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March 29, 2024, 08:27:44 am

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RuiAce

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Re: Mathematics Question Thread
« Reply #4140 on: April 08, 2019, 06:30:57 pm »
+1
Hi,

With this question, when I'm working it out, what am I meant to do to know whether the square root of y is + or - in the graph?

Thanks!
If you had to go from \(x = y^2\) to \(y = \pm \sqrt{x}\), you'd identify that the top branch is \(y=\sqrt{x}\) and the bottom branch is \(y=-\sqrt{x}\). This is because above the \(x\)-axis, the \(y\)-coordinates are positive, and below the \(x\)-axis, the \(y\)-coordinates are negative.

Similarly, if you had to go from \(y=x^2\) to \(x=\pm \sqrt{y}\), you'd identify that the left branch is \(x=\sqrt{y}\) and the right branch is \(x=-\sqrt{y}\). This is because to the right of the \(y\)-axis, the \(x\)-coordinates are positive, and to the left of the \(x\)-axis, the \(y\)-coordinates are negative.

Same goes here. You would take \(\boxed{x-2 = -\sqrt{y}}\) and hence \(x = 2-\sqrt{y} \), because the expression \(x-2\) will be negative for values to the left of the line \(x=2\).
hey, having a bit of a 'moment'...

Find the arc length, correct to 2 decimal places, given radius is 5.9cm and angle subtended is 23degrees 12minutes...

I understand the process of l=r.theta , but when I punch in the degrees bit into my calculator, the answer is way off!
That formula only works when \(\theta\) is radians. Write the angle as \( \left( 23 + \frac{12}{60} \right) \) degrees, and hence convert it to radians by writing it as \( \left( 23 + \frac{12}{60} \right) \times \frac\pi{180} \).
« Last Edit: April 08, 2019, 06:39:47 pm by RuiAce »

emilyyyyyyy

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Re: Mathematics Question Thread
« Reply #4141 on: April 08, 2019, 07:37:46 pm »
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Hi I need help again :)

How do I do part iii?

thanks!
« Last Edit: April 08, 2019, 07:39:17 pm by emilyyyyyyy »

fun_jirachi

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Re: Mathematics Question Thread
« Reply #4142 on: April 08, 2019, 07:47:54 pm »
+1
You could think about the probability of getting 45 or more in the first two sums, then subtracting that probability to obtain the answer. This basically uses the idea of the complement to make your life a lot easier.

There are two cases; getting a 45 and getting a 50.

There's only one way of getting a 50; rolling 5 all four times. The chance of that happening is (1/6)^4, or 1/1296.
There's four ways of getting a 45; if you think about it, a 45 is made of a 4 and 5, and a 5 and 5. What you should notice is that the four appears once and once only, and can appear anywhere, as long as all the other rolls are fives ie. 5 and 4 + 5 and 5 is different to 4 and 5 + 5 and 5. ie. the probability of this happening is 4 x (1/6)^4, or 4/1296.
Add these up, and subtract them from 1, and you get 1291/1296.

Hope this helps :)
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emilyyyyyyy

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Re: Mathematics Question Thread
« Reply #4143 on: April 08, 2019, 07:54:11 pm »
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You could think about the probability of getting 45 or more in the first two sums, then subtracting that probability to obtain the answer. This basically uses the idea of the complement to make your life a lot easier.

There are two cases; getting a 45 and getting a 50.

There's only one way of getting a 50; rolling 5 all four times. The chance of that happening is (1/6)^4, or 1/1296.
There's four ways of getting a 45; if you think about it, a 45 is made of a 4 and 5, and a 5 and 5. What you should notice is that the four appears once and once only, and can appear anywhere, as long as all the other rolls are fives ie. 5 and 4 + 5 and 5 is different to 4 and 5 + 5 and 5. ie. the probability of this happening is 4 x (1/6)^4, or 4/1296.
Add these up, and subtract them from 1, and you get 1291/1296.

Hope this helps :)

thank you! i understand the part about the 50 and how there are 4 chances of that happening - so to get a 50 it's (1/6)^4
But I'm still a little lost on the part about getting 45, is there another way you might be able to explain it to m (if not don't worry about it hahha)


emilyyyyyyy

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Re: Mathematics Question Thread
« Reply #4144 on: April 08, 2019, 09:46:15 pm »
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Hi,

Is someone able to tell me where I've gone wrong in working out the area of this graph? I cant get the right answer

thanks :))

jamonwindeyer

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Re: Mathematics Question Thread
« Reply #4145 on: April 08, 2019, 09:59:17 pm »
+1
thank you! i understand the part about the 50 and how there are 4 chances of that happening - so to get a 50 it's (1/6)^4
But I'm still a little lost on the part about getting 45, is there another way you might be able to explain it to m (if not don't worry about it hahha)

I always revert to listing the possible outcomes when it gets confusing! Here are the four outcomes Fun_Jirachi is referencing:

