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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Momo.05 on January 09, 2010, 10:28:54 pm

Title: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 09, 2010, 10:28:54 pm: Hi, I need some help solving a somewhat simple question that i haven't been able to get my head around.

Express the total surface area, S, of a cube as a function of:

(b) the volume V of the cube

Thanks for all the help :D

p.s there probably be many more questions, so i figured posting them on this trend would be the best idea
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 09, 2010, 10:29:54 pm: Oh i forgot to add , "the length is x cm of an edge"
Title: Re: This is METH-ness! (H/w help trend).
Post by: GerrySly on January 09, 2010, 10:34:05 pm: $\begin{align*} S&=6x^2\\ V&=x^3\\ x&=V^{\frac{1}{3}}\\ S&=6V^{\frac{2}{3}} \end{align*}$
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 09, 2010, 10:35:19 pm: Bleugh!!! What have I done! :o
Thanks GerrySly!!! :P
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 09, 2010, 10:35:56 pm: *glass break* ..that was simple.
Thanks :)

Is right to by the way.
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 09, 2010, 10:36:25 pm: Its*
Title: Re: This is METH-ness! (H/w help trend).
Post by: GerrySly on January 09, 2010, 10:36:44 pm: Quote from: Momo.05 on January 09, 2010, 10:35:56 pm
Is right to by the way.
Well that's lucky haha
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 09, 2010, 10:52:02 pm: Ohh another question,

Express the area of an equilateral triangle as a function of ;
"with the length s of each side"

(b) the altitude h

Thanks in advance ! (:
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 09, 2010, 11:09:13 pm: My turn!

Area of equilateral triangle = 0.5*s*h

Split the triangle into two right-angled triangles and use Pythagoras to get an equation linking s, h:
(0.5s)² + h² = s²
s = 2h√3/3

sub this into the original:
Area = 0.5*(2h√3/3)*h
Area = h²√3/3

Hope I got this one right!!! ;D
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 09, 2010, 11:10:37 pm: You did well my good friend :P

I was about to post "nevermind" since i solved it but u were too fast in being right lol (Y)
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 09, 2010, 11:12:11 pm: Haha :P
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 09, 2010, 11:43:17 pm: Ahh another question, last one for the day woot !

A car travels half the distance of a journey at an average speed of 80km/h and half at an average speed of x km/h.
Define a function, S<, which gives the average speed for the totally journey as a function of x.

- Thanks all for the help tonight, I know its late and its strange to be up this late and doing maths :/
But yeah ..? Thanks for the help .
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 09, 2010, 11:43:38 pm: S*
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 10, 2010, 09:58:22 am: This is probs very late now and you've probs got the answer! But this is good practice for me! Anyway, here goes:

Let total distance = d,

Then total time = (d/2)/80 + (d/2)/x
= d/160 + d/2x
= (dx+80d)/160x
= d(x+80)/160x

Average speed = (total distance)/(total time)
= d/[d(x+80)/160x]
= 160x/(x+80)

I think this is right...
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 10, 2010, 10:37:57 am: Yep, so what you're looking for would be $S(x) = \frac{160x}{x+80}$ .
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 10, 2010, 10:42:06 am: Ah yep, my bad! I didn't answer the question!!! :P
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 11:44:30 am: Yup, its right (Y)
Glad my slowish is a useful for someone out there lol (:
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 12:30:45 pm: Ohh darnn it, i need help again with another question ==

A cylinder is inscribed in a sphere with a radius of length 6 cm and a height of h cm ( please note the cylinder is in the circle wiht those measurements).
For the cylinder;

Define a function, V1, which gives the volume of a cylinder as a function of the height (h)
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 10, 2010, 12:41:22 pm: Hmmm...

Draw a triangle connecting the edge of the base of the cylinder, the middle of the base, and the middle of the height (centre of circle).
This should be right-angled with the side-lengths: r, h/2 & 6 (the last is the hypotenuse).

Using Pythagoras, get r² in terms of h:
r² + (h/2)² = 6²
r² = 36 - h²/4

sub this into the volume equation of a cylinder:
V(x) = πr²h
= π*(36 - h²/4)*h

Btw, that funny thing's supposed to be pi :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 12:42:08 pm: Coolies, thank you heaps (Y)
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 10, 2010, 12:42:45 pm: No probs, good practice for me ;D!
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 02:42:39 pm: Woo, another question i need some help on please (:

Thank you heaps in advance, a *virtual hi-five and pat on the back* ,

Let f(x) = $\frac{px+q}{x+r}$ , where x is an element of R\{-r,r}

(a) If f(x) = f(-x) for all x, show that f(x) = p for x being an element of R\{-r,r}

(b) If f(-x) = -f(x) for all x not equaling to 0, find the rule for f(x) in terms of q.
Title: Re: This is METH-ness! (H/w help trend).
Post by: cipherpol on January 10, 2010, 02:51:10 pm: For a), http://vcenotes.com/forum/index.php/topic,20998.msg214604.html#msg214604
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 02:57:38 pm: Thank you, that was very helpful (:
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 10, 2010, 03:21:57 pm: Interesting question...lemme see how I go..don't know if this is right though.

a.

$f(x) = f(-x)$

$\therefore \frac{px+q}{r+x} = \frac{q-px}{r-x}$

$\implies (r-x)(px+q) = (r+x)(q-px)$

$\implies pxr-px^2+qr-xq = qr + xq - pxr - px^2$

$\implies pxr - xq = xq - pxr$

$\implies 2pxr - 2xq = 0$

$\implies 2x(pr-q) = 0$

Hence, by the null factor law, $2x = 0$ or $pr-q = 0$ .

When $pr-q = 0$ , $pr = q$ .

