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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No, this would not be a biased question unless the people asked were all Thai people or all Chinese people. Then the information would be skewed because naturally, Chinese people would favor Chinese food and same for Thai.
9/19 - 3/14
= 126/140 - 57/140
= 69/140
So i'm assuming your eqtn is h = -16t^2 + 27t + 10
and we're looking for t = ? when h = 0
=> 16t^2 - 27t - 10 = 0
(16t + 5)(t - 2) = 0
t = 2 and t = -5/16
answer 2 seconds
Answer:
a) x changes by 6.5 units
b) y changes by 13 units
Step-by-step explanation:
a. Over this interval, how much does x change by?
Initially, we have that x = 2.
In the end, we have that x = 8.5.
So x changes by 8.5-2 = 6.5 units
b. Over this interval, how much does y change by?
Initially, when x = 2, we have that y = 2x + 11 = 2*2 + 11 = 15
In the end, when x = 8.5, we have that y = 2*8.5 + 11 = 28
So y changes by 28 - 15 = 13 units
