October 16, 2019, 08:10:33 pm

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#### fun_jirachi

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« Reply #4260 on: June 12, 2019, 11:39:16 pm »
+1
Still not quite sure what you're asking but will do my best to answer

If ever in doubt, always plot some points. With sine curves (and trigonometric curves in general) multiples of pi/2 are always good x-values to test out, as they're usually the x-intercepts or turning points of the curve.

eg.
$\text{For} \ f(x)=4+3\sin 2x \\ f(0)=4, f\left(\frac{\pi}{2}\right)=7, f(\pi)=4, \ \text{etc.}$

The curve is also the original sine curve compressed by a factor of two, stretched vertically by a factor of 3, then shifted up by 4 units. If you like, the amplitude gets tripled, the period gets halved and it gets shifted up 4 units. ie. x-values will line up with triple the y-value on the curve sin2x shifted up 4.

I hope this makes some sort of sense
« Last Edit: June 12, 2019, 11:42:12 pm by fun_jirachi »
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

#### kector

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« Reply #4261 on: June 17, 2019, 10:01:38 pm »
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Hi I'm really confused on doing Part III and sketching in general... Thanks for any help. 2004 HSC 7(b) **hopefully the screenshot went thru**

#### Abhiram

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« Reply #4262 on: June 18, 2019, 02:03:10 pm »
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hey Jake i need help doing this question......(i have attached an image i that doesn't work the question is written below

Sarah borrows $450000 from a bank. The loan is to be repaid in 20 years. The interest rate is 6%p.a. compounded monthly. There is no repayment for the first three months. Let A[n] be the amount owing after n months and M be the monthly repayments. (i) Find an expression for A[4] (ii) Show that A[5] = 450000(1.005)^5 - M(1+1.005) (iii) Find the monthly repayments if the loan is to be repaid in 20 years. #### InnererSchweinehund • MOTM: JUNE 19 • Forum Regular • Posts: 72 • Respect: +17 ##### Re: Mathematics Question Thread « Reply #4263 on: June 18, 2019, 02:42:49 pm » +2 Sarah borrows$450000 from a bank. The loan is to be repaid in 20 years. The interest rate is 6%p.a. compounded monthly. There is no repayment for the first three months. Let A[n] be the amount owing after n months and M be the monthly repayments.
(i)          Find an expression for A[4]
(ii)         Show that A[5] = 450000(1.005)^5 - M(1+1.005)
(iii)        Find the monthly repayments if the loan is to be repaid in 20 years.

Hi!

Unfortunately I'm not Jake... but I did do the HSC last year so hopefully my memory is still up to scratch.

If it doesn't, or it's not right, let me know!!

Ps. Sorry if the image is bad quality - I had to resize it so it would attach. Also let me know if you need a better one!!

#### fun_jirachi

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« Reply #4264 on: June 18, 2019, 03:34:35 pm »
+1
Hi I'm really confused on doing Part III and sketching in general... Thanks for any help. 2004 HSC 7(b) **hopefully the screenshot went thru**

Remember that when you're sketching the acceleration, you're essentially sketching the derivative of velocity ie. f'(x) for the graph you already have. As a start, try looking at the graph between x=0 and x=1 + x=3 and x=5: in these restricted domains the derivative is zero since the velocity is constant. As the velocity becomes more negative, you want to make sure the acceleration is somewhere below the x-axis, and at any points of inflexion there must be a max/min for the acceleration

Have a go, and if you need more help ask away
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

#### Abhiram

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« Reply #4265 on: June 18, 2019, 06:30:34 pm »
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Hi!

Unfortunately I'm not Jake... but I did do the HSC last year so hopefully my memory is still up to scratch.

If it doesn't, or it's not right, let me know!!

Ps. Sorry if the image is bad quality - I had to resize it so it would attach. Also let me know if you need a better one!!

thanks for the reply the image is perfect.

#### Thankunext

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« Reply #4266 on: June 19, 2019, 06:21:03 pm »
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Hello, can someone help with this probability question please. I'm really stuck.
Tim plays a video game 3 times and the probability that he wins at least once is 37/64. What is Tim's probability of winning one game?

#### fun_jirachi

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« Reply #4267 on: June 19, 2019, 06:48:53 pm »
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$\text{P(Winning at least once in three games) is the same as 1-P(no wins in three games)} \\ \text{Therefore, P(no wins in three games)} = 1-\frac{37}{64} = \frac{27}{64} \\ \text{Hence, the chance he doesn't win one game is} \ \frac{3}{4} \\ \text{Therefore, the chance he wins is} \ \frac{1}{4}$

Hope this helps

EDIT: If P(lose) = x, then P(losing all n games) = xn. Conversely, if P(losing all n games) = x, then P(lose) is the nth root of x. In this case, we take the cube root of 27/64.
« Last Edit: June 19, 2019, 09:03:40 pm by fun_jirachi »
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

#### Thankunext

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« Reply #4268 on: June 19, 2019, 07:17:24 pm »
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Sorry but how did you get the 3/4?

#### RuiAce

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« Reply #4269 on: June 19, 2019, 09:04:05 pm »
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Sorry but how did you get the 3/4?
$P(\text{No wins in 3 games}) = [P(\text{Does not win one game})]^3,\\ \text{therefore }[P(\text{Does not win one game})]^3 = \frac{27}{64}.$
Hence he took cube roots.

#### Thankunext

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« Reply #4270 on: July 08, 2019, 07:20:15 pm »
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Can someone help with this question please? I think you need to use the derivative given to find the answer, but I'm unsure how. Thank you!
The population of a city is P(t) at any one time. The rate of decline in population is proportional to the population P(t), that is, dP(t)/dt = -kP(t). What will the percentage rate of decline in population be after 10 years?

#### Thankunext

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« Reply #4271 on: July 08, 2019, 09:12:15 pm »
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Can someone help with this one please. Thanks.
If dQ/kQ, prove that Q=Ae^kt satisfies this equation by integrating dQ/dt=kQ.

#### Kombmail

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« Reply #4272 on: July 08, 2019, 11:43:40 pm »
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Can someone help with this one please. Thanks.
If dQ/kQ, prove that Q=Ae^kt satisfies this equation by integrating dQ/dt=kQ.
I think it’s one of those were you differentiate the first given formula and break it down to the given answer ie. Ae^kt = kAe^kt = k(Ae^kt) = kQ since Q = Ae^kt
-KgkG-

#### LoneWolf

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« Reply #4273 on: July 11, 2019, 05:49:56 am »
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Hey, think this is the right forum.
Can you help me understand why the second derivative of dy/dx is d^2y/dx^2 not dy^2/dx^2
Im in year 11 and doing a journal assessment!
« Last Edit: July 11, 2019, 06:10:10 am by DrewN20 »

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