Answer:
Option (C)
Step-by-step explanation:
The value of y exceeds the multiplication of x by 2.
The equation that describes this condition is:
y = 2x + 1
Thus, by inserting x values into this equation, we can produce a table reflecting the corresponding input and output values:
x 0 1 2 3 4
y 1 3 5 7 9
Consequently, the table provided in Option (C) is indeed the correct choice.
If rational roots exist, they would be factors of the constant term (9) over factors of the leading coefficient (12).
To determine the rates at which the inlet and outlet pipes fill and empty the reservoir, we remember that work done equals rate multiplied by time. Let’s denote the inlet rate as i and for the outlet pipe as 0. Therefore,
i(24) = 1
o(28) = 1
In this context, the '1' represents the total number of reservoirs, since the problem states the time needed for each pipe to either fill or empty a singular reservoir. Solving for rates yields:
i = 1/24 reservoirs/hour
o = 1/28 reservoirs/hour
Over the first six hours, the inlet pipe fills (1/24)(6) = 1/4 reservoirs and during the same period, the outlet pipe empties (1/28)(6) = 3/14 reservoirs. To calculate the net volume of the reservoir filled, we subtract the emptying total from the filling total:
1/4 - 3/14 = 1/28 reservoirs (note that if emptying exceeds filling, a negative value results. In such cases, treat that negative value as zero, indicating that the outlet rate surpasses the inlet rate, leading to an empty reservoir).
Now we need to find out how long it will take to fill up one reservoir since we’ve already partially filled 1/28 of it, after closing the outlet pipe. In simpler terms, we need to determine the time required for the inlet pipe to finish filling the remaining 27/28 of the reservoir. Fortunately, we have already established the filling rate for the inlet pipe, leading to the equation:
(1/24)t = 27/28
Solving for t gives us 23.14 hours. Remember to add the initial 6 hours to this result since the question seeks the total time. Thus, the final total is 29.14 hours.
Please ask me any questions you may have!
Response:
Step-by-step explanation:
Shift the decimal points in both the divisor and the dividend.
Transform the divisor (the number you're dividing by) into a whole number by moving its decimal to the furthest right. Simultaneously, adjust the dividend's decimal (the number being divided) the same number of places to the right.
In the quotient (the result), place a decimal point directly over where the decimal point is now located in the dividend.
Proceed with the division as normal, ensuring proper alignment so the decimal point appears correctly.
Align each digit in the quotient directly over the last digit of the dividend utilized in that step.
It's false due to the squares being reduced to their minimum values.
