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Advanced Maths Predictions/Thoughts (6 Viewers)

kittyc4tkittycar

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hi guys
i have a random kinda pedantic question
when using the trapezoidal rule or estimates in general, can we use squiggly equal sign instead of the dotted equal sign?
I always use the squiggly one and I'm used to that one, it's on the reference sheet for trapezoidal but all HSC answers use the dotted one.
I remember my math teacher saying somethign about how the squiggly ones aren't reccomended in HSC. Does it matter?
1761004540669.png
 

bigupsanky

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hi guys
i have a random kinda pedantic question
when using the trapezoidal rule or estimates in general, can we use squiggly equal sign instead of the dotted equal sign?
I always use the squiggly one and I'm used to that one, it's on the reference sheet for trapezoidal but all HSC answers use the dotted one.
I remember my math teacher saying somethign about how the squiggly ones aren't reccomended in HSC. Does it matter?
View attachment 49698
no not really i use squiggly lines and theyre fine
 

bigupsanky

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Also, how do I calculate return on interest for different scenarios, like for e.g. continuously adding money into the savings account every month or so, or like withdrawing money from that account every month or so. I always get questions like those wrong.
 

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zodiacocean

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Also, how do I calculate return on interest for different scenarios, like for e.g. continuously adding money into the savings account every month or so, or like withdrawing money from that account every month or so. I always get questions like those wrong.
wait ok so this is only three months so afaik you have to do it manually. so for the first month, $200000 earns interest of 0.5% for one month to make $201000, then $2000 is withdrawn to make $199000. for the second month $199000 earns interest of 0.5% for one month to make $199995, then $2000 is subtracted to make $197995. for the third month $197995 earns interest of 0.5% for one month to make $198984.98, then $2000 is subtracted to make $196984.98. does that make sense
 

Trial&Error

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Also, how do I calculate return on interest for different scenarios, like for e.g. continuously adding money into the savings account every month or so, or like withdrawing money from that account every month or so. I always get questions like those wrong.
a)
R = 1+ 0.06/12 = 1.005

A1 = 200,000R - 2000

A2 = 200,000R^2 - 2000R - 2000

A3 = 200,000R^3 - 2000R^2 - 2000R - 2000

= $196,984.98

This question was straight forward since the numner of periods was small but if its largher then you need to find a general equation for "An" for exampe if you wanted to find the money left after 7 months:

An = 200,000R^n - 2000R^n-1 -... 2000R - 2000
so that = 200,000R^n - 2000 (1 + ... R^n-1)

so for 7 months: 200,000R^7 - 2000 ( 1 +...R^6)

then use the sum formula: 200,000R^7 - 2000 (1(1-R^7) / (1-R))
= 192,894,12
 

bigupsanky

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h
a)
R = 1+ 0.06/12 = 1.005

A1 = 200,000R - 2000

A2 = 200,000R^2 - 2000R - 2000

A3 = 200,000R^3 - 2000R^2 - 2000R - 2000

= $196,984.98

This question was straight forward since the numner of periods was small but if its largher then you need to find a general equation for "An" for exampe if you wanted to find the money left after 7 months:

An = 200,000R^n - 2000R^n-1 -... 2000R - 2000
so that = 200,000R^n - 2000 (1 + ... R^n-1)

so for 7 months: 200,000R^7 - 2000 ( 1 +...R^6)

then use the sum formula: 200,000R^7 - 2000 (1(1-R^7) / (1-R))
= 192,894,12
no i got part a i got part b wrong my fault should've explained it clearer
 

Trial&Error

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Also, how do I calculate return on interest for different scenarios, like for e.g. continuously adding money into the savings account every month or so, or like withdrawing money from that account every month or so. I always get questions like those wrong.
for part b)

withdrawals = 3x2000 = $6000

If no interest was added the money left would be: 200,000-6000 = $194,000

So the difference between the interest added amount from part a and the hypothetical non-intrest amount is
196,984.98 - 194,000 = $2984.98
That's the interest added over the 3 periods.
 

bigupsanky

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for part b)

withdrawals = 3x2000 = $6000

If no interest was added the money left would be: 200,000-6000 = $194,000

So the difference between the interest added amount from part a and the hypothetical non-intrest amount is
196,984.98 - 194,000 = $2984.98
That's the interest added over the 3 periods.
ohh thats how u work it out thanks a lot
 

bigupsanky

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This is a rlly dumb question but why did you chose 0 and a as the two y values to check? Like I understand everyhting but is there a reason? So I can know and use it in a similar question where the values tested are variables... :redface:
basically she's checking how the gradient sign changes when the value is less than sqrt(a)/2 but u really don't know the value of a, so u can use 0 and a instead because sqrt(a)/2 is greater than 0 and a is greater than sqrt(a)/2
 

bigupsanky

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basically she's checking how the gradient sign changes when the value is less than sqrt(a)/2 but u really don't know the value of a, so u can use 0 and a instead because sqrt(a)/2 is greater than 0 and a is greater than sqrt(a)/2
i meant a/sqrt(2)
 

kittyc4tkittycar

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i meant a/sqrt(2)
ohhh i see. so it's relying on the fact that a>0 and k>0 completely (?) :pold: but you couldn't check with a number such as -1, right? Because of the variables in the question.
(I just worked it out with a and 0 and I understand it! :tongue: )
Also would it be better to check w second derivative via concavity?
 

bigupsanky

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ohhh i see. so it's relying on the fact that a>0 and k>0 completely (?) :pold: but you couldn't check with a number such as -1, right? Because of the variables in the question.
(I just worked it out with a and 0 and I understand it! :tongue: )
Also would it be better to check w second derivative via concavity?
1. no you can't check with -1 cuz dimensions cant be negative so they will deduct marks for that.
2. You can go for a second derivative but it will become too complex since the first derivative is complex.
3. Im not sure whether it relies on a > 0 and k>0 completely.
 

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