Solving a system of equations, we will see that there are 6000 kg of coal on the pile.
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How many kilograms of coal are in the pile of coal?</h3>
Let's say that there are N kilograms of coal, and it is planned to burn it totally in D days, these are our variables.
We know that:
- If the factory burns 1500 kilograms per day, it will finish burning the coal one day ahead of planned.
- If it burns 1000 kilograms per day, it will take an additional day than planned.
Then we can write the system of equations:
(1500)*(D - 1) = N
(1000)*(D + 1) = N
Because N is already isolated in both sides, we can write this as:
(1500)*(D - 1) = N = (1000)*(D + 1)
Then we can solve for D:
(1500)*(D - 1) = (1000)*(D + 1)
1500*D - 1500 = 1000*D + 1000
500*D = 2500
D = 2500/500 = 5
Now that we know the value of D, we can find N by replacing it in one of the two equations:
(1500)*(D - 1) = N
(1500)*(5 - 1) = N
(1500)*4= N = 6000
This means that there are 6000 kilograms of coal on the pile.
If you want to learn more about systems of equations:
brainly.com/question/13729904
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Answer:
18x^2+6
Step-by-step explanation:
2(9x^2+3)
multiply each term in parentheses by 2
2 x 9x^2+2 x 3
calculate product
18x^2+2 x 3
multiply numbers
18x^2+6
Answer:
The next time both will leave the bus station at the same time is 7:15 a.m.
Step-by-step explanation:
Since 15 and 75 are multiples to each other.
So, L.C.M. of 15 and 75 is 75.
So, after 75 minutes both buses will leave the station at the same time.
Also since buses begin their routes at 6 a.m.
So, next time they will meet at
6 hours 0 minutes + 75 minutes = 7 hours 15 minutes = 7:15 a.m.
Answer:
A: 8
B: 96
I'm not sure on C, sorry.