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vazorg [7]
3 years ago
10

A square sheet of art paper has an area of 625 square inches. What is the minimum side length of an easel that supports the whol

e sheet of paper?
Mathematics
2 answers:
Mila [183]3 years ago
7 0
If the area of the paper is 625 sq in, the length of one side of this square is sqrt(625 sq in), or 25 inches.  That's the min. side length of an easel that is to support the whole sheet of paper along its lowest side.
svetlana [45]3 years ago
4 0

area of a square = side^2

 to find the length of a side using area find the square root of the area

sqrt(625) = 25

so the easel has to be at least 25 inches

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Step-by-step explanation:

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Parameterize S by the vector function

\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k

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\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k

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In this case, the normal vector is

\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k

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\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx

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\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}

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Y=f(x)=(1/6)^x find f(x) when x=1/4
SashulF [63]
I think it is 0.64..
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4 years ago
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75% of what equals 48?
Alika [10]

Answer:

75 percent of 64 = 48

Step-by-step explanation:

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3 years ago
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