Answer what you can please and show work. I will mark you brainliest.

part 1:

1) AB = BC = 10

2) DE = sqrt(10² - 9²) = sqrt(19)

3) CED = 90⁰

4) BCD = (90⁰ - 53⁰).2 = 74⁰

part 2:

because FGHJ is a rhombus

=> GK = KJ

<=> 3x + 1 = x + 9

<=> 2x = 8

<=> x = 4

and we have:

GHK + FGK = 90⁰

<=> (4y - 8)/2 + y + 4 = 90⁰

<=> 2y - 4 + y + 4 = 90⁰

<=> 3y = 90⁰

<=> y = 30

5) with x = 4 => GJ = 2KJ = 2.(4 + 9) = 26

6) JK = 4 + 9 = 13

7) with y = 30, HFJ = 90 - FGK = 90 - 34 = 56⁰

8) HJF = 2.FGK = 2.34 = 68⁰

part 3:

because MNLO is rectangle

=> MP = PO

<=> 5a - 7 = 2a + 8

<=> 3a = 15

<=> a = 5

9) with a = 5, MO = 2.PO = 2.(2.5 + 8) = 2.18 = 36

10) NP = PO = 2.5 + 8 = 18

11) LM² = LN² - MN² = MO² - 26² = 36² - 26² = 620

=> LM = 2sqrt(156) = 24,9

12) tan NMO = NO/MN = LM/MN = 24.9/26

=> NMO = 43,8⁰

part 4:

because RSTU is square

=> SR = RU

=> 2k + 7 = 3k - 2

<=> k = 9

with k = 9 =>RU = SR = 2.9 + 7 = 25

13) RV = 1/2.SU = 1/2.sqrt(25² + 25²) = 17,7

14) SU = 2RV = 2.17,7 = 35,4

15) perimeter RSTU = 25.4 = 100

16) area RSTU = 25² = 625

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

24

step-by-step explanation: