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Fantom [35]
3 years ago
14

What is the surface area of this rectangular prism?

Mathematics
1 answer:
Lesechka [4]3 years ago
4 0

Answer:

160cm^2

Step-by-step explanation:

So the surface area is basicly the sum of all the areas of the rectangles in this prism.

So the formula can be written as  2(wl + hl+hw).

The w stands for width, the h stands for height, and the l stands for length.

5cm is the width, 10 cm is the length, and 2 cm is the height.

So inserting all those values for the variabls the formula now looks like this.

2((5)(10)+(2)(10)+(2)(5)).

So 5*10 is <u><em>50</em></u>

2*10 is <u><em>20</em></u>

2*5 is <em><u>10</u></em>

So adding all that up we get 80.

Then multiplyong that by 2 we get 160cm^2.

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What is if g(x,y,z) = x + y and S is the first octant portion of the plane 2x + 3y + z = 6 ?
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The question asks for the value of I=\int\int_Sx+y\textrm{ }dS where S=\{(x,y,z)\mid2x+3z+y=6,x\ge0,y\ge0,z\ge0\}.

First let's look at what that surface looks like.

Letting y=z=0 yields x=3
<span>Letting x=z=0 yields y=2
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</span>
Therefore S is the area of the triangle defined by the three points (3,0,0),(0,2,0),(0,0,6).

We can thus reformulate the integral as I=\int_{z=0}^6\int_{x=0}^{6-z}x+ydxdz.

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</span>I=\int_{z=0}^6\left[2x+\frac{x^2}6-\frac{zx}3\right]_{x=0}^{6-z}dz=\int_{z=0}^62(6-z)+\frac{(6-z)^2}6-\frac{z(6-z)}3\right]dz

<span>I=\int_{z=0}^6\frac{z^2}2-6z+18=\left[\frac{z^ 3}6-3z^2+18z\right]_{z=0}^6=36-108+108</span>

Hence \boxed{I=\int\int_Sx+y\textrm{ }dS=36}

<span>


</span>
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