CPhill

1: If the diameter of a right cylindrical can with circular bases is increased by 25%, by what percent should the height be increased in order to double the volume of the original can?

For convenience sake...let the original diameter of the can  =2

So....the original radius  = 1

So...the original volume is

V = pi * (1)^2 * original height

V = pi * original height

original height = V /pi

When the original diameter is increased by 25%, so is the original radius...so the new radius is 1.25

And we have that

2V  = pi * (1.25)^2 * new height

[2V ] / pi * (1.25)^2  =  new height

So

new height

____________   =

original height

[2V] / [ pi *(1.25)^2 ]

_________________      =

   V /pi

      2V                           pi

__________    *     _________

pi * (1.25)^2                V

2 / (1.25)^2  =   

1.28

So....the original height should be increased by [ 1.28  - 1]  =  .28  = 28%    to double the volume

2)Find the distance between the graphs of 5x - 12y + 45 = 0 and 5x - 12y - 46 = 0.

The first line has a y intercept  of 

-12 y = -45

y = 45 / 12

y = 15/4  = 3.75

So....the point  (0, 3.75)  is on the first line

And the distance betwen this point and the second line will be the perpendicular distance between the two lines....so...we have

l 5 (0)  - 12 (3.75) - 46  l             l 91 l                    91

____________________  =    ___________  =   ___   =   7   units

   √[ 5^2 + 12^2]                         √169                    13

1)Find the distance between (3,4) and the line 4x + 3y + 7 = 0

We can use this to find the answer.....the distance between  (a, b)  and the line  AX + BY + C  = 0  is given by

l A(a)  +  B(b)  +  C  l   

_________________

  √ [ A^2 + B^2 ]

So we have

  l  4(3) + 3(4)  l         l  24  l               24

  ___________  =     _______  =      ___  =    4.8 units

  √ [ 4^2 + 3^2]           √25                  5

2sin x   cos x  + √3 cos x  = 0        factor out  cos x

cos x  (2 sinx + √3 )   = 0

Set each factor to 0 and solve for x

cos x  = 0    and this happens at   x = pi/2  and 3pi/2

2sinx + √3  =  0     subtract  √3 from both sides

2 sin x  = -√3       divide both sides by 2

sin x  = -√3 / 2     and this happens at  4pi/3 and  5pi/3

So...the correct answer is

pi/2, 4pi/3, 3pi/2, 5pi/3

This is the sum of an arithmetic series

The first term  is   3(1)  - 1  = 2

The eighth term is 3(8)  - 1 = 23

The sum is

[first term + eigth term]  * number of terms  / 2  =

[ 2 + 23  ] * 8  / 2  =

[25] * 4  =

100

This  will be the sum of a geometric series where the first term is 4

The common ratio  is -5

And we will have 6 terms.....so....the sum is

first term   [ 1  -  common ratio^number of terms ] / [ 1 - common ratio ]  =

4  [1 - (-5)^6]  /  [ 1  - -5 ]  =

4 [ 1 - 15625 ] / 6  =

(2/3) ( -15624 )  =

-10416

a_n  = 7n  - 2

a_n   =1500 (1.15)^{n -1}   ⇒ last answer

The  amount of chocolate  contained  in the large hemisphere  is just

(2/3)pi (1)^3  =  (2/3) pi ft^3

So....we want to solve this for r  to find the radius of each of the smaller hemispheres

(2/3) pi   =  27  (2/3) pi (r)^3

1  = 27 * r^3

1/27 =  r^3

1/3  = r   

So....r = 1/3 ft  

This seems to be harder than it really is......

Notice that if we "unwrapped"  one  complete  revolution  of the string around the post, we would have a rectangle with a "width"  of  4 ft  [the circumference of the post ].....and the height of this rectangle would be just 12 / 4  =  3ft

And the string would run  from the top left of this rectangle to the bottom right, thus comprising the diagonal of this rectangle .......so....the length of this string would be the hypotenuse of a right triangle with legs of 3 and 4.....and its length  =  sqrt  [3^2 + 4^2]  = sqrt [ 25]  = 5

And we have 3 more  similar rectangles in the next 3 revolutions.....so....the string length  is just

1 revolution * 5 ft   +   3 revoultions * 5 ft  = 

4 * 5 ft  =

20 ft