Ask question

Ask question

All categories

2 years ago

11

15 Points!! Need help fast!!!

1 answer:

Lyrx [107]2 years ago

7 0

Answer:

its 18 units what ever your units is

You might be interested in

I’m not sure how to do this

Cerrena [4.2K]

It's b because you have to round up

5 0

3 years ago

Read 2 more answers

Rationalize the denominator of square root of negative 9 over open parentheses 4 minus 7 i close parentheses minus open parenthe

AlekseyPX

The answer is
$\frac{-6i-3}{5}$ .

Before we rationalize the denominator, we must simplify the numerator and denominator. We evaluate the square root on the top, and combine like terms on the bottom:
$\frac{\sqrt{-9}}{(4-7i)-(6-6i)}
\\
\\=\frac{\sqrt{9i^2}}{4-6-7i--6i}
\\
\\=\frac{3i}{-2-i}$

To rationalize the denominator, we multiply by the conjugate. The conjugate is found by making the imaginary term of the complex number, the i part, the opposite sign. The conjugate of -2-i is -2+i:

$\frac{3i}{-2-i}\times \frac{-2+i}{-2+i}
\\
\\=\frac{3i(-2+i)}{(-2-i)(-2+i)}
\\
\\=\frac{3i\times -2 + 3i\times i}{-2\times -2 + -2\times i + -2\times -i + -i\times i}
\\
\\=\frac{-6i+3i^2}{4-2i+2i-i^2}
\\
\\=\frac{-6i+3(-1)}{4-i^2}=\frac{-6i-3}{4-(-1)}=\frac{-6i-3}{5}$

7 0

3 years ago

Which equation is correct

BARSIC [14]

The last one, v = -2,000y + 20,000

3 0

3 years ago

Read 2 more answers

Simplify the expression

QveST [7]

Answer:

Step-by-step explanation:

It might be wrong but i beieve that it is 4 to the 5th multiplied 4 to the 2nd

8 0

2 years ago

Read 2 more answers

Hey Bestie! 50 pts Pls help! Real answers only pls <3

frosja888 [35]

Answer:

$1. \: 3i$

2. option D

3. option C

4. option D

5. option C

6. option B

7. option C

8. option D

9. option C

10. option C

Step-by-step explanation:

<h2> $1. \: \sqrt{ - 9}$ </h2>

$\sqrt{ - 1(9)}$

$\sqrt{ - 1} \times \sqrt{ 9}$

$i \times \sqrt{9}$

$i \times \sqrt{ {3}^{2} }$

$i \times 3$

$3i$

<h2> $2. \: \sqrt{ - 8}$ </h2>

$\sqrt{ - 1(8)}$

$\sqrt{ - 1} \times \sqrt{8}$

$i \times \sqrt{8}$

$i \times \sqrt{ {2}^{2} \times 2 }$

$2i \sqrt{2}$

<h2> $3. \: \sqrt{ - 80}$ </h2>

$\sqrt{ - 1} \times \sqrt{80}$

$i \times \sqrt{80}$

$4i \: \sqrt{5}$

<h2> $4. \: \sqrt{ - 75}$ </h2>

$\sqrt{ - 1} \times \sqrt{75}$

$i \times \sqrt{75}$

$i \times \sqrt{ {5}^{2} \times 3 }$

$5i \sqrt{3}$

<h2> $5. \: \sqrt{ - 72}$ </h2>

$\sqrt{ - 1} \times \sqrt{72}$

$i \times \sqrt{72}$

$i \times ( {6}^{2} \times 2)$

$6i \sqrt{2}$

<h2> $6. \sqrt{ - 20}$ </h2>

$\sqrt{ - 1} \times \sqrt{20}$

$i \times \sqrt{20}$

$i \times \sqrt{ {2}^{2} \times 5}$

$2i \sqrt{5}$

<h2> $7. \: \sqrt{ - 27}$ </h2>

$\sqrt{ - 1} \times \sqrt{27}$

$i \times \sqrt{27}$

$i \times \sqrt{ {3}^{2} \times 3}$

$3i \sqrt{3}$

<h2> $8. \: \sqrt{ - 12}$ </h2>

$\sqrt{ - 1 \times 12}$

$i \times \sqrt{12}$

$i \times \sqrt{4(3)}$

$2i \sqrt{3}$

<h2> $9. \: \sqrt{ - 125}$ </h2>

$\sqrt{ - 1} \times \sqrt{125}$

$i \times \sqrt{ {5}^{2} \times 5 }$

$5i \sqrt{5}$

<h2> $10. \: \sqrt{ - 180}$ </h2>

$\sqrt{ - 1} \times \sqrt{180}$

$i \times \sqrt{ {6}^{2} \times 5 }$

$6i \sqrt{5}$

<h3>Hope it is helpful...</h3>

5 0

2 years ago