What is the inverse of the equation y=x²-4, x≤ 0

Mathematics

1 answer:

Lina20 [59]1 year ago

6 0

The inverse of the equation y = x² - 4, x≤ 0 is $y = \sqrt{x + 4$

<h3>How to determine the inverse of the equation?</h3>

The equation is given as:

y = x² - 4, x≤ 0

Swap the x and y values

x = y² - 4

Add 4 to both sides

y² = x + 4

Take the square root of both sides

$y = \sqrt{x + 4$

Hence, the inverse of the equation y = x² - 4, x≤ 0 is $y = \sqrt{x + 4$

Read more about inverse equations at:

brainly.com/question/11302699

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Which number line shows the solution to −6 − (−2)? A number line is shown from negative 10 to 0 to positive 10. There are increm

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Answer:

The correct option is A) A number line is shown from negative 10 to 0 to positive 10. There are increments of 2 on either side of the number line. The even numbers are labeled on either side of the number line. An arrow pointing from 0 to negative 6 is shown. Above this, another arrow pointing from negative 6 to negative 4 is shown. A vertical bar is shown at the tip of the arrowhead of the top arrow.

Step-by-step explanation:

Consider the provided expression.

−6 − (−2)

Open the parentheses and change the sign.

−6 − (−2)

−6 + 2

Subtract the numbers.

−4

Now draw this on number line.

First draw a number line is shown from −10 to 0 to 10. with scale of 2 unit on either side of the number line. Draw an arrow pointing from 0 to −6 Which show −6. Above this, another arrow pointing from −6 to −4 which shows −6 − (−2) = −4. A vertical bar is shown at the tip of the arrowhead of the top arrow.

The required number line is shown in the figure 1.

Hence, the correct option is A) A number line is shown from negative 10 to 0 to positive 10. There are increments of 2 on either side of the number line. The even numbers are labeled on either side of the number line. An arrow pointing from 0 to negative 6 is shown. Above this, another arrow pointing from negative 6 to negative 4 is shown. A vertical bar is shown at the tip of the arrowhead of the top arrow.