Here is some thoughts on this:
cos 57 = sin (90-57) = sin 33
so cos 57 . cosec 33 = cos 57 / sin 33 = sin 33 / sin33 = 1
2 cos 60 = 2 * 1/2 = 1
so the last 2 parts work out to 0
now we have to find cos 70 / sin 20
sin 20 = cos 70 so this comes to 1
so finally the answer is 1
Yes u have to and as long are you do than you would be right
Standard for would be 6,670,404,013.19

Length of each side of each of these three equilateral triangle is :
Solution is in attachment ~
Answer:
± 
, 
Step-by-step explanation:
See the attached image
This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are:
that correspond to the force of resistance on the mass by the action of the spring and
that is an external force with unknown direction (that does not specify in the enounce).
For determinate
we can use Hooke's Law given by the formula
where
correspond to the elastic constant of the spring and
correspond to the relative displacement of the mass-spring system with respect of his rest state.
We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

Now we apply Newton's 2nd Law and obtaint that...
±
= 




Finally...
± 
We know from the problem that there's not initial displacement and initial velocity, so...
and 
Finally the Initial Value Problem that models the situation describe by the problem is
