CPhill

Not possible......to see why let x =  a/b  and let y  = c/d   where a,b,c,d   are integers

Then we have that

(a/b)^2   =  2(c/d)^2   rearrange as

(a/b)^2 /(c/d)^2   = 2

(a^2 *d^2)  /  (b^2 *c^2)   = 2         take the square root of both sides

(a * d) / (b * c)  =  √2

But  a*d  , b*c     are both integers.......let a*d  = m  and b*c  = n

m / n  = √2

But......the square root of two is irrational....so.....we're led to a false conclusion.....thus, x and y  are not both rational

We have that

S  = first term ( 1 - r^n) / (1 - r)  ......so.....

93   =  3  (1 - 2^n) ( 1 - 2)         divide both sides by 3

31  = (1- 2^n) / -1      multiply both sides by -1

-31  =  (1 - 2^n)        multiply both sides by -1 again

31  =  2^n  - 1      add 1 to both sides

32  = 2^n         →    n  = 5   terms

We have that the total vertical distance traveled  is :

              8

35 + 70 Σ .8^n     =   268.024 m  =  268 m 

              1

Oops.....thanks, hectictar, for catching my error....!!!

The lower limit of what???

What quadratic  ????

We are using The Law of Sines here

sinx / 38   = sin35 / 22      multiply both sides by 38

sin x   =   (38/22)sin35

Take the sine inverse

arcsin [   (38/22)sin35 ]   = x  ≈  82.2°

For the second one, we have

g / sin (74)  =  54/sin (58)      multiply both sides by sin(74)

g  = sin(74)* 58 / sin (58)   ≈  65.7

Let's go from here :

110  =  73(1.0235)^x     divide both sides by 73

110/ 73   = (1.0235)^x      take the log of both sides

log (110 / 73)  =  log (1.0235)^x

And, by a log property,   log(a)^x  =   x *log(a)...so.....the right side becomes

log (110 / 73)  =  x * log(1.0235)       divide both sides by log(1.0235)

log (110 / 73)   / log (1.0235)   =  x  ≈  17.65  years

So.....2006 + 17.65   =   2024  [rounded to the closest year ]

If we waned to stretch things.......based on the pattern, the nth term is given by :

1 /  [  (2n+ 1)(-1)ⁿ ]

Both of these can be solved by manipulating the Law of Cosines  as follows :

arccos [ (a^2 - b^2 - c^2) / [ -2(b)(c)] = A  

Where  "a" is the side opposite the angle we are trying to find [Angle "A"] and "b" and "c" are the other two sides

For the first, we have that

arccos  [ ( 4^2 - 7^2 - 7^2) / [ -2(7)(7)]   = A  ≈  33.2°  = 33°

For the secomd, we have

arccos  [ ( 6^2 - 7^2 - 10^2) / [ -2(7)(10)]   = E  ≈  36.18°   =  36° 

Based on these....see if you can do the last one....if you get stuck.....let us know.....