Author Topic: Mathematics Question Thread (Read 1304200 times) Tweet Share

lozil · « **Reply #420 on:** September 30, 2016, 09:54:51 pm »

Quote from: jakesilove on September 30, 2016, 09:15:11 pm

Hey! So, essentially, we need to find a value for k such that the area under the curve from -6 to 4 equals the area under the curve from 4 to k (that way, the negative and positive areas cancel out!

You can find the area under the first section in many ways; there's a formula, which I don't know, so I'll divide it into a triangle and a rectangle. The rectangle has area 10*2, and the triangle has area 10*(1/2)*3, so the total area will be 35 units squared.

The area of the second section can be found in the same way; the rectangle will have area (k-4)(4) and the triangle will have area (1/2)(k-4)(2). So, the total area will be (k-4)(4)+(1/2)(k-4)(2). Now, we need these two areas to be the same, so

$35=4(k-4)+(k-4)$
$35=5(k-4)$
$7=k-4$
$k=11$

As required.

Jake

ok, thanks

lozil · « **Reply #421 on:** October 01, 2016, 12:28:20 pm »

Hi again

So concerning the attachment (the question is express with a rational denominator), for my answer i got the second last line: 8(2-sqrt7)/3 (which is just factorised from the final answer). If i left it in the factorised form, would i still get full marks?

RuiAce · « **Reply #422 on:** October 01, 2016, 12:29:46 pm »

Quote from: lozil on October 01, 2016, 12:28:20 pm

Hi again
So concerning the attachment (the question is express with a rational denominator), for my answer i got the second last line: 8(2-sqrt7)/3 (which is just factorised from the final answer). If i left it in the factorised form, would i still get full marks?

Yes that last step wouldn't be required. Still a rational denominator.

lozil · « **Reply #423 on:** October 01, 2016, 03:12:49 pm »

Quote from: RuiAce on October 01, 2016, 12:29:46 pm

Yes that last step wouldn't be required. Still a rational denominator.

ok, thanks!

olivercutbill · « **Reply #424 on:** October 03, 2016, 04:32:06 pm »

Hey guys,

How does (d-6)^2 - 4 < 0

become 4 < d < 8 ?

im confused with the individual steps

RuiAce · « **Reply #425 on:** October 03, 2016, 04:41:55 pm »

Quote from: olivercutbill on October 03, 2016, 04:32:06 pm

Hey guys,

How does (d-6)^2 - 4 < 0

become 4 < d < 8 ?

im confused with the individual steps

$\text{We are interested in a quadratic inequality here. The more straightforward approach is}\\ \begin{align*}&&(d-6)^2&< 4\\ &&-\sqrt{4} &< d-6 < \sqrt{4}&&\\ &&4 &< d < 8\end{align*}$
$\text{Alternatively, expanding and refactorisation also suffices}\\ \begin{align*}d^2-12d+32&< 0\\ (d-4)(d-8)&< 0\\ 4&< d < 8\end{align*}$
$\text{For the second method, the last line }\textbf{IS NOT this mistake:}\\ d< 4\text{ and }d<8\\ \textbf{A quadratic inequality can NEVER be treated as though it was a linear inequality.}$
$\text{To observe why the second method works, consider }y=(d-4)(d-8)\\ \text{Sketch the graph of }y=(d-4)(d-8)$

$\text{Because }y=(d-4)(d-8) \\ (d-4)(d-8)< 0\text{ corresponds to }y< 0\\ \text{From the graph, clearly this is when }4< d< 8$

jakesilove · « **Reply #426 on:** October 03, 2016, 04:43:22 pm »

Quote from: olivercutbill on October 03, 2016, 04:32:06 pm

Hey guys,

How does (d-6)^2 - 4 < 0

become 4 < d < 8 ?

im confused with the individual steps

Hey! There are heaps of ways of going about this, but this is how I generally work through these kinds of questions. Let me know if you don't understand any steps!

$(d-6)^2-4<0$

Now, I usually work out inequalities by first solving an equation. As such,

$(d-6)^2-4=0$
$(d-6)^2=4$
$d-6=\pm 2$
$d=8, d=4$

So, we now have two solutions for d. The range will either be OUTSIDE (ie. d>8, d<4) or INSIDE (ie. 4<d<8 ). You can figure this out in a number of ways; the easiest is to sub in a point to the original equation. Let's say that d is equal to 5.

$(5-6)^2-4=(1)^2-4=-3<0$

Clearly, the relationship is true for d=5, which is between 4 and 8. Therefore, the solution for d is

$4 < d < 8$

As required!

Jake

kevin217 · « **Reply #427 on:** October 03, 2016, 04:45:09 pm »

Quote from: olivercutbill on October 03, 2016, 04:32:06 pm

Hey guys,

How does (d-6)^2 - 4 < 0

become 4 < d < 8 ?

im confused with the individual steps

Rearrange the equation to get:
(d-6)² < 4
(d-6) < 2 or (d-6) > - 2
d < 8 or d > 4
therefore 4 < d < 8
I hope this helped

Vinhtran · « **Reply #428 on:** October 03, 2016, 09:29:54 pm »

Please help me on this question. So I managed to get the mean and standard deviation correct, but i didn't understand why they made certain choices.

The probability of a seventeen year old boy having a part-time job while at school is 0.4
In a sample of 80 seventeen year old boys, what is the probability that between 20 and 25 of them have a part-time job.

The answer key says the answer is 0.07

RuiAce · « **Reply #429 on:** October 03, 2016, 09:43:12 pm »

Quote from: Vinhtran on October 03, 2016, 09:29:54 pm

Please help me on this question. So I managed to get the mean and standard deviation correct, but i didn't understand why they made certain choices.

The probability of a seventeen year old boy having a part-time job while at school is 0.4
In a sample of 80 seventeen year old boys, what is the probability that between 20 and 25 of them have a part-time job.

The answer key says the answer is 0.07

Statistics is not in the HSC mathematics course.
$\text{Let }X\text{ be the number of 17 year old boys out of 80 with part time jobs}\\ \text{Then, }X \sim \mathcal{B}(80,0.4)\\ \text{i.e. }X\text{ follows a binomial distribution of sample size 80 with probability }0.4$
$\text{Then we simply perform a ton of computations}\\ \begin{align*}&\> Pr(20 \le X \le 25) \\&=Pr(X=20)+Pr(X=21)+Pr(X=22)+Pr(X=23)+Pr(X=24)+Pr(X=25)\\ &= \binom{80}{20}0.6^{60}0.4^{20}+\binom{80}{21}0.6^{59}0.4^{21}+\binom{80}{22}0.6^{58}0.4^{22}+\binom{80}{23}0.6^{57}0.4^{23}+\binom{80}{24}0.6^{56}0.4^{24}+\binom{80}{25}0.6^{55}0.4^{25}\\ &= 0.06580277\dots \\ &\approx 0.07\end{align*}$

olivercutbill · « **Reply #430 on:** October 04, 2016, 09:59:45 pm »

Quote from: jakesilove on October 03, 2016, 04:43:22 pm

Hey! There are heaps of ways of going about this, but this is how I generally work through these kinds of questions. Let me know if you don't understand any steps!

$(d-6)^2-4<0$

Now, I usually work out inequalities by first solving an equation. As such,

$(d-6)^2-4=0$
$(d-6)^2=4$
$d-6=\pm 2$
$d=8, d=4$

So, we now have two solutions for d. The range will either be OUTSIDE (ie. d>8, d<4) or INSIDE (ie. 4<d<8 ). You can figure this out in a number of ways; the easiest is to sub in a point to the original equation. Let's say that d is equal to 5.

