∠2 =∠1 which is 180° -65° =115<span>°
So </span>∠2=115<span>° </span>
Left at 3 p.m
traveled 1 hr 30 min......now its 4:30 p.m.
stopped for 15 min....now its 4:45 p.m.
continued for 2 hrs 10 min....now its 6:55 p.m
so the train arrived at 6:55 p.m.
Let L be the low copper alloy and H be the high copper alloy.
We need 0.15L + 0.60H = 0.42 and L + H = 100
L=100 - H
Substituting this into the second equation for L, we get:
0.15(100-H) + 0.60H = 0.42(H+L)
15 - 0.15H + 0.60H = 0.42(H+100-H)
15 + 0.45H = 0.42(100)
0.45H = 42 - 15 = 27
H = 27/.45 = 60 lbs
Back substitution, L = 40 lbs.
Answer:
y+5 = 3/4(x+8)
Step-by-step explanation:
The point slope form of the equation of a line is
y - y1 = m(x-x1) where m is the slope and ( x1,y1) is a point on the line
The slope is 3/4 and the point is (-8,-5)
y - -5= 3/4( x - -8)
y+5 = 3/4(x+8)
