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katrin2010 [14]
3 years ago
13

Find the area of the polygon to the nearest square inch.

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
5 0

Answer:

137 in.^2

Step-by-step explanation:

A = nsa/2

where

n = number of sides = 9

s = length of side = 4.7 in.

a = length of apothem = 6.5 in.

A = (9)(4.7 in.)(6.5 in.)/2

A = 137.475 in.^2

Answer: 137 in.^2

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B is the answer because you have to math it
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Is a dodecahedron a 3 dimensional or a dodecagon a 3 dimensional ?
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What percent of the first 10 natural numbers are prime numbers? Answer in a percent please.<br> :)
saul85 [17]

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3 years ago
2 y + 4 x = 8, Solve for y
mash [69]

Answer:

y=−2x+4

Step-by-step:

2y+4x=8

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4x+2y+−4x=8+−4x

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2y /2 = −4x+8 /2

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3 years ago
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EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =
geniusboy [140]

The Lagrangian,

L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)

has critical points where its partial derivatives vanish:

L_x=1-\lambda-2\mu x=0

L_y=2+\lambda-2\mu y=0

L_z=9-\lambda=0

L_\lambda=x-y+z-1=0

L_\mu=x^2+y^2-1=0

L_z=0 tells us \lambda=9, so that

L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu

L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}

Then with L_\mu=0, we get

x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2

and L_\lambda=0 tells us

x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}

Then there are two critical points, \left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right). The critical point with the negative x-coordinates gives the maximum value, 9+\sqrt{185}.

8 0
3 years ago
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