Substitute $pr = q$ into original equation, where x is an element of R\{-r,r}.

$\implies \frac{px + (pr)}{r+x}$

$\implies \frac {p(r+x)}{r+x}$

$f(x) = p$ , where x is an element of R\{-r,r}.

b.

$f(-x)=-f(x)$

$\therefore \frac{q-px}{r-x} = -\frac{px+q}{r+x}$

$\implies \frac{q-px}{r-x} = \frac {-px-q}{r+x}$

$\implies (q-px)(r+x) = (-px-q)(r-x)$

$\implies qr+qx-pxr-px^2=-pxr+px^2-qr+qx$

$\implies qr - px^2 = px^2 - qr$

$\implies 2qr-2px^2 = 0$

$\implies qr = px^2$

$\implies r = \frac {px^2}{q}$

Substitute $r = \frac {px^2}{q}$ back into original equation.

$\frac {px + q}{\frac{px^2}{q} + x}$

$\implies \frac {px+q}{\frac {px^2+xq}{q}}$

$\implies \frac {q(px+q)}{x(px+q)}$

$f(x) = \frac {q}{x}$

Edit: Is this what you got? :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 03:34:37 pm: Its okay , i solved both of them questions :)

All u need to do is solve for r in your last step and then sub r =px^2/q into f(x) and wowlaaa :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 10, 2010, 03:37:35 pm: Aha! Smart...
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 04:06:47 pm: Can someone help me with question please,

Find the rule for the area, A(t), enclosed by the graph of the function:

f(x) = { 3x, 0 < x < 1 , equal to signs as well.. like the one with the line underneath
= {3, x >1

the x-axis, the y-aix and the vertical line x=t (t equal to or greater than 0). State the domain and the range of the function.

Thank you. Also can someone show me how to (if possible) to show >,< but where there's a line underneath it indicating the equal to thing ... ? (lol the thingy makes a face that describes what im feeling atm LOL)

P.s sorry for the f(x) thingy doesnt look right since i failed at using the latex thing
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 10, 2010, 04:11:17 pm: Have a read of this, you'll become a pro latex-er in no time!
http://vcenotes.com/viki/index.php/Help:LaTeX
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 04:13:56 pm: Thanks :)

I ahaha i tired using latex for the above question and my thing turned out in codes.. i didnt realize that u have to actually include , "\end{cases}" ==
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 04:15:08 pm: OHH thanks !

geqsplant.. i shall now go re write the question so the reader would find it more understand-able and get help on it too :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 10, 2010, 04:15:47 pm: Let's give this a shot!

$A(t) = \begin{cases} \frac{1*t*3t}{2}, & \mbox{if }0<t<1 \\ \frac{1*t*3t}{2} + 3(t-1), & \mbox{if }t\geqslant1 \end{cases}$

$\Longrightarrow A(t) = \begin{cases} \frac{3t^2}{2}, & \mbox{if }0<t<1 \\ \frac{3t^2}{2} + 3(t-1), & \mbox{if }t\geqslant1 \end{cases}$

I think this is it! This is the area of the triangle (0<x<1) plus the rest.
If it is wrong, LET ME KNOW!!!
Title: Re: This is METH-ness! (H/w help trend).
Post by: TrueTears on January 10, 2010, 04:16:14 pm: For greater than or equal to:

Code: [Select]
\geq
For less than or equal to:

Code: [Select]
\leq
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 10, 2010, 04:18:05 pm: Quote from: TrueTears on January 10, 2010, 04:16:14 pm
For greater than or equal to:

Code: [Select]
\geq
For less than or equal to:

Code: [Select]
\leq
Listen to TT. I suck at Latex!
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 04:18:51 pm: Find the rule for the area, A(t), enclosed by the graph of the function:

f(x) =
\begin{cases}
3x, & \mbox{if }x\mbox{ 0 <\geqslant x 1\geqslant} \\
3, & \mbox{if }x\mbox{ x>1}
\end{cases}

the x-axis, the y-axis and the vertical line x = t ( t >\geqslant 0). State the domain and range of the function.
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 04:20:03 pm: F.A.I.L ==
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 10, 2010, 04:21:31 pm: LOL it's my first time too! :D
Title: Re: This is METH-ness! (H/w help trend).
Post by: TrueTears on January 10, 2010, 04:21:45 pm: Code: [Select]
[tex] ... [/tex]
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 10, 2010, 04:21:55 pm: hahahahaha, practice makes perfect!! :D
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 04:23:45 pm: @ the.watchman
the answer is

A(t) = { 3t^2/2, 0 < (including) t < (including) 1
= { 3t - 3/2 , t >1
:)

(Im not using latex for a while until i get the hand of it properly)
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 10, 2010, 04:27:05 pm: For the question, roughly draw the graph. When $0 \le t \le 1$ the area is a triangle. When $t > 1$ , the area is a rectangle plus a triangle. Hope I'm not stating the obvious. ><
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 10, 2010, 04:30:54 pm: OOPS, my bad!

