CPhill

i'm using x in place of theta

cos (x)  + 4 sec(x)  =  8

cos (x)  + 4 / cos(x)  = 8         multiply through by cos (x)

cos^2x  + 4  = 8cos(x)

cos^2 (x) - 8cos(x) + 4  =  0

Let cos(x)  = m       and we have that

m^2 - 8m  + 4  = 0

m^2  - 8m  =   - 4      complete the square on m

m^2  - 8m + 16  =  -4 + 16

(m - 4)^2  =  12        take both roots

m - 4  =  ±2√3

m =  ±2√3 + 4

Only the negative root makes sense

So

m  =  cos x  =  (-2√3  + 4)

Take the cosine inverse

arcos ( -2√3 + 4)  =  x  ≈  1.005 rads     and  (2pi - 1.005)  ≈ 5.28 rads  

So....the sum of these  ≈  6.285 rads

6 tan ^2  ( 5x - 3)  = 2

tan^2 ( 5x - 3)  = 1/3      take both roots

tan (5x - 3)  =  ±√[1/3]

So  we have the following possibilities

arctan ( 1 / √3) =  5x - 3         or         arctan (-1/√3) = 5x - 3

150 = 5x - 3                                         330 = = 5x - 3

150 +3  = 5x                                        330 + 3  = 5x

153  = 5x                                              333  = 5x                                    

x = 153/5                                              333/5 = x

x = 30.6                                                  x = 66.6

510 = 5x - 3

510 + 3  = 5x

513 = 5x

513/5  = x

x = 102. 6°

Here's my attempt

x^3 -  x^2  =

x^2 * (x - 1)

We will have a perfect square whenever  (x - 1)  is a perfect square......so.....

x  = 1, 2,  5, 10, 17, 26, 37, 50, 65, 82

.5625  =

5625  / 10000  =

9 / 16

LOL!!!!!

f(5)  =  3f(5-2) - 2f(5 - 1)

f(5)  = 3f(3) - 2f(4)

f(5)  = 3 [ 3f(3-2) - 2f(3 - 1) ] - 2 [ 3f(4-2) -2f(4 - 1)]

f(5)  = 3 [ 3f(1) - 2f(2)] - 2[ 3f(2) - 2f(3) ]

f(5)  =  3 [ 3(-1) - 2(3) ]  - 2 [ 3(3) - 2f(3) ]

f(5) = 3[ -9] - 2 [ 9 ] + 4f(3)

f(5) = -45 + 4 [3f(3 - 2) - 2f(3 - 1)]

f(5) = -45 + 4 [ 3f(1) - 2f(2) ] 

f(5)  = -45 + 4 [ 3(-1) - 2(3) ]

f(5)  = -45 +4 [ -3 - 6 ]

f(5)  = -45 + 4 (-9)

f(5)  = -45 - 36

f(5)  = -81

We have the form  

ax^3 + bx^2 + cx +d

I think that we should have  f(-1)  = 15

Since (0,0)  is on the graph, then d  =  0

So  we have this system of equations

-a   +  b  -  c   = 15    (1)

a  +  b  +   c  = -5      (2)

8a  + 4b  + 2c   = 12    ⇒   4a + 2b + c  =  6  (3)

Add (1) and (2) and we get that  2b = 10  ⇒  b  = 5

Sub this into  (2) and (3) and we get that

a + 5 + c  =  -5  ⇒   a + c  =  -10   ⇒  -a - c  =  10   (4) 

4a + 2(5)  + c  = 6  →  4a + c  =  -4     (5)

Add (4)  and (5)  and we get that 

3a  = 6

a  = 2

And

-2 + 5 - c   = 15

-c  = 12

c  = -12

So....the function is

f(x)  = 2x^3  + 5x^2 - 12x

Factoring we have that

x (2x^2 + 5x - 12)  = 0

x (2x - 3)(x + 4)  = 0

So....settting  each factor to 0   and solving for x gives the  x intercepts  of  x  = 0 , x = 3/2 and x  = -4

If you did all the homework  and understood all the concepts.....no worries.....you'll do fine....

[125 * 6 ] * 2  /100  =

125 * 12  /100  =

1500 / 100 =

15 hrs

Center  =  ( 6, -3)

Endpoints of the major axis  are 10 units from the center

Endpoints of the minor axis  are 6 units  from the center

The final answer is just the next to the last answer box given