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____ [38]
3 years ago
13

During the cold winter months, a sheet of ice covers a lake near the Arctic Circle. At the beginning of spring, the ice starts t

o melt. The variable sss models the ice sheet's thickness (in meters) ttt weeks after the beginning of spring. s=-0.25t+4s=−0.25t+4s, equals, minus, 0, point, 25, t, plus, 4 By how much does the ice sheet's thickness decrease every 666 weeks? meters
Mathematics
1 answer:
Masja [62]3 years ago
3 0

Answer: The ice sheet's thickness decrease every 6 weeks= 1.5 meters

Step-by-step explanation:

GIven: At the beginning of spring, the ice starts to melt. The variable sss models the ice sheet's thickness (in meters) t weeks after the beginning of spring. s=-0.25t+4

At t=0, s=4

So, Thickness of ice sheet at the beginning = 4 meters

Now at t= 6, we get

s=-0.25(6)+4=-1.5+4=2.5

Thickness of ice sheet after 6 weeks = 2.5 meters

Decrease in thickness = 4 meters - 2.5 meters = 1.5 meters

Hence, the ice sheet's thickness decrease every 6 weeks= 1.5 meters

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3 years ago
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
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= (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}&#10;\\ = \frac{-2(l + w) \pm  \sqrt{(2(l + w))^2 - 4(4)(-lw)}}{2(4)} &#10;\\ = \frac{-2(l + w) \pm  \sqrt{(4(l + w)^2) + 16lw}}{8} &#10;\\ = \frac{-2(l + w) \pm  \sqrt{(4(l^2 + 2lw + w^2) + 4(4lw)}}{8}&#10;\\ = \frac{-2(l + w) \pm  \sqrt{(4(l^2 + 2lw + w^2 + 4lw)}}{8}&#10;\\ = \frac{-2(l + w) \pm  \sqrt{(4(l^2 + 6lw + w^2)}}{8}
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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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A certain type of cereal costs $3.79 for 19.5 ounces. What is the unit rate?
Radda [10]

Step-by-step explanation:

$3.79 ÷ 19.5 = .194

answer = $.19 per oz

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2 years ago
Read 2 more answers
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