$\Longrightarrow A(t) = \begin{cases} \frac{3t^2}{2}, & \mbox{if }0<t<1 \\ \frac{((t-1)+t)*3)}{2}, & \mbox{if }t\geqslant1 \end{cases}$

$\Longrightarrow A(t) = \begin{cases} \frac{3t^2}{2}, & \mbox{if }0<t<1 \\ 3t-\frac{3}{2}, & \mbox{if }t\geqslant1 \end{cases}$
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 04:32:16 pm: Yup .
I posted the answer for the.watchman to tell him its all good. Since he posted let me know if the answer is wrong or not .. Thought i post it up anyways for the sake of him being quite smart.. so far he answered everything i posted up (which im extremely grateful for lol)
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 04:43:41 pm: [tex]\begin{align}
f(x) & = (m+o)^2 \\
& = m^2+2o+0.5^2 \\
\end{align}[tex\]

Please ignore this.
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 10, 2010, 04:44:01 pm: [/tex] Momo. :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 04:44:32 pm: Urgh screw this. Ahaha, better get back on to finishing the question. --
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 10, 2010, 04:44:58 pm: Good luck learning latex! :D
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 04:45:42 pm: $\begin{align} f(x) & = (m+o)^2 \\ & = m^2+2o+0.5^2 \\ \end{align}$
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 04:46:57 pm: OHHH :D:D:D IT WORKED ! IT WORKED ! IT WORKED !
Thank you brightsky and yeah ahaha lesson learnt the.watchman

Plus i also finish this chapter WOOT ! Two wins !
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 10, 2010, 04:55:24 pm: Ok, back to the question.

When $0 \le t \le 1$ (remember that t is an x-value), the area is a triangle.

Draw any line $x = t$ , where t is between 0 and 1 inclusive.

The "base" of the triangle would have length $t$ whilst the "height" of the triangle would have length $3t$ as the line $f(x) = 3x$ has a positive 3 gradient.

Hence, when t is between 0 and 1 (inclusive), the area under the function would be $3t \times t \times \frac {1}{2} = \frac {3t^2}{2}$ .

Now, when the hybrid function progresses into $t > 1$ , the area would be a triangle plus a rectangle.

Draw any line $x = t$ , where t greater than 1.

In this region, the function has already covered the "whole triangle", which has a total area of $\frac {3}{2}$ . So we know that the area in this second region would be $\frac {3}{2} + C$ .

To find $C$ , we need to look back at our graph. In the rectangular portion, the "base" of the rectangle would have a length of $t - 1$ , whilst the "height" of the rectangle would have a length of $3$ . Hence, $C = 3(t-1) = 3t - 3$ .

Hence the whole area under the function when $t > 1$ would be equal to:

$3t - 3 + \frac {3}{2} = 3t - \frac {3}{2}$
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 10, 2010, 04:59:06 pm: Ahhh, smart! :)

I was using a trapezium for $t>1$ .
So the area is $\frac{t+(t-1)}{2}*3 = 3t-\frac{3}{2}$ .
I prefer your way, even though it's longer.
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 10, 2010, 05:18:15 pm: .. And the domain and range of this function is both equal to [0, infinity) =D (Y)
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 02:17:25 pm: Long time no assistance, but once again im stuck on another question

Three points have coordinates A(1,7) , B(7,5) and C(0,-2). Find:

(a) the equation of the perpendicular bisector of AB
(b) the point of intersection of this perpendicular bisector and BC .

Thank you in advance for the help :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 19, 2010, 02:35:04 pm: a) Gradient of AB = $\frac {5 - 7}{7-1} = -\frac{2}{6} = -\frac{1}{3}$

Equation of AB = $y - y_1 = m (x - x_1) \implies {y - 7} = -\frac{1}{3} (x - 1)$

$\implies y = -\frac{1}{3}x + 6\frac{2}{3}$

The point at half of AB is $(\frac{7+1}{2}, \frac{5+7}{2}) = (4, 6)$

Let the gradient of perpendicular bisector = $m_1$

$m_1 \times m = -1 \implies m_1 = 3$

The equation of the perpendicular bisector of AB is $y = 3x + b$ .

Substitute in the coordinates $(4,6)$ :

$6 = 3(4) + b$

$6 = 12 + b$

$b = -6$

Hence the equation is $y = 3x - 6$ . Is that right?
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 19, 2010, 02:41:40 pm: b) The gradient of the line BC = $\frac{-2-5}{0-7} = \frac{-7}{-7} = 1$

Equation of BC = $y - y_1 = m(x-x_1) \implies y - 5 = x - 7 \implies y = x - 2$

We've gathered that the equation of the perpendicular bisector of AB = $y = 3x - 6$

So we have two equations:

$y = x - 2$ .....(1)

$y = 3x -6$ ......(2)

Substitute equation (1) into equation (2):

$x - 2 = 3x - 6$

$x - 3x = 2 - 6$

$2x = 4$

$x = 2$

Substituting that back into equation (1):

$y = 0$

So the point of intersection is at (2,0).
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 02:44:09 pm: Yup, the answers are right :)
Thank you for the help, its very clear :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 19, 2010, 02:45:05 pm: An alternate move to the second step of each, i.e. when using the point-gradient formula $y - y_1 = m (x - x_1)$ is to substitute in the two x and y coordinates into the equation $y = -\frac{1}{3}x + c$ in the first one and $y = x + c$ for the second one and solving simultaneously to find $c$ .
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 04:44:38 pm: Another question ,

In the rectangle ABCD, A and B are the points (4,2) and (2,8) respectively. Given that the equation of AC is y = x - 2, find ;

(a) the equation of BC

Thank you :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: cipherpol on January 19, 2010, 05:00:56 pm: Quote from: Momo.05 on January 19, 2010, 04:44:38 pm
Another question ,

In the rectangle ABCD, A and B are the points (4,2) and (2,8) respectively. Given that the equation of AC is y = x - 2, find ;

(a) the equation of BC

Thank you :)

Gradient of AB = -3

BC is perpendicular to AB, so gradient of BC is $\frac{1}{3}$

$y-8=\frac{1}{3}(x-2)$
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 05:06:18 pm: Thank you
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 07:18:42 pm: OHHH last question in this exercise please help ! :)

ABCD is a parallelogram , lettered anitclockwise , such that A and C are the points (-1,5) and (5,1) respectively.

(b) Given that BD is parallel to the line whose equation is y + 5x = 2, find the equation of BD.

