Answer:
6kg pure copper and 30 kg 10% copper was mixed to give 36kg of 25% alloy
Step-by-step explanation:
Here, we want to produce 36 kg of 25% alloy
Let the Pure copper be x kg while 10% alloy be y kg
Pure copper is simply 100% copper
Thus;
x + y = 36 •••••(i)
Then;
100% of x + 10% of y = 25% of 36
= x + 0.1y = 9 •••••• ii)
From i x = 36-y
from ii, x = 9-0.1y
Equate both x
36-y = 9-0.1y
36-9 = 0.1y + y
0.9y = 27
y = 27/0.9
y = 30
x = 36-y
x = 36-30
x = 6 kg
<h3>
Answer: x(x+1)(5x+9) </h3>
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Work Shown:
5x^3 + 14x^2 + 9x
x( 5x^2 + 14x + 9 )
To factor 5x^2 + 14x + 9, we could use the AC method and guess and check our way to getting the correct result.
A better way in my opinion is to solve 5x^2 + 14x + 9 = 0 through the quadratic formula
Then use those two solutions to find the factorization
x = -1 or x = -9/5
x+1 = 0 or 5x = -9
x+1 = 0 or 5x+9 = 0
(x+1)(5x+9) = 0
So we have shown that 5x^2 + 14x + 9 factors to (x+1)(5x+9)
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Overall,
5x^3 + 14x^2 + 9x
factors to
x(x+1)(5x+9)
Answer:
ai) 73000 gal/yr
aii) $730 per year
b) 1000 days
Step-by-step explanation:
ai) The water usage per day is ...
(4 showers/day)(10 min/shower)(5 gal/min) = 200 gal/day
Then the usage per year is ...
(200 gal/day)(365 days/yr) = 73,000 gal/yr
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aii) The cost of electricity is the cost of heating half the water usage, so is ...
(0.20 kWh/gal)(1/2)(73,000 gal/yr)($0.10/kWh) = $730/yr
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b) The current daily cost of electricity for heating water is ...
($730/yr)/(365 days/yr) = $2/day
If the cost is cut in half, it will be $1 per day. That means the savings is $1 per day, so it will take 1000 days to recover the $1000 initial cost. (Effectively, the average cost for the first 1000 days is the same as if the water heater had not been replaced.)
It is x=-3 just use Photomath bro