This weeks challenge

Wow, there seem to be some math genius' in here. I really struggled with math back then. I remember some assignments that I did not understand when in 9th grade. I have now found my old math book, and I would like to see if anyone here can solve them. Maybe you Omar

Opg. 11 Naturfag 2.g

The number of horses in a country decreased after 1945 "eksponentially" (Don't know how to spell it ) with the prefix f(x) = 1.3 mill. * 0.92(lifted x)
a) How many horses disappeared pr year?
b) How many horses were there in 1950
c) In the period after 1950 the number of horses decreased wíth 7% pr year, how many horses were there in 1980?

Opg. 23 Naturfag 2.g

An eksponentially growing function "f" is given by f(x) = 12,5 * 1,17(lifted x)
a) By how many percent is f(x) increased, when x gets increased with 4
b) What growth shall x have, if f(x) is increasing 60%


Be my guest

 

The trickle-down effect of Affirmative Action. Back in the 60's you had to qualify to take honors.

 

a)
f(x)=1300000*(.92)^x
f '(x) also known as the rate of change per unit time (year)=1300000*(.92)^x*ln(.92)
b)
x=5 (5 years since 1945)
f(5)=1300000*(.92)^5
c)
A decrease of 7% means every year, there was only .93 of the last year (100% - 7%=93% or .93)

Thus:

Population in Year N+1 = (Population in year N) * (.93)

So say in 1952, it would be this: Pop in '51 (.93), where the pop in 51 was the pop of 50(.93). Thus overal the pop in 52 would be the pop in 50 times .93^2. It appears the exponent is the same as the number of years since 50 (ie 52-50 = 2)

Thus in 30 years (50 to 80) you will have only to plug in everything:
Population in Year 30 = (year 50 population) * (.93)^30
=(1300000*(.92)^5) * (.93)^30

I do not have a calculator and cant do the numbers out, but you get the idea.

 

a)
Normally (we will call it fx):
fx=12.5 * 1.17^x
After the change (we will call it fx2):
fx2=12.5 * 1.17^(x+4)

Now using the rule for exponents (add expo when they are multiplied and have same base) you know the following:

fx2=12.5 * (1.17^x) * (1.17)^4

Now for precent increase you ask 'what precent of the original is the new number' which is:
precent*fx=fx2
precent=fx2/fx

=[12.5 * (1.17^x) * (1.17)^4] / [12.5 * 1.17^x]
=(1.17)^4


b)12.5*1.60 (a 60% increase in size) is what f(x) equals, thus:
12.5*1.60 = 12,5 * 1,176^x
1.60=1,176^x

To solve, rearrange:
log base 1.176 (1.6) = x

Use one of the law of logs (dont remember which one, but it says the log base any number of A is the same as lnA/ln(any number). ie log10(35) = ln(35)/ln(10)

Thus: ln(1.6)/ln(1.176) = x

Your answer is ln(1.6)/ln(1.176)

 

omar, are u a math major.

 

I used to be, but to graduate as one would have meant another year at Hopkins. I still like math though, I taught myself calculus when I was in 10th grade in HS.

 

Omar, you are pretty good at that

I understood it when I came to college, and never back in 9th grade. I actually think the standard is set a little to high, but maybe that's because I never was into math Hated it.

 

WoW. There was a kid in my class last year (10th grade) who was taking math courses at St. Joes Univ. He taught himself calc in 9th grade. His name was Omar as well, hmmmm.

 

Awesome.

Is your school a private school? If they need a substitute that would work for free, I can do it in January.

 

yea its private, we dont do substitutes, if a teacher is absent u just have a procter and u can do whatever u want. Its all guys as well, so u have really nothing to look at during the day.

 

Ouch! Im so sorry!

 

its better than u think, i get better grades, i roll out of bed in the morning no shower no brushing of hair put on whatever is on the floor, their are plenty of all girls schools around and they all like us b/c we are the rich hot guys. You can fart in class and be congratulated. umm u have long discussions about sports and porn. And we are the champions in every sport except baseball. Heck, we have won the Football title 3 years in a row, and we are on our way to the 4th one this year. so yea dont be sorry.

 

Omar, I think something may be wrong in the above, although I don't have the time to read through it thoroughly (I'm writing this after 2 am Singapore time).

Very simple problem, I did most of it in my head, then scribbled on paper for the final step. Goes like this :

The question :

Equations are :
x^2 = 7x + 3y ----(1)
y^2 = 3x + 7y ----(2)

Take (1) - (2) :

x^2 - y^2 = 4(x-y)
LHS (left hand side) factorises easily giving :

(x+y)(x-y) = 4(x-y)

Rearranging,

(x+y-4)(x-y) = 0

So either x-y = 0 --- (A) OR x+y-4 = 0 --- (B) will yield valid solutions.

