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nika2105 [10]
3 years ago
7

Express the radical using the imaginary unit i

Mathematics
1 answer:
valentinak56 [21]3 years ago
7 0

Answer:

=\pm i\sqrt{66}

Step-by-step explanation:

So we have:

\pm\sqrt{-66}

This is the same as:

=\pm\sqrt{66\cdot-1}

Separate:

=\pm\sqrt{66}\cdot\sqrt{-1}

Replace the second term with i:

=\pm i\sqrt{66}

The square root of 66 cannot be simplified.

So, this is our answer :)

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Which inequality is shown in this graph
Katena32 [7]

ANSWER

y \leqslant   - 2x - 2

EXPLANATION

The boundary line passes through (-2,2) and (0,-2).

The slope of this line is

m =  \frac{ - 2 - 2}{0 -  - 2}

m =  \frac{ - 4}{2}  =  - 2

The y-intercept is , c=-2.

The slope-intercept form of this line is given by;

y = mx + c

We substitute values to obtain;

y =  - 2x - 2

Since the lower half-plane is shaded the required inequality is

y \leqslant   - 2x - 2

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3 years ago
HELP!!! It’s due is 20 minutes
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Answer:

B

Step-by-step explanation:

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Use Pythagorean theorem :)

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10, because then meg would have 5. 
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Marisa recorded the number of hours she babysat each day last summer. She plotted the data in the box plot. A number line goes f
il63 [147K]

Answer:

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(ASAP PICTURE ADDED) What is the simplified form of the following expression?
bonufazy [111]

Answer:

option c is correct.

Step-by-step explanation:

7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{16x}\right)-3\left(\sqrt[3]{8x}\right)

WE need to simplify this equation.

Solve the parenthesis of each term.

=7\left\sqrt[3]{2x}\right-3\left\sqrt[3]{16x}\right-3\left\sqrt[3]{8x}\right

Now, We will find factors of the terms inside the square root

factors of 2: 2

factors of 16 : 2x2x2x2

factors of 8: 2x2x2

Putting these values in our equation:=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2X2 x}\right)-3\left(\sqrt[3]{2X2X2 x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3*2\left(\sqrt[3] {2 x}\right)-3*2\left(\sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)

Adding like terms we get:

=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right\\=(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\

(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\can\,\,be \,\, written\,\, as\,\,\\(\sqrt[3] {2x})-6\left(\sqrt[3]{x}\right)

So, option c is correct

5 0
3 years ago
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