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1 year ago

A rectangular parking lot is 253.5 feet long and 176.5 feet wide. a 55-gallon drum of asphalt sealer covers

Mathematics

1 answer:

sashaice [31]1 year ago

4 0

Answer:

$1,194

Step-by-step explanation:

The area of the parking lot is 253.5(176.5) which equals 44,472.75. You then divide 44,472.75 by 4000 to see how many gallons you need to cover the parking lot. The answer to that was about 11.2 and since you can only purchase full drums you have to round it up to 12 to have enough. Finally, you multiply 12 by 99.50 and get your answer.

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Write the equation of the line that passes through the point (2,-9) and has a slope of -5.

balu736 [363]

Step-by-step explanation:

Find equation

y - y1 = m(x - x1)

y + 9 = -5(x - 2)

y = -5x + 10 - 9

y = -5x + 1 → Equaton

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2 years ago

Work out the surface area of the solid cuboid H=2cm W=4cm L=5cm

Nitella [24]

Answer:

total surface area = 76 sq.cm

Step-by-step explanation:

total surface area of cuboid = 2 ( lb + lh + bh )

( where l is length , b is width and h is height )

so, surface area :

= 2[ (5×4) + (5×2) + (2×4) ]

= 2 [ 20 + 10 + 8 ]

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3 years ago

X2 + y when x = -2 and y = 3.

OlgaM077 [116]

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3 years ago

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yKpoI14uk [10]

I hope this it

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2 years ago

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Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

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

so that the normal vector to $S$ is given by

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

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

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

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

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

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

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

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3 years ago