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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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If m∠agh = 4x and m∠ghd = 3x + 40, what is the value of x?
Mariulka [41]
I am attaching the full question...

Because AB is parallel to CD, the transversal line EF creates the same angles in AB and CD.

This means that 3x+40 = 4x

..........==>> x = 40 degrees

7 0
3 years ago
6x-5y=15<br> x=y+3 <br> what does x= and what does y=
Zepler [3.9K]

Answer:

(0,-3)

Step-by-step explanation:

hi, with this we can just do substitution (substitute x for y+3)

6(y+3)-5y=15

evaluate

6y+18-5y=15

y+18=15

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then we plug that in

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4 0
3 years ago
If -xy – 5+ y^2+ x^2= 0 and it is known that dy/dx= y-2x/-x+2y, find all
sleet_krkn [62]

Answer:

There is no point of the form (-1, y) on the curve where the tangent is horizontal

Step-by-step explanation:

Notice that when x = - 1. then dy/dx becomes:

dy/dx= (y+2) / (2y+1)

therefore, to request that the tangent is horizontal we ask for the y values that make dy/dx equal to ZERO:

0 = ( y + 2) / (2 y + 1)

And we obtain y = -2 as the answer.

But if we try the point (-1, -2) in the original equation, we find that it DOESN'T belong to the curve because it doesn't satisfy the equation as shown below:

(-1)^2 + (-2)^2 - (-1)*(-2) - 5 = 1 + 4 + 2 - 5 = 2  (instead of zero)

Then, we conclude that there is no horizontal tangent to the curve for x = -1.

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2 years ago
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