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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Adam is incorrect because if you substitute the value of y (which is 2) on the expression 52-3y=20, the statement is false since 46 is obviously not equal to 20.
52-3(2)=20
52-6=20
46 = 20
FALSE
Answer:
(1,155, 1.545).
Step-by-step explanation:
From Normal distribution tables the z score for 95% CI = 1.96.
Confidence Interval
= 1.35 +/- 1.96 * 0.63/√40
= 1.35 +/- 0.195
= (1,155, 1.545).
Yeah it’s 33.3 because there 3 feet in a yard and 100 divided by 3 is 33.3 repeating.
Step-by-step explanation:
I am not sure what your problem here is.
you understand the inequality signs ?
anyway, to get
6×f(-2) + 3×g(1)
we can calculate every part of the expression separately, and then combine all the results into one final result.
f(-2)
we look at the definition.
into what category is -2 falling ? the one with x<-2, or the one with x>=-2 ?
is -2 < -2 ? no.
is -2 >= -2 ? yes, because -2 = -2. therefore, it is also >= -2.
so, we have to use
1/3 x³
for x = -2 that is
1/3 × (-2)³ = 1/3 × -8 = -8/3
g(1)
again, we look at the definition.
into what category is 1 falling ? the one with x > 2 ? or the one with x <= 1 ?
is 1 > 2 ? no.
is 1 <= 1 ? yes, because 1=1. therefore it is also <= 1.
so we have to use
2×|x - 1| + 3
for x = 1 we get
2×0 + 3 = 3
6×f(-2) = 6 × -8/3 = 2× -8 = -16
3×g(1) = 3× 3 = 9
and so in total we get
6×f(-2) + 3×g(1) = -16 + 9 = -7
