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zvonat [6]
3 years ago
5

Which is equivalent to -√10 3/4^x?

Mathematics
2 answers:
Nat2105 [25]3 years ago
8 0

Answer:

\Rightarrow -\sqrt[8]{10^{3x}}

D is correct

Step-by-step explanation:

Given: -(\sqrt{10^{3/4}})^x

It is complex radical equation.

As we know,

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

or viceversa

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

In the given expression

\Rightarrow -(\sqrt{10^{3/4}})^x

\Rightarrow -10^{\frac{1}{2}\cdot\frac{3}{4}\cdot x}

simplifying the exponent

\Rightarrow -10^{\frac{3x}{8}}

Here, the radical is 8 and change exponent to radical form

\Rightarrow -\sqrt[8]{10^{3x}}

Hence, -(\sqrt{10^{3/4}})^x is equivalent to \Rightarrow -\sqrt[8]{10^{3x}}

Galina-37 [17]3 years ago
4 0

3√10^4x is equivalent to -√10 3/4^x and -4√10^3x they are all equivalent

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Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.
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Answer:

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b.

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Step-by-step explanation:

Remember that for any curve      r(t)  

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a.

For this case, using the chain rule

r'(t) = (  -10*2tsin(t^2) ,   102t cos(t^2)   )\\

And also remember that

|r'(t)| = \sqrt{(-10*2tsin(t^2))^2  +  ( 10*2t cos(t^2) )^2} \\\\       = \sqrt{400 t^2*(  sin(t^2)^2  +  cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t

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T(t) = r'(t) / |r'(t) | =  (  -10*2tsin(t^2) ,   10*2t cos(t^2)   )/ 20t\\\\ = (  -10*2tsin(t^2)/ 20t ,   10*2t cos(t^2) / 20t  )\\= ( -sin(t^2), cos(t^2) )

Similarly, using the quotient rule and the chain rule

T'(t) = ( -2t cos(t^2) , -2t sin(t^2))

And also

|T'(t)| = \sqrt{  ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t

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N(t) = T'(t) / |T'(t) |  =   (-cos(t^2) , -sin(t^2))

Notice that

1.   |N(t)| = |T(t) | = \sqrt{ cos(t^2)^2  + sin(t^2)^2 } = \sqrt{1} =  1

2.   N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0

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T(t) =r'(t)/|r'(t)| =  (2t/ \sqrt{4t^2   + 36} , -6/\sqrt{4t^2   + 36},0)

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And since

N(t) =T(t)/|T'(t)| =  (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)

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