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:
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Answer:
0.0625 of the students recorded the show.
Step-by-step explanation:
2/3×3/8×1/4= the fraction of people who watched and recorded the show (0.0625)
Answer:
44
Step-by-step explanation:
22 for every six kids
12 is 6 doubled so you double 22 to get 44
Answer:
Step-by-step explanation:
x+165 = 180
x = 180 - 165
x = 15
Answer : Option A
Explanation:
we can clearly see that from x+ y=5 we can substitute y = 5-x in y = 9x^2
Option B is clearly incorrect x can not be y -5
Option C is incorrect because y can not be 5 +x
Option D is incorrect because its y = 9x^2 not y^2 =9x^2 so y can not be 3x
if its y^2 = 9x^2 then only implies y = 3x