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Alchen [17]
3 years ago
8

In parts ​(a) through ​(e)​ below, mark the given statement as True or False. Justify each answer. All vectors are in set of rea

l numbers R Superscript nℝn. a. vtimes•vequals=Bold left norm v right normvsquared2 Choose the correct answer below. A. The given statement is false. It is not possible for it to be true because vtimes•v simplifies to a ​vector, whereas Bold left norm v right normvsquared2 simplifies to a scalar. B. The given statement is true. By the definition of the length of a vector v​, Bold left norm v right normvequals=StartRoot Bold v times Bold v EndRootv•v. C. The given statement is false. By the definition of the length of a vector ​v, Bold left norm v right normvequals=vtimes•v. It follows that Bold left norm v right normvsquared2equals=​(vtimes•v​)squared2. D. The given statement is true. By the definition of the length of a vector ​v, Bold left norm v right normvequals=vtimes•v. It follows that Bold left norm v right normvsquared2equals=​(vtimes•v​)squared2.
Mathematics
1 answer:
Oxana [17]3 years ago
5 0

Answer:

(c)

"The given statement is true, by definition of length of a vector v, ||v|| = \sqrt{v\bullet v}"

Step-by-step explanation:

(a) v  \bullet v = || v ||^2

That is completely correct Remember that if  v = (x_1,x_2,x_3)\\ then

v \bullet v = x_1*x_1+x_2*x_2+x_3*x_3 = x_1^2+x_2^2+x_3^2 = ||v||^2

Therefore the correct answer would be (c).

"The given statement is true, by definition of length of a vector v, ||v|| = \sqrt{v\bullet v}"

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<h3>Answer: Choice C</h3>

RootIndex 12 StartRoot 8 EndRoot Superscript x

12th root of 8^x = (12th root of 8)^x

\sqrt[12]{8^{x}} = \left(\sqrt[12]{8}\right)^{x}

=========================================

Explanation:

The general rule is

\sqrt[n]{x} = x^{1/n}

so any nth root is the same as having a fractional exponent 1/n.

Using that rule we can say the cube root of 8 is equivalent to 8^(1/3)

\sqrt[3]{8} = 8^{1/3}

-----

Raising this to the power of (1/4)x will have us multiply the exponents of 1/3 and (1/4)x like so

(1/3)*(1/4)x = (1/12)x

In other words,

\left(8^{1/3}\right)^{(1/4)x} = 8^{(1/3)*(1/4)x}

\left(8^{1/3}\right)^{(1/4)x} = 8^{(1/12)x}

-----

From here, we rewrite the fractional exponent 1/12 as a 12th root. which leads us to this

8^{(1/12)x} = \sqrt[12]{8^{x}}

8^{(1/12)x} = \left(\sqrt[12]{8}\right)^{x}

4 0
3 years ago
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3 years ago
Explaining your answer can either include a counterexample, a diagram or a written explanation to support your answer.
anzhelika [568]

Answer:

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