Author Topic: Mathematics Question Thread (Read 1625941 times) Tweet Share

RuiAce · « **Reply #2220 on:** July 19, 2017, 12:06:51 am »

I can't read. Yes it is A (fixing in original post now)

bigsweetpotato2000 · « **Reply #2221 on:** July 19, 2017, 12:10:02 am »

Quote from: RuiAce on July 19, 2017, 12:06:51 am

I can't read. Yes it is A (fixing in original post now)

Need to double the glasses power

itssona · « **Reply #2222 on:** July 19, 2017, 09:46:15 pm »

Um so i was doing past papers for my school exam andddd i need clarification

Whats domain of underroot (x-2)
Our school says x greater than and equal to 2 but in another exam a similar question had the answer of GREATER than only

Thank you

RuiAce · « **Reply #2223 on:** July 19, 2017, 09:48:13 pm »

Quote from: sssona09 on July 19, 2017, 09:46:15 pm

Um so i was doing past papers for my school exam andddd i need clarification

Whats domain of underroot (x-2)
Our school says x greater than and equal to 2 but in another exam a similar question had the answer of GREATER than only

Thank you

It's greater than or equal to. If we sub in x=2 we don't run into problems.
$\text{It may be worth mentioning that if we chucked the square root on the BOTTOM}\\ \text{i.e. }f(x)=\frac{1}{\sqrt{x-2}}\\ \text{Then the domain is }x>2$

itssona · « **Reply #2224 on:** July 19, 2017, 09:49:13 pm »

Quote from: RuiAce on July 19, 2017, 09:48:13 pm

It's greater than or equal to. If we sub in x=2 we don't run into problems.
$\text{It may be worth mentioning that if we chucked the square root on the BOTTOM}\\ \text{i.e. }f(x)=\frac{1}{\sqrt{x-2}}\\ \text{Then the domain is }x>2$

Ohhh THANK YOU RUI

Youssk · « **Reply #2225 on:** July 19, 2017, 09:57:20 pm »

Hi, Could I have some help with this question.
The acceleration of a particle is given by
a=1-2t,
where a is measured in cm/second and t is measured in seconds.
a) At what time is the particle at rest?
b) Where is the particle at this time?

RuiAce · « **Reply #2226 on:** July 19, 2017, 09:59:09 pm »

Quote from: Youssk on July 19, 2017, 09:57:20 pm

Hi, Could I have some help with this question.
The acceleration of a particle is given by
a=1-2t,
where a is measured in cm/second and t is measured in seconds.
a) At what time is the particle at rest?
b) Where is the particle at this time?

This question lacks information and cannot be done. Information needs to be given about initial conditions.

Youssk · « **Reply #2227 on:** July 19, 2017, 10:02:12 pm »

Initially the particle is at rest 2cm to the right of the origin.

RuiAce · « **Reply #2228 on:** July 19, 2017, 10:05:49 pm »

georgiia · « **Reply #2229 on:** July 20, 2017, 02:47:15 pm »

Could someone please help me for this? Its probably very simple but I have no clue??

Thanks!

jamonwindeyer · « **Reply #2230 on:** July 20, 2017, 02:52:47 pm »

Quote from: georgiia on July 20, 2017, 02:47:15 pm

Could someone please help me for this? Its probably very simple but I have no clue??

Thanks!

Hey! This is a quotient rule problem, definitely not simple! Helpful to differentiate $\ln{x^2}$ separately so we'll do $u$ and $v$ as well as their derivatives:

$u=x\implies u'=1\\v=\ln{x^2}\implies v'=\frac{2x}{x^2}=\frac{2}{x}$

Note you can also do that last one with log laws

put this into the quotient rule:

$\frac{u'v-v'u}{v^2}=\frac{\ln{x^2}-2}{\left(\ln{x^2}\right)^2}$

georgiia · « **Reply #2231 on:** July 20, 2017, 02:58:01 pm »

SORRY I ATTATCHED THE WRONG SCREENSHOT!!!!!!!

RuiAce · « **Reply #2232 on:** July 20, 2017, 03:01:06 pm »

Quote from: georgiia on July 20, 2017, 02:47:15 pm

Could someone please help me for this? Its probably very simple but I have no clue??

Thanks!

Presumably the attachment changed.
$\text{Let }KJ =x\\ \text{so note that }KI = x+16$
$\text{Applying trigonometry in:}\\ \triangle HKJ: \tan 74^\circ = \frac{HK}{x} \implies\boxed{ x = \frac{HK}{\tan 74^\circ}}\\ \triangle HKI: \tan 30^\circ = \frac{HK}{x+16}\implies \boxed{x+16 = \frac{HK}{\tan 30^\circ}}$
$\text{This is now a pair of simultaneous equations}\\ \text{Upon substituting the first into the second:}\\ \begin{align*}\frac{HK}{\tan 74^\circ}+16&= \frac{HK}{\tan 30^\circ}\\ 16&=HK\left(\cot 30^\circ - \cot 74^\circ\right)\\ HK&= \frac{16}{\frac1{\tan 30^\circ} - \frac1{\tan 74^\circ}}\\ &= 11.07032\dots\end{align*}$
$\therefore B$

kylesara · « **Reply #2233 on:** July 20, 2017, 04:03:58 pm »

Hi could i please have help with this question.

The cubic y=ax^3+bx^2+cx+d has a point of inflexion at x=p.
Show that p = − b . 3a

Thanks:)

RuiAce · « **Reply #2234 on:** July 20, 2017, 04:08:31 pm »

Quote from: kylesara on July 20, 2017, 04:03:58 pm

Hi could i please have help with this question.

The cubic y=ax^3+bx^2+cx+d has a point of inflexion at x=p.
Show that p = − b . 3a

Thanks:)

A typo was assumed.
$\frac{d^2y}{dx^2}=6ax+2b$
$\text{If an inflexion point occurs at }x=p\\ \text{then when }x=p, \frac{d^2y}{dx^2}=0\\ \text{Hence }6ap-2b=0 \implies \boxed{p=-\frac{b}{3a}}$

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