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SVETLANKA909090 [29]

1 year ago

8

What is the domain of the function y = RootIndex 3 StartRoot x minus 1 EndRoot?

1 answer:

finlep [7]1 year ago

4 0

By definition of cubic roots and power properties, we conclude that the domain of the cubic root function is the set of all real numbers.

<h3>What is the domain of the function?</h3>

The domain of the function is the set of all values of x such that the function exists.

In this problem we find a cubic root function, whose domain comprise the set of all real numbers based on the properties of power with negative bases, which shows that a power up to an odd exponent always brings out a negative result.

<h3>Remark</h3>

The statement is poorly formatted. Correct form is shown below:

<em>¿What is the domain of the function </em> $y = \sqrt[3]{x - 1}$ <em>?</em>

<em />

To learn more on domain and range of functions: brainly.com/question/28135761

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3 years ago

Is this correct btw I put 1400

Oxana [17]

Answer: Yes

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5 0

2 years ago

Read 2 more answers

Why is it preferable to measure a penny in millimeters rather than centimeters or meters?

andrey2020 [161]

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3 years ago

Read 2 more answers

(1 point) Linear System - Three Variables Solve the following system of equations using Gaussian elimination method. If there ar

EastWind [94]

Answer:

$x=\frac{-363}{70}$ ≈ -5.1857

$y=\frac{-192}{35}$ ≈ -5.4857

$z=\frac{313}{84}$ ≈ 3.7262

Step-by-step explanation:

Rewrite the equation system as:

$6x+8y=-75$

$-3x+6y+6z=5$

$2x-9y=39$

Now, write the system in its augmented matrix form:

$\left[\begin{array}{cccc}6&8&0&-75\\-3&6&6&5\\2&-9&0&39\end{array}\right]$

applying row reduction process to its associated augmented matrix:

Swap R1 and R3, and then Swap R1 and R2:

$\left[\begin{array}{cccc}-3&6&6&5\\2&-9&0&39\\6&8&0&-75\end{array}\right]$

R3+2R1

$\left[\begin{array}{cccc}-3&6&6&5\\2&-9&0&39\\0&20&12&-65\end{array}\right]$

3R2+2R1

$\left[\begin{array}{cccc}-3&6&6&5\\0&-15&12&127\\0&20&12&-65\end{array}\right]$

15R3+20R2

$\left[\begin{array}{cccc}-3&6&6&5\\0&-15&12&127\\0&0&420&1565\end{array}\right]$

Now we have a simplified system:

$-3x+6y+6z=5\\0-15y+12z=127\\0+0+420z=1565$

$-3x+6y+6z=5\hspace{5 mm}(1)\\0-15y+12z=127\hspace{3 mm}(2)\\0+0+420z=1565\hspace{3 mm}(3)$

From (3):

$z=\frac{313}{84}$ (4)

Replacing (4) in (2)

$y=\frac{-192}{35}$ (5)

Finally replacing (5) and (4) in (1)

$x=\frac{-363}{70}$

7 0

3 years ago