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3 years ago

Which of the following expressions are equivalent to this expression:

Mathematics

2 answers:

tatyana61 [14]3 years ago

8 0

Answer:

A: -1[4+(-4)]

Gekata [30.6K]3 years ago

6 0

The original expression is equal to 0 because anything multiplied by 0 is equal to 0. Solve inside the brackets for the possible answer choices to find what will equal 0.

Start with the first expression. Add 4 and negative 4 will become 0, and -1 times 0 is equal to 0. Let's solve for the others just to be sure.

In the second expression, solving inside the brackets gives you 8. -1 times 8 is equal to -8.

Adding 4 and negative 4 in the third expression leaves you with 0. But, 1 + 0 is equal to 1.

Adding negative 4 and negative 4 gives you the answer of -8, and -1 times -8 is equal to 8.

Your answer is the first expression, or A.

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Elaine has three types of nuts, peanuts, cashews and almonds, that she combines and stores in a large container: There are three

wlad13 [49]

Answer:

There are 9 ounces of peanuts

Make the ounces of cashews equal to three, and this works! Don’t know the math otherwise :|

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2 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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