Consider (A) first :

x = y

Substitute into eqn (1) :

x^2 = 7x + 3x ----(1)

x^2 - 10x = 0

therefore x = 0 or x = 10

The respective values for y will be 0 and 10.

So we get the solutions (0,0) and (10,10)

Putting (A) into eqn (2) will give the same results for y (0 and 10) and the same solution pairs as the above.

Now consider (B) :

x + y = 4

x = 4 - y

Put that into (2) :

y^2 = 3x + 7y ----(2)

y^2 = 3(4-y) + 7y

Rearranging, etc. :

y^2 - 4y - 12 = 0

(y-6)(y+2) = 0

Hence y = 6 or y = -2.

The respective values for x (= 4-y) are -2 and 6.

So we get the solution pairs (-2,6) and (6,-2).

Again, putting (B) into (1) instead will give the same pairs of solutions.

So the final solution set in terms of ordered pairs (x,y) is :

(0,0) ; (10,10) ; (-2,6) ; (6, -2)

and that's the required answer.

 

eh, I still liked it when I had something to look at in all my boring classes.

 

And before I turn in tonight, the fastest way to do this :

is this, IMHO :

2K-2 = 4M

Halve the whole thing :

K-1 = 2M

Rearrange,

K-M = M + 1

Since you're given that (K-M), which is the same as the left hand side of the above, is equal to 502,

M + 1 = 502

M = 501 and

4M = 2004.

I thought this way was fast and simple, because I managed the whole thing in my head in around 7 seconds.

 

Doc, I am going to go through your problem 3..looks right to me! I figured there was an easier way I wasnt thinking of.
I didnt show the work for speed, I illustrated each step so they knew what I was doing. I could do all of them, save 3 in my head.

 

Sorry, but your answer is INCORRECT.

If X=4 (a square), the answer should be "pi." Your
formula gets some weird result. You even got the
units wrong. (X/pi and pi do not have same units)

 

oops, I flipped the first the wrong way...it should be 360/X or 2pi/X for the first angle. Not the other way around.

a1=360/X
a2 = 1/2(180 - a1) = 1/2(180 - 360/X)

Note that a3 = 180 - 2(a2) =180 - (180 - 360/X) = 0 + 360/X or just 360/X.

Thus the angle we want is simply 180 - 2(360/X) or 2(90-360/X).

BTW, for n = 4, the answer would NOT be pi, as the 2 lines will never meet..and thus be zero.
Image Unavailable, Please Login

 

OK, wide awake now.

For question 2, Omar, you've posted the right expression in your latest post. This is also very easy, this is how I did it :

(Sorry to be a stickler, but I would prefer that the question said : Let ABCDE.... be a regular, plane, convex polygon for rigor. I would also prefer that some stipulation be placed on the number of sides [e.g. it should be at least a 5-gon, but really, the formula will work for 3 and 4-gons if interpreted correctly, so it's not such an issue]).

Anyway, making that assumption and moving on,

Let the polygon be x-sided. Produce side AB indefinitely beyond the polygon (sorry, no figures). First note that the acute angle subtended between the produced line AB and BC is equal to 360/x in degrees. This is easy to see, since you can think of a vector formed by the produced line AB that needs to be rotated x times to bring it back in line with itself. This is also a fast way to "see" the internal angle sum formula for polygons.

If the same is done for DC (produce it so that it intersects the produced line AB), then the analogous angle there will also be 360/x. Let the intersection point b/w the produced lines be called O. We now have a triangle BOC with 2 base angles equal to 360/x each.

Since angle sum of triangle is 180 degrees, the apical angle is :

angle BOC = 180 - 2*(360/x) = 180 - 720/x = 180*[1-(4/x)] in degrees.

which is the required answer. For radians, replace the 180 with pi.

Looks longwinded when you write it down but really a mental problem.

(Oh, yeah, for a square, the answer is 0, as Omar rightly pointed out (The lines in the case of a square are parallel and "meet at infinity" in an infinitesimally small angle (zero). For a triangle the answer is negative 60 degrees, because the lines diverge after they're produced and meet on the "other side" of the figure, so the answer should be expected to be negative. No answer exists for x <= 2).

 

Good point - it's all about communication anyway.

I was class of '80 Oakton - back when we went to class inside the school and not in trailers like it appears now (damn, I'm old). My wife went to Madison - hence the dig. Glad we moved to North Carolina where the yung-uns kan git a gooder Edukashun. Go Cougars. (Did you at least beat Madison in football this year?)

 

Just for completeness, since I've done the rest, I thought I'd answer this (even though Omar has posted the correct solution).

Rearrange,

(a-b)(x-y) = 0

Since (a-b) is nonzero by the conditions of the question, (x-y) = 0

Therefore x = y.

 

You are WRONG again!

 

On this point, he is surely right, although the answer of zero cannot be demonstrated in reality since parallel lines "never" meet.