Secondary Maths questions, give me your maths questions

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the population of a town increases by 6% every year. in january 2006 the population was 5500.
find population of the town...
A) in january 2007
B)after n years

find the year the population will reach 11000
 
showie said:
the population of a town increases by 6% every year. in january 2006 the population was 5500.
find population of the town...
A) in january 2007
B)after n years

find the year the population will reach 11000
Click to expand...
Seems easy.

P(0) = 5500
g = 6%

a) P(n) = P(0) x (1 + g)^n

P(1) = 5500(1.06)^1 = 5830

b) P(n) = P(0) x (1 + g)^n = 5500(1.06)^n

11000 = 5500(1.06)^n

2 = (1.06)^n

log(2) = n x log(1.06)

n = log(2)/log(1.06) = 11.90 years
 

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There's actually also a neat trick you can use to do a rough check.

A population doubles roughly once every 70/r years where i is the rate of growth. 70/6 = 11.67, and so the above answer is likely correct.
 
S "Thinks He's Daicos" J said:
No particular reason. I think it is purely by convention. If everyone did it the other way around it would work as well.
Click to expand...

No it wouldn't.


5 * 4 + 3
= 20 + 3
= 23

If we worked it out by adding first and multiplying second it'd become

5 * 4 + 3
= 5 * 7
= 35

We do it because it works.

A better answer then that? I don't have one. :)
 
No you are missing the point.

Currently, the operations of multiplication and division are given priority to addition and subtraction.

If we have four groups of three oranges and then take away two, we can write it as so:

4 x 3 - 2 = 10 (a)

Note, by convention, this is equivalent to:

(4 x 3) - 2 = 10 (b)

If we had four groups of three oranges, and took away two from each group, it must be written as:

4 x (3 - 2) = 4 (c)

This "works" because of the convention that is set.

However if it were the other way around, where addition and subtraction were given priority to multiplication and division, things would change.

No longer would the equation in (a) be equivalent to (b), but instead to (c).

Under these constraints, in order to express a situation of four groups of three oranges, before subtracting two, the expression must be written as (b).
 
SJ, you're effectively changing the definition of multiplication there.

We multiply first by definition, not by convention.

I.e. 4 x 5 - 2 = (5 + 5 + 5 + 5) - 2 by definition of the x operator.
 
mdc said:
SJ, you're effectively changing the definition of multiplication there.

We multiply first by definition, not by convention.

I.e. 4 x 5 - 2 = (5 + 5 + 5 + 5) - 2 by definition of the x operator.
Click to expand...
Why is it by definition? Why couldn't it be around the other way, if it was convention to do so?

Your example is written that way because that is the rule. That is what everyone does.

The question was asked why we do it that way.

Carl Spackler said:
Why do you times/multiply before you add/subtract?
Click to expand...

From my knowledge, there is no specific reason.

If addition had priority over multiplication in the order of operations, then using your example:

4 x 5 - 2 = 4 x (5 - 2) = 12

To write it the current way, we would have the place the parenthesis around the multiplication:

(4 x 5) - 2 = (5 + 5 + 5 + 5) - 2 = 18

I may be wrong. However then I'd like to ask why the definition of multiplication states it is first, if it isn't solely done by convention.
 
S "Thinks He's Daicos" J said:
Why is it by definition? Why couldn't it be around the other way, if it was convention to do so?
Click to expand...

Convention implies and arbitrary choice. That's not the case here.

Perhaps my use of brackets was unfortunate. To put it more clearly.

4 x 5 - 2 = 5 + 5 + 5 + 5 - 2 = (4 x 5) - 2 but doesn't = 4 x (5 - 2)

I.e. The first part is by definition, and we can check 5 + 5 + 5 + 5 - 2 = 18. Then we can also check that (4 x 5) -2 gives us 18 but 4 x (5-2) doesn't.
 
mdc said:
Convention implies and arbitrary choice. That's not the case here.

Perhaps my use of brackets was unfortunate. To put it more clearly.

4 x 5 - 2 = 5 + 5 + 5 + 5 - 2 = (4 x 5) - 2 but doesn't = 4 x (5 - 2)

I.e. The first part is by definition, and we can check 5 + 5 + 5 + 5 - 2 = 18. Then we can also check that (4 x 5) -2 gives us 18 but 4 x (5-2) doesn't.
Click to expand...
Isn't it?

The thing is, I understand all of what you write. Yes we do multiplication before addition. Yes placing the parentheses around a subtraction creates a different equation.

That's all primary school stuff.

But why is multiplication and division given priority over addition and subtraction?
 
Because multiplication is defined in terms of addition, and not the other way around. Without a definition for "add" we cannot define "multiply", and we'd need to for "the opposite way" to work.
 

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This actually explains it somewhat. I can see why it might be easier to do exponents, then multiplication, then addition in that order as they are all related to the next.

In saying that, I can't see why it wouldn't work the other way around. Just might not be as clean.

Edit: Just saw the above. Same thing, thanks.
 
