Alan

With x = 6 as one solution we must have:

(x-6)(x² + ax + b) = x³ - 37x + 6

or x³ + ax² + bx - 6x² - 6ax - 6b = x³ - 37x + 6

or x³ + (a-6)x² + (b-6a)x - 6b = x³ - 37x + 6

So we must have a = 6 and b = -1 for the LHS to equal the RHS.

The depressed equation is thus the quadratic: x² + 6x - 1 = 0

Can you take it from here?
 

Possible

primes.        Die 1.     Die 2.

     2.              1.            1

     3.              1.            2

                      2.            1

     5.              1.            4

                      4.            1

                      2.            3

                      3.            2

     7.              1.            6

                      6.            1

                      2.            5

                      5.            2

                      3.            4

                      4.            3

     11.            5.            6

                      6.            5

 There are 4 ways to make 5 out of 15 ways to make a prime.

Write this as tanθ - 2cosθtanθ = 0  or tanθ(1 - 2cosθ) = 0

So, either tanθ = 0  or  1 - 2cosθ = 0

Can you take it from here?

These might help:

Use arctan (or atan on many calcuators).

Here are parts 2 and 3:

Here's Part 1)

Look at the answer to https://web2.0calc.com/questions/pre-calc-help_9 and try the same technique here.

As follows:

a^b = b^a.     b = 9a

a^(9a) = (9a)^a

a*log(a^9) = a*log(9a)

log(a^9) = log(9a)

a^9 = 9a

a^8 = 9

a = 9^(1/8)

m = 8, n = 9

"A box contains 81 coins, only dimes and nickels. The amount of money in the box is $5.30. How many dimes and how many nickels are in the box?"

d + n = 81   (1)

10d + 5n = 530 divide by 5 to get 2d + n = 106   (2)

Subtract (1) from (2):   d = 106 - 81 = 25   (3)

Use (3) in (1):  25 + n = 81 or n = 81 - 25 = 56

There are 25 dimes and 56 nickels.