Let's define variables:
s = original speed
s + 12 = faster speed
The time for the half of the route is:
60 / s
The time for the second half of the route is:
60 / (s + 12)
The equation for the time of the trip is:
60 / s + 60 / (s + 12) + 1/6 = 120 / s
Where,
1/6: held up for 10 minutes (in hours).
Rewriting the equation we have:
6s (60) + s (s + 12) = 60 * 6 (s + 12)
360s + s ^ 2 + 12s = 360s + 4320
s ^ 2 + 12s = 4320
s ^ 2 + 12s - 4320 = 0
We factor the equation:
(s + 72) (s-60) = 0
We take the positive root so that the problem makes physical sense.
s = 60 Km / h
Answer:
The original speed of the train before it was held up is:
s = 60 Km / h
1) 20x+5y=120
Stp1: 5y=120-20x
Stp2: 5y=(120/5 and -20x/5)= [y=24-4x]
finding x:
20x+5(24-4x)=120
Stp1: 5*24=120,-4x*5=-20
20x+120-20x=120
Stp2: subtract 120 on both sides of = sign
20x-20x=0
Stp3: subtract the variables:
[x=0]
finding y:
20(0)+5y=120
stp1: multiply 20*0
5y=120
stp2: divide 120 by 5
[y=24]
2) 10x+7.5y=80
Stp1: move 7.5y to other sign of =
10x=80-7.5y
Stp2: divide by 10
[x=8-.75y]
substitute x and re do original:
stp1: 10(8-.75y)+7.5y=80
stp2: 80-7.5y+7.5y=80
stp3: subtract variabled num. with variabled num. and whole num. with whole number: [y=0]
solve for x:
10x-7.5(0)=80
10x=80
[x=8]
check:
10(8)+7.5(0)=80
80+0=80
[80=80]
Answer: 9. 54; 10. 220; 11. 71; 12. 262
Step-by-step explanation: Chord and tangent intersection angle is half the arc measure or circle, and inscribed angles are also half the included arc measure
Draw a diagram
the diagram is a right triangle
use the tangent function
tan=opposite/adjacent
call the angle "θ"
tanθ=15/57
tanθ=0.2631579
use the arctan function on your calculator
tan-1θ=0.2631579
θ=14.74356312º or 14º 44' 36.827"
second-period class average:
55+70+450+480+170+270+95 = 1590
1590/20 = 79.5
sixth-period class average:
65+225+480+595+270+190 = 1825
1825/20 = 91.25
on average, students in the sixth period class scored higher
