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March 29, 2024, 02:41:07 am

Author Topic: Mathematics Question Thread  (Read 1296762 times)  Share 

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emilyyyyyyy

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Re: Mathematics Question Thread
« Reply #4005 on: February 15, 2019, 07:43:55 pm »
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hello can someone please help me how to find the enclosed area between the y-axis and the curve x=y(y-2)??  I'm just not sure what the y-values im meant to be integrating are.
thanks!

meerae

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Re: Mathematics Question Thread
« Reply #4006 on: February 15, 2019, 08:28:33 pm »
+1
hello can someone please help me how to find the enclosed area between the y-axis and the curve x=y(y-2)??  I'm just not sure what the y-values im meant to be integrating are.
thanks!

Hey!
You'd be integrating between y=0 and y=2

Hope this helps!
meerae :)
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jamonwindeyer

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Re: Mathematics Question Thread
« Reply #4007 on: February 15, 2019, 08:49:51 pm »
+1
Can someone help with this question? Particularly stuck on how to write the equation for the second bit.
The sum of the first 4 terms of an arithmetic series is 42 and the sum of the 3rd and 7th term is 46. Find the sum of the first 20 terms.

Hey! So your first equation is:



The next one is tougher, but it is really just the sum of \(T_3\) and \(T_7\):



Solving those simultaneously will give you your first term and common difference! And that can be used to find \(S_{20}\) - Is that enough to get you rolling? :)
« Last Edit: February 15, 2019, 09:44:07 pm by jamonwindeyer »

jamonwindeyer

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Re: Mathematics Question Thread
« Reply #4008 on: February 15, 2019, 09:00:38 pm »
+1
hello can someone please help me how to find the enclosed area between the y-axis and the curve x=y(y-2)??  I'm just not sure what the y-values im meant to be integrating are.
thanks!

Hey! As meerae says, the values are \(y=0\) and \(y=2\). This is because these are the two points where the curve cuts the y-axis (sideways parabola!), so this defines the start and stop point for the area. The graph is here:



Also notice the area is to the left of the y-axis, meaning you'll need to compensate for the negative as well! Hope this helps :)

goodluck

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Re: Mathematics Question Thread
« Reply #4009 on: February 15, 2019, 09:22:06 pm »
+1
Hey how would I solve this? (it's absolute values if the writing doesn't make it clear!)

|(x-1)/(x+1)| <1

jamonwindeyer

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Re: Mathematics Question Thread
« Reply #4010 on: February 15, 2019, 09:44:30 pm »
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Ah yep thank you so much! :)
Though I think you wrote T4 instead of T3 by accident??

I did indeed, sorry!! Just tidied it up but the general idea is the same :)

RuiAce

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Re: Mathematics Question Thread
« Reply #4011 on: February 15, 2019, 09:50:39 pm »
+3
Hey how would I solve this? (it's absolute values if the writing doesn't make it clear!)

|(x-1)/(x+1)| <1
\[ \text{The solution is equivalent to that of}\\ |x-1| < |x+1| \]
(However, note that \(x \neq -1\) because otherwise in the original equation we have dividing by 0 issues.)
Recall
The definition of the absolute value states that
\[ |x| = \begin{cases} x & \text{if }x\geq 0\\ -x & \text{if }x < 0\end{cases} \]
Therefore we have
\begin{align*} |x-1| &= \begin{cases} x-1 & \text{if }x-1 \geq 0\\ -(x-1) & \text{if }x-1 < 0\end{cases} \\ &= \begin{cases} x-1 & \text{if }x \geq 1\\ -x+1 &\text{if }x < 1\end{cases} \end{align*}
And similarly
\[ |x+1| = \begin{cases} x+1 &\text{ if }x \geq -1\\ -x-1 &\text{if }x < -1\end{cases} \]
\[ \text{For this inequality, noting the above}\\ \text{we can accordingly split the following cases.} \]
\[ \text{Case 1: } x < -1.\\ \text{For this case the inequality becomes}\\ \begin{align*} -x+1 &< -x-1\\ 1 &< -1\end{align*}\\ \text{This obviously has no solution}\\ \text{so no values of }x\text{ in this case belong in the solution.} \]
\[ \text{Case 2: }-1\leq x < 1.\\ \text{For this case the inequality becomes}\\ \begin{align*}-x+1 &< x+1\\ 0 &< 2x\\ x &> 0 \end{align*}\]
\[ \text{So overlapping }x > 0\text{ with }-1\leq x < 1\\ \text{we see that }\boxed{0 < x < 1}\text{ is a part of the solution.}\]
\[ \text{Case 3: }x\geq 1.\\ \text{For this case the inequality becomes}\\ \begin{align*}x-1 &< x+1\\ -1 &< 1 \end{align*}\]
\[ \text{This is obviously true for every value of }x.\\ \text{So all values of }x\text{ in this case, i.e. }\\\boxed{ x\geq 1}\text{, is a part of the solution.}\]
\[ \text{Combining all parts, our solution is thus}\\ \boxed{x > 0}.\]

