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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To see what are the factors use, we write each of the numbers, as product of prime factors:
As we can see, the 4 factors used to produce the numbers in the list are {2, 3, 5, 7}
Answer: {2, 3, 5, 7}
Pay attention to the even numbers
Answer:
- 17900 + 700
answer to the equation would be -17,200
Answer: Ruth have to make 7 hats to break even.
Step-by-step explanation:
Break even point is a point where Cost = Revenue.
Let x be the number of hats sold.
Given: Cost of booth = $63
Each hat cost = $10
Selling price for each cost = $19
As per given,
Total cost = Cost of booth+ (Cost of each hat)(Number of hats)
= 63+10x
Total selling price = 19x
For break even,
63+10x=19x
9x=63
x=7
Hence, Ruth have to make 7 hats to break even.