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vova2212 [387]
1 year ago
11

it takes 126 cubes that have edge lengths of 1/3feet to completely fill this plastic bin what is the area of the base of bin

Mathematics
2 answers:
8_murik_8 [283]1 year ago
6 0

Answer: 1) Each of the 126 cubes has a volume = (1/3)3 = 1/3 × 1/3 × 1/3 = 1/27 cubic feet.

2) It takes 126 cubes to fill the bin, so the bin has a volume of 126 × 1/27 = 126/27 cubic feet. (Don't do this division yet - leave it as an improper fraction)

3) The volume of the bin is also given by the formula Volume = Area of Base × Height. Plus we know the volume is 126/27 cubic feet and the height is 2 1/3 = 7/3 of a foot, so:

Step-by-step explanation: Volume = Area of Base × Height

126/27 = Area of Base × 7/3

126/27 × 3/7 = Area of Base

Arada [10]1 year ago
4 0

The  area of the base of bin is 2 feet².

<h3>What is Volume of Cube?</h3>

The formula of volume of the cube is given by: Volume = a³, where a is the length of its sides or edges.

Given:

Edge = 1/3 feet

Volume of 1 cube= l³= (1/3)³ = 1/27

Totals cubes= 126

So, Volume of 126 cube= 126 x 1/27 = 126/ 27

Then, Volume of Plastic Bin= Area of base x height

126 / 27 =  Area of base x 7/3

Area of base = 126 /27 x 3 /7

Area of base = 2 ft²

Learn more about volume of cube here:

brainly.com/question/11168779

#SPJ2

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---------{NOTICE}----------

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3 years ago
Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

y'(x)= x^2y(x)-\frac12y^2(x)

Putting the value of y'(x) in equation (1)

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

Substituting x =0 and h= 0.2

y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

Substituting x =0.2 and h= 0.2

y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]

\Rightarrow y(0.4)\approx  2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]

\Rightarrow y(0.4)\approx 1.9926

Substituting x =0.4 and h= 0.2

y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]

\Rightarrow y(0.6)\approx  1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]

\Rightarrow y(0.6)\approx 1.6593

Substituting x =0.6 and h= 0.2

y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

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Substituting x =0.8 and h= 0.2

y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]

\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

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