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andrezito [222]
3 years ago
13

Robinson ' s square shaped air conditioner covers 144 feet of area of his room. what is the length of any side?

Mathematics
2 answers:
Flauer [41]3 years ago
4 0
If it's a square, then the sides must be equal. So, find the square root of 144, which is 12. To make sure you are correct, you can find the area of a square with 12 feet sides, which is 144. 12 is correct. 
astraxan [27]3 years ago
4 0
Hey there,
Area = 144 square feet
Length = square root 144 square feet 
            = 12 cm

Hope this helps :))

<em>~Top♥</em>
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I do not need an explanation, just a simple angle number!
Murljashka [212]

Answer:

Step-by-step explanation:

a + 143 = 180    {Linear pair}

a = 180 - 143

a = 37

37 + 93 + b = 180 {Angle sum property of triangle}

130 + b = 180

         b = 180 - 130

         b = 40

c + b = 90

c  + 40 = 90

c = 90 - 40

c = 50

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d = 180 - 120

d = 60

? = 60 + 40         {Exterior angle property of triangle}

? = 100

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The mean of the following data values is 32. <br> 19, 23, 35, 41, 42<br> A. True <br> B. False
wariber [46]

Answer:

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Step-by-step explanation:

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3 years ago
3f + 9 &lt; 21<br> Help please
fiasKO [112]

Answer:

f < 4

Step-by-step explanation:

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Complete the table to show the steps for combining like terms
Fantom [35]

Answer:

check the illustration I provided

Step-by-step explanation:

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2 years ago
For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f
butalik [34]

\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k

Let

\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k

The curl is

\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)

where \partial_\xi denotes the partial derivative operator with respect to \xi. Recall that

\vec\imath\times\vec\jmath=\vec k

\vec\jmath\times\vec k=\vec i

\vec k\times\vec\imath=\vec\jmath

and that for any two vectors \vec a and \vec b, \vec a\times\vec b=-\vec b\times\vec a, and \vec a\times\vec a=\vec0.

The cross product reduces to

\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

\nabla\times\vec f=\vec0

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Integrate both sides of

\dfrac{\partial f}{\partial y}=2xze^{2xyz}

with respect to y and

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Differentiate both sides with respect to x and

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Now

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)

and differentiating with respect to z gives

\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}

2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}

\dfrac{\mathrm dh}{\mathrm dz}=0

\implies h(z)=C

for some constant C. So

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C

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4 years ago
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