Mathematics Question Thread

[RuiAce]

Intriguing... there are a few similar questions with cubic equations in the textbook exercise that I am working on...Is there a certain way to approach these types of questions?

Unless you wanna try guessing (which is fine), there really isn't. Which textbook is this?

[kauac]

Which textbook is this?

Maths in Focus - Margaret Grove

[RuiAce]

Maths in Focus - Margaret Grove

Questions like Q13 and Q6 in that exercise look fine to me. Occasionally, they can be solved because the expression should be easy to factorise.

However, that particular one disgusts me and just further lowered my opinion of the Maths in Focus textbook. It is definitely not expected out of a 2U student and it's beyond unfair to throw it at them.

(There are 3U cheats around it. But a 2U student is not expected to know of these.)

[kauac]

Questions like Q13 and Q6 in that exercise look fine to me. Occasionally, they can be solved because the expression should be easy to factorise.

However, that particular one disgusts me and just further lowered my opinion of the Maths in Focus textbook. It is definitely not expected out of a 2U student and it's beyond unfair to throw it at them.

(There are 3U cheats around it. But a 2U student is not expected to know of these.)

Alright... Thanks for looking into it!

[jamonwindeyer]

However, that particular one disgusts me

Rui is disgusted! 

[brooksykait]

helloooo, i’m having some difficulties, could someone please help?

Find the sum of all integers between 1 and 200 that are not multiples of 9

(from Magaret Grove textbook - Challenge Exercise

i know this is supposed to be an easy question but i cannot think of a way to do this without literally counting the numbers up. any help would be appreciated

[EEEEEEP]

helloooo, i’m having some difficulties, could someone please help?

Find the sum of all integers between 1 and 200 that are not multiples of 9


(from Magaret Grove textbook - Challenge Exercise

i know this is supposed to be an easy question but i cannot think of a way to do this without literally counting the numbers up. any help would be appreciated

Hint:

Try using pairs. For numbers between 1 and 200, 2 numbers at each ends will go together to make 200, e.g.
(I excluded 199 for now)
2, 198
3, 197
4, 196
5, 195

Once you get the right amount of pairs , you can do 200 * X (which denotes the number of pairs), minus and plus some other things =)

[RuiAce]

Hint:

Try using pairs. For numbers between 1 and 200, 2 numbers at each ends will go together to make 200, e.g.
(I excluded 199 for now)
2, 198
3, 197
4, 196
5, 195

Once you get the right amount of pairs , you can do 200 * X (which denotes the number of pairs), minus and plus some other things =)

That's a bit overkill, especially with the whole "not multiples of 9" issue

helloooo, i’m having some difficulties, could someone please help?

Find the sum of all integers between 1 and 200 that are not multiples of 9

(from Magaret Grove textbook - Challenge Exercise

i know this is supposed to be an easy question but i cannot think of a way to do this without literally counting the numbers up. any help would be appreciated

[brooksykait]

[/quote]

Thank you, that worked. Except I think your calculation was wrong. My answer was 17823, and so was the textbook. Doesn't matter but thank you 

[RuiAce]

Thank you, that worked. Except I think your calculation was wrong. My answer was 17823, and so was the textbook. Doesn't matter but thank you 

I fixed it. I accidentally punched a + into the calculator instead of a -. Check updated post

[mirakhiralla]

Hey! Yeah not quite, but that's all good, let me show you!

So we start with $20,000, let that be \(A_0\). After 1 year (so, January 2009), we add 6% interest and then pay back $3000. That looks like this:



The next year, we take that amount, and do the same thing. Add 6%, subtract $3000:



If you expand, you'll get what I've got above! And if you do it again, you should get:



See the pattern? If you're just starting this topic it might look strange, but do a few of these and this is what they all look like, more or less. After 5 years (2013), we have:



Pop that in your calculator, I get $9853.23!! For Part (b), you need to instead consider a general version of the expression, after \(n\) years:



We need \(A_n=0\) -> See if you can manipulate that expression to find \(n\)! If you've never done a question like this before let me know and I'd be happy to show you, or perhaps read this guide which steps through it for you! Happy to help if anything above was confusing as well - I'm assuming you've seen something similar to it before but can definitely go slower if you haven't

So I got a right after doing it again, thank you.
For b, do I use the sum formula and then log? I get the same answer but I feel like my sum is wrong...

[RuiAce]

So I got a right after doing it again, thank you.
For b, do I use the sum formula and then log? I get the same answer but I feel like my sum is wrong...

\begin{align*}20000(1.06)^n - 3000\left(\frac{1.06^n - 1}{0.06} \right)&= 0\\ 20000(1.06)^n -50000(1.06^n - 1)&=0 \tag{from calculator}\\ 20000(1.06)^n - 50000(1.06)^n + 50000 &= 0\\ 30000(1.06)^n &= 50000\\ 1.06^n &= \frac{5}{3}\\ n \log 1.06 &= \log \frac{5}{3}\\ n &= \frac{\log \frac53}{\log 1.06}\\ &\approx 8.767\end{align*}

[gilliesb18]

Hello,

Just need help on a question; Find the equation of the normal to the curve, y=e^x at the point where x=3, in exact form.
I just can't remember how to do this!!
Any help is appreciated.....

Thanks heaps:)

[jazzycab]

Hello,

Just need help on a question; Find the equation of the normal to the curve, y=e^x at the point where x=3, in exact form.
I just can't remember how to do this!!
Any help is appreciated.....

Thanks heaps:)

The normal is perpendicular to the curve, so you can find it by first finding the gradient of the tangent at that point (by differentiating), then evaluating the gradient of the normal (the product of perpendicular gradients is -1).
You will also need the coordinates of a point on the line, but the line passes through \(y=e^x\) at \(x=3\), so it passes through \(\left(3,e^3\right)\)

[gilliesb18]

Ok cool! Thanks heaps...