I assume you mean that A moves directly toward B at 14 cm/s and B moves directly toward A at 25 cm/s. Then the distance between them is reducing at 15+ 25= 40 cm/s. The time required cross 480 cm at 40 cm/s is 480/40= 12 seconds.

! If an exterior angle is x degrees then the corresponding exterior angle is x- 180 degrees. So if the three angles are "x", "y", and "z" degrees, the corresponding interior angles are 180- x, 180- y, and 180- z. The sum of those is 180- x+ 180- y+ 180- z= 3(180)- (x+ y+ z) and since x+ y+ z= 200, that is 3(180)- 200= 540- 200= 340 degrees.5) The sum of 3 of the exterior angles of a 7-sided polygon is 200 deg. Find

i) the sum of the three interior angles corresponding to the three exterior angles

Given any polygon of n sides, draw a line from one vertex to each of the remaining vertices. That will divide the polygon into n-2 triangles (the "-2" is because the lines connecting the vertex to the two adjacent vertices are already sides of the polygon). Since a the angles in a triangle sum to 180 degrees, the angle in a polygon with n sides sum to 180(n- 2). Here n= 7 so the sum of all angles is 180(5)= 900 degrees. The first three angles sum to 340 degrees so the other four sum to 900- 340= ?ii) The sum of the remaining four interior angles

A variation of the "chinese remainder theorem! However, I have a question for this. When you say "there is one pupil short" do you mean that marching by 3s the last row has only two pupils, when marching by 5s there are 4 pupils in the last row, and when marching by 7s there are 6 pupils in the last row, or do you mean that there is always one pupil left over- i.e. only one pupil in the last row for all these? I would call that "one pupil extra" or "one pupil left over".6) A group of pupils is practising their march past.

If they march in pairs, one pupil is without a partner.

If they march in 3s, 5s, or 7s, there is one pupil short.

Calculate the smallest possible number of pupils in this group.

Thankssss!