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9)  Not as tough as 8

1st term = a  = 63      (1) 

4th term  = ar^3 = 122     (2)

Sub (1) into (2)

63r^3 = 122

r^3  = 122/63

r = (122/63)^(1/3)  ≈ 1.246 ≈  1.25

Gotcha thx

8)   Let the first term  =  a

Then we know that the second term = ar = 42     where r is the common difference

And the sixth term  = 6302.625  = ar^5

So.....we have the following equations

ar  =  42

ar^5 = 6302.25          divide the ssecond equation by the first and we have

r^4 = 6302.625 / 42  

r = ( 6302.625  / 42 )^1/4  = 3.5

To find the first term we can solve this

ar = 42

a (3.5) = 42

a = 42/3.5  = 12  = first term

See here, KeyLime :

https://web2.0calc.com/questions/help_53845

First one.....the area of the inscribed equilateral triangle  =

3(1/2)r^2 sin  120  =  ( 3√3/4) r^2

The Height of the circumscribed equilateral triangle = 3r

And we  can find the base as

tan 60 = 3r /(1/2) Base

√3 =6r / Base

Base = 6r/√3  = 2√3r

So....the area of the circumscribed eqilateral triangle is  (1/2)Base* Height  = (1/2)2√3r * 3r =  3√3r^2

So we have that

Area of circumscribed triangle - Area of inscribed triangle =  36

(3√3  - 3√3/4) r^2  = 36

3(√3 -  √3/4) r^2 = 36

  (√3 -  √3/4)r^2  = 12

r^2  =  12 / ( √3 - √3/4 )  

r = √ [  12 / ( √3 - √3/4 ) ] ≈  3.04 cm

 ∛[ 64* 2x^3 * x ]

 ∛ 64  *  ∛2   *   ∛x^4  =

4 *  ∛2 * x ∛x  =

4x  ∛ [ 2x ]

l a - b l  will be maximized when a is large as possible and b is as small as possible

So....suppose that   a > c > b

Then we have that

(a - b) +( c - b) + ( a - c)  = 20

2a - 2b  =  20            divide through by 2

a - b  =  10

So

l a - b l  =  10

sqrt3(64*2*x to the 3'rd* x)

- √ [ 128x^4 ]  =

- √ [( 64 * 2)  * x^4 ]  =

- √64*√2 *√ x^4  =

- 8 x^2 √2     

z +  1/z  = 1

z^2 + 1  = z

z^2 - z + 1   = 0

Complete the square on z

z^2  - z + 1/4  =   -1 + 1/4

(z - 1/2)^2  = -3/4

z - 1/2  =  ± i √3 / 2

z =  1/2 +   (√3/2) i        or    z   = 1/2  -  ( √3/2) i

In each case

z^3  =  - 1

x^2 - x - 2  factors as  (x -2) ( x + 1)

Multiply through by this factorization and we have that

1x + 7  = A( x + 1) + B( x - 2)    simplify

1x + 7  =  Ax + A  + Bx - 2B     equate coefficients

1x + 7  =  (A + B) x  + (A - 2B)

This implies that

A + B  =  1   ⇒  - A - B   = - 1    (1)

A - 2B = 7   (2)

Add (1) and (2)  and we have that

-3B =  6

B = -2

And

A + B  = 1

A - 2  = 1

A = 3

So  (A, B)  = ( 3, -2)

(a - bi)^2 =

a^2 - 2abi - b^2    =   8 - 6i

(a^2 - b^2) = 8     (1)

(-2ab)i  = - 6i

2ab i = 6i

ab  = 3      ⇒   b = 3/a      (2)

Sub (2) into (1)

a^2 - (3/a)^2  = 8

a^2 - 9/a^2  = 8

a^4 - 9  = 8a^2

a^4 - 8a^2 - 9  = 0

(a^2 - 9) (a^2 + 1)   = 0

The second factor produces a non-real for a

a^2 - 9  ⇒  a  = 3

And b = 3/a = 3/3  = 1

So  (a, b)  = (3, 1)

