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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V = Bh
.. = (20 ft^2)*(5 ft)
.. = 100 ft^3
The volume of this rectangular prism is 100 ft^3.
Answer:
c.-10
Step-by-step explanation:
4x+2=11
x=9/4...(1)
now,
-4x-1=-4x(9/4)-1=-10
(5x)^2 + x^2 = c^2
25x^2 + x^2 =c^2
sqrt 26 x ^ 2 = sqrt c^2
x sqrt 26= c
so answer is 6x + x sqrt 26
Answer:
p(r+1)
Step-by-step explanation:
Hope this helps
