Let be:Speed of the wind: WSpeed of the plane in still air: P
Against the wind the plane flew:Distance: d=175 milesTime: ta=1 hour 10 minutesta=1 hour (10 minutes)*(1 hour/60 minutes)ta=1 hour + 1/6 hourta=(6+1)/6 hourta=7/6 hourSpeed against the wind: Sa=d/taSa=(175 miles) / (7/6 hour)Sa=175*(6/7) miles/hourSa=1,050/7 miles per hourSa=150 mph
(1) P-W=Sa(1) P-W=150
The return trip only took 50 minutesDistance: d=175 milesTime: tr=50 minutestr=(50 minutes)*(1 hour/60 minutes)tr=5/6 hour
Speed retur trip: Sr=d/trSr=(175 miles) / (5/6 hour)Sr=175*(6/5) miles/hourSr=1,050/5 miles per hourSr=210 mph
(2) P+W=Sr(2) P+W=210
We have a system of 2 equations and 2 unknows:(1) P-W=150(2) P+W=210
Adding the equations:P-W+P+W=150+2102P=360Solving for P:2P/2=360/2P=180
Replacing P by 180 in equation (2):(2) P+W=210180+W=210
Solving for W:180+W-180=210-180W=30
Answers:The speed of the plane in still air was 180 mphThe speed of the wind was 30 mph
Follow the steps
106=x+79+x+45
106=2x+124
-124= -124
-18=2x
( -18)/2=(2x)/2 <----- solving for x
-9=x
then
angle LMF= -9+79
=70
angle FMN= -9+45
= 36
(when you add 70 and 36 together you get 106 which is the measurement of angle LMN or in other word the whole thing)
In conclusion the answer to the problem is:
Angle LMF is equal to 70.
17 is the constant term because it does not have any variables.
Answer:
16
Step-by-step explanation:
Answer:
Step-by-step explanation:
5x=y+6 --------- equ 1
2x-3y=4 -------- equ 2
5x - 6 = y
y = 5x - 6
substitute in equ 2
2x-3(5x - 6) =4
2x - 3*5x + 3*6 = 4
2x - 15x + 18 = 4
- 13x = 4 - 18
-13x = -14
x = -14/-13
x = 14/13
substitute in equ 1
5*14/13 = y+6
70/13 - 6 = y
y = ( 70 - 78)/13
y = -8/13
