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3 years ago

FG ∥ CB, A ∈ FG, D ∈ AB, E ∈ AC

Mathematics

1 answer:

andrey2020 [161]3 years ago

3 0

Answer:

The value of x is $31^{\circ}$ .

Step-by-step explanation:

Please look at the figure attached to get more clear solution.

We have given:

FG||CB

And the line that cut the parallel line is transversal so, here BA is transversal

And alternate interior angles on transverse line are equal

So, ∠1=∠4

And ∠4= $28^{\circ}$

Hence, ∠1=∠4= $28^{\circ}$

And On FG the sum of angles will be $180^{\circ}$

∠3+∠2+∠1= $180^{\circ}$

$90^{\circ}$ +∠2+ $28^{\circ}$ = $180^{\circ}$

Hence, ∠2= $62^{\circ}$

Now, we know that the sum of interior angles is equal to the exterior angle:

Therefore, ∠2+∠5=∠6+∠7

$62^{\circ}+3x=x+4x$

On simplification we get:

$2x=62^{\circ}$

$x=31^{\circ}$

Hence, the value of x is $31^{\circ}$ .

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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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