D midpoint of EC -----------------> FD parallel to AC and FD=AC/2=14/2=7
<span>2-EB=EA E midpoint of AB </span>
<span>DB=DC D midpoint BC ...............> ED=AC/2=2 </span>
<span>3-T midpoint of SR </span>
<span>U midpoint of QR ---------> TU = QS/2 </span>
<span>QS=2 TU = 4.4 </span>
<span>4- The same steps SR=2 UV=9 </span>
<span>5-N midpoint of KM </span>
<span>O midpoint of ML </span>
<span>* NO parallel to Kl</span>
<span>△PQR is similar to △STU
</span>m∠R = m∠U = 96°
m∠Q = m∠T = 6
m∠P = 180 - ( 96 + 6)
m∠P = 180 - 102
m∠P = 78
answer
m∠P = 78
Step-by-step explanation:
always draw triangle b first. for triangle b it is reflected across the x-axis which means all y valies become negative. then for triangle c, you swap the x and y values around.
Answer:
If what the sin 90º
Step-by-step explanation:
Answer:
y = -x + 9
Step-by-step explanation:
The line that passes through the points (-2,4) and (0,6) has a slope of 1, and a y intercept of 6. The equation to the first line is y = x + 6. and perpendicular lines always have the opposite, reciprocal slope of the other line. So the slope for the second line would be -1. A line with the slope of -1 and a point of (5,4) would contain the points (5,4) , (4,5) , (3,6) , (2,7) , (1,8) , and (0,9) , which is the y intercept for the second line. So the equation for the second line would be y = -x + 9
