Active Maths 4 Bk1 - Q & A

VirtualWorld

I'm self-studying LC HL Maths using this book as a base. To say that there are gaps in it is an understatement. It ought to be added I haven't done maths in years and therefore this could account for these gaps in my case.
As this book seems to be one of two options out there at the moment for LC, I was hoping that if I started this thread for the bits I can't seem to understand, that some of you might go about putting in examples or explanations. In doing so, hopefully you will benefit from some revision & help others too.

So my first question relates to:

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PAGE 16 Order of Magnitude.

I don't understand what that last green box is trying to say, or how to go about answering the related questions 1.4) 10) (i) to (v) on the next page.

Any help would be appreciated.
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VirtualWorld

I've added photos of the two pieces in question.

Canard

First, you write them in that notation - you can only have 1 number before the decimal point. The power beside the 10 is the number of digits *after* the point; 5 in the first case, and 8 in the second. The idea there is that the multiplication by 10 to the power of whatever will give you the original number again. You round any <5 to 1, and >/= 5 up to 10 (presumably because it's not small enough to be negligible once it gets that high).

Any number to the power of 0 is equal to 1 (I think it's kind of important to understand that for this box) - so 1 = 10^0. When you multiply powers you add them together, and that's how those two sums are being found (10^5 and 10^9).

Then to find the magnitude you put them over each other, meaning you subtract, and voila.

I hope that's all correct, I havent done that kind of stuff in a while because I don't remember it being on our course last year but I know I did it at some point. :pac: For the questions you listed you just substitute the numbers - e.g. 868 would be 8.68 x 10^2; 1 x 10^2; 10^1 x 10^2 = 10^3. With the ones like 0.01 they take 10 to a negative power.

Hope I helped a bit, maybe someone else can add to that / correct any inaccuracies.

VirtualWorld

that's how those two sums are being found (10^5 and 10^9).

Then to find the magnitude you put them over each other, meaning you subtract, and voila

Thank you so much. I knew it couldn't be difficult but I couldn't see what they were saying in those couple of lines you have here.

Much appreciated.

VirtualWorld

Hi again,

I'm attaching two images, the question and the answers to questions I can't fathom. I was able to answer Q.2 but am lost on 1 & 3 so far.

Any help would be appreciated. Thanks in advance.

VW

Moody_mona

Square out the brackets on the left hand side and then compare it to the expression on the right hand side. If you end up with x^2 +5x - 6 =ax^2 +bx + c then compare the coefficients of x^2, x and the constant. So a=1, b=5 and c=-6.

VirtualWorld

Thank you so much Mona. Very much appreciated.