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2 years ago

Volume of a sphere with a diameter of 8 meters? Use 3.14 for pi.

1 answer:

4 0

Answer: " 267.95 m³ " .
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Explanation:
________________________________________________________

V = (4/3) * $\pi$ * r³ ;

Given: diameter, "d" = "8 m" ;

radius, "r" = 8 m / 2 = 4 m "

V = (4/3) * $\pi$ * r³

V = (4/3) * $\pi$ *(4m)³

V = (4/3) * (3.14) * 4³ * m³ ;

V = (4/3) * (3.14) * (4*4*4) * m³ ;

V = (4/3) * (3.14) * (64) * m³ ;

V = 267.9466666666666667 m³ ;

round to: " 267.95 m³ " .
_________________________________________________________

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Dennis_Churaev [7]

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2 years ago

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2 years ago

Multiply. Check picture.

VashaNatasha [74]

The answer is $3x^4-13x^3-x^2-11x+6$ .

Solution:

Use algebraic identity: $a^m\times a^n=a^{m+n}$

For example: $x^2\times x=x^{2+1}=x^3$

Given expression $(x^2-5x+2)$ and $(3x^2+2x+3)$ .

To multiply these equations.

$(x^2-5x+2)\times(3x^2+2x+3)$

$=x^2(3x^2+2x+3)-5x(3x^2+2x+3)+2(3x^2+2x+3)$

$=(3x^4+2x^3+3x^2)+(-15x^3-10x^2-15x)+(6x^2+4x+6)$

$=3x^4+2x^3+3x^2-15x^3-10x^2-15x+6x^2+4x+6$

Combine like terms together.

$=3x^4+(2x^3-15x^3)+(3x^2-10x^2+6x^2)-15x+4x+6$

$=3x^4-13x^3-x^2-11x+6$

$(x^2-5x+2)\times(3x^2+2x+3)=3x^4-13x^3-x^2-11x+6$

Hence the answer is $3x^4-13x^3-x^2-11x+6$ .

3 0

2 years ago

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3241004551 [841]

Answer:

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Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

g A manufacturer is making cylindrical cans that hold 300 cm3. The dimensions of the can are not mandated, so to save manufactur

sdas [7]

Answer:

The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

Step-by-step explanation:

A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.

Recall that the volume for a cylinder is given by:

$\displaystyle V = \pi r^2h$

Substitute:

$\displaystyle (300) = \pi r^2 h$

Solve for <em>h: </em>

$\displaystyle \frac{300}{\pi r^2} = h$

Recall that the surface area of a cylinder is given by:

$\displaystyle A = 2\pi r^2 + 2\pi rh$

We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.

First, substitute for <em>h</em>.

$\displaystyle \begin{aligned} A &= 2\pi r^2 + 2\pi r\left(\frac{300}{\pi r^2}\right) \\ \\ &=2\pi r^2 + \frac{600}{ r} \end{aligned}$

Find its derivative:

$\displaystyle A' = 4\pi r - \frac{600}{r^2}$

Solve for its zero(s):

$\displaystyle \begin{aligned} (0) &= 4\pi r - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}$

Hence, the radius that minimizes the surface area will be about 3.628 centimeters.

Then the height will be:

$\displaystyle \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2} \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm} \end{aligned}$

In conclusion, the dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

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2 years ago