S "Thinks He's Daicos" J said:
That's right. Maybe "why" isn't the correct question, as I understand the algebra. I haven't studied mathematics since second year - how does e^(x*i) = cos(x)+isin(x) again?
Click to expand...

Hey, I'd forgotten about this thread. My own attempt at explaining this:

exp(a) = lim x->inf (1 + a/n)^n

with a = 1 this is used to define e.

So exp(i) = lim x-> inf (1 + i/n) ^ n = -1
 
Can someone explain transposing equations to me?

I have a lot of trouble with some of the semi-complex types.

For example

s = θ/360 x 2pir

Arc Length = Angle Size/360 x 2 x Pi x Radius Length

I can apply that formula, no worries.

When given the formula to find Angle Size or Radius Length I can apply those as well.

But what I want to know, is given the formula for Arc Length above, how would I actually transpose the equation to get either Radius Length or Angel Size by themselves.
 
superfraser said:
how would I actually transpose the equation to get either Radius Length or Angel Size by themselves.
Click to expand...

You want to "reverse" the operations. To get r on its own, for example, think about what's happening to r on the Right Hand Side; namely, it's being multiplied by (θ/360 x 2Pi). So to reverse that, we simply want to divide both sides by that term. So;

s/(θ/360 x 2Pi) = [strike](θ/360 x 2Pi)[/strike]r/[strike](θ/360 x 2Pi)[/strike]

s/(θ/360 x 2Pi) = r

If you want to re-arrange for neatness, this gives r = 360 x s/(2Piθ)

θ is found similarly, by dividing both sides by (1/360 x 2Pir)
 
mdc said:
You want to "reverse" the operations. To get r on its own, for example, think about what's happening to r on the Right Hand Side; namely, it's being multiplied by (θ/360 x 2Pi). So to reverse that, we simply want to divide both sides by that term. So;

s/(θ/360 x 2Pi) = [strike](θ/360 x 2Pi)[/strike]r/[strike](θ/360 x 2Pi)[/strike]

s/(θ/360 x 2Pi) = r

If you want to re-arrange for neatness, this gives r = 360 x s/(2Piθ)

θ is found similarly, by dividing both sides by (1/360 x 2Pir)
Click to expand...

Yup.

That's exactly how I thought you did it..

..So it's actually the re-arraging for neatness that I need to understand.

Would you mind explaining that part?

Cheers.. :D
 
superfraser said:
Yup.

That's exactly how I thought you did it..

..So it's actually the re-arraging for neatness that I need to understand.

Would you mind explaining that part?

Cheers.. :D
Click to expand...

This is something that a lot of people have trouble with, the thing to remember is that dividing is the same as multiplying by the inverse (to get the inverse just swap the top and bottom of the fraction). So s/(2Piθ/360) = s ÷(2Piθ/360) = s x (360/(2Piθ) = 360s/(2Piθ)
 
I need to sketch a parabola for the area of a rectangle,
where p = 2l + 2w, and
a = w*l.

length = x
I needed to find the width, and by rearranging the perimet formula i got

width=1/2p-x

By subsituting w = 1/2p - x and l = x into a = w*l, i get...
Area = x(1/2p-x)

I need to sketch this as a parabola with A as the y axis and X as the x axis.

I know how to do it when I know what the perimeter is but I can't do it otherwise. Sorry for the long confusing explanation I've given you but if anyone can help it would be great.
 
Not sure if your rearranged width equation (with respect to length and perimeter) is correct.

p = 2l + 2w
2w = p - 2l
w = p/2 - l = p/2 - l (where x = l)

In terms of the question, by definition, the area of a rectangle is the function of two variables. You can't find area in terms of width only (unless the rectangle is a square).
 
-TAZZA- said:
I just changed p/2 to 1/2*p just to make it read a bit better. Does this change the whole equation completely?
Click to expand...
That's correct, I thought you had rearranged it to 1/(2p) rather than (1/2)p.

I edited the post above to share my thoughts on the actual problem.
 
Yes, I'm still a bit confused as to how I actually sketch the graph without knowing the value of x, A, or P.

I did a quick google search and found this.

2(L+W) = P
L + W = P/2
W = P/2 - L

Then the area will be

LW = L(P/2 - L)
A = (P/2)L - L^2

Although I'm not really sure if that helps me sketch it. Its a bit hard to explain this question over the internet, I'll probably need to ask my teacher to explain it visually.
 
But again it's the same issue. You have an equation with two unkowns.

You could sketch it without numerical figures though:

A = (P/2)L - L^2 = -L(L - P/2)

Therefore intercepts on the horizontal axis at L = 0 and L = P/2.

Then you just need the turning point:

A = (P/2)L - L^2

dA/dL = P/2 - 2L = 0

L = P/4

Then:

A = (P/2)L - L^2

A = (P/2)(P/4) - (P/4)^2 = P^2/8 - P^2/16 = P^2/16

Therefore the turning point at (P/4, P^2/16)

You could draw it like that.
 
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