CPhill

Very nice, Rom....I like this problem.....!!!!

s    and r  are both perpendicular to t

So....they must be parallel

So

s ll  r

In the form   f(x) = Acos (Bx) + D

B =   pi  = the number of periods in 2pi

So

B = 2pi / period

pi  =  2pi / period        rearrange  as

period =  2pi / pi    =  2

And the frequency  =   1 /period  =   1/2

Equation of midline  ...  y =  [ max value of function + min value of function ]  / 2

Note that the max value of the cosine = 1    and the min value  = -1

So....max value of function =  3(1) - 2.5  =   .5

And the min value of the function is  3(-1) - 2.5 = -5.5

So the equation of the midline is

y =   [ .5  +    -5.5] / 2   =  [ -5  ] / 2   =   - 2.5

See the graph here :  https://www.desmos.com/calculator/lesls9edag

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We have the form

g(x)  = Asin (x) + D

A = the vertical stretch = 2

D = the shift up

So...the equation is

g(x) = 2sin (x) + 3

We have the form

y = A sin (Bx + C ) + D

A = the amplitude =  4

B =  the number of periods in 2pi....so    ...B =   2pi / period  =  2pi/pi   = 2

C is the phase shift....there is none....so....C = 0

D = the shift up / down   =  -3

So.....the equation is

y = 4sin (2x) - 3

Here is the graph :  https://www.desmos.com/calculator/67jvpq3xip

Thanks, Mclovin.....here's another way...

5x ^2 - 3x = 25     divide through by 5

x^2  - (3/5)x  =  5         

Take (1/2) of (3/5) = 3/10....square it = 9/100.....add to both sides

x^2 - (3/5)x + 9/100 =  5 + 9/100      factor the right side

(x - 3/10)^2 =   509/100    take both roots

x - 3/10 =  ±√ [509/100]

x - 3/10 = ±√(509) /10     add 3/10 to both sides

x =     [ 3 ±√509 ] / 10

Very nice, Melody!!!!

I didn't see this one, at all....!!!!

Not possible....the integral does not converge....[ it is undefined at x = 0 ]