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jasenka [17]
2 years ago
11

(a+3b)(2c+d+4e) simplify​

Mathematics
1 answer:
Savatey [412]2 years ago
6 0

Answer:

2ac + ad + 4ea + 6bc + 3bd + 12be (Simplified)

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Please someone help me !
brilliants [131]

Step-by-step explanation:

a. If x is the total numbers of students in school, 35%x = 140.

0.35x = 140

x = 140/0.35 = 400

b. Since there are 400 kids in the school, 15% of them take the bus which is 0.15 * 400 = 60 kids.

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4 years ago
Help me on question. 5
Volgvan
Because he was wrong
8 0
4 years ago
Read 2 more answers
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
For every person who walked, there were forty-four who drove. If 270 made the trip, how many walked? PLEASE HELP! NO SPAMMING OR
svp [43]
For every person that walked there were 12 who drove and if the 270 made the trip then 87 walked
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3 years ago
True or false? The sum of the exterior angles depends on how many sides there are in the polygon?
kap26 [50]

Answer: true i believe

Step-by-step explanation:

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3 years ago
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