Prove: every point of s=(0, 1) is an interior point of s.

Proved

Step-by-step explanation:

To prove that every point in the open interval (0,1) is an interior point of S

This we can prove by contradiction method.

Let, if possible c be a point in the interval which is not an interior point.

Then c has a neighbourhood which contains atleast one point not in (0,1)

Let d be the point which is in neighbourhood of c but not in S(0,1)

Then the points between c and d would be either in (0,1) or not in (0,1)

If out of all points say d1,d2 we find that dn is a point which is in (0,1) and dn+1 is not in (0,1) however large n is.

Then we find that dn is a boundary point of S

But since S is an open interval there is no boundary point hence we get a contradiction. Our assumption was wrong.

Every point of S=(0, 1) is an interior point of S.

p ÷ r = tan(x)

step-by-step explanation:

given that there is a figure below shows a right triangle: a right triangle is shown with hypotenuse equal to q units and the length of the legs equal to p units and r units.

graph is missing so i will create the graph as per given information.

the angle between the legs having lengths p units and q units is y degrees. the angle between the legs having lengths r units and q units is x degrees.

now we need to find about what is p ÷ r equal to.

so we can use trigonometric ratio tan(x)=[altitude]/[base] to find the value of p ÷ r, because p is the altitude and r is the base.

p ÷ r=tan(x)

hence final answer is p ÷ r=tan(x).

answer: im gonna go with milli

step-by-step explanation: