Author Topic: Mathematics Question Thread (Read 1615319 times) Tweet Share

RuiAce · « **Reply #3375 on:** March 31, 2018, 02:34:56 pm »

Quote from: username2000 on March 31, 2018, 02:23:48 pm

Hey Jamon,
more specifically...ive attached some questions i need help with, just get a bit confused with the e's, and whether to leave it as decimal places or what.
thanks heaps!
cheers

$\text{Take for example, }Q3.\\ y = xe^{3x} \implies \frac{dy}{dx} = e^{3x} + 3xe^{3x}\\ \text{from the product rule.}$
$\text{When }x=1, y = e^3.\\ \text{Also }\frac{dy}{dx} = e^3 + 3e^3 = 4e^3.$
$\text{So plugging into the point-gradient form gives}\\ \begin{align*}y-e^3 &= 4e^3(x - 1)\\ y &= 4e^3 x - 4e^3 + e^3\\ y &= 4e^3 x - 3e^3\end{align*}$
Also note that when differentiating in Q4, you should use log laws to rewrite $ \ln \sqrt{x} $ as $ \frac12\ln x$ before differentiating. Note that you should never round unless explicitly told to do so.

If you have further problems, you should post any relevant progress.

Mate2425 · « **Reply #3376 on:** March 31, 2018, 03:11:51 pm »

Hey could someone please help me with this question.
Q. Solve 7Sin3x = 2x-1 for .... 0 <= x <= 2Pi
Where: < = is x is greater than/ equal to
<= is X is less than/ equal to
Thanks heaps

RuiAce · « **Reply #3377 on:** March 31, 2018, 03:14:37 pm »

Quote from: Mate2425 on March 31, 2018, 03:11:51 pm

Hey could someone please help me with this question.
Q. Solve 7Sin3x = 2x-1 for .... 0 <= x <= 2Pi
Where: < = is x is greater than/ equal to
<= is X is less than/ equal to
Thanks heaps

You don't. Either you approximate the solution by drawing two graphs or there's a mistake.

jamonwindeyer · « **Reply #3378 on:** March 31, 2018, 03:14:41 pm »

Quote from: Mate2425 on March 31, 2018, 03:11:51 pm

Hey could someone please help me with this question.
Q. Solve 7Sin3x = 2x-1 for .... 0 <= x <= 2Pi
Where: < = is x is greater than/ equal to
<= is X is less than/ equal to
Thanks heaps

Hey! There's no way to solve this precisely, so what you'll want to do is draw graphs of both $y=7\sin{3x}$ and $y=2x-1$ to scale, on the same axes. Then, use the x-coordinates of the points of intersection as estimates for your answer (since those x-coordinates are where the LHS equals the RHS, where the two curves cross)

username2000 · « **Reply #3379 on:** March 31, 2018, 05:26:06 pm »

Quote from: RuiAce on March 31, 2018, 02:34:56 pm

$\text{Take for example, }Q3.\\ y = xe^{3x} \implies \frac{dy}{dx} = e^{3x} + 3xe^{3x}\\ \text{from the product rule.}$
$\text{When }x=1, y = e^3.\\ \text{Also }\frac{dy}{dx} = e^3 + 3e^3 = 4e^3.$
$\text{So plugging into the point-gradient form gives}\\ \begin{align*}y-e^3 &= 4e^3(x - 1)\\ y &= 4e^3 x - 4e^3 + e^3\\ y &= 4e^3 x - 3e^3\end{align*}$
Also note that when differentiating in Q4, you should use log laws to rewrite $ \ln \sqrt{x} $ as $ \frac12\ln x$ before differentiating. Note that you should never round unless explicitly told to do so.

If you have further problems, you should post any relevant progress.

thanks, v helpful RuiAce

but i cant seem to find that answer on the sheet...well it is P when i chnaged it to general form...but when i put in in it says its wrong. its from mathsonline

RuiAce · « **Reply #3380 on:** March 31, 2018, 05:58:26 pm »

Quote from: username2000 on March 31, 2018, 05:26:06 pm

thanks, v helpful RuiAce
but i cant seem to find that answer on the sheet...well it is P when i chnaged it to general form...but when i put in in it says its wrong. its from mathsonline

Erm, not sure what you're trying to tell me here. Are these from mathsonline or are they a worksheet but without answers, or something else?

username2000 · « **Reply #3381 on:** March 31, 2018, 06:49:44 pm »

sorry...forgot i hadnt posted the full thing...my bad
its a worksheet from mathsonline and the thing i sent before was just a snip..
thanks again for your help...especially when its easter

atar notes is sooo good

RuiAce · « **Reply #3382 on:** March 31, 2018, 06:54:15 pm »

I'm looking at their "answers" and my answer to Q3 looks like the same as their answer K)

username2000 · « **Reply #3383 on:** March 31, 2018, 07:02:30 pm »

Quote from: RuiAce on March 31, 2018, 06:54:15 pm

I'm looking at their "answers" and my answer to Q3 looks like the same as their answer K)

is that the same as p or am i seeing things...i might be doing something wrong but it still says thats wrong...unless their answers are incorrect

unless it can be factorised or simplified further...maybe im just wasting ur time sorry $:-\$

RuiAce · « **Reply #3384 on:** March 31, 2018, 07:17:20 pm »

Quote from: username2000 on March 31, 2018, 07:02:30 pm

is that the same as p or am i seeing things...i might be doing something wrong but it still says thats wrong...unless their answers are incorrect
unless it can be factorised or simplified further...maybe im just wasting ur time sorry $:-\$

Nope you're right. That's definitely the same as P) there.

(Nor would I know why, rip)

username2000 · « **Reply #3385 on:** March 31, 2018, 07:19:47 pm »

Quote from: RuiAce on March 31, 2018, 07:17:20 pm

Nope you're right. That's definitely the same as P) there.

(Nor would I know why, rip)

so do u agreee your answers right

RuiAce · « **Reply #3386 on:** March 31, 2018, 07:22:49 pm »

I know I'm right when I right

sady123 · « **Reply #3387 on:** April 01, 2018, 01:20:29 pm »

Hi can someone help me with this question?? Thanks!

RuiAce · « **Reply #3388 on:** April 01, 2018, 01:25:46 pm »

Quote from: sady123 on April 01, 2018, 01:20:29 pm

Hi can someone help me with this question?? Thanks!

$\text{The area of }\triangle BCD\text{ is simply}\\A = \frac{1}{2}\times 10 \times 12 \times \sin 120^\circ = 30\sqrt{3}\text{ m}.$
$\text{Taking the hint on board, }\triangle BCD \equiv \triangle BAD.\\ \text{Since congruent triangles have the same area, it follows that}\\ \text{the area of }\triangle BAD\text{ is also }30\sqrt{3}\text{ m}.$
$\text{So by combining areas, the area of parallelogram }ABCD\\ \text{will just be }A = 60\sqrt{3}\text{ m}.$

Never.Give.Up · « **Reply #3389 on:** April 01, 2018, 10:27:28 pm »

can i pls have some help with...

Find the exact area bounded by the curve y=e^2x, the x-axis and the lines x=2 and x=5.

sorry, i know this is really easy but i've had a mind blank

thanks heaps

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