Author Topic: Mathematics Question Thread (Read 1627384 times) Tweet Share

owidjaja · « **Reply #3450 on:** May 03, 2018, 07:41:30 pm »

Hey guys,
I'm not entirely sure where I went wrong in my working out- could someone help me find my mistake?

Here's the question (the answer as well) and here's my working out.

RuiAce · « **Reply #3451 on:** May 03, 2018, 07:45:50 pm »

Quote from: owidjaja on May 03, 2018, 07:41:30 pm

Hey guys,
I'm not entirely sure where I went wrong in my working out- could someone help me find my mistake?

Here's the question (the answer as well) and here's my working out.

\[ T_n = a r^{n-1} =4 \left(\frac12\right)^{n-1}.\\ \text{You can't throw the 4 under the bracket here.} \]
\begin{align*}4 \times \frac{1}{2^{n-1}} &= \frac1{128}\\ \frac{1}{2^{n-1}}&=\frac1{512}\end{align*}

owidjaja · « **Reply #3452 on:** May 03, 2018, 08:07:14 pm »

I'm also not sure where I went wrong in this other question. Here is the question and here is my working out.

RuiAce · « **Reply #3453 on:** May 03, 2018, 08:30:57 pm »

Quote from: owidjaja on May 03, 2018, 08:07:14 pm

I'm also not sure where I went wrong in this other question. Here is the question and here is my working out.

Your working out is fine. You just haven't finished the question

owidjaja · « **Reply #3454 on:** May 07, 2018, 06:59:43 pm »

Hey guys,
I'm not entirely sure how to figure out this question, so it'll be great to get some help.

Thanks in advance

RuiAce · « **Reply #3455 on:** May 07, 2018, 07:07:32 pm »

Quote from: owidjaja on May 07, 2018, 06:59:43 pm

Hey guys,
I'm not entirely sure how to figure out this question, so it'll be great to get some help.

Thanks in advance

$T_2 = 2\text{, so }\boxed{ar = 2}$
$\text{The limiting sum is 9, so }\boxed{\frac{a}{1-r} = 9}.$
$\text{This is now a pair of simultaneous equations.}\\ \text{The second rearranges to }\boxed{a = 9(1-r)}\\ \text{which, upon subbing into the first, gives }\boxed{9r(1-r) = 2}\\ \text{which rearranges to }9r - 9r^2 - 2 = 0$
$\text{This is now a quadratic, whose solutions are }r = \frac13\text{ and }r=\frac23.\\ \text{I leave the details to fill in as your exercise.}$

owidjaja · « **Reply #3456 on:** May 09, 2018, 08:53:51 pm »

Hey guys,
I'm not entirely sure how to do the following question, so help is appreciated.

Thanks in advance

RuiAce · « **Reply #3457 on:** May 09, 2018, 09:04:46 pm »

Quote from: owidjaja on May 09, 2018, 08:53:51 pm

Hey guys,
I'm not entirely sure how to do the following question, so help is appreciated.

Thanks in advance

(Shows how shit MIF is as a textbook with that horrendous typo...)
$\text{What they meant was that it first jumps 0.5, then it jumps 0.1,}\\ \text{and it keeps jumping 0.2 }\textbf{times}\text{ that of the previous distance.}$
$\text{i.e. on the third jump, it jumps 0.02 m.}\\ \text{On the foruth jump, it jumps 0.004 m.}\\ \text{(And so on).}$
$\text{So it's just an infinite geometric series with }a=0.5\text{ and }r=0.2$

Mate2425 · « **Reply #3458 on:** May 11, 2018, 11:00:21 am »

Hi could i please have some help with the following AP Question.

Q. Find the sum of all integers between 1 and 100 that are not multiples of 6.

RuiAce · « **Reply #3459 on:** May 11, 2018, 12:04:18 pm »

Quote from: Mate2425 on May 11, 2018, 11:00:21 am

Hi could i please have some help with the following AP Question.

Q. Find the sum of all integers between 1 and 100 that are not multiples of 6.

$\text{The question can be reinterpreted}\\ \text{as the sum of all integers between 1 and 100,}\\ \text{but then with the multiples of 6 }\textbf{subtracted back out}.$
$\text{i.e. we can consider}\\ (1 + 2 + 3 + \dots + 100) - (6 + 12 + 18 + \dots + 96).$
$\text{Using the A.P. partial sum formula, this gives us}\\ \frac{100}{2} ( 1 + 100) - \frac{16}{2} (6 + 96)$

Mate2425 · « **Reply #3460 on:** May 16, 2018, 07:40:52 pm »

Quote from: RuiAce on May 11, 2018, 12:04:18 pm

$\text{The question can be reinterpreted}\\ \text{as the sum of all integers between 1 and 100,}\\ \text{but then with the multiples of 6 }\textbf{subtracted back out}.$
$\text{i.e. we can consider}\\ (1 + 2 + 3 + \dots + 100) - (6 + 12 + 18 + \dots + 96).$
$\text{Using the A.P. partial sum formula, this gives us}\\ \frac{100}{2} ( 1 + 100) - \frac{16}{2} (6 + 96)$

Thanks RuiAce

LaraC · « **Reply #3461 on:** May 22, 2018, 12:14:09 pm »

Hello

Could someone please help me with the following question...I'm a little confused as I used the limiting sum formula with a= 1.5m and r=2/5....but that gives me 2.5m and the answer says 3.5m?!

Mary bounces a ball, dropping it from 1.5m on its first bounce. It then rises up to 2/5 of its height on each bounce. Find the distance through which the ball travels.

Thanks!

RuiAce · « **Reply #3462 on:** May 22, 2018, 12:32:50 pm »

Quote from: LaraC on May 22, 2018, 12:14:09 pm

Hello

Could someone please help me with the following question...I'm a little confused as I used the limiting sum formula with a= 1.5m and r=2/5....but that gives me 2.5m and the answer says 3.5m?!

Mary bounces a ball, dropping it from 1.5m on its first bounce. It then rises up to 2/5 of its height on each bounce. Find the distance through which the ball travels.

Thanks!

Some similar questions were asked here and here. Take a look at the first.

The idea is that you can’t treat these as just one geometric series; you actually have two geometric series added together. In your case, you’ll have $(1.5 + 1.5(0.4) + 1.5(0.4)^2 + \dots) + (1.5(0.4) + 1.5(0.4)^2 + 1.5(0.4)^3 + \dots)$ which will give $ \frac{1.5}{1-0.4} + \frac{1.5(0.4)}{1 - 0.4} $.

LaraC · « **Reply #3463 on:** May 22, 2018, 12:40:49 pm »

Oh yep - that makes sense now! Thanks Rui!

LaraC · « **Reply #3464 on:** May 23, 2018, 10:13:19 am »

Hello again,

This isn't exactly a maths question, but more a calculator qu

I have a series of questions that include things like "find correct to 2 decimal place cos 0.589", and " find correct to 2 decimal places tan 0.056" etc. They are all along the same lines. Punching that into my calculator is easy enough, but it says it has to be done using the radian mode on my calculator (clearly giving a different answer).....how do I convert my calculator into radian mode?
It is a Casio fx-82AU PLUS.

Sorry I've looked up online manuals but still a bit confused!!

Thanks!

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