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soldi70 [24.7K]
3 years ago
9

Given that a cubit is equivalent to 17.7 in., find the volume of the ark in cubic feet.

Mathematics
1 answer:
Daniel [21]3 years ago
8 0

Since an ark is 2.5 cubits in length and 1.5 cubits in width and height.

It is given that a cubit is equivalent to 17.7 in.

We have to determine the volume of the ark in cubic feet.

Since 1 feet = 12 inches.

Since, length = 2.5 cubits

= 2.5 \times 17.7 inches

= 44.25 inches

= \frac{44.25}{12}

= 3.7 feet

Now, width = 1.5 cubits

= 1.5 \times 17.7 inches

= 26.55 inches

= \frac{26.55}{12}

= 2.2 feet

Now, height = 1.5 cubits

= 1.5 \times 17.7 inches

= 26.55 inches

= \frac{26.55}{12}

= 2.2 feet

SO, Volume of rectangular prism = length \times width \times height

So, Volume of ark= 3.7 \times 2.2 \times 2.2

= 17.9

= 18 cubic feet.

Therefore, the volume of an ark is 18 cubic feet.

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The equation in spherical coordinates will be a constant, as we are describing a spherical shell.

r(φ, θ) = 8 units.

<h3>How to rewrite the equation in spherical coordinates?</h3>

The equation:

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Defines a sphere of radius R.

Then the equation:

x^2 + y^2 + z^2 = 64

Defines a sphere of radius √64 = 8.

Then we will have that the radius is a constant for any given angle, then we can write r, the radius, as a constant function of θ and φ, the equation will be:

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brainly.com/question/10171109

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Step-by-step explanation:

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Answer:

5x^2+7x+30+\frac{102}{x-3}

Step-by-step explanation:

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}5x^3-8x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{5x^3}{x}=5x^2\\\mathrm{Quotient}=5x^2\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}5x^2:\:5x^3-15x^2\\\mathrm{Subtract\:}5x^3-15x^2\mathrm{\:from\:}5x^3-8x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=7x^2+9x+12\\\mathrm{Therefore}\\=5x^2+\frac{7x^2+9x+12}{x-3}

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{7x^2}{x}=7x\\\mathrm{Quotient}=7x\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}7x:\:7x^2-21x\\\mathrm{Subtract\:}7x^2-21x\mathrm{\:from\:}7x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=30x+12\\\mathrm{Therefore}\\=5x^2+7x+\frac{30x+12}{x-3}\\

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}30x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{30x}{x}=30\\\mathrm{Quotient}=30\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}30:\:30x-90\\\mathrm{Subtract\:}30x-90\mathrm{\:from\:}30x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=102\\\mathrm{Therefore}\\=5x^2+7x+30+\frac{102}{x-3}

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Simplify the parentheses:

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