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.
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Each side is 6 feet.
let s be 1 side of the square
24 = 4s
6 = s
15ft because 12x.25=3 so 3ft is 25% of the 12ft, add the 3+12=15
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Answer: 
</h3>
n starts at 1, and n is a positive whole number (1,2,3,...)
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Explanation:
The sequence is arithmetic with first term 40 and common difference 10. Meaning we add 10 to each term to get the next one.
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a1 = 40 = first term
d = 10 = common difference
is the general nth term of this arithmetic sequence
Plug in n = 1 and you should get 
Plug in n = 2 and you should get 
and so on
The two expressions that are comparable to one another for the total cost of the cloth are and
(7.99 × 7/3) and 18.64
This is further explained below.
<h3>What is
expressions ?</h3>
In most cases, the two expressions that are comparable to one another for the total cost of the cloth are
(7.99 × 7/3) and 18.64
Total cost = Cost of fabric per yard * Number of fabrics
= 7.99 × 2 1/3
= (7.99 × 7/3)
= 55.93 / 3
= 18.64
As a result, the two expressions that are comparable to one another for the total cost of the cloth are and
(7.99 × 7/3) and 18.64
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