Solving a system of equations, we will see that there are 6000 kg of coal on the pile.
<h3>
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
#SPJ1
Answer:
6x-12=-48
6x=-36
x=-6
Step-by-step explanation:
Answer:
(0, -4)
Step-by-step explanation:
The coordinates of the points from which the directed line segment extends = (-6, -6) to (9, -1)
The ratio the required point partitions the line = 2 to 3
The formula for finding the coordinate of a point that partitions a line AB into a ratio 'a' to 'b', where the coordinates of, A = (x₁, y₁) and B = (x₂, y₂) is given as follows;
Therefore, the required point is located as follows;
The coordinates of the point is (0, -4)
Supply is the amount of product a seller is able to make
