# 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:

I'm just struggling to find the link between

#### InteGrand

##### Well-Known Member

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

I'm just struggling to find the link between

#### davidgoes4wce

##### Well-Known Member

OK I get it now.

I treated the origin at the base of the cliff.

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

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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.

(treating anything downwards as positive g=+10)

#### HeroicPandas

##### Heroic!
• AfroNerd

#### 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. Last edited:

#### 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 #### eyeseeyou

##### Well-Known Member

How do I do this #### eyeseeyou

##### Well-Known Member

Also #### Drongoski

##### Well-Known Member

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• kashkow

#### 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-

Also 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

• eyeseeyou