Guys me to do im tryin’ my best but it’s so hard

Step-by-step explanation:

1.

∠ACB=6 x°

∠DCE=30°

6x=30°divide both sides by 6

x=30°/6= 5°

∠BCE=180°-30°=150°sum of angles on a straight line

2.

∠ABD=(4+5x)°

∠CBD=(x+2)°

(4+5x)°+(x+2)°=180°sum of angles on a straight line

4+5x+x+2=180°collect like terms

6x+6=180°

6x=180°-6°

6x=174°divide by 6 both sides

x=174°÷6= 29°

∠ABD=4+5x=4+(5*29)= 4+145=149°

∠CBD= x+2=29°+2°=31°

3.

BE bisects ∠ABE=(3x+1)°

m∠DBA=(8x-14)°

1/2 (m∠DBA)° = m∠ABE

1/2(8x-14)°=(3x+1)°

(4x-7)°=(3x+1)°

4x-7=3x+1

4x-3x=7+1collecting like terms

x=8°

m∠ABE=(3x+1)°=(3*8+1)=24+1=25°

m∠DBA=(8x-14)°= (8*8-14)°=(64-14)=50°

4.

∠ADE=∠CDG

50+3x-y=x+2x-16collect like terms

50+16-y=x+2x-3x

66-y=3x-3x

66°=y

∠ADC +∠ADB+∠BDE

90°+50°+(3x-y)°=180°

140°+(3x-y)°=180°

3x-66°=180°-140°

3x=40°+66°

3x=106°divide both sides by 3

x=106°÷3= 35.33°

∠FDG=(2x-16)°= (2×35.33° - 16° )= 54.67°

∠BDE = (3x-y)°= (3×35.33°-66°)= 105.99-66=39.99°

5.

∠ABD+∠DBC=90°

(6x+4)°+32°=90°

(6x+4)°=90°-32°

6x+4=58°

6x=58°-4°

6x=54°

x=54°÷6=9°

∠ABD= (6x+4)°+32°

∠ABD= (6×9 +4)°+32°

∠ABD= (54°+4°) + 32°

∠ABD=58°+32°=90°

6.

∠AED=∠CEB

(3x+5)°=(4y-15)°form equation of equality

3x+5=4y-15

5+15=4y-3x

20=4y-3x(1)

∠AEC=∠DEB

(y+20)°=(x+15)°

y+20=x+15

20-15=x-y

5°=x-y

5+y=x(2)

Use equation (2) in equation (1)

20=4y-3x

20=4y-3(5+y)

20=4y-15-3y

20+15=4y-3y

35°=y

Solve for x

x=5+y=5°+35°=40°

∠AED=(3x+5)°=(3×40 +5 )=120+5=125°

∠AEC= (y+20)°= 35° + 20° =55°

c. 4 1/4 pounds

1.true 2.false? ? i think. .

step-by-step explanation: