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igomit [66]
3 years ago
11

The ymca lap pool is a right rectangular prism 36.8\, \text{m}36.8m36, point, 8, space, m long by 20 \,\text{m}20m20, space, m w

ide. the pool contains 1472 \text{ m}^31472 m 3 1472, space, m, start superscript, 3, end superscript of water. how deep is the water in the pool?
Mathematics
2 answers:
Jlenok [28]3 years ago
7 0
The volume of a rectangular prism is given by:
 V = (w) * (l) * (h)
 Where,
 w: width
 l: length
 h: height
 Substituting values we have:
 1472 = (20) * (36.8) * (h)
 Clearing h we have:
 h = (1472) / ((20) * (36.8))
 h = 2 m
 Answer:
 
the water in the pool is 2m deep
erastovalidia [21]3 years ago
4 0

Answer:

2 m deep

Step-by-step explanation:

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Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0
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So, the definite integral  \int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74

Given that

\int\limits^1_0 {x^{2} } \, dx = 13

We find

\int\limits^1_0 {(4 - 6x^{2} )} \, dx

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx

So, \int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\=  4[x]^{1}_{0}    - \int\limits^1_0 {6x^{2} } \, dx \\=  4[x]^{1}_{0}    - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0]    - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1]    - 6\int\limits^1_0 {x^{2} } \, dx\\= 4    - 6\int\limits^1_0 {x^{2} } \, dx

Since

\int\limits^1_0 {x^{2} } \, dx = 13,

Substituting this into the equation the equation, we have

\int\limits^1_0 {(4 - 6x^{2} )} \, dx = 4 - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6 X 13 \\= 4 - 78\\= -74

So, \int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74

Learn more about definite integrals here:

brainly.com/question/17074932

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Find the surface area of this net.
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Answer:

<h2>231cm²</h2>

Step-by-step explanation:

First, let's find the surface area of both the triangles

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So, the surface area of the triangles is 15 sq.cm

Now, let's find the surface area of the base (large rectangle in the middle)

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