Btw can someone explain to what the Karma:*insert number* mean ? And thank you for helping me out with this question :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: simonhu81292 on January 19, 2010, 07:42:57 pm: alright . .
first if BD is parralel to the y=-5x+2
the gradient for BD would be -5x
we then get the midpoint of a and c ...
so for x, 5+-1 /2 = 2
and for y, 5+1 /2 =3
y-3+-5(x-2)
y = -5x+13
hope this is right ... also correct me if i am wrong
thanks :angel:
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 07:45:33 pm: Nope your right, thank you . :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 19, 2010, 07:46:13 pm: The Karma thing is just that: that "karma" given to you by other members of VN.
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 07:55:39 pm: Ohh i see.
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 09:57:11 pm: Hrmm i stumble on another question ,

A car journey of o300 km lasts for 4 hours. Part of this journey is on a freeway at an average speed of 90 km/h. The rest is on a country roads at an average speed of 70 km/h.

Find "T"

Okay, so this what i did ..

Let d be the distance traveled on the freeway

SO d = 90T (1)

Let d be the distance traveled on country roads

SO d = 70(4 -T) (2)

So let (1)=(2)

.. 90T = 70 ( 4 - T)

... Which gives me an answer of T = 1.75, which is wrong .. since B.O.B says T only equals to 1?

How is this possible? Did i over look something and mess up badly ?
All help would be much appreciated (:
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 09:59:10 pm: Wait i think i got what i was doing wrong !
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 10:00:49 pm: No actually , i dont know what im doing ..
BTW T = time in hours spent on the freeway :/
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 19, 2010, 10:04:25 pm: Didn't you double-define 'd'?
The distance on each type of road isn't the same...

You want d and (300-d), I think
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 10:05:56 pm: OHh true
so it should be "300 -d " or something yeah ?
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 19, 2010, 10:06:57 pm: Quote from: Momo.05 on January 19, 2010, 10:05:56 pm
OHh true
so it should be "300 -d " or something yeah ?

Yeh, that should give the right answer

It gives T=1, d=90
Title: Re: This is METH-ness! (H/w help trend).
Post by: Studyinghard on January 19, 2010, 10:07:09 pm: Well I can see how it can be 1 hr because...

1hr at 90km/hr will be 90km done and 3 hrs at 70km/hr will be 210 km.

210 + 90 = 300km
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 10:07:39 pm: Yup thats it ... GOT IT ! WOOOT :D
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 19, 2010, 10:08:47 pm: I didnt notice i let double defined "d" to equal the same thing .. didnt take into consideration the total distance of 300 km
==
Title: Re: This is METH-ness! (H/w help trend).
Post by: Studyinghard on January 19, 2010, 10:12:48 pm: Can someone just do the working out for this one. I get the logic behind this but thats never enough in an exam :P
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 19, 2010, 10:17:23 pm: Ok

Let d be the distance on freeway

$d=90t$ (1)

AND $300-d=70(4-t)$ (where $300-d$ is the distance on country roads and $4-t$ is the time on country roads)

$\Longrightarrow d=70t+20$ (2)

sub (1) into (2):

$90t=70t+20$

$\Longrightarrow 20t=20$ , $t=1$
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 19, 2010, 10:24:52 pm: Let the distance spent on the freeway be $d$ km.

So journey spent on country roads would be $(300 - d)$ km.

Let $t$ hours be the time spent on the freeway.

So the time spent on country roads would be $(4 - t)$ hours.

$t = \frac{d}{90}$ (1)

$4 - t = \frac {300 - d}{70}$ (2)

Substitute (1) into (2):

$4 - \frac{d}{90} = \frac{300-d}{70} \implies \frac{360-d}{90} = \frac{300-d}{70} \implies 70(360 - d) = 90 (300 - d) \implies 7(360-d) = 9(300- d) \implies 2520 - 7d = 2700 - 9d \implies 2d = 180 \implies d = 90$

Substitute $d = 90$ back into (1):

$t = \frac{90}{90} = 1$

EDIT: Beaten! :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 19, 2010, 10:26:25 pm: LOL, two things:

1. We don't need to find d :P

2. If you find t first, I assure you it's less working :D

Good job anyway!
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 19, 2010, 10:28:45 pm: Quote from: the.watchman on January 19, 2010, 10:26:25 pm
LOL, two things:

1. We don't need to find d :P

2. If you find t first, I assure you it's less working :D

Good job anyway!

LOL, hehe, woops! Blame the time! It's 10:28 pm and I'm tired! :p
Title: Re: This is METH-ness! (H/w help trend).
Post by: Ilovemathsmeth on January 20, 2010, 12:03:17 am: Actually that question only appears confusing. It's solution is so simple that I thought it was a trick question :P
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 21, 2010, 04:32:10 pm: Hrmm, this is a matrix related question Im quite stuck on, please help !

Find the value of m for which the simultaneous equations;

(m+3)x + my = 12
(m-1)x + (m-3)y = 7

have no solutions .

Thank-you :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: cipherpol on January 21, 2010, 04:43:04 pm: Ok, so if we put it into a matrix we get something like:

m+3 m
m-1 m-3

So for no solutions we require that the determinant=0.

