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photoshop1234 [79]
3 years ago
10

It takes 8 minutes to fill an empty aquarium to a depth of

Mathematics
2 answers:
krek1111 [17]3 years ago
6 0
<h3>Answer:   20 minutes per meter</h3>

===================================================

Work Shown:

8 minutes = (2/5) meters

5*(8 minutes) = 5*( (2/5) meters )

40 minutes = 2 meters

40/2 minutes = 2/2 meters

20 minutes = 1 meter

The unit rate in minutes per meter is 20 minutes per meter.

In other words, it takes 20 minutes to raise the level 1 meter.

Sidana [21]3 years ago
4 0

Answer:

Step-by-step explanation:

Answer

0

andrew8253

Virtuoso

237 answers

42.1K people helped

Answer:

The rate is 15 minutes per meter.

Step-by-step explanation:

It takes 9 minutes to fill a depth of  meters.

Filling   meters take 9 minutes time.

Filling 3 meters of the aquarium takes  = 45 minutes time (Multiplying by 5 both the numbers).

In order to fill 1 meter of the aquarium it will take  minutes (Dividing by 3 both the numerical values).

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Help help i need help anyone ?
REY [17]
-9-6i/-3-2i
Factor -3 out of expression
-3(3+2i)/-3-2i
Then extract the negative sign out of the expression
-3(3+2i)/-(3+2i)
Reduce the fraction with -(3+2i)
-3/-1
-3*-1=3
Hope this helps
3 0
3 years ago
Over what interval is the growth rate of the function decreasing?
aleksklad [387]
I think the rate is decreasing from:
(-infinity, ln9/3)
6 0
3 years ago
Anyone down to zo0m with this chaotic idi0t? I've got nothin to do lol
gladu [14]

Answer:

I wouldddd but I can’t sorry

Step-by-step explanation:

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5 0
3 years ago
Read 2 more answers
How do you do this question?
alina1380 [7]

Answer:

(8√2) / 15

Step-by-step explanation:

A curve bounded by the y-axis is represented by in terms of dy;

\int \:x\:dt

When the curve crosses the y-axis, x will be 0. In this case x is the function of t, so we have to solve for x(t) = 0;

0 = t^2 + 2t --- (1)

Solution(s) => t = 0, t = 2

dy = (1/2 * 1/√t)dt --- (2)

Our solutions (0, 2) are our limits. The area of the curve is in the form A\:=\:\int _b^a\:f\left(t\right)g'\left(t\right)dt , so now let's introduce the limits of integration, x(t) and dy/dt. Remember, dy/dt = (1/2 * 1/√t) (second equation). 1/2 * 1/√t can be rewritten as 1/2 * t^(-1/2)....

A\:=\:\int _2^0\:\left(t^2-2t\right)\left(\frac{1}{2}t^{-\frac{1}{2}}\right)dt\\\\= \int _2^0\:\left(\frac{1}{2}t^{\frac{3}{2}}-t^{\frac{1}{2}}\right)dt\\\\= \left[\frac{t^{\frac{5}{2}}}{5}-\frac{2t^{\frac{3}{2}}}{3}\right]_2^0\\\\= 0\:-\:\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\\\\= \frac{8\sqrt{2}}{15}

Your solution is 8√2 / 15

7 0
2 years ago
V=Bh over 3 for B=15 inches two squared .and h=28inches
Leni [432]
Explain what u need help with cuz i think i know this

3 0
3 years ago
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