goodluck

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Re: Mathematics Question Thread
« Reply #4012 on: February 16, 2019, 06:11:59 pm »
0
Hey, thank you so much for your help, what's the maximal domain and range of y= (1-2sinx)^1/2?  I think the max range is 0<y< sqrt3 but was confused on how you express the domain

RuiAce

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Re: Mathematics Question Thread
« Reply #4013 on: February 16, 2019, 06:41:31 pm »
+1
The groundwork
\[ \text{Essentially we are solving }1-2\sin x \geq 0\\ \text{which rearranges to }\sin x \leq \frac12. \]
\[ \text{First note that the solutions to }\sin x = \frac12 \text{ for }0\leq x \leq 2\pi\\ \text{are }x=\frac\pi6\text{ or } \frac{5\pi}6.\]
\[ \text{Following the periodicity of the trigonometric functions, the other solutions will be at}\\ x = 2k\pi+\frac\pi6\text{ or }x=2k\pi + \frac{5\pi}{6}\\ \text{where }k\text{ is an integer.}\]
_____________________________
\[\text{Now for the inequality, we can sketch the curve }y=\sin x\\ \text{along with the line }y=\frac12.\\ \text{The above computations track down each point of intersection for us.}\]
\[ \text{As we are interested in }\sin x \leq \frac12,\\ \text{we're interested in the }x\text{ values for which}\\ y=\sin x\text{ lies }\textbf{below}\text{ the line }y=\frac12.\]
\[ \text{We see that we'll have }-\frac{7\pi}{6} \leq x \leq \frac\pi6, \, \frac{5\pi}{6}\leq x \leq \frac{13\pi}6\\ \text{and so on.}\\ \text{All of these intervals will form our answer.} \]
So expressing it in a nice way is reasonably hard, because we have to express an infinite number of intervals. But we can always work around this by just setting \(k\) to be an arbitrary integer. One possible answer is \( -\frac{7\pi}{6}+2k\pi \leq x \leq \frac\pi6+2k\pi \), where \(k\) is an integer. Having this arbitrary integer placeholder lets us get around this issue, because \(k\) can then be substituted for every integer possible.

Note that I purposely avoided "general solutions" as this was posted in the 2U section.

Edit: fixed typos
« Last Edit: February 16, 2019, 07:14:37 pm by RuiAce »

goodluck

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Re: Mathematics Question Thread
« Reply #4014 on: February 16, 2019, 07:01:58 pm »
0
Oh you can express it via general solutions?  (maybe should have posted it in the 3U section then!)
but how does the possible answer work if k is like 10 for example since it will be bigger than pi/6? would we just do general solution there?