$det=ad-bc$

$\implies det=(m+3)(m-3)-m(m-1)=0$

$\implies m-9=0$

$m=9$ for no solutions.
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 21, 2010, 04:45:25 pm: OH, the det has to equal 0 for no solutions ..?
Guess i missed that
THANK YOU :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 21, 2010, 06:36:23 pm: Quote from: Momo.05 on January 21, 2010, 04:45:25 pm
OH, the det has to equal 0 for no solutions ..?
Guess i missed that
THANK YOU :)

When the determinant = 0, there can be both no solutions or infinitely many solutions. Just need to sub values back in to check which one it is.
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 21, 2010, 06:55:52 pm: Ohhh, i see.
Thanks for the heads up, i thought when the det = 0 it just mean no solutions ..
But its all done now, just going to put that note down :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 21, 2010, 08:08:15 pm: Graphically speaking, the two lines have the same gradient when the determinant is equal to 0

Therefore, if their y-intercepts are the same, then there are infinite solutions.
Or if they are different, no solutions.
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 21, 2010, 08:14:59 pm: Hey, random question, what is the difference between differentiating and differentiating with respect to x?
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 21, 2010, 08:16:15 pm: Quote from: the.watchman on January 21, 2010, 08:08:15 pm
Graphically speaking, the two lines have the same gradient when the determinant is equal to 0

Therefore, if their y-intercepts are the same, then there are infinite solutions.
Or if they are different, no solutions.

Yep, so imagine two parallel lines. If the two parallel lines have the same equation, then the two lines would be the same, and hence there are infinite solutions. But if the parallel lines are apart, they will never intersect, hence no solutions.
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 21, 2010, 08:17:59 pm: Quote from: brightsky on January 21, 2010, 08:14:59 pm
Hey, random question, what is the difference between differentiating and differentiating with respect to x?

If you differentiate, I think is has to be with respect to a variable, eg. x

Hence why the joke referring to $\frac{d}{dy} e^x\neq e^x$ , it actually equals 0
Title: Re: This is METH-ness! (H/w help trend).
Post by: TrueTears on January 21, 2010, 08:18:30 pm: Not if y = x
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 21, 2010, 08:18:53 pm: Quote from: TrueTears on January 21, 2010, 08:18:30 pm
Not if y = x

LOL, good point :P
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 21, 2010, 08:19:31 pm: Quote from: the.watchman on January 21, 2010, 08:17:59 pm
Quote from: brightsky on January 21, 2010, 08:14:59 pm
Hey, random question, what is the difference between differentiating and differentiating with respect to x?

If you differentiate, I think is has to be with respect to a variable, eg. x

Hence why the joke referring to $\frac{d}{dy} e^x\neq e^x$ , it actually equals 0

Still don't get it. Like how would you differentiate $ax^n$ with respect to x?
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 21, 2010, 08:21:32 pm: Quote from: brightsky on January 21, 2010, 08:19:31 pm
Quote from: the.watchman on January 21, 2010, 08:17:59 pm
Quote from: brightsky on January 21, 2010, 08:14:59 pm
Hey, random question, what is the difference between differentiating and differentiating with respect to x?

If you differentiate, I think is has to be with respect to a variable, eg. x

Hence why the joke referring to $\frac{d}{dy} e^x\neq e^x$ , it actually equals 0

Still don't get it. Like how would you differentiate $ax^n$ with respect to x?

With respect to x, just differentiate it as normal

With respect to a, $x^n$ is the co-efficient of a, so the derivative is $x^n$ I think

With respect to another variable ( $\neq x$ , $\neq a$ ), it is 0, as $ax^n$ does not have the variable in question
Title: Re: This is METH-ness! (H/w help trend).
Post by: TrueTears on January 21, 2010, 08:23:31 pm: $\int \sqrt{1+x^3}dx$

Find the anti derivative and I will worship you forever.
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 21, 2010, 08:24:04 pm: Take the example:

Example
Differentiate a^x with respect to x.

You might be tempted to write xa^(x-1) as the answer. This is wrong. That would be the answer if we were differentiating with respect to a not x.

Put y = a^x .

Then, taking logarithms of both sides, we get:

ln y = ln (a^x)
so ln y = x lna

So, differentiating implicitly, we get: (1/y) (dy/dx) = lna
and so dy/dx = y lna = a^x lna

Someone explain it to me. :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 21, 2010, 08:32:10 pm: This is some shocking working!

$y=a^x=e^{xln(a)}$

so the derivative is $[xln(a)]'\times e^{xln(a)}=ln(a)\times a^x$
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 21, 2010, 08:39:41 pm: Quote from: the.watchman on January 21, 2010, 08:32:10 pm
This is some shocking working!

$y=a^x=e^{xln(a)}$

so the derivative is $[xln(a)]'\times e^{xln(a)}=ln(a)\times a^x$

But why is it wrong to say the derivative in respects to x is $xa^{n-1}$ ?
Title: Re: This is METH-ness! (H/w help trend).
Post by: stonecold on January 21, 2010, 08:42:29 pm: Quote from: Momo.05 on January 21, 2010, 06:55:52 pm
Ohhh, i see.
Thanks for the heads up, i thought when the det = 0 it just mean no solutions ..
But its all done now, just going to put that note down :)

It means one of two things:
-You have two parallel lines (same gradient, different y-intercept), and they will never intersect ,or
-Your equations are actually the same line. (e.g. x+2y=4 is the same as 2x+4y=8)

So yeah, you put them into a matrix and calculate the determinant. If it is 0, then the matrix is said to be 'singular' and depending on whether the lines are parallel or the same, there are either no solutions or an infinite number of solutions.
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 21, 2010, 08:58:03 pm: Quote from: brightsky on January 21, 2010, 08:39:41 pm
Quote from: the.watchman on January 21, 2010, 08:32:10 pm
This is some shocking working!

$y=a^x=e^{xln(a)}$

so the derivative is $[xln(a)]'\times e^{xln(a)}=ln(a)\times a^x$

But why is it wrong to say the derivative in respects to x is $xa^{n-1}$ ?

