CPhill

Excellent, RP  !!!

An insribed angle  measures  1/2  of its intercepted arc  ...

So

(1/2) *DG  =   55      multiply both sides by 2

DG  =  110°

9.  Yes,  2/3

10.  

10th term   =  81(1/3)^{n -1}   =  81(1/3)^10-1  =  81(1/3)⁹  = 1/243

They are the answers  .....I just used "?"  to stand for a variable

6. Not arithmetic because the difference between terms is not constant

7.  12th term  =

15  - 4 (12 - 1)  =

15 - 4(11)  =

15  -  44 =

-29

8.  To find the missing term

[ First term + Third term ]  /  2   =

[ 20 + 32 ]  / 2

52  / 2  =

26

8.

16 / 8  =   ? / 20

2  = ? / 20           multiply both sides by 20

40  = ?

9.

25/ 20  = ? / 8

5/4  = ? / 8     multiply both sides by 8

8 * 5/4  =  40/4  =  10  =  ?

10.  Note that the missing side of A  =  1/6 * 48    =  8

4.  After Third Bounce 

6 (.80)^3  = 3.072 ft  =  3.1 ft  

5.   The number of bacterium increases by 22 every hour....so we have

a_n  =  22n 

tan (30°)  =  1 / √3  =  √3 / 3

Scale factor  =  40 / 24  =  10/6  =  5/3

Last one

We have 8 congruent triangles each with a base  of 5.8 cm

We can use the Pythagorean Theorem to find the slant height of each  =

√[ 7^2 + 7^2]  =  √ [2 * 7^2]   =  7√2

So.....the lateral area  is

8 (1/2) (base) (slant height)  =

4 (5.8) (7√2 )  =  229. 7 cm^2