Friday

October 31, 2014

October 31, 2014

Posted by **anna** on Sunday, November 25, 2012 at 5:36pm.

720 = 11.895+2.545sin [2pi/366(t-80.5)]

- trig -
**Steve**, Sunday, November 25, 2012 at 8:32pmsame way you would solve for any equation. collect t stuff on one side, and work your way inside out the expressions till t ia all alone:

708.105 = 2.545sin [2pi/366(t-80.5)]

278.2338 = sin [2pi/366(t-80.5)]

at this point you are stuck. sin(x) is never more than 1, so there is no way it can be 278.2338

I suspect a typo somewhere.

- trig -
**Reiny**, Sunday, November 25, 2012 at 8:35pmIt is intuitively obvious that this equation cannot have a solution the way it was written

the sin(anything) has to be a number between -1 and +1

so if 720 = 11.895 + x , the value of x cannot possible fall between - 1 and +1

check your typing, or if correctly typed, there is no solution

- trig -
**anna**, Monday, November 26, 2012 at 2:51ami think i figured out my mistake. it should be

720 = 713.7+152.7 sin [(2pi/366)(t-80.75)]

this problem is modelling minutes of daylight in a town using the form y = d+a sin[b(t-c)]. let y equal 720 minutes of daylight and let t=the number of days in 2012

**Answer this Question**

**Related Questions**

Math - How would I solve for t if y equals 720 minutes of daylight and t=the ...

trig - determine the solutions for tanxcos^2x-tanx=0 in the interval XE[-2pi,2pi...

bobpursly, check please,math - do you mean like this: 720/(V-30) + 720/(V+30) = ...

graphing trig functions - given y=(3x+2pi) how would i graph one period this ...

trig - Solve cos x-1 = sin^2 x Find all solutions on the interval [0,2pi) a. x=...

Trig - 2sin^2(x) = 2 + cos(x) interval [0,2pi) How do I solve this? Help would ...

Trig help - 1. What is the period of the function? y = 4 cos pi x A. 1 B. 2 C. ...

Trig - Solve:cos^2x=1 for 0 < x is less than or equal to 2pi

Trig - Solve for x in the interval [0,2pi) sin^2x+2cosx=2

help,math - can some set the equation up for me. Science and medicine. A plane ...