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3 years ago

How to rewrite in simplest radical form 1/-3/x^6

Mathematics

1 answer:

Vilka [71]3 years ago

8 0

$\dfrac{\frac{1}{-3}}{x^6}=\dfrac{1}{-3}\cdot\dfrac{1}{x^6}=\boxed{-\frac{1}{3x^6}}$

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Is -6.333 greater less equal to -6.375

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6.375 is greater than 6.333

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What is 24.5 as a fraction.

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24 1/2. 0.5 is equal to 1/2

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A right triangle is removed from a rectangle to create the shaded region shown below. Find the area of the shaded region. Be sur

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The sides of a right triangle are
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Which statement describes the inverse of m(x) = x^2 – 17x?

DochEvi [55]

Given:

The function is

$m(x)=x^2-17x$

To find:

The inverse of the given function.

Solution:

We have,

$m(x)=x^2-17x$

Substitute m(x)=y.

$y=x^2-17x$

Interchange x and y.

$x=y^2-17y$

Add square of half of coefficient of y , i.e., $\left(\dfrac{-17}{2}\right)^2$ on both sides,

$x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2$ $[\because (a-b)^2=a^2-2ab+b^2]$

Taking square root on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}$

Add $\dfrac{17}{2}$ on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y$

Substitute $y=m^{-1}(x)$ .

$m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$

We know that, negative term inside the root is not real number. So,

$x+\left(\dfrac{17}{2}\right)^2\geq 0$

$x\geq -\left(\dfrac{17}{2}\right)^2$

Therefore, the restricted domain is $x\geq -\left(\dfrac{17}{2}\right)^2$ and the inverse function is $m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$ .

Hence, option D is correct.

Note: In all the options square of $\dfrac{17}{2}$ is missing in restricted domain.

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3 years ago

Find the value of 15 - 8 ÷ 2.

Mamont248 [21]

8÷2= 4

15-4= 11

First you divide 8 by 2.
The you subtract 15 minus 4 and the answer will be 11.

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