The rate at which the water was drained from the tank is 6 gallons of water per minute (option A)
To know the speed at which the tank was drained, we must carry out the following operations:
Taking into account the information provided by the table we establish how much water has been drained from the tank in 20 minutes.
450 gallons - 330 gallons = 120 gallons.
Subsequently, we must divide the amount of water (120 gallons) that was drained by the time it was drained (20 minutes).
120 gallons ÷ 20 minutes = 6 gallons / minute.
According to the above, we can establish that the bath drain rate is 6 gallons / minute (option A).
Note: This question is incomplete because there is some missing information. Here is the complete information:
What is the rate at which the water was drained from the tank?
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6 gallons of water per minute.
- 11 gallons of water per minute.
- 45 gallons of water per minute.
- 120 gallons of water per minute.
Learn more about linear equation in: brainly.com/question/11897796
18.2 divided by 18 is 1.011
23. The answer is B) x-6.
23.) The answer is D) x + 3
First one = 9x3=27
Second one 12x9=108
3rd one = 3x4=12
Idk what that last one is
Part A. You have the correct first and second derivative.
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Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.
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Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out
To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0.
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well.
