Answer1:an international agreement, usually regarding routine administrative matters not warranting a formal treaty, made by the executive branch of the US government without ratification by the Senate Answer2: An executive agreement is an agreement between the heads of government of two or more nations that has not been ratified by the legislature as treaties are ratified. Executive agreements are considered politically binding to distinguish them from treaties which are legally binding
As the payroll manager the specific items that relates to vacation leave are:
- Employees getting their leave after they have been in the organization for up to 5 years.
- The duration of the vacation, this should be about three weeks.
- The schedule of vacation time.
- Compliance with labor standards and employment standards.
<h3>The elements related to vacation pay</h3>
- The pay for vacation should be equivalent of 4 percent of salary.
- This amount is a lump sum that has to be paid before the period.
- The employer has to receive gross pay as well as commissions and bonuses.
- The payment should be through the use of checks and all payment must be in compliance to employment standards.
Read more on vacation here:
brainly.com/question/96862
Answer:
Concentric circles are circles with a common center. The region between two concentric circles of different radii is called an annulus. Any two circles can be made concentric by inversion by picking the inversion center as one of the limiting points.
1. Picking any two points on the outer circle and connecting them gives 1/3.
2. Picking any random point on a diagonal and then picking the chord that perpendicularly bisects it gives 1/2.
3. Picking any point on the large circle, drawing a line to the center, and then drawing the perpendicularly bisected chord gives 1/4.
So some care is obviously needed in specifying what is meant by "random" in this problem.
Given an arbitrary chord BB^' to the larger of two concentric circles centered on O, the distance between inner and outer intersections is equal on both sides (AB=A^'B^'). To prove this, take the perpendicular to BB^' passing through O and crossing at P. By symmetry, it must be true that PA and PA^' are equal. Similarly, PB and PB^' must be equal. Therefore, PB-PA=AB equals PB^'-PA^'=A^'B^'. Incidentally, this is also true for homeoids, but the proof is nontrivial.
