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:
![\sqrt[5]{2^4}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B2%5E4%7D)
Step-by-step explanation:
Maybe you want 2^(4/5) in radical form.
The denominator of the fractional power is the index of the root. Either the inside or the outside can be raised to the power of the numerator.
![2^{\frac{4}{5}}=\boxed{\sqrt[5]{2^4}=(\sqrt[5]{2})^4}](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%3D%5Cboxed%7B%5Csqrt%5B5%5D%7B2%5E4%7D%3D%28%5Csqrt%5B5%5D%7B2%7D%29%5E4%7D)
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In many cases, it is preferred to keep the power inside the radical symbol.
Answer:
20×(5×(-16) = (20×5)×(-16)
Step-by-step explanation:
They both equal -1600
Answer:
y=25
Step-by-step explanation:
y=4+4x+x^2
x=3
y=4+4(3)+(3)^2
y=4+12+(3)^2
y=4+12+(3x3)
y=4+12+(9)
y=16+9
y=25
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