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RuiAce

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Re: Mathematics Question Thread
« Reply #990 on: December 03, 2016, 11:40:36 am »
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What topic is it taught in? As I have only done Integration, Exp and Logs, Geometry 2, and Trig Functions?
The last topic. Applications of calculus.

Happy Physics Land

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Re: Mathematics Question Thread
« Reply #991 on: December 03, 2016, 11:40:47 am »
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What topic is it taught in? As I have only done Integration, Exp and Logs, Geometry 2, and Trig Functions?

It will be taught later in the 2 unit course in physical application of calculus to the real world. But it should be general knowledge that:

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jamonwindeyer

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Re: Mathematics Question Thread
« Reply #992 on: December 03, 2016, 11:41:56 am »
+1
It will be taught later in the 2 unit course in physical application of calculus to the real world. But it should be general knowledge that:



Only for Physicists is this general knowledge ;)

J.B

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Re: Mathematics Question Thread
« Reply #993 on: December 03, 2016, 11:42:46 am »
+1
Thank you.

anotherworld2b

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Re: Mathematics Question Thread
« Reply #994 on: December 06, 2016, 01:29:40 am »
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Hi i was wondering if i could please get help with these two questions. I have attempted to do q8 but im not sure whether i am right.
In regards to q12 i was particularly confused on how to do parts a and b
« Last Edit: December 06, 2016, 01:31:12 am by anotherworld2b »

jamonwindeyer

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Re: Mathematics Question Thread
« Reply #995 on: December 06, 2016, 02:18:11 am »
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Hi i was wondering if i could please get help with these two questions. I have attempted to do q8 but im not sure whether i am right.
In regards to q12 i was particularly confused on how to do parts a and b

Hey!

For Question 8, I think parts of your working are correct. Let me give you a hand interpreting!

Part A: The two lines are each parallel to \(x+2y=0\). Now if you rearrange this line, you get \(y=\frac{-x}{2}\), you did this yourself. So this line has a gradient of \(m=\frac{-1}{2}\). The lines L1 and L2 just need to be any lines with a gradient of this, aka, any line of the form \(y=\frac{-x}{2}+c\). Just pick two random values for \(c\) ;D

Part B: Here, it is now easier to consider two general equations in gradient intercept form. Before we do though, the fact that the lines are perpendicular means that we can set the two gradients to be \(m\), and \(\frac{-1}{m}\). This is because, when two lines are perpendicular, \(m_1=\frac{-1}{m_2}\). So we can write the lines like this:

L1: \(y=mx+b_1\)
L2: \(y=\frac{-x}{m}+b_2\)

Note that the two intercepts are different, at least for now. In fact, both lines meet at the point (0, 4), which sits on the y-axis. So in actuality, both lines have a y-intercept of 4, so the equations are now:

L1: \(y=mx+4\)
L2: \(y=\frac{-x}{m}+4\)

Both of our conditions are now satisfied, so you can pick any value of \(m\) you like and substitute it here to get your answer :)

Part C: Just do a cheeky answer; any two parallel lines will do. Do y=0 and y=1, those will definitely not intersect! ;D

For Question 12 your equations look fine for Part B, that answer is correct! For Part A, your equation should be \(y=80x+100\), the $80 is the half-hourly rate and the $100 is the constant call out fee. Be careful, you have defined \(x\) as half an hour here but one hour in Part B. This is okay as long as you are aware of it!

Try subbing with this new equation, you should get $740.00 as your answer for Part A ;D

anotherworld2b

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Re: Mathematics Question Thread
« Reply #996 on: December 07, 2016, 12:28:14 am »
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thank you very much for your help :D
I was wondering what would be the best way to do q12 d)?

Hey!

For Question 8, I think parts of your working are correct. Let me give you a hand interpreting!

Part A: The two lines are each parallel to \(x+2y=0\). Now if you rearrange this line, you get \(y=\frac{-x}{2}\), you did this yourself. So this line has a gradient of \(m=\frac{-1}{2}\). The lines L1 and L2 just need to be any lines with a gradient of this, aka, any line of the form \(y=\frac{-x}{2}+c\). Just pick two random values for \(c\) ;D

Part B: Here, it is now easier to consider two general equations in gradient intercept form. Before we do though, the fact that the lines are perpendicular means that we can set the two gradients to be \(m\), and \(\frac{-1}{m}\). This is because, when two lines are perpendicular, \(m_1=\frac{-1}{m_2}\). So we can write the lines like this:

L1: \(y=mx+b_1\)
L2: \(y=\frac{-x}{m}+b_2\)

Note that the two intercepts are different, at least for now. In fact, both lines meet at the point (0, 4), which sits on the y-axis. So in actuality, both lines have a y-intercept of 4, so the equations are now:

L1: \(y=mx+4\)
L2: \(y=\frac{-x}{m}+4\)

Both of our conditions are now satisfied, so you can pick any value of \(m\) you like and substitute it here to get your answer :)

Part C: Just do a cheeky answer; any two parallel lines will do. Do y=0 and y=1, those will definitely not intersect! ;D

For Question 12 your equations look fine for Part B, that answer is correct! For Part A, your equation should be \(y=80x+100\), the $80 is the half-hourly rate and the $100 is the constant call out fee. Be careful, you have defined \(x\) as half an hour here but one hour in Part B. This is okay as long as you are aware of it!

