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2 years ago

Marking brainlest if done in 5 mins

Mathematics

2 answers:

MA_775_DIABLO [31]2 years ago

8 0

Answer:

392in² would be my guess

Step-by-step explanation:

The area of the parallelogram is equal to,

$A=bh$

where b is base and h is the length of the height.

Area of parallelogram = 14 inches× 10 inches = 140 in²

There are two parallelograms, the total area of two parallelograms is 280 in².

The area of the triangle is calculated Below

$Area=BH/2$

$Area = (28 inches)-18 inches - 10 inches- / 2 = 112 in^{2}$

Adding the areas we get --> $392in^{2}$

Therefore the Answer will be $392in^{2}$

Good Luck with your Assignment!

Have a great day!

Luden [163]2 years ago

7 0

Answer: The answer is 392 squared

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Please see attachment

Dafna11 [192]

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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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