This formula only works for equations of the form $y=ax^n$
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 21, 2010, 09:00:33 pm: Oh...was meant to say $xa^{x-1}$ . :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 21, 2010, 09:01:59 pm: Quote from: brightsky on January 21, 2010, 09:00:33 pm
Oh...was meant to say $xa^{x-1}$ . :)

Yep I figured :P

The variable which a diff is 'in respect to' is important in figuring the final answer

PS. 400 posts!!! (200th MM post!!!) :D
Title: Re: This is METH-ness! (H/w help trend).
Post by: TrueTears on January 21, 2010, 09:02:55 pm: It is wrong because of what the.watchman just showed.

Do you want me to rigorise it?
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 21, 2010, 09:05:47 pm: Quote from: TrueTears on January 21, 2010, 08:23:31 pm
$\int \sqrt{1+x^3}dx$

Find the anti derivative and I will worship you forever.

Apparently you need to use "elliptical integrals"...I have no idea what that is..
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 21, 2010, 09:06:12 pm: Quote from: TrueTears on January 21, 2010, 09:02:55 pm
It is wrong because of what the.watchman just showed.

Do you want me to rigorise it?

Yes, please.
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 21, 2010, 09:11:18 pm: Quote from: brightsky on January 21, 2010, 09:06:12 pm
Quote from: TrueTears on January 21, 2010, 09:02:55 pm
It is wrong because of what the.watchman just showed.

Do you want me to rigorise it?

Yes, please.

$\frac{d}{dx}x^n=nx^{n-1}$ (because of the rule $x^n \Longrightarrow nx^{n-1}$ )

$\frac{d}{da}a^n=na^{n-1}$ (because of the SAME rule $x^n \Longrightarrow nx^{n-1}$ , x is interchangeable with any other variable)

$\frac{d}{da}a^x=xa^{x-1}$ (because of the rule $x^n \Longrightarrow nx^{n-1}$ , a is not the variable in question, it is x)

BUT $\frac{d}{dx}a^x \neq xa^{x-1}$ (because of the rule $x^n \Longrightarrow nx^{n-1}$ , a is not the variable in question, it is x)

$\frac{d}{dx}a^x = ln(a)a^x$ , as shown earlier (EDIT: And by TT :D)
Title: Re: This is METH-ness! (H/w help trend).
Post by: TrueTears on January 21, 2010, 09:12:01 pm: pretty boring and trivial but watever

let $f(x) = a^x$

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{a^{x+h}-a^x}{h} = a^x \lim_{h \to 0} \frac{a^h-1}{h} = a^xf'(0)$

isn't that $f'(0)$ annoying? That's why e is so special since e is defined as $\lim_{h \to 0} \frac{e^h-1}{h} = 1$

Good, so all other exponential except for e have inexact values for f'(0), so wouldn't it be nice to convert to base e always :smitten:

Now the.watchman post is continuation of this.

Quote from: the.watchman on January 21, 2010, 08:32:10 pm
This is some shocking working!

$y=a^x=e^{xln(a)}$

so the derivative is $[xln(a)]'\times e^{xln(a)}=ln(a)\times a^x$
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 22, 2010, 03:37:11 pm: Okay, i dont need help with anything atm, but i got a question out there ..

Im assuming everyone here is doing methods 3/4 right ? I was wondering what are you guys using as your summary book at the end of the year.. Like someone of my friends are using actual books with notes in them, others are just gathering up a lot of notes on loose leaf and then binding them together at the end of the year ...

So i was hoping what is the best way to have a "summary book" and any advice on what to include and what not should be included.

Also are you allowed to colour code anything in it ?

PS sorry about posting some random question
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 22, 2010, 03:42:20 pm: Yeh, I'll probs write my own, it's better than trawling through a massive textbook :P
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 22, 2010, 03:55:33 pm: Ahaha yeah me too ...
But i you can go office works i think to get alot of loose leaf binded together to make a proper book. That sounds tempting too ..LOL
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on January 22, 2010, 03:56:20 pm: Talking about officeworks...is the Back-to-school sale still on? (random question I know..:))
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 22, 2010, 03:59:08 pm: I think so :/
I think it goes till feb or something ? Google it :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 22, 2010, 04:09:36 pm: Quote from: Momo.05 on January 22, 2010, 03:55:33 pm
Ahaha yeah me too ...
But i you can go office works i think to get alot of loose leaf binded together to make a proper book. That sounds tempting too ..LOL

Good idea! I might do that too :)
Title: Re: This is METH-ness! (H/w help trend).
Post by: m@tty on January 22, 2010, 04:14:04 pm: Quote from: Momo.05 on January 22, 2010, 03:37:11 pm
Okay, i dont need help with anything atm, but i got a question out there ..

Im assuming everyone here is doing methods 3/4 right ? I was wondering what are you guys using as your summary book at the end of the year.. Like someone of my friends are using actual books with notes in them, others are just gathering up a lot of notes on loose leaf and then binding them together at the end of the year ...

So i was hoping what is the best way to have a "summary book" and any advice on what to include and what not should be included.

Also are you allowed to colour code anything in it ?

PS sorry about posting some random question
I just used the one I obtained at a Derrick Ha's lecture. One thing to remember: You most likely won't even look at it during the exam. I certainly didn't. By all means, write your own, but don't expect to use it.
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on January 22, 2010, 04:17:02 pm: Quote from: m@tty on January 22, 2010, 04:14:04 pm
Quote from: Momo.05 on January 22, 2010, 03:37:11 pm
Okay, i dont need help with anything atm, but i got a question out there ..

Im assuming everyone here is doing methods 3/4 right ? I was wondering what are you guys using as your summary book at the end of the year.. Like someone of my friends are using actual books with notes in them, others are just gathering up a lot of notes on loose leaf and then binding them together at the end of the year ...

So i was hoping what is the best way to have a "summary book" and any advice on what to include and what not should be included.

Also are you allowed to colour code anything in it ?