And

a - bi  =

3  - i

OK...note that  

mAngle ACB  + mAngle  ACD  = 180°     (1)

And

mAngle CAD + mAngle ADC + mAngle ACD = 180°    (2)

So

(1) = (2)  so

mAngle ACB + mAngle ACD  =  mAngle CAD + mAngle ADC + mAngle ACD

Subtract  mACD from each side

So

mAngle ACB = mAngle CAD + mAngle ADC

Thank you for this but I have one question. How is your step 3 the Alternate Exterior Angles Theorem? I do not understand how you got that and I was wondering if you could elaborate.

(1) AB = AC                                      Given

(2) Angle ABC = Angle ACB             In a triangle, angles opposite equal sides are also equal

(3) Measure of Angle ACB =            Exterior Angle Theorem

Measure of Angle CAD +

Measure of  Angle ADC

(4) Measure of Angle ACB >         Consequence of (3)

Measure of Angle ADC

(5) Measure of Angle ABC >        Substitution 

Measure of Angle ADC

(6) Measure Angle B >                By (5)

Measure Angle D

On Monday, the produce manager stocked the display case with eighty heads of lettuce. By the end of the day some of the lettuce had been sold.

On Tuesday, the manager surveyed the display case and counted the number of heads of lettuce left. He decided to add an equal number of heads of lettuce. (He doubled the leftovers.) By the end of the day he had sold the same number of heads of lettuce as on Monday.

On Wednesday, the manager decided to triple the number of heads of lettuce that had been left in the case Tuesday night. He sold the same number of heads of lettuce that day, too. However, at the end of the day there were no heads of lettuce left. How many heads of lettuce were sold each day?

I only know one method

Call N the number sold on Monday.......so...80 - N  were left

On Tuesday  (80 - N ) were added to the (80-N) and N   were sold...so  2(80 - N)  - N  = 160 - 2N - N = (160 - 3N) were left

On Wed 2(160 - 3N) were added to (160 - N) and N were sold....and none were left ...so...

3(160 - 3N) - N = 0

480 - 9N - N = 0

480  -10N = 0

480 = 10N

N = 48   = number sold each day

22/24  =  11/12

So

10  11/12

Sorry, but this is incorrect!

3z  +4i (z conjugate) = 1 - 81

3 ( a + bi)  - 4 i ( a - bi)  = 1 - 8i

3a + 3bi  - 4a i - 4b  = 1 - 8i

(3a - 4b) + (3b - 4a) i  = 1 - 8i

Equate coefficients  on real , imaginary parts

3a - 4b  = 1       ⇒  12a - 16b  = 4   

-4a + 3b  = - 8  ⇒  -12a + 9b  = -24          add the last two equations

-7 b = - 20      ⇒  b = 20/7

And 

3a - 4(20/7)  = 1

3a - 80/7 = 7/7

3a  = 87/7

a = 87/21  = 29/7

So

z   =    29/7  + (20/7) i

-3i (z conjugate)  =

-3 i ( a - bi)  =

-3i a   + 3b i^2  =

-3b - 3 i a

Thank you!

CPhill: Did you make a mistake in the 2nd line?

(3 * 5) - (5 * 3)  =

[3*5 + 5^2]  - [ 5*3 - 3^2]  =Souldn't this read: [3*5 + 5^2] - [5*3 + 3^2]

[ 15 + 25 ] - [ 15 - 9 ]  = [15 + 25] - [15 + 9] =40 - 24 = 16 ???

The reciprocal of every prime p (other than two and five) has a period, that is the decimal expansion of 1/p repeats in blocks of some set length. This period is the period of p. ... For example: 3, 11, 37 and 101 are the only primes with periods one, two, three and four respectively−so they are unique-primes.

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.693.5325&rep=rep1&type=pdf