someone answer pls
armys out there,....helpppp!!!!
Answers
Answer:Answer:
Answer:let the angle ZYQ =x;
Answer:let the angle ZYQ =x;angle PYQ=x
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight line
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2→x=58°
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2→x=58°♦ anglePYQ=58°
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2→x=58°♦ anglePYQ=58°♦ angle QYZ=58°
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2→x=58°♦ anglePYQ=58°♦ angle QYZ=58°→angleXYQ=angle XYZ+angleZYQ
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2→x=58°♦ anglePYQ=58°♦ angle QYZ=58°→angleXYQ=angle XYZ+angleZYQ→ angleXYQ =64°+58°
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2→x=58°♦ anglePYQ=58°♦ angle QYZ=58°→angleXYQ=angle XYZ+angleZYQ→ angleXYQ =64°+58°→angleXYQ=122°
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2→x=58°♦ anglePYQ=58°♦ angle QYZ=58°→angleXYQ=angle XYZ+angleZYQ→ angleXYQ =64°+58°→angleXYQ=122°Reflex angleQYP=360°- 58°
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2→x=58°♦ anglePYQ=58°♦ angle QYZ=58°→angleXYQ=angle XYZ+angleZYQ→ angleXYQ =64°+58°→angleXYQ=122°Reflex angleQYP=360°- 58°=302°
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2→x=58°♦ anglePYQ=58°♦ angle QYZ=58°→angleXYQ=angle XYZ+angleZYQ→ angleXYQ =64°+58°→angleXYQ=122°Reflex angleQYP=360°- 58°=302°Therefore, answer = 58°,122°,
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2→x=58°♦ anglePYQ=58°♦ angle QYZ=58°→angleXYQ=angle XYZ+angleZYQ→ angleXYQ =64°+58°→angleXYQ=122°Reflex angleQYP=360°- 58°=302°Therefore, answer = 58°,122°,302°.
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2→x=58°♦ anglePYQ=58°♦ angle QYZ=58°→angleXYQ=angle XYZ+angleZYQ→ angleXYQ =64°+58°→angleXYQ=122°Reflex angleQYP=360°- 58°=302°Therefore, answer = 58°,122°,302°.HOPE IT HELPS
Answer:let the angle ZYQ =x;angle PYQ=xPYX is a straight lineangle PQY + angle QYZ+ angle XYZ=100°→x+x+64°=180°→2x+64°=180°→ 2x= 180 -64°→x=116°/2→x=58°♦ anglePYQ=58°♦ angle QYZ=58°→angleXYQ=angle XYZ+angleZYQ→ angleXYQ =64°+58°→angleXYQ=122°Reflex angleQYP=360°- 58°=302°Therefore, answer = 58°,122°,302°.HOPE IT HELPS♠ANSWER BY YASHAWINI♠
Answer:
assertion is true but reason is false