Try subbing with this new equation, you should get $740.00 as your answer for Part A ;D

Jakeybaby

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Re: Mathematics Question Thread
« Reply #997 on: December 07, 2016, 12:45:56 am »
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thank you very much for your help :D
I was wondering what would be the best way to do q12 d)?
Please correct me if I am wrong, but your equation for Bill seems incorrect to me personally.
If Bill charges a call-out fee of $100, initially, the cost will be $100 after 0 half-hours of work. Therefore, the y-intercept of your equation should be $100.
As he charges $80 per half-hour, if we allow x to be the number of hours worked, therefore:
 y = 80(2x) + 100
We shall use 2x, as there are obviously two half hours in one full hour, meaning that per hour, Bill costs $160.
Therefore: y = 160x + 100
So for a):
For a job that takes exactly 4 hours,
 y = 160(4) + 80

B is correct.

c)
A job that takes 3 hours and 20 minutes, this means that the number of hours worked is 3.33333.... hours, (i.e. 10/3 hours), as 20 minutes is one-third of an hour.
Therefore:
For Ian: y = 180(10/3)
Bill: y = 160(10/3) + 100
You can justify which one is cheaper for the job.

d)
I would find the point on the lines where they intersect, this shows you the length of job where they both cost the same amount of money.
Sketching the two equations with a graphics calculator will be best.
You can see both the point of intersection, one employee is cheaper than the other, and then after the point of intersection, the one that was cheaper, is now more expensive for the job.
(Hint: You need to state the domain (hours) of the job needed for Bill to be cheaper than Ian.)

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jamonwindeyer

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Re: Mathematics Question Thread
« Reply #998 on: December 07, 2016, 12:51:34 am »
+1
thank you very much for your help :D
I was wondering what would be the best way to do q12 d)?

Just a warning that D is more complicated than what is shown above, it's a tricky one. Thanks for your post Jake! ;D

You could use an inequality, but the different way that the things change is tricky. To do it, you would have to consider both men in terms of their price per hour. That turns Bill's equation into \(y=100+160x\). Now, \(x\) represents hours for both men. So we want when Bill is cheaper:



Indeed, at any time above 5 hours, Bill costs less than Ian (you could check this with substitution). However, since Ian charges hourly and Bill charges half hourly, you have some extra options.

Consider a simple example.

For a single hour, Bill costs $260.00 ($100.00 plus $80.00 each for two half hour periods) and Ian costs $180.00. For anything above that, both plumbers have a price jump, since they charge for any part of an hour/half an hour. So into the next half hour, Ian charges $360.00 and Bill charges $340.00; Bill is cheaper between 1 and 1.5 hours.

However, say we jump once more to a period between 1.5-2 hours. Ian's price would stay the same, but Bill's would jump to $420.00; he is now more expensive. Once we exceed 2 hours, Bill jumps to $500.00 and Ian jumps to $540.00, Bill is again cheaper. Exceed two and a half, and Bill jumps to $580 and is again more expensive.

So, the full answer is that Bill is cheaper for any work exceeding 5 hours, or, any work exceeding 1 hour that is an odd number of half hour periods :) honestly, the intuitive approach is best here, because this is quite a tricky little problem :)

anotherworld2b

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Re: Mathematics Question Thread
« Reply #999 on: December 07, 2016, 12:34:03 pm »
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Hi i was wondering if i could get help with these questions please.

I am not sure how to find the exact x intercepts without using a calculator.
I am also not sure how to find the equation of a parabola given the diagram. There are 3 different forms?  :-*

jamonwindeyer

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Re: Mathematics Question Thread
« Reply #1000 on: December 07, 2016, 01:04:45 pm »
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Hi i was wondering if i could get help with these questions please.

I am not sure how to find the exact x intercepts without using a calculator.
I am also not sure how to find the equation of a parabola given the diagram. There are 3 different forms?  :-*

The two questions you have about finding x-intercepts can be answered with the quadratic formula:



You may need to expand the quadratic first (the first question), but from there, just substituting into this formula is all you need! Do you learn it in WACE?

As for determining a formula given a diagram, you have a few options. In this case we have intercepts, so it is probably best to think of the parabola like this:



It needs to be in this form, because the roots need to be -1 and 3 (from the diagram). We can only multiply by some constant \(a\), so if we expand and then differentiate:



Now we need our turning point to be at (1,7). The first derivative definitely allows this, but we need the equation to yield \(y=7\) when \(x=1\);



So the equation of the parabola is \(y=-10(x+1)(x-3)\) ;D

There are heaps of ways to do this question, this is just my approach. I'm not sure what forms you learn, this is a way to do it from scratch :)

RuiAce

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Re: Mathematics Question Thread
« Reply #1001 on: December 07, 2016, 01:15:43 pm »
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Hi i was wondering if i could get help with these questions please.

I am not sure how to find the exact x intercepts without using a calculator.
I am also not sure how to find the equation of a parabola given the diagram. There are 3 different forms?  :-*

« Last Edit: December 07, 2016, 01:19:25 pm by RuiAce »

anotherworld2b

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Re: Mathematics Question Thread
« Reply #1002 on: December 09, 2016, 04:44:54 pm »
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Could i please get help with these two questions?
I am not sure how to draw the diagrams  :-*

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Re: Mathematics Question Thread
« Reply #1003 on: December 10, 2016, 05:19:59 pm »
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how to do this without guess and check

find the two numbers whose sum is 28 and whose product is a maximum
get me out of here

RuiAce

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Re: Mathematics Question Thread
« Reply #1004 on: December 10, 2016, 05:28:16 pm »
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how to do this without guess and check

find the two numbers whose sum is 28 and whose product is a maximum



Incidentally, the correct answer is the answer where a=b. It can be proven that in general, if the perimeter of any rectangle is fixed, the area is maximised when the rectangle is a square. This is just the geometry put into numbers only.
« Last Edit: December 10, 2016, 05:40:16 pm by RuiAce »