CPhill

Your hypothesis is true only if this triangle is either equilateral or isosceles [ with angle K = N ]....   X would lie on the angle bisector of angle S........

1.   2500(1 + .04)^2   = $2704

2.  2500(1.05)^N  = 3000     divide both sides by 2500

1.05^N = 6/5       take the log of both sides

log (1.05 )^N   = log(6/5)     and we can write

N * log (1.05)  = log(6/5)      divide both sides by log (1.05)

N =  log(6/5) / log(1.05)   =  about 3.74  years

3.  2500(1 + r)^3  = 2755      divide both sides by 2500

(1 + r)^3  =  2755/2500          take the cube root of both sides

1 + r =  1.033    subtract 1 from both sides

r = .033  =  about  3.3%

x+4y=-22     (1)

-2x-2y=14    →   -x - y  = 7   (2)

Add (1) and (2)   and we get

x+4y=-22

-x - y  = 7

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     3y = - 15        divide both sides by 3     and y = -5

And using (2) .......  -x - y = 7  ......   -x - (- 5 ) = 7   ......-x + 5 = 7    ........ -x  =  2  ..........x  =  -2

Infinity is a concept that doesn't obey "normal" math operations

Infinity^3  * 2   =   infinity     ..... suprisingly....  !!!!

Based on the sample, 75%  of any number of students would have the same preference .....so....

125 * .75  = 93.75  = about 94 students

tan^-1 (-3)   =  -71.565051177078°

So

sin (-71.565051177078°)  =  -0.948683298051

V = [pi/3] * 10^2  * 9  =  942.48 cm^3

RPM   =  Pages read / number of minutes  ....so we have

[ 8 + 1/3]  / 5   =

[25/3] / [5 / 1]   =

25/3  *  1/5   =

25/5 * 1/3  =

5 * 1/3   =

5/3   = 1 + 2/3  pages per minute

V  = [ pi/3 ] * r^2 * h  =    [pi/3] * 1.2^2 * 5   = about 7.54 units^3

This problem comes from one of the greatest American puzzleists of all time.....Sam Lloyd.....it seems more difficult than it really is......

Let W = the river's width

When they first meet, they have traveled the same amout of time.....one boat has traveled 720 yards and the other has traveled W-720 yards......let the rate of the first boat = R1  and the rate of the second boat = R2  ....and we have that.....

T1     =   T2    ....or......

[720]/R1  = [W - 720]/R2   →   R2/ R1   = [W- 720] / [720]    (1)

Wnen they next meet, they have again traveled the same amount of time......The first boat has traveled the width of the river, (W), plus 400 yards and the second boat has traveled 2W - 400....again  we have that

T3   = T4   ....or.....

[W +  400] /R1  = [ 2W - 400] / R2  →  R2/R1  = [2W - 400] / [W + 400]  (2)

Equating (1) with (2), we have

 [ W - 720] / 720=  [2W - 400] [W + 400]

720[2W - 400]  = [W - 720] [W + 400]

1440W - 288000 = W^2 - 320W - 288000

W^2 - 1760W  = 0

W(W - 1760)  = 0

Reject W = 0......and the solution  is.....W = 1760 yards.....exactly 1 mile wide  !!!!