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Excuse me for not understanding.  If the sequence consists of all positive multiples of 3 that contain at least one digit that is a 1, shouldn't numbers like 111, 114, 141, 117, 171, etc. belong to the sequence as well?
 

The probability is 4/51.

The length of arc YZ is 6*pi.

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.

(a) See the graph below.

(b) g(x) = 2/3*f(3x - 1).

(c) We stretch the graph horiztonally by a factor of 3, then stretch the graph vertically by a factor of 1/2, then shift down 3 units.

The value of x is 115 (degrees).

c = 4 + 4*sqrt(5)/5.

The area of triangle BCD is 20/3.

5-12-13  right triangle

A

            10 

5                   D

                                 3

C             12                      B

sin  ABC  =  5/13

Area  of   BCD   =    (1/2)  ( BC * DB  sin (ABC)  =   

(1/2)  (12)  (3)  ( 5/13)   =

180  /  26   =

90   /  13   units^2

See   here  :    https://web2.0calc.com/questions/help_39113

Interior  angles sum to  360°

128  + ( 128 - d  )  + (128 - 2d )  +(128 - 3d )  =  360

512  -  6d  =   360

512  -  360   =   6d

152  =  6d

152 /  6  =  d    =   76/3

Smallest  angle  =   128  - 3 ( 76/3)   =  128  -76  =   52°

z =  a +  bi        z (complement)   =  a  - bi               i^2  =  -1

3 (a + bi) +  4i (a - bi)   =  3 - 3i

3a  +  3bi  +  4ai  - 4bi^2   =  3  -  3i

3a   +  (4a  + 3b)i  + 4b  =  3  - 3i

(3a + 4b)   +  (4a  + 3b) i  =   3   - 3i               implies  that

3a + 4b  =    3    ⇒   12a  + 16b  =  12      (1)

4a + 3b   =  -3   ⇒  -12a  -9b  =    9        (2)

Add ( 1)  and (2)

7b   = 21

b   =  3

3a   +  4 (3)  =  3

3a  +  12  =  3

3a  =  3 - 12

3a  =    -9

3a  =   -9 / 3

a  =  -3

z =   -3    +   3 i

With replacement   =   (1/13)^4  =  1  / 28561

( 23 +  16  + 30 +  33  + x  +  x + 2)  /  6  =  27

(104  +  2x)  = 27*6

104  +  2x   =  162

2x =   162   - 104

2x  =  58

x  = 58 / 2  =   29               x + 2  = 31

So.......16  is discarded

(23 +  29 +  30  +  31  +  33)  / 5   =  146 / 5   =    29.2  = new mean

See here  :    https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1047022.html

Note that when   f(x) =    y  =   1

We  have  the following  

f(-3)   = 1   f(-2)  =  1    f (-1)  = 1    f(1)  = 1     f(2)  =  1    f(3)  = 1

Sum of all  c^s =     6

Using Binet's formula, c = 2 + sqrt(5)/5.

The number of possible ways is 484.

The interior angle that is missing is 58 degrees.

Using coordinates, the area of triangle BCD is 120/7.

slope = (11 - (-17))/(5 - (-2)) = 28/7 = 4

y + 17 = 4(x + 2)

y + 17 = 4x + 8

y = 4x - 9