CPhill

Excellent, Hugo!!!

A "Best Answer"

OK.....no prob......this one is confusing   !!!!

Number  of integers divisible  by  5  from 1 to 1000 =  1000 / 5  = 200

Number of integers  divisible by 7  from 1 to 1000 =   integer part (1000/7)  =   142

But we  must subtract out  those integers  which are multiples of 5 and 7, ie., all the  multiples of 35

So   integer part (1000/35)  =  28

So  ....number of integers that are multiples of 5 or 7  =  200 + 142  - 28  =   314

Consider this, hugo

[ Remember  that tangents drawn to a circle  from an exterior point have equal lemgths ]

If we draw  radii  to the  midpoints  of AD . DC, AB,BC  we will create  congruent quadrilaterals

Therefore....the midpoint  of BC to B  = 4

And this is a tangent

And the midpoint of AB to B  is  a tangent with the same length

But  since we have congruent quadrilaterals then the tangents drawn from D  to the circle will have the same length as those drawn from B to the circle =  4

And the tangents drawn from A  have the same length  = 3

And the tangents drawn from C have the same length = 5

So...the perimeter  is just  2 [ 4 + 4 + 5 + 3]  =  32 cm

Yeah....that's a little confusing

Note  that if angle B  = angle D, than the tangents drawn from D  have  the same length as the tangents drawn from B

The tangents from B both have the length of 4 cm

So....the two tangents drawn from D have the same length

And the tangents drawn from A  = 3cm

And the tangents drawn from C  = 5cm

So

[ tangents drawn from B + tangents drawn from D  + tangents drawn from A  + tangents drawn from C] =

[  (4 + 4)   + (4 + 4)  + (3 + 3)  +(5 + 5) ]  =

[ 8  +  8   +  6  +  10]  =

32 cm

V  =  product of the dimensions shown =   (36 * 12 * 15)  = 6480 in^3

Perimeter  = 2 (3 + 5 + 4 + 4)  = 2 ( 16)  =  32 cm

r  =   sqrt [ (4√3)^2 + (-4)^2 ]  =  sqrt [ 48 + 16]  = sqrt (64)  =  8

θ   =  arctan (-4 / [4sqrt(3)]  =  arctan ( - 1/sqrt (3) )  =  -pi/6  = 11pi/6

So

( r, θ)  =  (8, 11pi/6)

Each tangent length has a counterpart with the same length

So....the perimeter  = 2  (3 + 4 + 4 + 2)  =  2 ( 13)  =  26 cm

[We  used the fact that  tangents drawn to a circle from an exterior point have the same length ]

See the following image :

Using some trig

tan CAB = 3/2

tan CAB  = S / AD          where S  = side of the square...let AD  = x

3/2 = S/x

x = (2/3)S

And

tan CBA  =  2/3

tan CBA   = S/ (BC - (S + x))                  BC = sqrt (13)

(2/3)  = S / (sqrt (13) - (S + (2/3)S)            sub in for x

(2/3)  = S / ( sqrt (13) - (5/3)S)

(2/3) (sqrt (13) -(5/3)S)  =  S

(2/3)sqrt (13) - (10/9)S  = S      multiply through by 9

6sqrt (13) - 10S  = 9S

6sqrt (13)  =  19S

6sqrt (13) / 19   =   S