Mathematics Question Thread

[samthemanfromacan]

2009 Chatswood Girls Paper

[RuiAce]

2009 Chatswood Girls Paper

(Image removed from quote.)

Numbers subject to inaccuracy as they were typed up rapidly. Method is all there.

[BarnesK01]

Just need a help with question 7 if I could. For part i) I am using SAS, so far I have used Angle EBC = Angle CDB (base angle of an isoscles triangle) and BC (common).

[RuiAce]

Just need a help with question 7 if I could. For part i) I am using SAS, so far I have used Angle EBC = Angle CDB (base angle of an isoscles triangle) and BC (common).

To show that BE = DC, consider the following facts:
- BA = CA
- BE = 1/2 BA
- DC = 1/2 CA

Remember, the question tells you information about midpoints.

[Wales]

A tad embarrassed to be asking something here.. Friend asked me a question and I cannot seem to get the answer.

i) Show that e^1-ln2=e/2
ii) Find the equation of the tangent to the curve y=e^1-4x at x=ln2/4

I've had a look at the solutions but cant for the love of God figure out why x=e/2/. I get x=ln2/4.

http://puu.sh/whQDT/8cba8b73fd.jpg - Solution

[RuiAce]

A tad embarrassed to be asking something here.. Friend asked me a question and I cannot seem to get the answer.

i) Show that e^1-ln2=e/2
ii) Find the equation of the tangent to the curve y=e^1-4x at x=ln2/4

I've had a look at the solutions but cant for the love of God figure out why x=e/2/. I get x=ln2/4.

http://puu.sh/whQDT/8cba8b73fd.jpg - Solution

For the sake of clarity, please use brackets if you intend to describe e^(1-4x)

[Wales]

For the sake of clarity, please use brackets if you intend to describe e^(1-4x)

I understand that. What I don't understand is that in the final solution using the point gradient formula the X value is e/2.  I got the same X value as you stated. 

Apologies for the lack of brackets. I told my friend off for the same thing but forgot to correct it when copying the question over.

[jamonwindeyer]

I understand that. What I don't understand is that in the final solution using the point gradient formula the X value is e/2.  I got the same X value as you stated. 

Apologies for the lack of brackets. I told my friend off for the same thing but forgot to correct it when copying the question over.

It's an error - You are finding the equation of the tangent at \(x=\frac{\ln{2}}{4}\), so that has to be the x-value!

[RuiAce]

Oh. Yeah, couldn't tell what you meant. Once I saw the point gradient form I just said no

[Wales]

It's an error - You are finding the equation of the tangent at \(x=\frac{\ln{2}}{4}\), so that has to be the x-value!

At last I can rest Goddamn I spent far too long trying to figure that out,

Cheers

[IronBark]

Hi RuiAce,

I am really sorry for this but I need help with these two questions. My maths exam is tomorrow, so on the bright side, after today you will never hear from me again  .


a) I evaluated that 30(deg)=1/4 though I dont know what to do with the squaring...
b) I graphed but dont know how to solve or how to answer it.

Thanks Rui,

Im so sorry for all the trouble mate.

[jamonwindeyer]

Hi RuiAce,

I am really sorry for this but I need help with these two questions. My maths exam is tomorrow, so on the bright side, after today you will never hear from me again  .
(Image removed from quote.)

a) I evaluated that 30(deg)=1/4 though I dont know what to do with the squaring...
b) I graphed but dont know how to solve or how to answer it.

Thanks Rui,

Im so sorry for all the trouble mate.

Don't be sorry friend! We'd love to keep hearing from you after your exam - We love having people around we start that first one by taking the square root of both sides:



So this is just now two regular trig equations, \(\sin{x}=\frac{1}{2}\) and \(\sin{x}=-\frac{1}{2}\), each of which has two solutions you can find the normal way

Let me draw a graph for that second one, just a sec...

[jamonwindeyer]

Okay, so the graph above should match yours. Now, a few things we notice:

- The graphs intersect at two points in the domain we care about. The x-coordinates of these are actually the two solutions to \(\sin{x}=\frac{1}{2}\) we found in the previous problem, namely: \(\frac{\pi}{6},\frac{5\pi}{6}\).
- Between those two points, the sine curve is above the line. Everywhere else, it is below.

And that is the answer. We want when \(\sin{x}\le\frac{1}{2}\), when the line is below the curve. So we define it with two regions:



Read this through once or twice, hopefully it makes sense - Let me know!

[Wales]

Don't be sorry friend! We'd love to keep hearing from you after your exam - We love having people around we start that first one by taking the square root of both sides:



So this is just now two regular trig equations, \(\sin{x}=\frac{1}{2}\) and \(\sin{x}=-\frac{1}{2}\), each of which has two solutions you can find the normal way

Let me draw a graph for that second one, just a sec...

*1 hour later*

Try out desmos to graph if you're struggling

http://puu.sh/wiORK/8df6de2f27.png

There's a sketch of y=1/2 and y=sinx between 0 and 2pi.

Post back if you need anymore help

[jamonwindeyer]

*1 hour later*
Try out desmos to graph if you're struggling
http://puu.sh/wiORK/8df6de2f27.png
There's a sketch of y=1/2 and y=sinx between 0 and 2pi.
Post back if you need anymore help

I think you might have missed the post I made just above yours, hidden on a new page love your work though Wales, as always