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3 years ago

Find the sum. 1. -7+(-5) O-12 O-2 0 2 0 12

Mathematics

1 answer:

AleksAgata [21]3 years ago

7 0

Answer:

-12

Step-by-step explanation:

-7+(-5)=

-7-5=

-12

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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:

Step(i):-

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

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

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

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

$2 x^{2} = 1$

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

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

Step(ii):-

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

Step(iii):-

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

Final answer:-

a) The value of absolute minimum value = - 0.3536

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

3 0

3 years ago

How would I do this

xz_007 [3.2K]

When,
correct is 0 then p is 0/20.
correct is 1 p is 1/20
correct is 2 p is 2/20 that is 1/10
correct is 3 p is 3/20.
correct is 4 p is 4/20 that is 1/5
correct is 5 p is 5/20 that is 1/4

5 0

3 years ago

If X plus 5 equals 8, then what is X?

katrin [286]

Answer:

X plus 5 equals 8, then whats is X. The answer would be X = 3

Step-by-step explanation: Hope it helps

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2 years ago

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Answer:

its the other won D

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3 years ago

Find the sum. 1. -7+(-5) O-12 O-2 0 2 0 12​

Find the sum. 1. -7+(-5) O-12 O-2 0 2 0 12