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3 years ago

Simplify 21v ^-6 divided by (-7v^-6)

Mathematics

2 answers:

lorasvet [3.4K]3 years ago

4 0

Answer:

hope this help you

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miskamm [114]3 years ago

3 0

The answer is -3. I just did it on a calculator

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Answer:

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Step-by-step explanation:

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3 years ago

Why is this scientific notation: 12,050,000 = 1.25×107 wrong

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3 years ago

Why is it not a system of linear equations?

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

3 0

3 years ago

HEEELP

Drupady [299]

Answer: (c)

Step-by-step explanation:

Given

$f(x)=\dfrac{1}{x-3}\\\\g(x)=\sqrt{x+5}$

Here, $\sqrt{x+5}\ \text{is always greater than equal to 0}\\\Rightarrow x+5\geq 0\\\Rightarrow x\geq -5\quad \ldots(i)$

To get $f\left(g(x)\right)$ , replace $x$ in $f(x)$ by $g(x)\ \text{i.e. by}\ \sqrt{x+5}$

$\Rightarrow f\left(g(x)\right)=\dfrac{1}{\sqrt{x+5}-3}\\\\\text{Denominator must not be equal to 0}\\\\\therefore \sqrt{x+5}-3\neq0\\\Rightarrow \sqrt{x+5}\neq 3\\\Rightarrow x+5\neq 9\\\Rightarrow x\neq 4\quad \ldots(ii)$

Using $(i)$ and $(ii)$ it can be concluded that the domain of $f\left(g(x)\right)$ is all real numbers except 0.

Therefore, its domain is given by

$x\in [-5,4)\cup (4,\infty)$

Option (c) is correct.

5 0

3 years ago