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:
X=17.9
Step-by-step explanation:
Answer:
A=4ft²
Answerrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr
Step-by-step explanation:
B is the answer b.converse
Answer:
(-2, -6)
Step-by-step explanation:
g(x) = (x+2)²-6
= x²+4x+4-6
= x²+4x-3
the vertex :
x = -4/2 = -2
y = (-2+2)²-6 =0-6 = -6
point of vertex : (-2, -6)
