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wariber [46]
3 years ago
7

The level of water in a dam was decreasing by 20% each day. If the level of water was 1500cm,what was the level after two days?

Mathematics
1 answer:
makkiz [27]3 years ago
6 0

Answer:

L_o= 1500 cm

And we know that the level is decreasing 20% each day so then we can use the following model for the problem

L_f = L_o (1-0.2)^t

Where t represent the number of days and L the level for this case we want to fnd the level after two days so then t =2 and replacing we got:

L_f = 1500cm(0.8)^2 = 960cm

Step-by-step explanation:

For this case we know that the initial volume of water is:

L_o= 1500 cm

And we know that the level is decreasing 20% each day so then we can use the following model for the problem

L_f = L_o (1-0.2)^t

Where t represent the number of days and L the level for this case we want to fnd the level after two days so then t =2 and replacing we got:

L_f = 1500cm(0.8)^2 = 960cm

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2 years ago
Eden just bought a trough in the shape of a rectangular prism for her horses. She needs to know what volume of water to add to t
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The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Let w represent the width, hence:

length = w + 33, height = w - 13

Volume (V) = w(w + 33)(w - 13) = w³ + 20w² - 429w

V(w) = w³ + 20w² - 429w

Rate of change = dV/dw = 3w² + 40w - 429

When w = 38, dV/dw = 3(38)² + 40(38) - 429 = 5423

When w = 53, dV/dw = 3(53)² + 40(53) - 429 = 10118

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The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in

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4 0
1 year ago
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From the given information,

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Read 2 more answers
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