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3 years ago

PLEASE HELP PICTURE SHOWN

Mathematics

2 answers:

Rainbow [258]3 years ago

4 0

Your answer is B. 2
This is because to rationalise the denominator, we need to multiply it by (3 - √7), so we get:
(3 + √7)(3 - √7)
3 × 3 = 9
3 × √7 = 3√7
3 × -√7 = -3√7
√7 × -√7 = -7
So all in all you get 9 - 7 which is 2.

I hope this helps!

Strike441 [17]3 years ago

3 0

I believe that would be B. 2

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Can you simplify 8/144

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One number is nine more than the other. Their sum is 33. Find the numbers. (Enter your answers as a comma-separated list.)

Bas_tet [7]

Answer:

The numbers are 12, 21.

Step-by-step explanation:

x+(x+9)=33

x+x+9=33

2x+9=33

2x=33-9

2x=24

x=24/2

x=12

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x+9=12+9=21

Please mark me as Brainliest if you're satisfied with the answer.

5 0

3 years ago

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

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lutik1710 [3]

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