What you don't want is the value of r(t) becoming negative. Surely that would represent water escaping the reservoir.
How big can (t) get before water actually starts escaping the reservoir?
Essentially, to figure this out r(t) would have to be equal to 0.
700 - 40t = 0
40t=700
t=700/40=17.5
So the first answer is 17.5 seconds. After this amount of time has elapsed the reservoir will start to lose water as r(t) would become negative.
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The reservoir had the least amount of water in it before it was being filled. That was when t=0. The volume of water in the reservoir wasn't negatively impacted as not enough water had escaped it during the 17.5 to 30 second period.
Answer:
Step-by-step explanation:
y = (x^2 + 4x) + 2
Take 1/2 of the linear term 4/2 = 2 and square that result. 2^2 = 4.
Put it after 4x
y = (x^2 + 4x + 4) +2 Subtract what you put inside the brackets on the outside.
y = (x^2 + 4x + 4) + 2 - 4 Combine the right.
y = (x^2 + 4x + 4) - 2 Express the brackets as a square.
y = (x + 2)^2 - 2
That's your answer
a = 1
h = 2
k = -2
Here is the information we have:
1. The perimeter is at most 130 cm.
2. The length of the rectangle is 4 times the width.
We can let l stand for length and w for width.
The formula for the perimeter of a rectangle is 2l + 2w.
We have to change the formula a bit.
The length of this rectangle is going to be 4 times the width
So, replace 2l with 2(4w).
Then make this equation equal to 130.
2(4w) + 2w = 130 ; Start
8w + 2w = 130 ; Distribute the 2 across the 4w
10w = 130 ; Combine the like terms 8w and 2w
w = 13 ; Divide both sides by the coefficient of w. Which is 10
Answer:
where is the question so that I can solve it
I think its the promise to not share it with anyone idk
