Math Past Paper help (1 Viewer)

[UTRazilBay]

I'm doing a James Ruse paper and I'm stuck on a couple of the questions






Q1: I cant find a relationship between the terms, its not a GP or an AP
Q2: I have no idea what that is, but I'm sure its not 2 unit (correct me if I'm wrong)
Q3: Need help with third part
Q4: How can dy/dt be a max or min when the question says the rate is constant??
Q5: For part ii) i keep getting powers of ^(n-1) instead of the required ^(n)


Thanks

 

[RishBonjour]

whats the answer for the first one?
its neither, but rather, you rationalise the denominator
and then
sub in r=1, 2 3.. and you see a pattern
they all cancel each other out except the 1 at the beginning and the last term, root 64 which is 8 therefore 8-1=7

thats the answer I got, makes sense, but I might have done a silly mistake, but thats how you do it.

number 2. yeah it is 2 unit, you can just look at it and do it, although they won't ask it in HSC, as it sorta does require "u-sub" but its -1/2cos(x^2)

ill do the rest later if no one replies.

 

[Timske]

5. i. y1=1000(1.06) - 72

ii. Find a pattern.
y1=1000(1.06) - 72
y2=1000(1.06)^2 - 72(1.06) - 72
y3=1000(1.06)^3 - 72(1.06)^2 - 72(1.06) - 72
.
.
.
yn=1000(1.06)^n - 72(1.06)^n-1 - ... - (1.06) - 72
yn=1000(1.06)^n - 72(1 + 1.06 + ... + 1.06^n-1)
Using sum of G.P
yn=1000(1.06)^n - 72[(1.06^n - 1)/(1.06-1)]

 

[Sy123]

Q3:



Q4:

The amount of water being put into the jug is in a constant rate, so dV/dt is constant. The meaning why this is so, is 3U (parameters), but its just there to tell you that there are no annomalies when they fill the jug. By observation and logic, the minimum dy/dt means the minimum rate at which DEPTH is increasing, since hte Volume of water is constantly increasing at a constant rate then the most widest areas will have the largest rate, hence 12cm.

Maximum dy/dt occurs at the thinnest areas. 4cm

To graph this, you need to know that at y=4 dy/dt is a maximum and that y=12 is a minimum. Note that these are the y-values not the t-values. What we are given is the rate at which y is increasing (hence we can deduce that y=f(t) is monotonic increasing).
We know that at y=4 dy/dt is a max hence its a point of inflection (d2y/dx2=0)
And at y=12 is a point of inflection
We know that the graph at t=8 is y=16 (thats how long it takes to fill up)

We also know that when y is concave down dy/dt is decreasing, y is concave up dy/dt is increasing. So using these 2 facts and the ones above we can sketch y=f(t) illustrating concavity changes (and the right concavities)

For example from y=0-> y=4 the graph is concave up since it is approaching maxima at y=4
Do the rest yourself etc.

Honestly not bothered to do finance lol but Ill do it I need the practise:

Q5:

We can see that this question is a loan repayment question, we are withdrawing 72 from the account each year. An account of initially 1000 dollars, which gains interest of 6% per annum and 72 is deducted each year after interest is added:








Just sub in the value for A_0 which is 841.83 and you get your answer

 

[Shadowless]

I did questions 1, 2 and 3 for you. For question 4, I'm not to sure about it so I'll let someone else answer it.

EDIT: -_- ppl beat me to it