Question

Picture of question is now up help with problems 3-5 :))

Answer 1

For problem #3:
\[
(4x-10)+8x+(2x+50)=180\implies\\
14x-10+50=14x+40=180\implies\\
14x=180-40=140\implies\\
x=10
\]Plugging this back in for \(\text m\angle 3\):
\[
\text m\angle 3=2(10)+50=20+50=70
\]

Answer 2

okay 3:

(4x-10) + (8x) + (2x+50) = 180

Answer 3

so ill do 4

Answer 4

Sounds good

Answer 5

3x-30<90 because its acute it less that 90

Answer 6

thanks @LolWolf can you help with 5?

Answer 7

solve for x

add 30 to both sides: 3x< 120
divide by 3 on both sides x<40

so x must be less than 40, that is the restriction

Answer 8

thanks @TAKEBACKMATH

Answer 9

angles ABD + angle dbc = 90
68 + X = 90
-68 -68
x = 22

Answer 10

no prob

Answer 11

thanks :)))

Answer 12

wait @TAKEBACKMATH im confused why did you leave alot of space there? did you want me to fill in the balnks?

Answer 13

for 5 21'40'34''

Answer 14

how'd you get that answer?

Answer 15

@frazerdon how'd you get that?

Answer 16

90'00'00'' - 68'19'26''

Answer 17

did you get 90'00'00" from angle ABC?

Answer 18

yes because it's a right angled triangle