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Could you explain how you got the formula for the constant term?

ElectricPavlov:

Good effort ElectrcPavlov but no cigar!!. Read the following carefully and try to undersrand it.

1) Since each child is to receive the same amount when they turn 21, then will have to find the PV of $1 in each case:

PV=FV [1 + R ]^-N

In the case of the 17-year old we have:

PV=1[1 + .08/4 ]^-(4*4)-He/she has 4 more years or 16 quartes to reach 21.

PV=.728445813714 

In the case of the 12-year old we have:

PV=1 [1 + .08/4 ]^-(9*4)--He/she has 9 more years or 36 qurters to reach 21

PV=.490223150352

Now, will add these two PV's together and will get:

=.728445813714 + .490223150352

=1.21866896406. Next, will divide $10,000 by this figure, which is what each child will receive:

$10,0000 / 1.21866896406

=$8,205.67--Which is the the amount that each child will receive when they turn 21.

2) Check:

The $10,000 will grow for 4 more years until the 17-year old turns 21. At the time, it will be worth:

=13,727.86. Then will subtract his/her inheritance of $8,205.67 from this amount and the balance is:

$13,727.86 - $8,205.67=$5,522.19.

This balance of $5,522.19 will continue to grow @ 8% comp. quarterly for another 5 years or 20 quarters until the 12-year old, who is now 16, turns 21.

FV=$5,522.19[1.02]^20

=$8,205.67. And that is what the 12-year old will receive at 21, which is the same amount that the other child received. And that is the end of this saga!!!!.

Of course not.

This question has been answered here:

http://web2.0calc.com/questions/we-have-10-standard-6-sided-dice-all-different-colors-in-how-many-ways-can-we-roll-them-to-get-a-sum-of-20#r7

Just copy and paste in the address bar of your Browser.

We will have to assume that they are both EXACTLY 12 and 17  today (they both had birthdays today)

then we have 10,000 invested at 2% per quarter and that 1/2 of the amount will be distributed in 16 quarters and the remaining will be distributed in 36 quarters and the amounts can be different?????     Well there is a whole bunch of possibilites depending on how it is split when the first kid turns 21 !     

  Wouldn't it be better to split the accumulated amount EQUALLY when the first kid turns 21 and then let the remainder continue to grow until final disbursement to the second child when she/he turns 21 ????     We'll need some more info to proceed.....BUT here is the way I would do it

10000(1+.02)^16 = 13727.86    when the 17 y/o turns 21......give half of the money to them  $6863.93

6863.93   will continue to grow for 20 more quarters before being disbursed to the younger child

6863.93(1.02)^20 = 10199.44   to the younger child when she/he turns 21    

This MAY seem unfair, but it is NOT....the FIRST child could leave the money deposited and get the same amount if she/he waits until the younger turns 21 to receive the money.

These will occur in pairs......

The first will occur at  C ( 4n -2, 2)   and the second will occur at C(4n - 1, 2)

So......there are 25 pairs of these from C(2, 2) to C(100, 2)  =  50 integers

Proof of the first :

[ 4n - 2] ! /   ( [4n - 2 - 2]! * 2! ]  =

[4n -2] ! / [(4n-4)! *2! ]  =

[4n -2] * [4n - 3] / 2 =

[4n -2]/2 * [4n -3]  =

[2n -1] * [4n -3]  ..........this is the product of two odds which always  = an  odd

Proof of the second

[4n - 1]!/[ (4n -1 -2)! * 2!] =

[ 4n -1]! / [(4n -3)! 2!]  =

[4n -1][4n-2] / 2  =

[4n -2] /  2  * [4n -1]  =

[2n -1] [4n -1]   =   odd

Egads !  I do not know that I could figure this one out without a little more info......but I will try....lets put the 250 pound force on the x-axis

Forces vertical =  200 sin x      &   150 sin y      Since the  250 pound will not exert any vertical force because it is strictly horizontal these two must be equal and opposite THEN   200 sin x = - 150 sin y    ... .(EQUATION 1)

Horizontal forces   200 cos x   &  150 cos y    and 250    I wil have to assume that the two smaller horizontal forces are balancing the larger 250 force

 so   200 cos x + 150 cos y = 250    ... . (EQUATION 2)

We now have two equations with two unknowns and should be able to solve for the angles x and y.....

I'm missing some trig rule to figure out these two equations, BUT by successive trial and elimination (i.e interprolation)  I come up with (approximately)  

250 horizontal to the right

200 @ 143.77 degrees

150 @  232     degrees     Sorry.....I got STUCK !

