Author Topic: Mathematics Question Thread (Read 1626895 times) Tweet Share

Dragomistress · « **Reply #2595 on:** September 24, 2017, 08:38:52 pm »

How does one solve this?

The point M divides the interval AB internally in the ratio 3:1. The point N divides the interval AB externally in the ratio 3:1. What is the value of NB:MB.

RuiAce · « **Reply #2596 on:** September 24, 2017, 08:41:38 pm »

Quote from: Dragomistress on September 24, 2017, 08:38:52 pm

How does one solve this?

The point M divides the interval AB internally in the ratio 3:1. The point N divides the interval AB externally in the ratio 3:1. What is the value of NB:MB.

This is an Extension 1 level question. If you need help with it, please post it in the more relevant thread.

Nuwanda_g · « **Reply #2597 on:** September 25, 2017, 02:04:22 pm »

Hey guys,
I need help for HSC 2016 Mathematics paper, question 16 (b) of i.

I dont get how to differentiate the inside bracket of (1+19e^-0.5t).

Thanks in advance.

Shadowxo · « **Reply #2598 on:** September 25, 2017, 03:02:09 pm »

Quote from: Nuwanda_g on September 25, 2017, 02:04:22 pm

Hey guys,
I need help for HSC 2016 Mathematics paper, question 16 (b) of i.

I dont get how to differentiate the inside bracket of (1+19e^-0.5t).

Thanks in advance.

$y=\frac{200}{1+19e^{-0.5t}} \\y=200(1+19e^{-0.5t})^{-1}\\\frac{dy}{dt}=-1*200*\frac{d}{dt}(1+19e^{-0.5t})*(1+19e^{-0.5t})^{-2}\\=-1*200*(19*-0.5*e^{-0.5t})*(1+19e^{-0.5t})^{-2}\\=1900e^{-0.5t}*(1+19e^{-0.5t})^{-2}\\=\frac{1900e^{-0.5t}}{(1+19e^{-0.5t})^{2}}$

Remember, the derivative of e^ax=a*e^ax and the derivative of a constant is 0
Let me know if you're still confused

_____ · « **Reply #2599 on:** September 26, 2017, 02:39:42 pm »

I've attached a question and its solution as a word document because it's too cumbersome for images/text.

I don't want it solved, I'm just wondering what's the reasoning behind the working for iv? Why is (where x is theta) sinx/(cosx+1) > 0 and tan2x < 0?

Thanks!

RuiAce · « **Reply #2600 on:** September 26, 2017, 02:53:20 pm »

Quote from: _____ on September 26, 2017, 02:39:42 pm

I've attached a question and its solution as a word document because it's too cumbersome for images/text.

I don't want it solved, I'm just wondering what's the reasoning behind the working for iv? Why is (where x is theta) sinx/(cosx+1) > 0 and tan2x < 0?

Thanks!

$\text{Because trigonometry said so.}$
$\text{In the event }45^\circ < \theta < 90^\circ\\ \text{we have }-\infty < \tan 2\theta < 0\\ \text{i.e. }\tan 2\theta < 0\\ \text{This can be seen by graphing }y=\tan 2x\text{ over }0^\circ < \theta < 180^\circ.$
$\text{Also, clearly }\boxed{\cos \theta + 1 > 0} \text{ for all }\theta\\ \text{This falls from finding the range of }f(\theta) = \cos\theta + 1$
$\text{Similarly, when }0^\circ < \theta < 90^\circ\text{, from another graph it is clear that }\boxed{\sin\theta > 0}$
$\text{Since the quotient of two positive numbers is still a positive number}\\ \text{it then follows that }\boxed{\frac{\sin\theta}{1+\cos\theta} > 0}\text{ as well.}$
__________________________________________________
$\text{This is just a question of discarding the nonsensical case }0^\circ < \theta < 45^\circ\\ \text{When this happens, the LHS is positive and the RHS is negative}\\ \text{which can't happen, since the LHS and RHS aren't both equal to 0.}$

Becky234 · « **Reply #2601 on:** September 27, 2017, 05:10:03 pm »

hey guys
can someone pls help me with these 2 questions:
1) Solve 2^(2x + 1) = 32.
2) Differentiate ln(5x + 2) with respect to x.

sophiegmaher · « **Reply #2602 on:** September 27, 2017, 05:13:27 pm »

How do you complete the attached question??

Natasha.97 · « **Reply #2603 on:** September 27, 2017, 05:18:32 pm »

Quote from: Becky234 on September 27, 2017, 05:10:03 pm

hey guys
can someone pls help me with these 2 questions:
1) Solve 2^(2x + 1) = 32.
2) Differentiate ln(5x + 2) with respect to x.

Hi!

Hope this helps

RuiAce · « **Reply #2604 on:** September 27, 2017, 05:22:54 pm »

Quote from: sophiegmaher on September 27, 2017, 05:13:27 pm

How do you complete the attached question??

$\text{There used to be the standard integral}\\ \int \sec ax \tan ax \, dx = \frac1a \sec ax + C\\ \text{So here the answer is }\frac13 \sec 3x +C$
$\text{Because the table is gone, you don't need to worry about it either}$

sophiegmaher · « **Reply #2605 on:** September 27, 2017, 05:25:50 pm »

Quote from: RuiAce on September 27, 2017, 05:22:54 pm

$\text{There used to be the standard integral}\\ \int \sec ax \tan ax \, dx = \frac1a \sec ax + C\\ \text{So here the answer is }\frac13 \sec 3x +C$
$\text{Because the table is gone, you don't need to worry about it either}$

ohhhh okay awesome thank you!

Becky234 · « **Reply #2606 on:** September 27, 2017, 05:33:34 pm »

Quote from: Jess1113 on September 27, 2017, 05:18:32 pm

Hi!

(Image removed from quote.)

Hope this helps

for part 1) can u pls explain how u got 5

Mathew587 · « **Reply #2607 on:** September 27, 2017, 05:38:27 pm »

How do we integrate y=ln(x-2) ?

Mathew587 · « **Reply #2608 on:** September 27, 2017, 05:40:58 pm »

Quote from: Becky234 on September 27, 2017, 05:33:34 pm

for part 1) can u pls explain how u got 5

for qs where its ln(f(x)), dy/dx = f'(x)/f(x)

so for ln (5x+2)
its
5/5x+2 + c

RuiAce · « **Reply #2609 on:** September 27, 2017, 05:41:51 pm »

Quote from: Mathew587 on September 27, 2017, 05:38:27 pm

How do we integrate y=ln(x-2) ?

You don't have any techniques for doing this in 2U. Please post the full question.

Quote from: Becky234 on September 27, 2017, 05:33:34 pm

for part 1) can u pls explain how u got 5

From basic arithmetic/the calculator, 2⁵ = 32.

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