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3 years ago

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muhammad, a shipping and receiving clerk uses 1 yard, 2 feet and 4 inches Of twine to wrap each carton he packs. he has to pack

29 cartons today. how much twine will he need ?

1 answer:

antiseptic1488 [7]3 years ago

8 0

Answer:

<em>He will need 51 yards, 1 foot and 8 inches twine to wrap 29 cartons. </em>

Step-by-step explanation:

Muhammad uses 1 yard, 2 feet and 4 inches of twine to wrap each carton he packs.

<u>We know that, 1 yard = 36 inches and 1 foot = 12 inches</u>.

So, 1 yard, 2 feet and 4 inches $=[(1*36)+(2*12)+4]inches = 64$ inches.

That means, he uses 64 inches of twine to wrap each carton.

He has to pack 29 cartons today. So, <u>the total length of the twine he will need</u> $=(64*29) inches= 1856$ inches.

Now....

1856 inches

$=\frac{1856}{36} yards \\ \\= 51 \frac{5}{9} yards \\ \\= 51 yards+(\frac{5}{9}*36)inches \\ \\= 51 yards, 20 inches \\ \\= 51 yards, (12+8)inches \\ \\=51 yards, 1 foot, 8 inches.$

Thus, he will need 51 yards, 1 foot and 8 inches twine to wrap 29 cartons.

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Answer:

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Step-by-step explanation:

a)

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

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On one hand we have,

$\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}$

On the other hand,

$\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}$

So

$\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

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The domain of definition of F is

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i.e., all the plane X-Y except the (0,0)

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