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3 years ago

What is 4,216 divided by 7

Mathematics

2 answers:

scZoUnD [109]3 years ago

5 0

The answer is 602.29

Akimi4 [234]3 years ago

5 0

If you want to do it step by step without a calculator then try doing long division :)

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Jim is riding his bicycle. He rides for 22.5 kilometers at a speed of 9 kilometers per hour. For how many hours does he ride?

Stells [14]

Answer:

It is 2.5 hours.

Step-by-step explanation:

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2 years ago

Ali’s cow gives 2 1/4 gallons of milk each day. How much is this in quarts? Write your answer as whole number or a mixed number

MrRissso [65]

Answer:

9 quarts

Step-by-step explanation:

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3 years ago

Please Help!!

Harlamova29_29 [7]

The answer is C

You would distribute the 5 so 5 times x= 5x

and 5 times positive 10 = 50

You are left with 5x=50

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4 years ago

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What are all the shapes that are used to create the net of a cylinder

Ksenya-84 [330]

2 circles and for the part in the middle if you look at it as a net it will be a rectangle just rolled up. That is called a lateral surface or a lateral face

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3 years ago

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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.

6 0

3 years ago