$(5-6)^2-4=(1)^2-4=-3<0$

Clearly, the relationship is true for d=5, which is between 4 and 8. Therefore, the solution for d is

$4 < d < 8$

As required!

Jake

This makes more sense to me than the solution posted above this. How are quadratic inequalities different from linear ones? In the first method posted by RuiAce, I don't understand the movement from the first line to the second.

Thanks guys!

RuiAce · « **Reply #431 on:** October 04, 2016, 10:10:22 pm »

Quote from: olivercutbill on October 04, 2016, 09:59:45 pm

Ignore method 1 if it makes no sense and focus on method 2. Because it is the exact same as Jake's method
________________________
$\text{Although if you really need to know how method 1 works, the idea is that}\\ (d-6)^2 < 4\\ \text{becomes }\\ |d-6| < 2\\ \text{upon square rooting.}$
$\text{Then the absolute value inequality is evaluated.}\\ -2 < d-6 < 2$
$\textbf{An extremely common misconception}\text{ however not usually problematic in 2U.}\\ \text{People tend to think that }\sqrt{x^2} = x\\ \text{False. }\sqrt{x^2}=|x|$
________________________
Quadratic inequalities (nothing's an identity here) are different from linear ones in that for starters, parabolas have a turning point. This messes things up. A linear function is just a straight line.

Secondly, every quadratic function has two roots. (You just don't get taught that in 2U because you don't know what complex numbers are)

IkeaandOfficeworks · « **Reply #432 on:** October 04, 2016, 10:30:12 pm »

Hi guys! I'm a bit confused with this question:

A piece of wire of length 50 cm is cut into two sections. One section is used to construct a rectangle whose
dimensions are in the ratio 3 : 1; the other section is used to construct a square. Find the dimensions of the
rectangle and the square so that the total enclosed area is a minimum.

I'm confused with the 3:1 ratio and how am I going to incorporate this in another similar question that talks about ratio? Thank you

RuiAce · « **Reply #433 on:** October 04, 2016, 10:48:45 pm »

Quote from: IkeaandOfficeworks on October 04, 2016, 10:30:12 pm

Hi guys! I'm a bit confused with this question:

A piece of wire of length 50 cm is cut into two sections. One section is used to construct a rectangle whose
dimensions are in the ratio 3 : 1; the other section is used to construct a square. Find the dimensions of the
rectangle and the square so that the total enclosed area is a minimum.

I'm confused with the 3:1 ratio and how am I going to incorporate this in another similar question that talks about ratio? Thank you

$\text{Let the breadth of the triangle be }\ell\\ \text{Then, from the 3:1 ratio, the length of the triangle is }3\ell$
$\text{Additionally, let the length of the square be }x$
$\text{We know that all of the lengths were made from a 50cm wire.}\\ \text{This means, if we add the lengths back up, we'd get 50.}\\ \text{i.e. }50=\underbrace{6\ell+2\ell}_{\text{perimeter of rectangle}}+\underbrace{4x}_{\text{square}}$
$\text{So we have }\frac{25}{2}=2\ell+x\\ \text{Now we need to consider the area.}$
$\text{The area of the rectangle is }3\ell^2\\ \text{The area of the square is }x^2\\ \text{Hence the total area is given by }A=3\ell^2+x^2$
$\text{Substituting in the above result we get }\\ A=3\ell^2+\left(\frac{25}{2}-2\ell\right)^2\\ \text{This is now your regular max/min question, so expand, compute }\frac{dA}{d\ell}\text{ and carry through.}$

fizzy.123 · « **Reply #434 on:** October 05, 2016, 01:21:24 am »

A particle is moving in a straight life, starting from the origin. At time t seconds the particle has a displacement of x metres from the origin and a velocity v m s^-1. the displacement is given by x=2t - 3ln(t+1). Find the distance travelled by the particle in the first 2 seconds. (HSC Paper 2000).
Can someone please help me out

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