# Cambridge HSC MX1 Textbook Marathon/Q&A (1 Viewer)

#### davidgoes4wce

##### Well-Known Member

I know from Physics formulae with the equations of motions we have:

$\bg_white v^2=u^2+2as$

$\bg_white s=ut+\frac{1}{2} at^2$

I'm just struggling to find the link between

$\bg_white 180= V \times t$

#### InteGrand

##### Well-Known Member

I know from Physics formulae with the equations of motions we have:

$\bg_white v^2=u^2+2as$

$\bg_white s=ut+\frac{1}{2} at^2$

I'm just struggling to find the link between

$\bg_white 180= V \times t$
$\bg_white \noindent Do you mean how did I arrive at 180 = Vt at the time of collision? This was done by equating the two equations of motion and cancelling common terms from each side.$

#### davidgoes4wce

##### Well-Known Member

OK I get it now.

I treated the origin at the base of the cliff.

$\bg_white y_B=180-5t^2$

Where I went wrong was I assumed the stone had an initial velocity of 0, the question stated it was V.

$\bg_white y_S=Vt-5t^2$

$\bg_white By equating y_S = y_B I can then get the equation of 180=Vt$

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#### davidgoes4wce

##### Well-Known Member

If I treated the top of the lookout as the mean line, my equations would obviously be different.

$\bg_white y_B=5t^2$

$\bg_white y_S=-Vt+5t^2+180$

(treating anything downwards as positive g=+10)

$\bg_white By equating y_B=y_S , 5t^2=-Vt+5t^2+180$

$\bg_white Vt=180$

#### AfroNerd

##### New Member

having trouble with this binomial question

#### HeroicPandas

##### Heroic!

having trouble with this binomial question
View attachment 33393
\bg_white (1+x)^{n-1} = \binom{n-1}{0} + \binom{n-1}{1}x + \dots + \binom{n-1}{n-2} x^{n-2} + \binom{n-1}{n-1}x^{n-1} \\ Put x = 1, \\\\ 2^{n-1} = \binom{n-1}{0} + \underbrace{\binom{n-1}{1} + \dots + \binom{n-1}{n-2}}_{\text{isolate this}} + \binom{n-1}{n-1} \\\\ \begin{align*} \binom{n-1}{1} + \dots + \binom{n-1}{n-2} &= 2^{n-1} - \binom{n-1}{0} - \binom{n-1}{n-1} \\&= 2^{n-1} - 1 -1 \end{align*}

Then multiply both sides by n

#### si2136

##### Well-Known Member

I think there is a formula for this but I'm not sure what it's called.

Q: Four out of the seven letters of the word COCONUT are selected and arranged in a row. How many different arranges are possible?

#### InteGrand

##### Well-Known Member

I think there is a formula for this but I'm not sure what it's called.

Q: Four out of the seven letters of the word COCONUT are selected and arranged in a row. How many different arranges are possible?
You may wish to consider cases based on things like how many ''double-letters'' are selected.

#### si2136

##### Well-Known Member

You may wish to consider cases based on things like how many ''double-letters'' are selected.
I know how to do it if it was all of the letters, but not 4 out of the 7

#### InteGrand

##### Well-Known Member

I know how to do it if it was all of the letters, but not 4 out of the 7
Yeah, that's why you may find it helpful to consider cases. E.g. The cases could be: no pairs of double-letters; exactly one pair of double-letters; exactly two pairs of double-letters.

Then count the no. of possibilities in each case and sum them up.

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#### Mongoose528

##### Member

Isn't the formula this?

For the number of ways to arrange n objects taken r at at a time

n!/(n-r)!

So it would be

7!/(7-4)! = 7!/3! = 7*6*5*4*3!/3! = 7*6*5*4=840

But since we have 2 c's and 2 o's, the answer would be 840/2!2! = 840/4 =210

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#### si2136

##### Well-Known Member

Isn't the formula this?

For the number of ways to arrange n objects taken r at at a time

n!/(n-r)!

So it would be

7!/(7-4)! = 7!/3! = 7*6*5*4*3!/3! = 7*6*5*4=840

But since we have 2 c's and 2 o's, the answer would be 840/2!2! = 840/4 =210
Wrong, you would need to take it by cases or Pork's Insertion Method

How do I do this

Also

#### infiniteeee

##### New Member

anyone who can help me with question 12b and 13a of exercise 4F?

