Author Topic: VCE Methods Question Thread! (Read 5666109 times) Tweet Share

KevinooBz · « **Reply #1515 on:** February 02, 2013, 11:04:00 pm »

Quote from: Will T on February 02, 2013, 10:45:55 pm

Consider where g(x) := x^2-5 > 0.

Also, I believe the answer to Exercise 3J question 2f. of the Essentials is incorrect. Verification appreciated.

Inverse is 1/(x-2)^4/3 + 1. So you solve for (x-2)^4/3>2. So domain is (2,∞).

Jaswinder · « **Reply #1516 on:** February 03, 2013, 11:35:00 am »

express the area of an equilateral triangle as a function of the length s of each side.

I keep getting $\frac{\sqrt{3}}{3}s^2$

pi · « **Reply #1517 on:** February 03, 2013, 11:45:47 am »

A = sh/2
h^2 + (s/2)^2 = s^2 as equilateral
h^2 = 3s^2/4
h = sqrt(3)s/2, h>0
Therefore A = s(sqrt(3)s/2)/2 = (s^2)sqrt(3)/4

Jaswinder · « **Reply #1518 on:** February 03, 2013, 11:55:00 am »

so the base of the triangle is the length (s)?

pi · « **Reply #1519 on:** February 03, 2013, 11:58:38 am »

Yeah, sorry I should have explained that. s = base (or any side, as it's equilateral) and h = height (from the traditional formula of A = (1/2)bh)

Jaswinder · « **Reply #1520 on:** February 03, 2013, 12:04:37 pm »

alrighttt! i have another one :/

A car travels half the distance at 80km/h, and the other half at x km/h, whats the average speed of the total journey

b^3 · « **Reply #1521 on:** February 03, 2013, 01:03:46 pm »

This question keeps getting asked year after year

(and multiple times in this thread)

Quote from: b^3 on January 02, 2013, 09:00:21 pm

Let $d$ be the total distance of the journey.
$\begin{alignedat}{1}s & =\frac{d}{t} \\ t & =\frac{d}{s } \\ t_{1} & =\frac{d/2}{80} \\ & =\frac{d}{160} \\ t_{2} & =\frac{d/2}{x} \\ & =\frac{d}{2x} \\ t_{total} & =t_{1}+t_{2} \\ & =\frac{d}{160}+\frac{d}{2x} \\ & =\frac{dx}{160x}+\frac{80d}{160x} \\ & =\frac{d\left(x+80\right)}{160x} \\ S_{avg} & =\frac{d_{total}}{t_{total}} \\ & =d\div\frac{d\left(x+80\right)}{160x} \\ & =d\times\frac{160x}{d\left(x+80\right)} \\ & =\frac{160x}{x+80} \end{alignedat} $

brightsky · « **Reply #1522 on:** February 03, 2013, 01:04:04 pm »

average speed = total distance/time spent = d/[d/2/80 + d/2/x] = 160x/(x+80)

abcdqdxD · « **Reply #1523 on:** February 03, 2013, 02:35:06 pm »

Hey, can you help me with this:

If f(x) = 2x+5 then f(x) + f(y) is: ?

b^3 · « **Reply #1524 on:** February 03, 2013, 02:43:40 pm »

1) If $f\left(-\frac{b}{a}\right)=0$ , then $ax+b$ is a factor of the equation (factor theorem). Now we can rearrange what we have to get a cubic expression
$\begin{alignedat}{1}3x^{3}+ax^{2} & =33x+b \\ 3x^{3}+ax^{2}-33x-b & =0 \end{alignedat}$
Now we know that since -2 and 5 are solutions, that $f(-2)=0$ and $f(5)=0$ .

Spoiler

So
$\begin{alignedat}{1}f\left(-2\right) & =0 \\ 3\left(-2\right)^{3}+a\left(-2\right)^{2}-33\left(-2\right)-b & =0 \\ -4a+b & =42...[1] \\ f\left(5\right) & =0 \\ 3\left(5\right)^{3}+a\left(5\right)^{2}-33\left(5\right)-b & =0 \\ 375+25a-165-b & =0 \\ -25a+b & =210...[2] \end{alignedat}$
Then solve the simultaneous equations.

2) Just algebraic manipulation, for f(y), where ever we see x in f(x), we put y

Spoiler

$\begin{alignedat}{1}f\left(x\right)+f\left(y\right) & =\left(2x+5\right)+\left(2y+5\right) \\ & =2x+2y+10 \\ & =2\left(x+y+5\right) \end{alignedat}$

abcdqdxD · « **Reply #1525 on:** February 03, 2013, 04:00:39 pm »

cheers

ashoni · « **Reply #1526 on:** February 04, 2013, 08:58:42 pm »

In the expansion of (2a -1)^n, the coefficient of the second term is -192. Find the value of n.
help pleasee! D:

saba.ay · « **Reply #1527 on:** February 04, 2013, 09:24:40 pm »

Consider the following linear simulataneous equations.

mx - 5y = 10
3x - (m-2)*y=6

a) find the values of m, where m belongs to R, for which there are infinitely many or no solutions.

values for m are :
m=5 (infinitely many solutions)
m=-3 (no solutions)

b) Find the unique solution for the equations in terms of m, specifying the restrictions on m.
??
Can someone please help with this part.

Thank you.

Homer · « **Reply #1528 on:** February 04, 2013, 09:32:54 pm »

Quote from: tony_123 on February 04, 2013, 08:58:42 pm

In the expansion of (2a -1)^n, the coefficient of the second term is -192. Find the value of n.
help pleasee! D:

second term would be

$n(2a)^{n-1}(-1)=-192$
$n(2a)^{n-1}=192$
$(2a)^{n-1}=\frac{192}{n}$

since we only need the coefficient we can get rid of a

$(2)^{n-1}=\frac{192}{n}$

Solve for n either on calculator or by hand and you should get n=6

ashoni · « **Reply #1529 on:** February 04, 2013, 09:41:59 pm »

Quote from: Homer on February 04, 2013, 09:32:54 pm

second term would be

$n(2a)^{n-1}(-1)=-192$
$n(2a)^{n-1}=192$
$(2a)^{n-1}=\frac{192}{n}$

since we only need the coefficient we can get rid of a

$(2)^{n-1}=\frac{192}{n}$

Solve for n either on calculator or by hand and you should get n=6

could you please show how you would work it out by hand? it'll really help me

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