Answer:
The speed of the first train is 45 mph and the speed of the second train is 75 mph
Step-by-step explanation:
Let x represent the speed of the first train in mph. Since the second train, is 30 mph faster then the first, therefore the speed of the second train is (x + 30).
The first train leaves at 1:00 pm, therefore at 6:00 pm, the time taken is 5 hours. Therefore the distance covered by the first train at 6:00 pm = x mph * 5 hours = 5x miles
The second train leaves at 3:00 pm, therefore at 6:00 pm, the time taken is 3 hours. Therefore the distance covered by the second train at 6:00 pm = (x + 30) mph * 3 hours = (3x + 90) miles
Since the second train overtakes the first at 6:00 pm, hence:
3x + 90 = 5x
2x = 90
x = 45
Therefore the speed of the first train is 45 mph and the speed of the second train is 75 mph (45 mph + 30 mph).
There is not enough information to answer this question.
Answer:
Is it clockwise rotation or counterclock wise?
If it’s clockwise rotation then C is the answer but if it’s counter lock wise rotation then the answer is A.
Hope this helps!
Hi.
Answer:

The answer should have positive sign but not negative.
Step-by-step explanation:
First, you add by 8 from both sides.

Then, simplify.

Next, you do is subtract by 4x from both sides.

Simplify.

Therefore, you divide by 2 from both sides.

Finally, you solve and simplify.


X=17 is the correct answer.
Hope this helps you!
Have a nice day! :)
Answer:
1. Volume of a cone: 1. V = (1/3)πr2h 2. Slant height of a cone: 1. s = √(r2 + h2) 3. Lateral surface area of a cone: 1. L = πrs = πr√(r2 + h2) 4. Base surface area of a cone (a circle): 1. B = πr2 5. Total surface area of a cone: 1. A = L + B = πrs + πr2 = πr(s + r) = πr(r + √(r2 + h2))
Step-by-step explanation:
1. Volume of a cone: 1. V = (1/3)πr2h 2. Slant height of a cone: 1. s = √(r2 + h2) 3. Lateral surface area of a cone: 1. L = πrs = πr√(r2 + h2) 4. Base surface area of a cone (a circle): 1. B = πr2 5. Total surface area of a cone: 1. A = L + B = πrs + πr2 = πr(s + r) = πr(r + √(r2 + h2))