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3 years ago

PLEASE HELP WITH THIS QUESITION WILL GIVE BRAINLIST TO BEST ANSWER

Mathematics

1 answer:

Elena L [17]3 years ago

3 0

Step-by-step explanation:

When dividing two exponents, you can use a formula of:

$\frac{x^a}{x^b} = x^{a - b}$

So your question would return:

$\frac{3^2}{3^6} = 3^{2-6} = 3^{-4}$

When an exponent is negative, it is equal to 1 divided by the positive number, so for this question, it would be:

$\frac{1}{3^4}$

Or in its expanded form:

$\frac{1}{81}$

Hope this helps!

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If point P is 3/7 the distance from A to B, then it is ____ the distance from B to A.

IrinaVladis [17]

The answer is 4/7, because if A to P is 3/7 of the way, going to B would be to the end. So, you have to go 4/7 of the way to get to the end.

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Use lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y = xyz; x^

Snezhnost [94]

I'm assuming the constraint involves some plus signs that aren't appearing for some reason, so that you're finding the extrema subject to $x^2+2y^2+3z^2=96$ .

Set $f(x,y,z)=xyz$ and $g(x,y,z)=x^2+2y^2+3z^2-96$ , so that the Lagrangian is

$L(x,y,z,\lambda)=xyz+\lambda(x^2+2y^2+3z^2-96)$

Take the partial derivatives and set them equal to zero.

$\begin{cases}L_x=yz+2\lambda x=0\\L_y=xz+4\lambda y=0\\L_z=xy+6\lambda z=0\\L_\lambda=x^2+2y^2+3z^2-96=0\end{cases}$

One way to find the possible critical points is to multiply the first three equations by the variable that is missing in the first term and dividing by 2. This gives

$\begin{cases}\dfrac{xyz}2+\lambda x^2=0\\\\\dfrac{xyz}2+2\lambda y^2=0\\\\\dfrac{xyz}2+3\lambda z^2=0\\\\x^2+2y^2+3y^2=96\end{cases}$

So by adding the first three equations together, you end up with

$\dfrac32xyz+\lambda(x^2+2y^2+3z^2)=0$

and the fourth equation allows you to write

$\dfrac32xyz+96\lambda=0\implies \dfrac{xyz}2=-32\lambda$

Now, substituting this into the first three equations in the most recent system yields

$\begin{cases}-32\lambda+\lambda x^2=0\\-32\lambda+2\lambda y^2=0\\-32\lambda+3\lambda z^2=0\end{cases}\implies\begin{cases}x=\pm4\sqrt2\\y=\pm4\\z=\pm4\sqrt{\dfrac23}\end{cases}$

So we found a grand total of 8 possible critical points. Evaluating $f(x,y,z)=xyz$ at each of these points, you find that $f(x,y,z)$ attains a maximum value of $\dfrac{128}{\sqrt3}$ whenever exactly none or two of the critical points' coordinates are negative (four cases of this), and a minimum value of $-\dfrac{128}{\sqrt3}$ whenever exactly one or all of the critical points' coordinates are negative.

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3 years ago

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Step-by-step explanation:

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3 years ago

Can someone help me with this

Jlenok [28]

Answer:

just add

32 plus 9

Step-by-step explanation:

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