PS sorry about posting some random question
I just used the one I obtained at a Derrick Ha's lecture. One thing to remember: You most likely won't even look at it during the exam. I certainly didn't. By all means, write your own, but don't expect to use it.

+1, if you know your stuff well enough, it won't be necessary
Title: Re: This is METH-ness! (H/w help trend).
Post by: superflya on January 22, 2010, 04:42:17 pm: i never end up using the cheat sheets i make, good to make sure u dont forget those little things.
Title: Re: This is METH-ness! (H/w help trend).
Post by: Momo.05 on January 22, 2010, 05:11:24 pm: Yeah thats what people keep saying as well, that i wont use it much but its good for revision :/

Ahh i shall go buy a book later to start on my notes :) - It shall become my back up brain if i (hopefully not) suffer from a blank mind during the exam =,="
Title: Re: This is METH-ness! (H/w help trend).
Post by: TrueTears on January 22, 2010, 08:58:07 pm: Yeah... no need to make notes for methods ... it's the one subject where notes are useless.
Title: Re: This is METH-ness! (H/w help trend).
Post by: Ilovemathsmeth on February 05, 2010, 07:32:58 pm: Don't take notes. Just understand the theory. The only time you make notes is for your bound book of notes. That's it. Do tonnes of qs.
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on February 05, 2010, 07:53:14 pm: Quote from: Ilovemathsmeth on February 05, 2010, 07:32:58 pm
Don't take notes. Just understand the theory. The only time you make notes is for your bound book of notes. That's it. Do tonnes of qs.

When it gets closer to exam time, it may be worthwhile you trying to answer all the MM questions on VN, they can be good practice questions as well (that's what I'm doing! :P)
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on February 05, 2010, 08:16:45 pm: Quote from: the.watchman on February 05, 2010, 07:53:14 pm
When it gets closer to exam time, it may be worthwhile you trying to answer all the MM questions on VN, they can be good practice questions as well (that's what I'm doing! :P)

Although you're a good year ahead. :p
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on February 05, 2010, 08:39:02 pm: Quote from: brightsky on February 05, 2010, 08:16:45 pm
Quote from: the.watchman on February 05, 2010, 07:53:14 pm
When it gets closer to exam time, it may be worthwhile you trying to answer all the MM questions on VN, they can be good practice questions as well (that's what I'm doing! :P)

Although you're a good year ahead. :p

Excuse me, brightsky! Who's the maths freak? :P
Title: Re: This is METH-ness! (H/w help trend).
Post by: Rockim on February 05, 2010, 08:41:52 pm: I need help with binomial expansion

Find the coefficient of x^3 in the expansion of:

(x-1/x)^10
Title: Re: This is METH-ness! (H/w help trend).
Post by: appianway on February 05, 2010, 08:42:30 pm: In terms of cheatsheets, they shouldn't be needed. They're a good back up, but you probably won't end up using them. I didn't use mine for physics, and I definitely wasn't alone... most people didn't!
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on February 05, 2010, 08:46:37 pm: There is no $x^3$ term, I think, so the co-efficient of $x^3$ is 0.

This is because there is no $r$ (between 0 and 10) such that $x^r * (-1/x)^{n-r}$ becomes $x^3$
Title: Re: This is METH-ness! (H/w help trend).
Post by: Rockim on February 05, 2010, 08:48:06 pm: alright thanks watchman
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on February 05, 2010, 09:03:01 pm: Quote from: Rockim on February 05, 2010, 08:41:52 pm
I need help with binomial expansion

Find the coefficient of x^3 in the expansion of:

(x-1/x)^10

Yep no $x^3$

$\left(x - \frac{1}{x}\right)^{10}$

The possible "bases" are $x^{10} \left(\frac{1}{x}\right)^0, x^9 \left(\frac{1}{x}\right)^1, x^8 \left(\frac{1}{x}\right)^2, \cdots$

As you can see, when we have a even power of $x$ , the power of $\frac{1}{x}$ takes an even number away (or zero) (e.g. $x^8 \left(\frac{1}{x}\right)^2 = \frac{x^8}{x^2} = x^6$ . If we have an odd power of $x$ , the power of $\frac{1}{x}$ , takes an odd number away (e.g. $x^9 \left(\frac{1}{x}\right)^1 = \frac{x^9}{x^1} = x^8$ . This is to say, no matter if we have an odd or even power of $x$ , the resulting "base" will always have an even power, or a "negative" even power (like -2). And because $x^3$ has an odd power, it will never exist...

Dodgy explanation I know...:p
Title: Re: This is METH-ness! (H/w help trend).
Post by: Rockim on February 05, 2010, 09:10:16 pm: What about this one for arrangements?

Find the number of ways the letters in the word ROOSTER can be arranged if each letter must be used and:

the R's must be together and the O's must be together.

All my classmates, including myself, have shown the working out differently. Can someone please clarify? the answer is 120
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on February 05, 2010, 09:13:14 pm: Quote from: Rockim on February 05, 2010, 08:48:06 pm
alright thanks watchman

No prob.

brightsky, a funny way to prove it would be by using simultaneous equations:

$n=10$ and $|(n-r)-r|=3$ , with only integer solutions and $0\leq r \leq10$ :P
Title: Re: This is METH-ness! (H/w help trend).
Post by: superflya on February 05, 2010, 09:15:22 pm: 7 letters and both O's and R's must be together.... $5!$ ??
Title: Re: This is METH-ness! (H/w help trend).
Post by: andy456 on February 05, 2010, 09:16:18 pm: Quote from: Rockim on February 05, 2010, 09:10:16 pm
What about this one for arrangements?

Find the number of ways the letters in the word ROOSTER can be arranged if each letter must be used and:

the R's must be together and the O's must be together.

