1)!k]
RHS = (n+1)Ck
= (n+1)!/(n+1 -k)!k!
= [n!(n+1)]/[(n-k+1)(n-k)!(k-1)!k]
LHS = RHS
therefore, nCk+nC(k-1)=(n+1)Ck
LHS =nCk+nC(k-1)
= n!/(n-k)!k! + n!(n-k+1)!(k-1)!
= n!/(n-k)!k(k-1)! + n!/(n-k+1)(n-k)!(k-1)!
= [n!(n-k+1) + n!(k)]/(n-k+1)(n-k)!(k-1)!k
=( n!n + n!)/(n-k+1)(n-k)!(k-1)!k
= [n!(n+1)]/[(n-k+1)(n-k)!(k-1)!k]
RHS = (n+1)Ck
= (n+1)!/(n+1 -k)!k!
= [n!(n+1)]/[(n-k+1)(n-k)!(k-1)!k]
LHS = RHS
therefore, nCk+nC(k-1)=(n+1)Ck
ninetypercent
ninety ninety ninety
new question
Give that f(x) = ax^3 + bx^2 + cx + d
has a double root at x = -1, and with a minimum value of -4 when x =1
Find the values of a, b, c and d
Give that f(x) = ax^3 + bx^2 + cx + d
has a double root at x = -1, and with a minimum value of -4 when x =1
Find the values of a, b, c and d
addikaye03
The A-Team
I still think it's a 4U concept. I think you would find students who do 2/3U wouldn't understand it. I'm not saying your teacher didn't teach you it in year 11, i'm just saying it's 4U content
My class mates didn't have much trouble understanding it.
In addition, this is definitely not 2U material because in 2U you don't deal with finding roots of equations that have a degree higher than quadratics.
addikaye03
The A-Team
Nah i wasn't saying it was 2U material at all... It's definitely 3U. Doesn't matter anyway, lets move on. Got a new Q Jm01? I can't think of one
hscishard
Active Member
Lol...are there any 2011 students thinking "wtf" when reading these questions? I know i am 
addikaye03
The A-Team
lol ofcourse, if you haven't studied the material, then it's hard to answer Q based upon it. You'll be right...when you learn it that is.Lol...are there any 2011 students thinking "wtf" when reading these questions? I know i am
ohexploitable
hey
Lol...are there any 2011 students thinking "wtf" when reading these questions? I know i am
New question:
(a) Using the result for tan(A+B), prove that
=\frac{tanA+tanB+tanC-tanAtanBtanC}{1-tanAtanB-tanBtanC-tanCtanA})
[2 m]
(b) Given that A, B and C are the angles of a triangle and
[2 m]
(c) Hence calculate the smallest of the angles to the nearest minute. [1 m]
(a) Using the result for tan(A+B), prove that
[2 m]
(b) Given that A, B and C are the angles of a triangle and
[2 m]
(c) Hence calculate the smallest of the angles to the nearest minute. [1 m]
Last edited:
ninetypercent
ninety ninety ninety
what is part 2 supposed to say?
hello-there
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SERIES APP.
Lets see who is smart here
Q. Benny Franklin left a will in which he established a fund of $1000 for the citizens of Boston. His instructions were that his money was to be invested at 5% interest, compounded annually. (ANS=$131 501
i) If Franklin's instructions were folllowed, how much money would have been in the fund 100 yrs after it was established?
ii) (here comes the tricky one) Suppose that at the beginning of each subsequent yr after establishment, a further $1000 had been added to the fund and had also earned 5% inteerest, compounded annually. How much would have been in the fund after 200yrs, just b4 the nest $1000 would have been added to the fund? (ANS=$363123197)
yes long and tedious
Lets see who is smart here
Q. Benny Franklin left a will in which he established a fund of $1000 for the citizens of Boston. His instructions were that his money was to be invested at 5% interest, compounded annually. (ANS=$131 501
i) If Franklin's instructions were folllowed, how much money would have been in the fund 100 yrs after it was established?
ii) (here comes the tricky one) Suppose that at the beginning of each subsequent yr after establishment, a further $1000 had been added to the fund and had also earned 5% inteerest, compounded annually. How much would have been in the fund after 200yrs, just b4 the nest $1000 would have been added to the fund? (ANS=$363123197)
yes long and tedious
xV1P3R
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i) A = 1000(1+0.05)^100
= 131,501.2578 with calculator
ii) Let (1+0.05) = 1.05 = i
k = 1000
1st Year
A = ki
2nd Year
A = ((k)i + k)i
A = (i + 1)ki
3rd Year
A = ((i + 1)ki + k)i
A = ((i + 1)i + 1)ki
A = (i^2 + i + 1)ki
4th Year
A = (((i + 1)i + 1)ki + k)i
A = (((i + 1)i + 1)i + 1)ki
A = ((i^2 + i + 1)i + 1)ki
A = (i^3 + i^2 + i + 1)ki
200th Year
%20\\(1-i)A%20=%20ki(1+i+i^{2}+i^{3}+...+i^{199}-i-i^{2}-i^{3}-i^{4}-...-i^{200})%20\\(1-i)A%20=%20ki(1-i^{200})%20\\A%20=%20ki(1+i+i^{2}+i^{3}+...+i^{199})%20\\(1-i)A%20=%20ki(1+i+i^{2}+i^{3}+...+i^{199}-i-i^{2}-i^{3}-i^{4}-...-i^{200})%20\\A%20=%20\frac{ki}{1-i}(1-i^{200})%20\\A%20=%20\frac{1050}{-0.05}(1-172925.58082)%20\\A%20=%20363,123,197.1)
New Question
Describe what processes happen to y = cos(x) when you transform it into y = cos(2x - pi/2) eg. Flip it, reflect off y-axis, translate this much to the left etc.
Hence or otherwise simplify y = cos(2x - pi/2)
= 131,501.2578 with calculator
ii) Let (1+0.05) = 1.05 = i
k = 1000
1st Year
A = ki
2nd Year
A = ((k)i + k)i
A = (i + 1)ki
3rd Year
A = ((i + 1)ki + k)i
A = ((i + 1)i + 1)ki
A = (i^2 + i + 1)ki
4th Year
A = (((i + 1)i + 1)ki + k)i
A = (((i + 1)i + 1)i + 1)ki
A = ((i^2 + i + 1)i + 1)ki
A = (i^3 + i^2 + i + 1)ki
200th Year
New Question
Describe what processes happen to y = cos(x) when you transform it into y = cos(2x - pi/2) eg. Flip it, reflect off y-axis, translate this much to the left etc.
Hence or otherwise simplify y = cos(2x - pi/2)