The groundwork
\[ \text{Essentially we are solving }1-2\sin x \geq 0\\ \text{which rearranges to }\sin x \leq \frac12. \]
\[ \text{First note that the solutions to }\sin x = \frac12 \text{ for }0\leq x \leq 2\pi\\ \text{are }x=\frac\pi6\text{ or } \frac{5\pi}6.\]
\[ \text{Following the periodicity of the trigonometric functions, the other solutions will be at}\\ x = 2k\pi+\frac\pi6\text{ or }x=2k\pi + \frac{5\pi}{6}\\ \text{where }k\text{ is an integer.}\]
_____________________________
\[\text{Now for the inequality, we can sketch the curve }y=\sin x\\ \text{along with the line }y=\frac12.\\ \text{The above computations track down each point of intersection for us.}\]
\[ \text{As we are interested in }\sin x \leq \frac12,\\ \text{we're interested in the }x\text{ values for which}\\ y=\sin x\text{ lies }\textbf{below}\text{ the line }y=\frac12.\]
\[ \text{We see that we'll have }-\frac{7\pi}{6} \leq x \leq \frac\pi6, \, \frac{5\pi}{6}\leq x \leq \frac{13\pi}6\\ \text{and so on.}\\ \text{All of these intervals will form our answer.} \]
So expressing it in a nice way is reasonably hard, because we have to express an infinite number of intervals. But we can always work around this by just setting \(k\) to be an arbitrary integer. One possible answer is \( -\frac{5\pi}{6}+2k\pi \leq x \leq \frac\pi6 \), where \(k\) is an integer. Having this arbitrary integer placeholder lets us get around this issue, because \(k\) can then be substituted for every integer possible.

Note that I purposely avoided "general solutions" as this was posted in the 2U section.

AlphaZero

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Re: Mathematics Question Thread
« Reply #4015 on: February 16, 2019, 07:11:06 pm »
+1
Oh you can express it via general solutions?  (maybe should have posted it in the 3U section then!)
but how does the possible answer work if k is like 10 for example since it will be bigger than pi/6? would we just do general solution there?

I think he meant to write  \[\frac{-7\pi}{6}+2\pi k\leq x\leq \frac{\pi}{6}+2\pi k,\quad \text{where }k\ \text{is an integer.}\]
« Last Edit: February 16, 2019, 07:17:43 pm by dantraicos »
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Youssefh_

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Re: Mathematics Question Thread
« Reply #4016 on: February 18, 2019, 07:48:57 pm »
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can someone please explain these two questions for me because i am getting confused at what to do, thank you

meerae

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Re: Mathematics Question Thread
« Reply #4017 on: February 18, 2019, 08:05:11 pm »
+3
can someone please explain these two questions for me because i am getting confused at what to do, thank you

Hey!

For the first question, the formula to find arc length is The question has given you the θ in radians, so now you have to find the radius.
Luckily, they gave you the circumference, so you would use the circumference formula and solve for r where C=300mm (keep in mind the units). After you find r, you would substitute it into the formula and solve for l. Hope this makes sense.

For the second question, you are required to find the angle. The question begins with telling you the area of the circle, so it would make sense to note this formula down as You would then use the area formula to solve for r. Using you would substitute l=8cm as that is the length of the arc, as well as the r you found. Then solve for  θ . Keeping in mind  θ is in radians and your length is in cm.

Hope this helps!
meerae :)
« Last Edit: February 18, 2019, 08:08:40 pm by meerae »
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Thankunext

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Re: Mathematics Question Thread
« Reply #4018 on: February 18, 2019, 09:34:10 pm »
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Can someone please help with the working out of these? I tried using log but I think I'm doing something wrong  :'( :'(
1. Which term of the series 8 - 4 + 2 - ... is 1/128?
2. Which term of 54 + 18 + 6 + ... is 2/243?
3. Find the value of n if the nth term of the series -2 + 3/2 - 9/8 + ... is -81/128.

fun_jirachi

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Re: Mathematics Question Thread
« Reply #4019 on: February 18, 2019, 09:49:03 pm »
+3
Using Tn=arn-1, a=8, r=-1/2, Tn=1/128




You can also use other log laws here to help you out, but in this case by knowing powers of two you can technically do it by inspection without logs and calculators.

A similar process follows for the second and third question, try them out!
« Last Edit: February 18, 2019, 09:56:00 pm by fun_jirachi »
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