All my classmates, including myself, have shown the working out differently. Can someone please clarify? the answer is 120

5! = 5x4x3x2x1
= 120
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on February 05, 2010, 09:16:39 pm: Quote from: Rockim on February 05, 2010, 09:10:16 pm
What about this one for arrangements?

Find the number of ways the letters in the word ROOSTER can be arranged if each letter must be used and:

the R's must be together and the O's must be together.

All my classmates, including myself, have shown the working out differently. Can someone please clarify? the answer is 120

Two groups: [R,R] & [O,O], plus S,T,E anywhere

So there are, sort of, six available positions (5!=120 possibilities)

I'm not so sure though...
Title: Re: This is METH-ness! (H/w help trend).
Post by: kyzoo on February 06, 2010, 02:45:49 pm: Quote from: the.watchman on February 05, 2010, 07:53:14 pm
When it gets closer to exam time, it may be worthwhile you trying to answer all the MM questions on VN, they can be good practice questions as well.

=/ Sounds very inefficient: The time you take in finding questions, and the fact (at least for me) that it's easy to be distracted on the Internet.
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on February 06, 2010, 02:58:12 pm: Quote from: kyzoo on February 06, 2010, 02:45:49 pm
Quote from: the.watchman on February 05, 2010, 07:53:14 pm
When it gets closer to exam time, it may be worthwhile you trying to answer all the MM questions on VN, they can be good practice questions as well.

=/ Sounds very inefficient: The time you take in finding questions, and the fact (at least for me) that it's easy to be distracted on the Internet.

Yeah, I guess, but I multitask (answer own questions & surf internet/VN) :P
Title: Re: This is METH-ness! (H/w help trend).
Post by: Rockim on February 06, 2010, 07:36:29 pm: i need help with this combination question:

5!/r!(5-r)!=10 or (5Cr=10)

find the pronumeral 'r'

sorry i still don't know how to show my maths questions in the right format
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on February 06, 2010, 08:48:49 pm: You can use Pascal's triangle (a bit cheap, I know), it's quicker than algebraically (answer is 2,3)
Title: Re: This is METH-ness! (H/w help trend).
Post by: Watery on February 10, 2010, 10:21:07 pm: A question from Essential Maths (cas). Help would be deeply appreciated :D

Consider the simultaneous equation
mx + 2y = 8
4x - (2-m)y = 2m

a) Find the value of m for which there are:
ii) no solutions ii) indefinitely many solutions

b) Solve the equation in terms of m, for suitable values of m

I have done section a but I don't know how to do (b), especially I dont know what they mean by 'for suitable values of m'
And when I solved it simultanously like normal I checked the answer and it wasn't correct.
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on February 10, 2010, 10:28:58 pm: So, solving simultaneously:

$4mx+8y = 32$ (1) (from the first eqn)

$4mx-m(2-m)y = 2m^2$ (2) (from the second eqn)

1-2: $8y+m(2-m)y = 32-2m^2$

$y=\frac{2(m+4)}{m+2}$

Then solve for x

Is this what you got?
Title: Re: This is METH-ness! (H/w help trend).
Post by: Biology on February 10, 2010, 10:30:02 pm: That's correct, thanks a lot :D
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on February 10, 2010, 10:31:25 pm: Quote from: Biology on February 10, 2010, 10:30:02 pm
That's correct, thanks a lot :D

No prob :D
Title: Re: This is METH-ness! (H/w help trend).
Post by: Watery on February 10, 2010, 10:39:35 pm: lol biology you're stuck on the same question?

And thanks watchman!!!
Title: Re: This is METH-ness! (H/w help trend).
Post by: the.watchman on February 10, 2010, 10:40:40 pm: No, it's just m: $m(4x-(2-m)y) = m(2m)$

So, $4mx-m(2-m)y = 2m^2$

EDIT: This post just became USELESS!!! :D
Title: Re: This is METH-ness! (H/w help trend).
Post by: Biology on February 10, 2010, 10:41:21 pm: ^
Yep I was wrong on that part. Doh :P
Title: Re: This is METH-ness! (H/w help trend).
Post by: Watery on February 10, 2010, 10:58:43 pm: Here's another simultaneous equation question which I have problem with.

Solve for x and y:
$(a+b)x+cy = bc$
$(b+c)y+ax = -ab$

The answer is supposed to be

$x=c$

$y=-a$

I got these mumbo jumbo answers which is far from the solution ><
Help please.
Title: Re: This is METH-ness! (H/w help trend).
Post by: brightsky on February 11, 2010, 12:08:38 am: $(a+b)x + cy = bc$ (1)

$(b+c)y + ax = -ab$ (2)

From (1):

$(a+b)x = bc - cy$

$\implies x = \frac{bc - cy}{a+b}$

Substitute that into equation (2):

$(b+c)y + a\left(\frac{bc-cy}{a+b}\right) = -ab$

Multiply everything by $a+b$ :

$(a+b)(b+c)y + a(bc-cy) = -ab(a+b)$

$(a+b)(b+c)y + abc - acy = -ab (a+b)$

$(a+b)(b+c)y - acy = -ab(a+b+c)$

$y[(a+b)(b+c) - ac] = -ab(a+b+c)$

$y = \frac{-ab(a+b+c)}{(a+b)(b+c) - ac}$

$\implies \frac{-a(ba + b^2 + bc)}{ab + ac + b^2 + bc - ac}$

$\implies \frac{-a(ba+b^2 + bc)}{ab + b^2 + bc}$

$\implies -a$

Substitute this into the equation to get $x$
Title: Re: This is METH-ness! (H/w help trend).
Post by: Watery on February 11, 2010, 12:10:26 am: Omg you're a legend, you just saved me from possible scold and detention lol
:smitten: Thank you so much!