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#### RuiAce

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« Reply #4305 on: July 23, 2019, 12:37:32 pm »
+2
Hey People,
I am learning the product rule atm however i cannot solve this question:

-2(x^2+x+1)^3  x.

\begin{align*}
u&=x^2+x+1\\
u^\prime &= 2x+1\\
v &= x\\
v^\prime &= 1
\end{align*}
\begin{align*}
\therefore \frac{d}{dx} -2(x^2+x+1)^3 x &= -2 \frac{d}{dx} (x^2+x+1)^3 x\\
&= -2 \left[x (2x+1) + (x^2+x+1) \right]
\end{align*}
You could probably simplify this further.
« Last Edit: July 23, 2019, 01:41:01 pm by RuiAce »

#### LoneWolf

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« Reply #4306 on: July 23, 2019, 01:40:36 pm »
0
Legend thanks!
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#### benneale

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« Reply #4307 on: August 03, 2019, 03:03:48 pm »
0
Hey guys!

A little confused as to how sinxcosx all over cos^2x is equal to tanx... can u show me like the proof or something?? thanks!

#### Jakeybaby

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« Reply #4308 on: August 03, 2019, 03:22:52 pm »
0
Hey guys!

A little confused as to how sinxcosx all over cos^2x is equal to tanx... can u show me like the proof or something?? thanks!
Some useful things to keep in mind!
$\frac{\sin{x}}{\cos{x}}=\tan{x} \\ \cos^2{x} = \cos{x}\cos{x}$
Use these 2 pieces of information to progress through the proof - let me know how you go!
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#### Kombmail

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« Reply #4309 on: August 05, 2019, 06:56:15 pm »
0
Guys does anyone now how to solve this:

Integrate 2/(5-x)^4dx ?
Ps any tips for index law help?
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#### fun_jirachi

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« Reply #4310 on: August 05, 2019, 07:58:26 pm »
0
Hey there!

All there is to the index law is on your reference sheet.

For reference, this is what it has:

$\frac{d}{dx} x^n = nx^{n-1} \\ \int (ax+b)^n \ dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C \\ \text{Hence,} \ \int \frac{2}{(5-x)^4} \ dx = \frac{2}{3(5-x)^3} + C$

Hope this helps
« Last Edit: August 05, 2019, 08:00:32 pm by fun_jirachi »
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#### mirakhiralla

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« Reply #4311 on: August 06, 2019, 03:35:41 pm »
0

Without finding the point of intersection, find the equation of the line which passes through the point of intersection of 7x+3y-13=0 and 3x+2y-12=0, and also passes through D(1.5, -1)

#### emmajb37

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« Reply #4312 on: August 07, 2019, 11:06:33 am »
0
Hi, with this question I understand the concept but I cannot seem to simplify the equation and the working of the answers isn't helping so i would love an explanation.
Find the value of p, where x=p is a vertical line that divides the area between y=√x, x=16 and the x-axis into 2 equal parts.

#### DrDusk

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« Reply #4313 on: August 07, 2019, 11:58:07 am »
0
Hi, with this question I understand the concept but I cannot seem to simplify the equation and the working of the answers isn't helping so i would love an explanation.
Find the value of p, where x=p is a vertical line that divides the area between y=√x, x=16 and the x-axis into 2 equal parts.

$A_{total} = \int_{0}^{16}\sqrt{x}dx$

$A_p = \int_{0}^{p}\sqrt{x}dx = \frac{1}{2}A_{Total}$

Solve for p from there because you know what the total area is from the first integral
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#### Coolmate

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« Reply #4314 on: August 07, 2019, 08:00:15 pm »
0
Hello everyone,

I am having a bit of trouble with these exponential equations (see attached) and was wondering if anyone could please step me through how to do them?
Also, do you times the front and the end numbers (outside of the brackets) together? Or do you just leave them?

Thankyou!

Coolmate  👍
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#### RuiAce

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« Reply #4315 on: August 09, 2019, 09:30:06 am »
+1

Without finding the point of intersection, find the equation of the line which passes through the point of intersection of 7x+3y-13=0 and 3x+2y-12=0, and also passes through D(1.5, -1)
If you're not allowed to do it the classic way, then you more or less have no choice but to use the mythical 'k'-method. Incredibly disgusting, but again, no choice.
$\text{The equation will take the form}\\ 7x+3y-13 + k(3x+2y-12) = 0.$
$\text{If the point }D\left( \frac32, -1\right) \text{lies on the line, then upon substituting in we obtain}\\ 7(1.5)+3(-1)-13 + k(3(1.5)+2(-1)-12) = 0 \implies \boxed{k=-\frac{11}{19}}.$
$\text{Hence the equation of the line will be}\\ 7x+3y-13 - \frac{11}{19}(3x+2y-12)=0\\ \text{which you can multiply by 19, then expand and simplify.}$

#### RuiAce

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« Reply #4316 on: August 09, 2019, 09:35:41 am »
+2
Hello everyone,

I am having a bit of trouble with these exponential equations (see attached) and was wondering if anyone could please step me through how to do them?
Also, do you times the front and the end numbers (outside of the brackets) together? Or do you just leave them?

Thankyou!

Coolmate  👍
You haven't specified what you want me to do for question j.

With that sketch in question a, you should first note that $f(x)=a^x$ has asymptote $y=0$. So in your case, $f(x) =4(3^x) + 1$ has asymptote $y=1$. (Which can be properly verified - as $x\to \infty$, it is clear that there is no asymptote. But as $x\to -\infty$, we have $4(3^x)+1 \to 0 + 1 = 1$.

Now the $4$ gets multiplied to the entire exponential term, i.e. the $3^x$ bit. So it forms some kind of a stretch (dilation) to just the exponential - it doesn't move the asymptote around. As an example, your $y$-intercept is now at $y=4(1)+1 = 5$, instead of at $y=2$.

In general, graphically showing the extra factor in front isn't exactly possible. Some points need to be plotted (and the graph drawn to scale) for it to have a noticeable impact. But if you try plotting on Desmos/GeoGebra, it will be 100% clear that some kind of stretching is happening.

#### Coolmate

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« Reply #4317 on: August 09, 2019, 01:05:34 pm »
0
Cheers Rui!
I have re looked over the question and it makes much more sense now, thanks for the thorough explanation!

With regards to Question j I have attached the original question of what to do (Sorry)

Thanks Again!

Coolmate
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#### fun_jirachi

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« Reply #4318 on: August 09, 2019, 05:30:44 pm »
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You should try following a similar process to what Rui previously outlined:
- Asymptotes, limiting value as x approaches positive and negative infinity
In this case, the horizontal asymptote is at x=4, and as x approaches positive infinity y approaches 4 from below.
- Dilation/Shift to the right/left, orientation
The dilation is likewise pretty hard to see, but basically, note that from the two negative signs (or by subbing in numbers) that the exponential will tend towards negative infinity as y approaches negative infinity.

Hope this helps
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#### Coolmate

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« Reply #4319 on: August 10, 2019, 01:19:34 pm »
0
You should try following a similar process to what Rui previously outlined:
- Asymptotes, limiting value as x approaches positive and negative infinity
In this case, the horizontal asymptote is at x=4, and as x approaches positive infinity y approaches 4 from below.
- Dilation/Shift to the right/left, orientation
The dilation is likewise pretty hard to see, but basically, note that from the two negative signs (or by subbing in numbers) that the exponential will tend towards negative infinity as y approaches negative infinity.

Hope this helps

Hi fun_jirachi!

This was extremely helpful and I understood the question perfectly, thankyou for your help!

Coolmate
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