Could you give me an example I could go threw the whole prosses with an example

Basically , if you solve the equation for 'y' it will be in the slope intercept form:  y = mx + b

9x+8y= -72

8y = -9x -72

y = -9/8 x - 9        m = -9/8 = slope   -9 = b = y intercept

Force perpindicular is found by the cosin of 15 degrees

cos15   = adjacent/hypotenuse   = adj/225    Where adj is the perpindicular force (f2)

225 cos 15 = adj = 217.333 = f2

You can use pythagorean theorem to find the friction force or use the sin function....let's use the sin function to find the answer and check it with the Pythag th.

sin 15 = opposite/hypotenuse = opp/225

225 sin 15 = opposite       where opposite is the force of friction (f1) being exerted to keep the box from sliding.

f1 = 58.234 

Use Pythagorean Theorem to check:    58.234^2 + 217.333^2 = 225^2        ????      Yep !

What is the equation that you are trying to solve?

Just use the calculator on this website!

If the slope is 25% then the road must RISE 25 feet for every 100 feet of horizontal distance

tan x = opposite/adjacent = rise/run = slope = 25/100 = .25

Arctan(.25) = x = 14 degrees

Subtract '1' from both sides of the equation to get:

-5x = 20     Now, divide by -5 to get 'x'

x = 20/(-5) = -4

Statics:   Since the object is hanging in equalibrium, all of the vertical forces have to balance(i.e. sum= o) and all of the horizontal forces have to balance.

Vertical forces:

32rope = tcos32

24rope=  t'tcos24

weight = -450          

SO they must balance:  tcos32 + t'cos24 - 450 =0

tcos32 +t'cos24 = 450    (FIRST equation)

Horizontal forces must cancel out too:   (the horizontal force in 32rope must equal horizontal force in 24rope)

32rope = t sin 32

24rope = t' sin 24       These forces are EQUAL (if they were not EQUAL something would MOVE)

tsin32 = t'sin24            we can solve this for t and substitute into the FIRST eqaution for VERTICAL forces to solve

t = t'  sin 24/sin32       (SECOND equation)

Substitute:  (t' sin24/sin32)  * cos32 + t'cos24 = 450

                   t' (.6509 + .9135) = 450       t' = 287.64 lbs   (tension in 24 degree rope)

SECOND equation  t = t' sin 24/sin32    t = 220.77  lbs   (tension in 34 degree rope)

Total temperature change = 16-(-33)= 49 degrees

Time change 2:00 to 9:00      7 hours                           

Average drop per hour     49/7 = 7 degrees/hr         

Thanks very much Alan   

Very nicely done Melody!

For the length use Pythagoras's theorem:  r = sqrt(x^2 + y^2)

For the direction, the angle to the horizontal is given by theta = tan^-1(y/x)

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I'm not sure this is actually simplifying anything, but you could do the following:

\(x=2^{1/3}3^{1/4} \\ x^{12}=432 \\ x=432^{1/12}\)

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No, in general you can't invert the function to get x explicitly in terms of y.  If you have a value for y, then you could numerically find a corresponding value of x depending on the range of valuea of interest.  See the graph below:

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I did this the hard way before I realised that it is really quite easy. (This is the easy version)

Any triangle in a semicircle is a right triangle. In this question the diameter will be AB.

The third point of the triangle is P

So if P is one the circumference then angle APB will be a rtight angle

If the point P lies inside the semicircle then angle APB will be obtuse and 

if the point P lies outside the semicircle then the triangle APB will be accute.

A(0,2)   B(4,0)

\(AB=\sqrt{4^2+(-2)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\\ \mbox{So the radius of the circle is }\sqrt{5}\\ \mbox{The centre of the circle is (2,1)}\)

The area of this semicircle is    

\(area=\frac{1}{2}*\pi*(\sqrt{5})^2\\ area=\frac{1}{2}*\pi*5\\ area=\frac{5\pi}{2}\;\;units ^2\\ \)

Any point P inside this semicircle will meet the criterion.

That is APB will be obtuse.

Now the area of the pentagon

\(Area\;of\; pentagon=4(2\pi+1)-\frac{1}{2}*2*4\\ Area\;of\; pentagon=8\pi+4-4\\ Area\;of\; pentagon=8\pi\\~\\ \mbox{P(APB is obtuse)}=\frac{5\pi}{2}\div8\pi=\frac{5\pi}{2*8\pi}=\frac{5}{16}\)

Here is a diagram to help you understand.

The volume is given  by  [pi * area of the base * height] / 3 =    [ pi *10^2 * 6] / 3   = about 200pi yd^3  = about 628.32 yds ^3