CPhill

4)Find the area of the region bounded by the lines y = 5 and 5x + 2y = 25 and the coordinate axes.

Look at the  following pic:

This are will be the difference  in the area  of triangle ADB  - area of triangle ACE

The area of triangle  ADB  =  (1/2) base (height)  = (1/2) (5)(12.5)  =  31.25 units^2

Since the triangles are similiar, the linear scale factor of ACE to ADB  = 3/5

So...the area of triangle  ACE  =  area of ADB * (3/5)^2  = 31.25 * (3/5)^2  = 11.25 units^2

So...the area of the bounded region is   31.25 - 11.25 =   20 units^2

3)A circle centered at the origin is tangent to the line x - y + 4 = 0. What is the area of the circle?

The distance from the centrer to the line will be the radius of the circle....so we have

l (0)  - (0) +  4  l                        4

______________  =              ____   =  2√2  =  √8  =  the radius 

   √[ 1^2 + 1^2]                        √2

So...the area of the circle  is

pi * (  √8)^2   =

8pi   units^2

The fact that  BC is parallel to DE  sets up the following relationship :

AC/CE  = AB /BD

AC / 15  = 8 /10     

AC / 15 =  4 /5    multiply both sides by 15

AC  = 15 * 4  /5   =   60 / 5   = 12

The bisection of the apex angle sets up the following relationship :

DB / DA    =  BC / AC

x / 2.25   =   4 / 3      multiply both sides by 2.25

x  = 2.25 *  4  /3 

x  =  9/ 3

x  = 3 

sin  ( x + pi)

_________

cos ( x - pi)

sinx * cos pi  + sin pi * cos x

_______________________ 

cosx *cos pi + sinx * sin pi

 sin x * (-1) + (0) * cos x

___________________

oos x * (-1)  = sin x * (0)

-sin x

_____

-cos x

tan x

44.

We  have this function

y  = A cos (Bθ )

A  = the amplitude = l 0.5  l   = 0.5

To find B....we have that

2pi  / period  = B

2pi / [ 2 pi / 3 ]  = B

2pi * 3  / 2pi  = B 

3  = B

So...the correct answer is

y  = 0.5cos (3θ )

We have that

x /  4   =  40 / 32

x / 4   =  5 / 4       multiply both sides by  4

x = 4 ( 5/4)   =   5  cm

Notice that the perimeter of the second figure  / perimeter of the first figure will be the linear scale factor  =

34 / 20  =   17/10   = 1.7

So....the area of the second figure will be =

area of first figure  * (linear scale factor)^2  =

19.6  * (1.7)^2  =

56.6  m^2

We will have this system of equations

a(2)^2  + b(2)  + c  = 28.4

a(3)^2  + b(3) + c  = 63.9

a(4)^2  + b(4)  + c  = 113.6       simplifying, we have

4a + 2b + c  = 28.4

9a +  3b + c = 63.9

16a  + 4b + c = 113.6

Subtract the first equation from both the second and third equations and we have this :

5a +  b  = 35.5

12a + 2b  = 85.2

Multiply the  first equation by  -2  and add it to the second equation

-10a -2b  = - 71

12a + 2b  = 85.2

______________

  2a  =  14.2            divide both sides by  2

a  =  7.1

And using 5a + b = 35.5,  we can solve for b as

5 (7.1) + b = 35.5

35.5 + b  =35.5

b = 0

And using  4a + 2b + c  =28.8  to find c

4(7.1) + 2(0) + c  =28.8

28.8  + c  =28.8

c  = 0

So...our function  is

y  = (7.1 )x^2           and  b  and c  = 0

2: A steel sphere with a 3-inch radius is made by removing metal from the corners of a cube that has the shortest possible side lengths. How many cubic inches are in the volume of the cube?

The sphere will be tangent to each face of the cube.....so....the side length of the cube must  be  6 in

So....the volume of the cube  is   just

side^3   =   6^3   =  216 in^3