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3 years ago

Should i use dy\dx (normal differentiation) or d²y\dx² (differentiation of the differentiation)?

Mathematics

1 answer:

11111nata11111 [884]3 years ago

7 0

Use both!

You want to minimize P, so differentiate P with respect to x and set the derivative equal to 0 and solve for any critical points.

P = 8/x + 2x

dP/dx = -8/x² + 2 = 0

8/x² = 2

x² = 8/2 = 4

x = ± √4 = ± 2

You can then use the second derivative to determine the concavity of P, and its sign at a given critical point decides whether it is a minimum or a maximum.

We have

d²P/dx² = 16/x³

When x = -2, the second derivative is negative, which means there's a relative maximum here.

When x = 2, the second derivative is positive, which means there's a relative minimum here.

So, P has a relative maximum value of 8/(-2) + 2(-2) = -8 when x = -2.

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3 years ago

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Crazy boy [7]

Answer:

see explanation

Step-by-step explanation:

∠ YXW = ∠ ZXV ( vertically opposite angles ) , so

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a = 25

Then

(7)

∠ ZXV = 2a + 65 = 2(25) + 65 = 50 + 65 = 115°

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∠ YXW = ∠ ZXV = 115°

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(9)

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3 years ago

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aev [14]

Considering the market options that an investor takes, Each scenario listed is attached to the right action that is shown on the tiles:

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<h3>Meaning of an Investor</h3>

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In conclusion, the list above are the required actions for each scenario that is encountered by an investor

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2 years ago

Perform the indicated operation. <img src="https://tex.z-dn.net/?f=x%20%2B%205%20%2F%203" id="TexFormula1" title="x + 5 / 3"

Salsk061 [2.6K]

Answer:

$x + \frac{5}{3}-x - \frac{3}{2} = \frac{1}{6}$

Step-by-step explanation:

Given

$x + \frac{5}{3}-x - \frac{3}{2}$

Required

Solve

$x + \frac{5}{3}-x - \frac{3}{2}$

Collect like terms

$x + \frac{5}{3}-x - \frac{3}{2} = x -x+ \frac{5}{3} - \frac{3}{2}$

$x + \frac{5}{3}-x - \frac{3}{2} = \frac{5}{3} - \frac{3}{2}$

Take LCM

$x + \frac{5}{3}-x - \frac{3}{2} = \frac{10 -9}{6}$

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3 years ago

Consider the two groups listed below. Which statement describes the sets?

antoniya [11.8K]

The correct answer is: 
D) neither the relation (length, volume) nor the relation (volume, length) is a function.

Explanation: 
A function is a relation in which each element of the domain is mapped to no more than 1 element of the range.
For (length, volume), the domain would be length and the range would be volume.
There could possibly be a length value that is mapped to more than one volume value, depending on the height of the pool. This means that (length, volume) is not always a function.
For (volume, length), volume is the domain and length is the range.
There could possibly be a volume that is mapped to more than one length, depending on the height of the pool. This means that (volume, length) is not always a function either.

6 0

4 years ago

Should i use dy\dx (normal differentiation) or d²y\dx² (differentiation of the differentiation)?​

Should i use dy\dx (normal differentiation) or d²y\dx² (differentiation of the differentiation)?