Author Topic: Mathematics Question Thread (Read 1627097 times) Tweet Share

Justinhales · « **Reply #3105 on:** October 23, 2017, 11:16:45 am »

Hey!
just a little confused about 13d) iii 2013 hsc pp! Can someone pls explain the step in the answers from
370 000(1.005)^n = 599 600 (1.005^n − 1) to
599 600 = (599 600 − 370 000)(1.005^n)

thanks!!

https://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf

https://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-marking-guide-maths.pdf

RuiAce · « **Reply #3106 on:** October 23, 2017, 11:19:49 am »

Quote from: Justinhales on October 23, 2017, 11:16:45 am

Hey!
just a little confused about 13d) iii 2013 hsc pp! Can someone pls explain the step in the answers from
370 000(1.005)^n = 599 600 (1.005^n − 1) to
599 600 = (599 600 − 370 000)(1.005^n)

thanks!!

https://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf

https://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-marking-guide-maths.pdf

\begin{align*}370000(1.005)^n &= 599600\left(1.005^n - 1\right)\\ 370000(1.005)^n &= 599600\left(1.005\right)^n - 599600 \tag{expanding brackets}\\ 599600 &= 599600(1.005)^n - 370000(1.005)^n \tag{rearranging}\\ 599600 &= \left(599600-370000\right)(1.005)^n\tag{factorising}\end{align*}

Justinhales · « **Reply #3107 on:** October 23, 2017, 11:22:11 am »

thanku!!!

taylorlucy · « **Reply #3108 on:** October 23, 2017, 12:00:58 pm »

A basic question, but would anyone be able to explain why the answer to 2016 HSC Q10 is D?

http://educationstandards.nsw.edu.au/wps/wcm/connect/47b7a5d2-8729-4d55-825a-7e840444d93f/2016-hsc-maths.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-47b7a5d2-8729-4d55-825a-7e840444d93f-lCmfMeF

Thanks in advance

RuiAce · « **Reply #3109 on:** October 23, 2017, 12:02:16 pm »

Quote from: taylorlucy on October 23, 2017, 12:00:58 pm

A basic question, but would anyone be able to explain why the answer to 2016 HSC Q10 is D?

http://educationstandards.nsw.edu.au/wps/wcm/connect/47b7a5d2-8729-4d55-825a-7e840444d93f/2016-hsc-maths.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-47b7a5d2-8729-4d55-825a-7e840444d93f-lCmfMeF

Thanks in advance

$\text{Since }4 = \log_2 16\text{, by using appropriate log laws}$
\begin{align*}4+\log_2 x &= \log_2 16 + \log_2 x \\ &= \log_2 (16x)\end{align*}

asd987 · « **Reply #3110 on:** October 23, 2017, 12:10:16 pm »

hey could someone help me with this:
the equation ax^2-x+b=0 has roots -2 and 4. find the value of a and b

RuiAce · « **Reply #3111 on:** October 23, 2017, 12:12:46 pm »

Quote from: asd987 on October 23, 2017, 12:10:16 pm

hey could someone help me with this:
the equation ax^2-x+b=0 has roots -2 and 4. find the value of a and b

$\text{From the sum of roots,}\\ \frac{1}{a} = -2 + 4\implies \boxed{a=-\frac12}$
$\text{From the product of roots,}\\ \frac{b}{a}=-2\times 4\implies \boxed{b=-4}$

gilliesb18 · « **Reply #3112 on:** October 23, 2017, 04:18:19 pm »

Hello!
I just need a small explanation on how to do this question... I bet its real simple for those of u out there but i'm new to this and just need help.
Question: The focal chord that cuts the parabola x^2= -6y at (6, -6) cuts the parabola again at X. Find the coordinates of X.

Thanks heaps in advance!!

