I think the equation is y= -1/3x+1. The slope is going down, so it’s negative. And the y intercept is 1. And how I got a fraction is by using rise/run.
Y = 125000
x = 0.25(125000) + 62500 = 31250 + 62500 = 93750
Answer: 93750
We want to know when we brought it back in to the room. at that time
the temperature difference is 80 - 42 = 38 so we have to solve
<span>
38=9<span>e<span>^−.3344t</span></span></span> for t
this time we get
<span><span>38 / 9</span>=<span>e^<span>−.3344t</span></span></span>
<span>t=<span><span>ln(<span>38 / 9</span>)</span><span> / −.3344</span></span></span>
<span>t=−4.3</span>
so 4.3 minutes before 2:10
False. All points on the graph of an equation satisfies the equation.
Solving a system of equations, we will see that there are 6000 kg of coal on the pile.
<h3>
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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