Answer:
10100=15700(1-0.075)^x
10100=15700(0.925)^x
divide both sides by 15700
0.925^x=101/157
convert decimal
(37/40)^x=101/157
take the log
x= log 37/40 (101/157)
x= 5.65824
X² + y² = 225
x - 7y = -75
x = 7y - 75
x² + y² = 225
(7y-75)² + y² = 225
(7y - 75)(7y - 75) + y² = 225
49y² - 525y - 525y + 5625 + y² = 225
50y² - 1050y + 5400 = 0
50(y² - 21y + 108) = 0
y² - 21y + 108 ⇒ (y - 12)(y - 9)
x = 7(12) - 75
x = 84 - 75
x = 9 ⇒ (9,12)
x = 7(9) - 75
x = 63 - 75
x = -12 ⇒ (-12,9)
So,
We are trying to figure out when Grandpa Lopez's age was twice that of Dad.
Let x represent the number of years before/after when G. Lopez's age was twice that of Dad.
66 + x = 2(37 + x)
Distribute.
66 + x = 74 + 2x
Subtract x from both sides.
66 = 74 + x
Subtract 74 from both sides.
-8 = x
So 8 years ago, G. Lopez was twice as old as Dad. Let's check that.
66 - 8 = 58
37 = 8 = 29
29 * 2 = 58
58 = 58
It checks.
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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