- (4x5) + (5x5)
- (5x4) + (5x5)
- (5x5) + (4x5)
- (5x5) + (5x4)

All of these give an answer of 45. Notice it is just all fives with a four in position #1, #2, #3 or #4 - Hence the four possible ways it can be done ;D

jamonwindeyer

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Re: Mathematics Question Thread
« Reply #4146 on: April 08, 2019, 10:01:44 pm »
+1
Hi,

Is someone able to tell me where I've gone wrong in working out the area of this graph? I cant get the right answer

thanks :))

Hi! If you just want the blue shaded area, then you only need the \(A_1\) from your working! That gives you the area between the curve and the x-axis for x=0 to x=1, which is exactly what you are looking for ;D the dotted line above is just the asymptote of the graph but it actually doesn't matter because the area never touches it!

emilyyyyyyy

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Re: Mathematics Question Thread
« Reply #4147 on: April 08, 2019, 10:06:37 pm »
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Hi! If you just want the blue shaded area, then you only need the \(A_1\) from your working! That gives you the area between the curve and the x-axis for x=0 to x=1, which is exactly what you are looking for ;D the dotted line above is just the asymptote of the graph but it actually doesn't matter because the area never touches it!

okay thank you! I think I was getting a bit confused bc when I was doing this question attached, I had to find the area under the curve and also of the rectangular shape? are you able to possibly explain why?

thanks!

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Re: Mathematics Question Thread
« Reply #4148 on: April 08, 2019, 10:12:45 pm »
+1
okay thank you! I think I was getting a bit confused bc when I was doing this question attached, I had to find the area under the curve and also of the rectangular shape? are you able to possibly explain why?

thanks!

So in this question, what you want (just straight up) is:



Problem is, we can't find that directly because we can't integrate logarithms. So, we instead rearrange it to \(x=2+e^y\) and find the area with respect to the y-axis, that's the area to the left of the curve.



That's not the one we want - The one we want is the area of the rectangle MINUS that area! And that's why we do it that way!

In summary, we need to use the rectangle for this question because we can't directly integrate a logarithm. So, we need to use tricks - We don't need to do that for exponentials ;D
« Last Edit: April 08, 2019, 10:15:53 pm by jamonwindeyer »

emilyyyyyyy

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Re: Mathematics Question Thread
« Reply #4149 on: April 09, 2019, 07:31:36 pm »
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Hi!

How do I find the area in this question? I've tried splitting it up into positive and negative areas, but my equations or something aren't working because I can't seem to get the answer.

Thanks! :)

r1ckworthy

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Re: Mathematics Question Thread
« Reply #4150 on: April 09, 2019, 08:50:25 pm »
+1
Hi!
How do I find the area in this question? I've tried splitting it up into positive and negative areas, but my equations or something aren't working because I can't seem to get the answer.
Thanks! :)
Hey,

Here is my solution of the problem. I don't think you need to split the curve up into positive and negative areas, as you are only working out the area between two curves. I think the time you do split the curve into positive and negative areas is when working out the area under a curve in a negative region. Hopefully someone can back this up, as I am not too sure on whether this is correct or not.

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emilyyyyyyy

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Re: Mathematics Question Thread
« Reply #4151 on: April 09, 2019, 08:55:39 pm »
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Hey,

Here is my solution of the problem. I don't think you need to split the curve up into positive and negative areas, as you are only working out the area between two curves. I think the time you do split the curve into positive and negative areas is when working out the area under a curve in a negative region. Hopefully someone can back this up, as I am not too sure on whether this is correct or not.

(Image removed from quote.)

Thanks! what you've done is actually what I did the first time, but for some reason I guess I put the numbers in my calculator wrong :)
I think I was getting confused with working out two separate areas on a curve - i.e. one that is above the x-axis and one that is below.

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Re: Mathematics Question Thread
« Reply #4152 on: April 09, 2019, 09:32:22 pm »
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Thanks! what you've done is actually what I did the first time, but for some reason I guess I put the numbers in my calculator wrong :)
I think I was getting confused with working out two separate areas on a curve - i.e. one that is above the x-axis and one that is below.

Cool, no worries! Yeah, I get what you mean, I got my numbers all wrong too until I checked it on an integration calculator online (where they calculate the answer to any integral- so useful).
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Re: Mathematics Question Thread
« Reply #4153 on: April 10, 2019, 07:07:23 pm »
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To clarify on the above question, the graph shown is y = f(x) and the question wants you to sketch y=f^-1(x)

RuiAce

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Re: Mathematics Question Thread
« Reply #4154 on: April 10, 2019, 07:27:53 pm »
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To clarify on the above question, the graph shown is y = f(x) and the question wants you to sketch y=f^-1(x)
In saying that, this is an MX1 question, so you should post in the appropriate MX1 thread if you require assistance.