ElectricPavlov

Using pythagorean theorem:  c^2 = a^2 + b^2    where c = diagonal and 'a' and 'b' are the sides...BUT 'a' and 'b' are equal because it is a square:

(7 sqrt2)^2 = 2 a^2        Take sqrt of both sides

7sqrt2 = a sqrt2              Divide both sides by sqrt2

7 = a     The sides are 7 inches

10 * sqrt10

I will assume that equation is  y = 3/5x-7     ONE perpindicular would be   - 5/3x-7

Works for me ..    .0191

ARCsin 1.1   will NOT WORK because sine only goes to +1 ---   -1

I'm NOT sure either BUT if you put Rachel in ANY car there are 11 people left who COULD be in the car with her and ONE of them is Tim for a 1/11 chance he will be in the same car.....but I do not know if this is correct.

What kind of 'compound interest' ???   Continous is what I will assume:

2000e^(.025) = 2050.63 at the beginning of year two

2050.63 e^(r*2) = 2124.46

e^(2r) = 2124.46/2050.63

ln (e^(2r) = ln (2124.46/2050.63)

2r=,03537

r= 1.768 % for years 2 and 3    CONTINOUS COMPOUNDING EXAMPLE

I agree with Guest1 ,....here is the calculation:

FV = PV e^(.034n) = 1500 e^(.034 x 30)= 4159.79

1^7  = 1x1x1x1x1x1x1 = 1

Volume of a sphere = 4/3 pi r^3    Now lets make r=2r  (doubling the radius of the baloon

     volume = 4/3 pi (2r)^3 = 4/3 pi 8 r^3

Now the second volume to the original volume ===   (4/3 pi 8 r^3)  / (4/3 pi r^3) = 8/1 = 8 times more volume!

...as a start, from the FIRST equation given  t2 cos 45 = 1000- t1 cos30    Substitute this into the SECOND equation given:

(1000- t1 cos30) - t1 cos 60 = 0     NOW, you should be able to solve this for t1......Right?

      Then you can use this value of t1 in the FIRST or SECOND equations to find t2    ... OK ???