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3 years ago

12

3^sqrt -64w^12

qrt[3]{ - 64w^{12}} " alt=" \sqrt[3]{ - 64w^{12}} " align="absmiddle" class="latex-formula">

1 answer:

Agata [3.3K]3 years ago

7 0

-64=(-4)³, w^12=(w^4)³
the result is -4w^4

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Find the directional derivative of the function at the given point in the direction of the vector v. h(r, s, t) = ln(3r + 6s + 9

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Step-by-step explanation:

First, let’s check to see if the direction vector is a unit vector:

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It’s not a unit vector. therefore let's divide the vector by its magnitude in order to convert it into a unit vector:

$\overrightarrow{\rm v}=(\frac{14}{49}\overrightarrow{\rm i})+(\frac{42}{49}\overrightarrow{\rm j})+(\frac{21}{49}\overrightarrow{\rm k})= (\frac{2}{7}\overrightarrow{\rm i})+(\frac{6}{7}\overrightarrow{\rm j})+(\frac{3}{7}\overrightarrow{\rm k})$

Now, the directional derivative is given by:

$D_\overrightarrow{\rm v}$ $h(r,s,t)=\frac{\partial h(r,s,t)}{\partial r}\overrightarrow{\rm i} + \frac{\partial h(r,s,t)}{\partial s}\overrightarrow{\rm j} + \frac{\partial h(r,s,t)}{\partial t}\overrightarrow{\rm k}$

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$\frac{\partial }{\partial r} ln(3r+6s+9t)=\frac{3}{3r+6s+9t}=\frac{1}{r+2s+3t}$

$\frac{\partial }{\partial s} ln(3r+6s+9t)=\frac{6}{3r+6s+9t}=\frac{2}{r+2s+3t}$

$\frac{\partial }{\partial t} ln(3r+6s+9t)=\frac{9}{3r+6s+9t}=\frac{3}{r+2s+3t}$

Therefore:

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4 years ago

7-x = -2 whats the awnser

Liono4ka [1.6K]

Add 2+x
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If you're wanting the value of x, the answer is ...
x = 9

_____
You can always solve an equation like this by subtracting one side, then dividing by the coefficient of x, then adding the opposite of the constant. If you choose to subtract the right side, these steps will give you ...
9 -x = 0 . . . . . . subtract -2 from both sides
-9 +x = 0 . . . . . divide by the coefficient of x, which was -1
x = 9 . . . . . . . . .add the opposite of the constant, so add 9

Whatever you do to one side of the equation, you must also do to the other side. When we say "subtract -2", we mean "subtract -2 from both sides of the equation."

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