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Ilya [14]
2 years ago
12

What is the radius of a sphere with a volume of 384\text{ ft}^3,384 ft

Mathematics
1 answer:
Alecsey [184]2 years ago
3 0

Answer:

r≈4.5 ft

Step-by-step explanation:

r≈4.5 ft

\text{Volume of a Sphere:}

Volume of a Sphere:

V=\frac{4}{3}\pi r^3

V=

3

4

​

πr

3

384=

384=

\,\,\left(\frac{4}{3}\pi\right) r^3

(

3

4

​

π)r

3

384=

384=

\,\,(4.1887902)r^3

(4.1887902)r

3

Evaluate 4/3pi in calc

\frac{384}{4.1887902}=

4.1887902

384

​

=

\,\,\frac{(4.1887902)r^3}{4.1887902}

4.1887902

(4.1887902)r

3

​

Evaluate \frac{4}{3}\pi

3

4

​

π in calc

91.6732472=

91.6732472=

\,\,r^3

r

3

\sqrt[3]{91.6732472}=

3

 

91.6732472

​

=

\,\,\sqrt[3]{r^3}

3

 

r

3

​

Cube root both sides

4.5090066=

4.5090066=

\,\,r

r

\text{Final Answer:}

Final Answer:

r\approx 4.5\text{ ft}

r≈4.5 ft

Round to nearest tenth

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There are several ways two triangles can be congruent.

  • \mathbf{AC = BD}<em> congruent by SAS</em>
  • \mathbf{\angle ABC \cong \angle BAD}<em> congruent by corresponding theorem</em>

In \mathbf{\triangle AOL} and \mathbf{\triangle BOK} (see attachment), we have the following observations

1.\ \mathbf{AO = DO} --- Because O is the midpoint of line segment AD

2.\ \mathbf{BO = CO} --- Because O is the midpoint of line segment BC

3.\ \mathbf{\angle AOB =\angle COD} ---- Because vertical angles are congruent

4.\ \mathbf{\angle AOC =\angle BOD} ---- Because vertical angles are congruent

Using the SAS (<em>side-angle-side</em>) postulate, we have:

\mathbf{AC = BD}

Using corresponding theorem,

\mathbf{\angle ABC \cong \angle BAD} ---- i.e. both triangles are congruent

The above congruence equation is true because:

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Applying Properties of Exponents In Exercise,use the properties of exponents to simplify the expression.
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Answer:

1.~e^{-2} \\2.~e^{\frac{7}{2}}\\3.~e^8\\4.~e^{\frac{-11}{2}}

Step-by-step explanation:

We have to simplify the given exponential exponents.

Exponential Properties:

e^0 =1\\e^a.e^b = e^{a+b}\\\\\displaystyle\frac{e^a}{e^b} = e^{a-b}\\\\(e^a)^b = e^{ab}\\\\e^{-a} = \frac{1}{e^a}

Simplification takes place in the following manner:

a)

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b)

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c)

(e^{-2})^{-4}\\(e^a)^b = e^{ab}\\= e^{-2\times -4}\\=e^8

d)

(e^{-4})(e^{\frac{-3}{2}})\\e^a.e^b = e^{a+b}\\=e^{(-4+\frac{-3}{2})} \\= e^{\frac{-11}{2}}

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