Answer:
<u>The Eastbound train speed is 76.5 miles per hour and the Westbound train speed is 79.5 miles per hour.</u>
Step-by-step explanation:
1. Let's review all the information provided for solving this question:
Eastbound train speed = x
Westbound train speed = x + 3 miles per hour (The speed of the westbound train is 3mph greater than the speed of the eastbound train)
Duration of trip = 3 hours
Distance between trains after 3 hours = 468 miles
2. Let's find the speed of each train, using the following equation:
3x + 3(x + 3) = 468
3x + 3x + 9 = 468
6x + 9 = 468
6x = 468 - 9 (Subtracting 9 at both sides)
6x = 459
x = 459/6 (Dividing by 6 at both sides)
x = 76.5 miles per hour
<u>The Eastbound train speed is 76.5 miles per hour and the Westbound train speed is 79.5 miles per hour.</u>
3. Proof that x = 76.5 is correct
3x + 3(x + 3) = 468
3 (76.5) + 3 (76.5 + 3) = 468
229.5 + 238.5 = 468
<u>468 = 468</u>
<u>The value of x = 76. 5 is correct. And now we know that the Eastbound train has traveled 229.5 miles since the departure from Cleveland and the Westbound train 238.5 miles.</u>
<span>Find the wind speed and the plane's airspeed.
:
Let s = speed of the plane in still air
Let w = speed of the wind
then
(s-w) = plane speed against the wind
and
(s+w) = plane speed with the wind
:
Change 3 3/8 hrs to 3.375 hrs
:
The trips there and back are equal distance, (1890 mi) write two distance equations
dist = time * speed
:
3.375(s-w) = 1890
3.0(s + w) = 1890
:
It is convenient that we can simplify both these equations:
divide the 1st by 3.375
divide the 2nd by 3
resulting in two simple equations that can be used for elimination of w
s - w = 560
s + w = 630
----------------adding eliminates w, find s
2s = 1190
s =
s = 595 mph is the plane speed in still air
Find w
595 + w = 630
w = 630 - 595
w = 35 mph is the wind spee</span>
Answer: 7500
Step-by-step explanation:
multiply 150 by 50
Answer:
7 Celsius
Step-by-step explanation:
Initial temperature at 12pm is 1 Celsius.
Rises by 6 (add by 6).
