For problems such as this, it can be handy to remember that
.. 1/(1/a +1/b) = ab/(a +b)
The time required to do a job is the inverse of the rate at which the job is done. Rates are added when people work together (in math problems, not in real life).
(2:44)*(3:14)/(2:44 +3:14) = 1:28:52.3
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2:44 = 41/15 hours = 82/30 hours
3:14 = 97/30 hours
Using the formula above, we have a "working together" time of
.. (82/30)*(97/30)/(82/30 +97/30) = 82*97/(30*179) = 3977/2685 . . . hours
.. = 1 hour 28 156/179 minutes
.. = 1 hour 28 minutes 52 52/179 seconds
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Learn to use the fraction functions on your calculator. They can be very helpful.
Answer:
THEY messed you upon this one and I don't see if you have to write or not!
More details please and thank you!
Step-by-step explanation:
Answer:
T maximum=T average -7.8 seconds
T minimum=T average +7.8 seconds
Step-by-step explanation:
Calculation for the equation that can be
use to find the maximum and minimum times for the track team
Using this equation to find the maximum times for the track team
T maximum=T average -7.8 seconds
T maximum=64.6 seconds-7.8 seconds
Using this equation to find the minimum times for the the track team
T minimum=T average +7.8 seconds
T minimum=64.6 seconds +7.8 seconds
Therefore the equation for the maximum and minimum times for the track team are :
T maximum=T average -7.8 seconds
T minimum=T average +7.8 seconds
Step 1. Simplify
7x^2 - 8x - 4 - 2x^2 + 3x + 5 + 5x^2 - 10x - 8
Step 2. Collect like terms
(7x^2 - 2x^2 + 5x^2) + (-8x + 3x - 10x) + (-4 + 5 - 8)
Step 3. Simplify
10x^2 - 15x - 7
12 / 3 so the answer would be 4
