CPhill

5.

Since 60 do drama and sports....then  85 - 60  = 25   are in drama alone

And 200 - 60  =  140   are in sports alone

So...the total  number of students that do  drama and sports  =    25  + 60 + 140  =  225

So  there are    330  - 225  =   105 who don't participate in drama or sports

So....the probability  is       105  / 330   =   7/22  ≈   31.8%

1.

Let the number  of voters  = V

Since we have 360°  in a circle, we have that

The number of   women voters  =   (360 - 130 )  / 360 V  =   (230 / 360) V  =   (23 / 36) V

And the number of  men  voters  is  (130 / 360) V  = (13/36) v

And we know that   (23/36) V  - (13/36)V  = 2880   simplify

(10/36)V  = 2800

(5/18)V  = 280      multiply both sides by  18/5

V  = 2800 ( 18/5)   = 10080 voters

The  volume of a regualr tetrahedron with edge length " a "   is given  by

a^3 /  √72

So...the volume of  the tetrahedron  is

1^3  /√72   = 

1  / √72   [units^3]

So

V^2  =    1 /72   

Not as hard as it looks....!!!!

Note that  25^x  - 9^x   can be written as   (5^x)^2  - (3^x)^2

And this will factor as a difference of squares as

( 5^x - 3^x)  (5^x  + 3^x)  

And since this equals 18  and (5^x - 3^x)  = 3

Then  (5^x + 3^x)   must  = 6

If AC  = AB ....then triangle  ABC is isosceles....with angles ABC  and ACB  being equal

And the measure of angle ABC  = (180 - 42) / 2   =  69°

And since angle ABC  is an inscribed angle, then minor arc AC  will  have twice the measure of angle ABC  =  2 * 69  = 138°

And since ED is a tangent to the circle at  C, then, by Euclid, angle ACD will have half the measure of minor arc AC  = (1/2)138  =   69°

If we  unwrapped the label, we would have a rectangle  whose dimensions woud be :

6  *   circumference of the can  = area of the label

So....we have that

6 * 2*pi*r  =  60 pi    ....   divide out  pi

12r  =  60      divide both sides by 12

r  = 5 in =  radius of the can

Let   M  be the midpoint of   AQ  and N be the midpoint of BQ

Thus....M and N will be the centers of  both of the smaller semi circles

Connect  M and N

By Euclid, a segment connecting the centers of two tangent circles or semi-circles will pass through the point of tangency

So  MQ   will be the radius  of the semicircle with the diameter AQ  = 14

So MQ  = 7

And call the radius of the smallest semi-circle, r 

So....MQN  will form  right triangle with  leg  MQ  = 7, leg  QN  = 14 - r  and

hypotenuse MN = ( 7 + r )

So we have that, by the Pythagorean Theorem

7^2  +  (14 - r)^2 = (7 + r)^2

49 + r^2 - 28r + 196  = r^2 + 14r + 49

196 = 42r

196/42  = r

14/3   = r = the radius of the smaller semi-circle

Call the radius  of the semi-circle and the circle, r

Call the side of the  small square, s

So...using the Pythagorean Theorem, we have that

(1/2)s^2  + s^2  = r^2

(1/2)s^2 + s^2  = r^2

(5/4)s^2  = r^2

√5/2 s  = r

s  =   2r / √5

So...the area of the small square = s^2  =   (4/5)r^2

And..in the larger circle....the side of the square is √2r

So....the area of the square in the larger circle is just    2r^2

So....the  area of the smaller square's area to  the larger square's   area is

(4/5)r^2  / 2 r^2   =

4/10  = 

2/5  of the large square's area

Call the side of the equilateral triangle , S

Angle ABC  = 60°

Angle DBC  = 45°

And angle  ABD  =  ABC  - DBC  =  60 - 45  = 15°

The area of triangle  CDB  =

(1/2) * S * BD * sin 45

And the area of triangle ADB =

(1/2) * S * BD * sin 15

So...the area  of  triangle ADB  to the area of triangle CDB  =

(1/2) * S * BD * sin 15           sin  15

_________________   =     ______

(1/2 * S * BD  * sin 45          sin  45

sin 15   = sin (45 - 30)  = 

sin45 cos30  - sin 30 cos 45  =

√2/2 * √3/2    -  (1/2)(√2/2) =

[ √6  - √2 ] / 4

And sin 45  = √2/2

So    sin 15  / sin 45   =

[ √6  - √2 ] / 4   *  2 /  √2   =

[ √6  - √2 ]  / [ 2√2 ] =

√6 / [ 2√2 ]  - √2 / [ 2√2] = 

√3 / 2  -  1/ 2  =

(√3  - 1 )

______

     2

Here's the third one

Orient the paper so that 22 is the width and 39 is the height

Note that  the number of retangles of (2 * 3)  that can  be cut from this sheet is

(39/3)  * ( 22/2)  = 

13  * 11  =

143

But each of the rectangles  can  be divided into two triangles with bases of 2 and heights of 3

So....the total number of triangles  is  143 * 2  =  286