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3 years ago

6

Solve. –355 + 355 = 0 355 710 –710

1 answer:

Verdich [7]3 years ago

3 0

Hey there,

<span>–355 + 355 =

Your correct answer is 0.

So,for example, -1+1+0.

The answer would be 0 because it used to be a negative number, but once there is exact number (but positive) as the same number as before, they would get you 0.

Hope this helps.
~Jurgen</span>

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A box contains 8 videotapes and 5 DVDs. I’m how many ways can one videotape and one DVD be selected?

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Answer:

40

Step-by-step explanation:

8x5=40

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3 years ago

Read 2 more answers

Please see attachment

Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

<u>Step(i)</u>:-

Given function

$f(x) = \frac{-x}{2x^{2} +1}$ ...(i)

Differentiating equation (i) with respective to 'x'

$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$2 x^{2}-1 = 0$

$2 x^{2} = 1$

$x^{2} = \frac{1}{2}$

$x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

<u><em>Step(ii):</em></u>-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

<u><em>Step(iii):</em></u>-

The value of absolute minimum value

$f(x) = \frac{-x}{2x^{2} +1}$

$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

<u><em>Final answer</em></u>:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

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3 years ago

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Answer:

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Step-by-step explanation:

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Gnoma [55]

Answer:

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2 years ago