CPhill

48. 

arctan (5.67)  =  79.99°   = 80°

49.   I just subbed  "x" for "theta"....same difference  

sin^2 x                1  - cos^2x         (1 + cos x) (1 - cos x)

________  =      _________  =    _________________  =       1  - cos x

1 + cos x             1 + cos x              1  +  cos x

√ [7x ]  ( √x  - 7√7 )

x√7  -  7√[49x]

x√7  - 7*7√x

x√7  - 49√x

He  can choose any 3 of the 12 shirts  =

C (12,3)   = 220 different choices

8)A circle centered at (3,2) is tangent to the line y = (x+1)/3. Find the area of the circle.

y  = (x + 1)  / 3      can be written as

y = (1/3)x + 1/3

3y = x + 1

x - 3y + 1   = 0

The distance from ( 3,2) to this line will be the radius of the circle

l 3 - 3(2) + 1 l               2             √10          √2

____________  =     ____   =      ___  =     ___

√[ 1^2 + 3^2 ]             √10              5            √5

So...the area  of the  circle is

pi * ( √ (2/5) ) ^2  =

2pi / 5   units^2

I worked  (9)   but my answer disappeared...it is a little lengthy to re-create...but....the answer ends up as  20/3  units^2

For convenience, position the lower left vertex of the equilateral triangle at (0,0)

The side of this triangle is twice the length of the side opposite the 30 degree angle in the other triangle  =  2 * 6   = 12

And the height of the other triangle  is  √3  times the side opposite the 30 degree angle  =   6√3  units =  √108  units

So...the vertex  at the top right of the figure has the coordinates  (12, √108 )

And the distance between this vertex  and  (0, 0)  is

√ [ 12^2  + (√108)^2  ]  =

√ [144  + 108 ] =

√252  =

√ [ 36 * 7 ]  =

6√7  units

g(x)  = 2x + 5

h(x)  = ax + b

If  h (3)  = 26     we have that   26  =a(3) + b  ⇒    3a + b  = 26   (1)

And   g^-1 (x)   is     [ x - 5] / 2

So  g^-1(21)   is    [ 21 - 5] /2   = 16 /2   =  8

So  g^-1(21)  = 8  =  h(1)   which implies that

8 = a(1) + b    ⇒  a + b  = 8    (2)

So...we have the system

3a + b  = 26

a + b  =  8          subtract the second equation from the first

2a  = 18

a = 9

And    a + b  = 8     so   9 + b  = 8   ⇒   b  = -1

So  the function is

y = 9x  - 1

And a = 9   and  b  = -1

7)We are given O = (0,0), A = (5,2) and B = (6, -3). Find [OAB].

We have triangle  OAB

Let AB   be the base

And its length is

√ [ (6 - 5)^2  + (-3 - 2)^2  ]  =  √ [ 1^2  + 5^2  ]  = √26

And the slope between these two points  is    [-3 -2 ] / [ 6 -5]  = --5

And the equation of the line containing AB  is

y  =-5(x - 5) + 2

y  = -5x + 25  + 2

y = -5x + 27

5x + 1y -27  = 0

And the distance from (0,0) to this line forms the height of OAB

l 5(0)  + 1(0)  -27]                 27

_______________  =        ____

√[ 5^2  + 1^2 ]                    √26

So....the area  of  OAB   is  (1/2)√26 * 27/√26  =   13.5 units^2

6)For some positive real number r, the line x + y = r is tangent to the circle x^2 + y^2 = r. Find r.

At the point of tangency,  x  = x^2  and  y  = y^2

And these will be true  at   0  and 1

But...x, y  cannot  = 0 since the center of the circle is (0,0)

So...x, y  = 1

And

x + y  =

1 + 1  =

2   = r

5)We are given A = (2,5), B = (-5, 2), C = (-2, -5), and D = (5,-2). Find [ABCD].

We can split this up  into  the area of two triangles.....ABD  and CBD

BD will serve as the base for both

And its length is  √ [ (-5-5)^2  + (-2 -2)^2 ] =√[ 10^2 + 4^2  ]  = √116 =  2√29 units

And the slope of BD  is  [ -2 - 2 ] / [ 5 - -5]  =  [-4] / [10]  = -2/5

And the equation of the line containing  BD  is

y = (-2/5)(x -5) - 2

5y = -2(x - 5) - 10

5y  = -2x + 10 - 10

5y + 2x = 0

2x + 5y  = 0

So...the distance from A to this line will be the height of triangle ABD  =

l 2(2) + 5(5) l                     29

___________   =           ____  =  √29  units

√ [2^2 + 5^2 ]                  √29

So...the area  of  triangle ADB  = (1/2) *2√29 *√29  =  29   units^2

And the distance  between C and this line will be the height of triangle CBD

l 2(-2)  + 5(-5)  l              29

_____________  =       _____  =  √29  units

 √29                                √29

And  triangle CBD will have the same area as triangle ABD

So...the total area  of  [ABCD ] =    58 units^2