geno3141

Since     log_b(a)  =  x     --->     b^x  =  a

log_{3x + 5}(9x² + 8x + 8)  > 2     ---->   (3x + 5)²  <  9x² + 8x + 8

Then continue ...

The probability of winning plus the probability of losing is 1.000.

So, if his probability of winning at Hoopla is 0.3, his probability of losing at Hoopla is 0.7.

And, if his probability of winning at Darts is 0.4, his probability of losing at Darts is 0.6.

[These numbers should help you complete the chart.]

The probability that he wins at both games

  =  the probability that he wins at Hoople  times  the probability that he wins at darts

  =                            0.3                                 x                           0.4                                    =  0.12

The probability that he wins at just one of the games

  =  the probability that he wins at Hoople  times  the probability that he loses at darts

  =                              0.3                               x                           0.6                                    =   0.18

plus

  =  the probability that he loses at Hoople  times  the probability that he wins at darts

  =                              0.7                                x                           0.4                                   =  0.28

Adding these together:  0.18 + 0.28  =  0.46

I'm going to use these identities:

1)  sin(A - B)  =  sin(A)cos(B) - cos(A)sin(B)

2)  sin(A + B)  =  sin(A)cos(B) + cos(A)sin(B)  

3)  sin(3A)  =  3sin(A) - 4sin²(A)

4 · sin(x) · sin(60 -x) · sin(60+x)

=  4 · sin(x) · [ sin(60 - x) ] · [ sin(60 + x) ]

=  4 · sin(x) · [ sin(60)cos(x) - cos(60)sin(x) ] · [ sin(60)cos(x) + cos(60)sin(x) ]

=  4 · sin(x) · [ sqrt(3)cos(x)/2 - sin(x)/2 ] · [ sqrt(3)cos(x)/2 + sin(x)/2 ]

=  4 · sin(x) · [ 3cos²(x)/4 - sin²(x)/4 ]

=  sin(x) · [ 3cos²(x) - sin²(x) ]

=  sin(x) · [ 3( 1 - sin²(x) ) - sin²(x) ]

=  sin(x) · [ 3 - 3sin²(x) - sin²(x) ]

=  sin(x) · [ 3 - 4sin²(x) ]

=  3sin(x) - 4sin³(x)

=  sin(3x) 

Questions 2, 3, and 7 are straight-forward questions. Have you tried these?

Question 1 can be tricky. What have you done, so far, in attempting this problem? 

Question 4:  What is the degree of f(x) + g(x)?  Can the constant 'a' change this degree?

Question 5:  What value for 'a' can change the degree of the sum?

Question 6 asks you to find the equation of a quadratic. Do you know how to do this?

cos(0)  cancels out  cos(180)

cos(5)  cancels out  cos(175)

...

cos(85)  cancels out cos(5)

This leaves you with cos(90); how much is cos(90)?

cos(355)  cancels out  cos(185)

cos(350)  cancels out  cos(190)

...

cos(275)  cancels out  cos(265)

This leaves you with cos(270); how much is cos(270)?

So, how much is  cos(90) + cos(270)?

Let O be the center of the circle.

Let CD be the chord of length 16

Let AOB be the diameter of the circle that is the perpendicular bisector of the chord; A is on minor arc(CD)

     and D is on major arc(CD).

Let the point of intersection of AOB and CD be point X.

Being a radius, OD = 10.

XD = 8.

Triangle(OXD) is a right triangle.     OX² + XD²  =  OD²  

                                                           OX² + 8²  =  10² 

                                                           OX² + 64  =  100

                                                                   OX²  =  36

                                                                     OX  =  6

OX is the distance from the center to the chord.

log₃(x + 7)  <  log₉(x² + 77)

Using the change-of-base formula:

                 [ log(x + 7) / log(3) ]  <  [ log(x² + 77) / log(9) ]

     [ log(x + 7) / log(3) ] · log(9)  <  log(x² + 77)                [multiply both sides by log(9)]

   [ log(x + 7) / log(3) ] · log(3²)  <  log(x² + 77)                [change 9 into 3²]

[ log(x + 7) / log(3) ] · 2 · log(3)  <  log(x² + 77)               [use property of logs]

                         [ log(x + 7) ] · 2  <  log(x² + 77)               [cancel the log(3) factors]

                             2 · log(x + 7)  <  log(x² + 77)               [rewrite]

                                log(x + 7)²  <  log(x² + 77)                [use property of logs]

                                     (x + 7)²  < x² + 77                        [raisee both sides to power of 10]

                            x² + 14x + 49  <  x² + 77                       [multiply out]

                                   14x + 49  <  77

                                           14x  <  28

                                               x  <  2

x² + 5 / (x - 3)  =  9 + 5 / (x - 3)

First:  note that x cannot = 3.

Subtract  5 / (x - 3)  from both sides:  x²  =  9 

--->  Either  x = 3  or  x = -3  (but x cannot = 3)

--->   x = -3

Simplify:        [ a(a + 1) / 2 ]²  -  [ a(a - 1) / 2 ]² 

  =                  [ (a² + a) / 2 ]²  -  [ (a² - a) / 2 ]²​ 

  =            (a⁴ + 2a³ + a²) / 4  -  (a⁴ - 2a³ + a²​) / 4 

   =              [ (a⁴ + 2a³ + a²)  -  (a⁴ - 2a³ + a²​) ]  /  4

   =              [ a⁴ + 2a³ + a² - a⁴ + 2a³ - a² ] / 4

   =                                     4a³ / 4

   =                                         a³