September 27, 2020, 12:04:05 am

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#### RuiAce

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« Reply #4290 on: July 22, 2019, 04:43:39 pm »
+1
Ahhhh I totally get it.
Thanks so much!
And if it was f(x) = f(-x) there would be a max turning point at (-3,7)??
Correct, because even function symmetry is about the $y$-axis, so it's just a reflection.

#### LoneWolf

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« Reply #4291 on: July 23, 2019, 12:28:07 pm »
0
Hey People,
I am learning the product rule atm however i cannot solve this question:

-2(x^2+x+1)^3  x.
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#### RuiAce

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« Reply #4292 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 #4293 on: July 23, 2019, 01:40:36 pm »
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Legend thanks!
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#### benneale

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« Reply #4294 on: August 03, 2019, 03:03:48 pm »
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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 #4295 on: August 03, 2019, 03:22:52 pm »
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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 #4296 on: August 05, 2019, 06:56:15 pm »
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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 #4297 on: August 05, 2019, 07:58:26 pm »
+1
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 #4298 on: August 06, 2019, 03:35:41 pm »
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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 #4299 on: August 07, 2019, 11:06:33 am »
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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 #4300 on: August 07, 2019, 11:58:07 am »
+1
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 #4301 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!

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#### RuiAce

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« Reply #4302 on: August 09, 2019, 09:30:06 am »
+2

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 #4303 on: August 09, 2019, 09:35:41 am »
+3
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 #4304 on: August 09, 2019, 01:05:34 pm »
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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!

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