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3 years ago

Select the quadrant or access where are each ordered pair is located on a coordinate plane

Mathematics

1 answer:

NISA [10]3 years ago

6 0

Answer:

Below.

Step-by-step explanation:

(-3, 9) - Quadrant II.

(6, -6) - Quadrant IV.

(0, -1 1/2) - x-axis 0, y-axis -1/1/2

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Solve for x: 8x + 10 - 5x = 15.

MrRa [10]

8x + 10 - 5x = 15
3x + 10 = 15
3x = 5
x = 5/3

hope it helps

7 0

3 years ago

Read 2 more answers

Help please!!! Thanks

tatiyna

Answer:

25.6

Explanation:

t = 2 seconds

h = 8 + 40.8(2) - 16(2)^2

Solve for h by simplifying both sides of the equation, then isolating the variable.

h = 25.6

3 0

3 years ago

SOMEONE PLEASE HELP ME ASAP PLEASE!!!

nadezda [96]

Answer:

$y=-4x+5$

Step-by-step explanation:

Parallel lines have the same slope, but different y-intecpets. They are lines that travel in the same directions but never intercept.

6 0

3 years ago

Read 2 more answers

What is the square root of -2i?

Studentka2010 [4]

Answer:

1-i and -1+i

Step-by-step explanation:

We are to find the square roots of $z=0-2i$ . First, convert from Cartesian to polar form:

$r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2$

$\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}$

$z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})$

Next, use the formula $\displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr]$ where $\displaystyle k=0,1,2,...\:,n-1$ to find the square roots:

When k=1

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]$

$\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i$

When k=0

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i$

Thus, the square roots of -2i are 1-i and -1+i

4 0

2 years ago

Help me please help

ololo11 [35]

Answer:

D.) Not enough information

Step-by-step explanation:

The marks are the same, but they aren't placed in the same position in the triangles.

8 0

3 years ago