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AURORKA [14]
1 year ago
11

Zoe is packaging a gift to send in the mail. A smaller gift box is placed inside a similar yet larger packaging box. The empty s

pace inside the packaging box is filled with tissue paper to keep the gift box from moving. The dimensions of the larger box are each 2.1 times longer than the corresponding dimensions of the smaller box. The volume of the smaller box is 24 cubic centimeters. What is the volume of the space filled with tissue paper in the packaging box? 81.84 cubic centimeters 105.84 cubic centimeters 198.264 cubic centimeters 222.264 cubic centimeters
Mathematics
1 answer:
oksian1 [2.3K]1 year ago
4 0

The volume of the space filled with tissue paper in the packaging box is 198.264 cubic centimeters

<h3>Volume</h3>

  • Volume if small, Vs = 24 cubic inches

Constant = k

= 2.1³

= 9.261

Volume of large : volume of small = k

Vl / Vs = k

Vl = Vs × k

= 24 × 9.261

= 222.264 cubic inches

Therefore,

Volume of tissue paper = volume of large - volume of small

= 222.264 cubic inches - 24 cubic inches

= 198.261 cubic inches

Learn more about volume:

brainly.com/question/3692276

#SPJ1

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Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices
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Step-by-step explanation:

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\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt

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\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy

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a) Directly

Let us break down C into 4 paths \large C_1,C_2,C_3,C_4 which represents the sides of the rectangle.

\large C_1 is the line segment from (0,0) to (5,0)

\large C_2 is the line segment from (5,0) to (5,1)

\large C_3 is the line segment from (5,1) to (0,1)

\large C_4 is the line segment from (0,1) to (0,0)

Then

\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize \large C_1 as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

\large \displaystyle\int_{C_1}xydx+x^2dy=0

 Now the second integral. We parametrize \large C_2 as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25

The third integral. We parametrize \large C_3 as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2

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r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

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