#### Drongoski

##### Well-Known Member

$\bg_white \therefore x^3 - Ax + 0.x + 3A = 0\\ \\ \therefore sum of roots = \alpha + \beta + (\alpha + \beta) = -(-A) = A \implies 2(\alpha + \beta) = A \implies \alpha + \beta = \frac {1}{2} A \\ \\ sum of roots taken 2 at a time = \alpha \beta +\alpha (\alpha+\beta) + \beta (\alpha+\beta) = 0 \\ \\ \therefore \alpha \beta = - (\alpha + \beta)^2 = - (\frac {1}{2} A)^2 = -\frac {1}{4} A^2 \\ \\ product of roots = \alpha \beta (\alpha + \beta) -3A \implies \alpha \beta = \frac {-3A}{\frac {1}{2}A} = -6 \implies -\frac {1}{4}A^2 = -6 \\ \\ \implies A = 2\sqrt 6$

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#### kashkow

##### Active Member

View attachment 33481

anyone who can help me with question 12b and 13a of exercise 4F?
The first question:

The "sum of the products of pairs of roots" refers to the x term which is absent, meaning the coefficient is 0. Therefore the "sum of the products of pairs of roots" = 0

Algebraically: ab + a(a+b) + b(a+b) = 0 [... a = alpha, b = beta]

Expanding: ab + a^2 + ab + ab + b^2 = 0

ab + a^2 + 2ab + b^2 = 0

Combining the perfect square:

ab + (a+b)^2 = 0

And you can use the results from the question before 12a what (a+b)^2 is.

That is substitute (a+b) = 1/2*A = A/2

ab + (A/2)^2 = 0

ab + (A^2)/4 = 0

Therefore: ab = -(A^2)/4 = -1/4 * A^2

Sorry if it's hard to read. Rats i was late again...

##### -insert title here-

Cosine Rule XOR Sine Rule

##### -insert title here-

How do I do this

...seriously just draw a diagram

#### davidgoes4wce

##### Well-Known Member

View attachment 33481

anyone who can help me with question 12b and 13a of exercise 4F?
EX 4F Q13 a

$\bg_white Let the three roots be \alpha, \alpha \ and \beta$

$\bg_white 2 \alpha + \beta =\frac{8}{4}=2 \ \textcircled{1}$

$\bg_white \alpha \beta +\alpha \beta +\alpha^2 =\frac{-3}{4} \implies 2 \alpha \beta +\alpha^2=\frac{-3}{4} \textcircled{2}$

$\bg_white \alpha^2 \beta =\frac{-9}{4} \textcircled{3}$

$\bg_white Rearranging \textcircled{1} \ \beta=2-2\alpha$

$\bg_white Substituting \textcircled{1} into \textcircled{2}$

$\bg_white 2 \alpha (2-2 \alpha)+\alpha^2=\frac{-3}{4}$

$\bg_white 4 \alpha -4 \alpha^2+\alpha^2=\frac{-3}{4}$

$\bg_white -3 \alpha^2+4\alpha=\frac{-3}{4}$

$\bg_white -12 \alpha^2+16 \alpha =-3 \implies 0=12\alpha^2-16\alpha-3$

$\bg_white By using the Quadratic Roots formula x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} we look to solve \alpha$

$\bg_white \alpha=\frac{16 \pm \sqrt{256-4(12)(-3)}}{2 \times 12}$

$\bg_white \alpha= \frac{3}{2} \ and \ \alpha=\frac{-1}{6}$

$\bg_white if \alpha =\frac{3}{2} \beta=-1$

$\bg_white Substitute \alpha =\frac{3}{2} \ into equation \textcircled{1} \ \beta=-1$

$\bg_white Substitute \alpha =\frac{3}{2} \ into equation \textcircled{3} \ \beta=-1$

$\bg_white Therefore, \alpha =\frac{3}{2} and \beta=-1 are consistent solutions$

$\bg_white if \alpha =\frac{-1}{6}$

$\bg_white Substitute \alpha =\frac{-1}{6} \ into equation \textcircled{1} \ \beta=\frac{7}{3}$

$\bg_white Substitute \alpha =\frac{-1}{6} \ into equation \textcircled{3} \ \beta=-81$

$\bg_white Therefore, \alpha =\frac{-1}{6} is not consistent and is not a solution to the cubic polynomial.$