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Anna [14]
3 years ago
9

Anton is approved for overtime at his job. Last week he worked a total of 45 hours , how many hours of overtime did he work

Mathematics
1 answer:
pochemuha3 years ago
8 0

Answer:

B. 5 Hours

Step-by-step explanation:

any work over 40 hours in a 168 hour period is counted as overtime. So take 45 (his total) and subtract it by 40 and you get 5. 45 - 40 = 5 hours.

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A rectangular swimming pool is bordered by a concrete patio. the width of the patio is the same on every side. the area of the s
andre [41]
Answer:

x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)

where

l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Explanation: 

Let 

x = width of the patio
l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Since the pool is bordered by a complete patio, 

Length of the pool (with the patio) 
= (length of the pool (w/o the patio)) + 2*(width of the patio)
Length of the pool (with the patio) = l + 2x

Width of the pool (with the patio) 
= (width of the pool (w/o the patio)) + 2*(width of the patio)
Width of the pool (with the patio) = w + 2x

Note that

Area of the pool (w/o the patio)
=  (length of the pool (w/o the patio))(width of the pool (w/o the patio))
Area of the pool (w/o the patio) = lw

Area of the pool (with the patio)
= (length of the pool (w/o the patio))(width of the pool (w/o the patio))
= (l + 2x)(w + 2x)
= w(l + 2x) + 2x(l + 2x)
= lw + 2xw + 2xl + 4x²
Area of the pool (with the patio) = 4x² + 2x(l + w) + lw

Area of the patio
= (Area of the pool (with the patio)) - (Area of the pool (w/o the patio))
= (4x² + 2x(l + w) + lw) - lw
Area of the patio = 4x² + 2x(l + w)

Since the area of the patio is equal to the area of the surface of the pool, the area of the patio is equal to the area of the pool without the patio. In terms of the equation,

Area of the patio = Area of the pool (w/o the patio)
4x² + 2x(l + w) = lw
4x² + 2x(l + w) - lw = 0    (1)

Let 

a = numerical coefficient of x² = 4
b = numerical coefficient of x = 2(l + w)
c = constant term = -lw

Then using quadratic formula, the roots of the equation 4x² + 2x(l + w) - lw = 0 is given by

x = \frac{-b \pm  \sqrt{b^2 - 4ac}}{2a}
\\ = \frac{-2(l + w) \pm  \sqrt{(2(l + w))^2 - 4(4)(-lw)}}{2(4)} 
\\ = \frac{-2(l + w) \pm  \sqrt{(4(l + w)^2) + 16lw}}{8} 
\\ = \frac{-2(l + w) \pm  \sqrt{(4(l^2 + 2lw + w^2) + 4(4lw)}}{8}
\\ = \frac{-2(l + w) \pm  \sqrt{(4(l^2 + 2lw + w^2 + 4lw)}}{8}
\\ = \frac{-2(l + w) \pm  \sqrt{(4(l^2 + 6lw + w^2)}}{8}
= \frac{-2(l + w) \pm 2\sqrt{l^2 + 6lw + w^2}}{8} \\= \frac{2}{8}(-(l + w) \pm \sqrt{l^2 + 6lw + w^2}) \\x = \frac{1}{4}(-(l + w) \pm \sqrt{l^2 + 6lw + w^2}) \\\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right) \text{ or }}
\\\boxed{x = -\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2} \right)}


Since (l + w) + \sqrt{l^2 + 6lw + w^2} \ \textgreater \  0, -\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2}\right) is negative. Since x represents the patio width, x cannot be negative. Hence, the patio width is given by 

\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)}




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