MisterNeo · « **Reply #3113 on:** October 23, 2017, 10:18:30 pm »

Quote from: gilliesb18 on October 23, 2017, 04:18:19 pm

Hello!
I just need a small explanation on how to do this question... I bet its real simple for those of u out there but i'm new to this and just need help.
Question: The focal chord that cuts the parabola x^2= -6y at (6, -6) cuts the parabola again at X. Find the coordinates of X.

Thanks heaps in advance!!

$\text{I would find the focus of the parabola first.}\\x^2=-4ay\\ a= \frac{-6}{-4}=\frac32\\ \text{The vertex is (0,0), so focus is (0,-3/2).}$ $\text{Find the gradient of the chord using the focus and (6,-6) to find equation of chord.}\\ m=\frac{-\frac{3}{2}+6}{0-6}=-\frac34\\y+6=-\frac{3}{4}(x-6)\\y=-\frac{3}{4}x-\frac32$ $\text{Solve this simultaneously with the parabola to obtain points of intersections.}\\ -\frac{1}{6}x^2=-\frac{3}{4}x-\frac32\\2x^2=9x+18\\0=2x^2-9x-18\\0=(2x+3)(x-6)\\x=6 \text{or} -\frac32$ $\text{We already know x=6, so sub x=-3/2 to get X}\\X=(-\frac32,-\frac38)$

gilliesb18 · « **Reply #3114 on:** October 25, 2017, 01:51:26 pm »

Ok great, thanks heaps for that!!

Also, just on a side note, I have just bought the ATARNotes HSC Mathematics complete course notes book and it is awesome!! I seriously recommend this book to anyone out there finding this course hard like I have been. It covers everything you need to know!!
Heres the link to it: https://atarnotes.com/product/hsc-mathematics-notes/
Everyone on here asking questions should really get one

(In fact anyone just doing this course should get one

)

itssona · « **Reply #3115 on:** October 25, 2017, 02:50:25 pm »

been stuck on how to do this in a clever way using arithmetic sequence

a woman is employed at $1600 a month but at the end of each month she receives an increase of $60.
what will be her annual income?

RuiAce · « **Reply #3116 on:** October 25, 2017, 06:06:43 pm »

Quote from: sssona09 on October 25, 2017, 02:50:25 pm

been stuck on how to do this in a clever way using arithmetic sequence

a woman is employed at $1600 a month but at the end of each month she receives an increase of $60.
what will be her annual income?

The question lacks information unless we assume she's employed in January. When that happens it's just an AP with a = 1600 and d = 60 (and you'd be calculation S₁₂)

itssona · « **Reply #3117 on:** October 25, 2017, 06:21:55 pm »

Quote from: RuiAce on October 25, 2017, 06:06:43 pm

The question lacks information unless we assume she's employed in January. When that happens it's just an AP with a = 1600 and d = 60 (and you'd be calculation S₁₂)

ohh yes thats right, thank you!!

Mate2425 · « **Reply #3118 on:** October 28, 2017, 06:58:01 pm »

Hi could i please get some help on this question...
The curve y=ax*3 + bx*2 - x + 5 has a point of inflexion at (1,-2). Find the values of 'a' and 'b'.

Thanks.

RuiAce · « **Reply #3119 on:** October 28, 2017, 07:34:17 pm »

Quote from: Mate2425 on October 28, 2017, 06:58:01 pm

Hi could i please get some help on this question...
The curve y=ax*3 + bx*2 - x + 5 has a point of inflexion at (1,-2). Find the values of 'a' and 'b'.

Thanks.

$\text{The point }(1,-2)\text{ lies on the curve, so subbing:}\\ -2 = a + b - 1 + 5 \implies \boxed{a+b=-6}$
$\text{Also }\frac{d^2y}{dx^2}=6ax+2b\\ \text{so since we have a point of inflexion at }(1,-2)\\ \text{when }x=1, \frac{d^2y}{dx^2}=0. \text{ So}\\ 0 = 6a+2b \implies \boxed{0=3a+b}$
$\text{This is now a system of simultaneous equations}\\ \text{with solution }a=3, b=-9$

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