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olga_2 [115]
3 years ago
7

Abandoned mines frequently fill with water. Before an abandoned mine can be reopened, the water must be pumped out. The size of

pump required depends on the depth of the mine. If pumping out a mine that is D feet deep requires a pump that pumps a minimum of + 4D – 250 gallons per minute, pumping out a mine that is 150 feet deep would require a pump that pumps a minimum of how many gallons per minute?
362


500


800


1,250


1,750
Mathematics
1 answer:
nekit [7.7K]3 years ago
5 0

Answer:

350

Step-by-step explanation:

using the formula given

4D -250 substitute 150 for D

4(150)-250 = 350.

it's not shown? but it's the answer. unless there are pieces missing to the question.

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Please help me with this!!!
guajiro [1.7K]

Answer: 41

Step-by-step explanation:

9*(-2)*(-2) - 4*(-2) - 3 = 9*4 + 8 - 3 = 36 + 5 = 41

3 0
3 years ago
Coefficient of x^2 in the expansion of (x-3)^3
Mila [183]

Answer:

-9

Step-by-step explanation:

(x-3)(x-3)(x-3)

(x^2-6x+9)(x-3)

x^3-9x^2+27x-27

Therefore, the coefficient of x^2 is -9.

7 0
3 years ago
Read 2 more answers
If g(x) = 3x^2+2x-1. Find g(-1).​
Mandarinka [93]

Step-by-step explanation:

g(x) = 3x² + 2x - 1

g(-1) = 3(-1)² + 2(-1) - 1

= 3(1) -2 -1

= 3 - 3

= 0

5 0
3 years ago
What is the product of 3y<br> product of (3y^)(274)
Dima020 [189]

3y^-4 × (2y^-4) = 6y^-8

= 6/1/y^8

= 6/y^8

.................................................

1) When x^1 multiply with x^2, 1 will add to 2 and become 3

= x^3

2) y^-1 = 1/y

3^-2 = 1/3^2 = 1/9

5^-2 = 1/5^2 = 1/25

5 0
3 years ago
In May you used 900 ​kilowatt-hours of energy for electricity.
Sloan [31]

Answer:

a) "The total electrical energy use was 3,240,000,000 joules."

b) "The average power use is 1210 watts."

c)

<u>Part 1:</u> "To generate the electricity you​ used, 810 L is needed."

<u>Part 2:</u> "To generate the electricity you​ used, 214.00 gal is needed."

Step-by-step explanation:

a)

1 kilowatt-hours = 3.6x10^6 joules

Converting 900 kilowatt-hours to joules, we have:

900*(3.6*10^6)=3.24*10^9 joules, or

Total Electrical Energy use = 3,240,000,000 joules

"The total electrical energy use was 3,240,000,000 joules."

b)

The month of May has 31 days and 1 day  is 24 hours. So May has:

31*24=744 hours

Now, we divide 900 kW-hr by the number of hours in the month (744 hrs) to get average power use:

\frac{900}{744}=1.21 kW. Since 1000 Watts = 1 kW, we multiply this by 1000 to get the answer in Watts:

1.21 * 1000 = 1210 Watts

"The average power use is 1210 watts."

c)

<u>Part 1:</u>

The conversion efficiency of most generating stations is 33%. So we need to multiply the total electrical energy use (in joules) by a factor of 3 to get the amount of joules required. So:

(3.24 * 10^9)*3 = 9.72 * 10^9 joules

We divide this by 12 million joules to get the number of liters:

\frac{9.72*10^9}{12*10^6}=810

So, 810 liters is needed

"To generate the electricity you​ used, 810 L is needed."

<u>Part 2:</u>

We know that 0.2642 gallons is equal to 1 liters. To get the number of gallons needed, we multiply 810 by 0.2642. So:

810 * 0.2642 = 214.00 gallons

"To generate the electricity you​ used, 214.00 gal is needed."

7 0
3 years ago
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