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3 years ago

What is the midpoint of the line segment graphed below?

Mathematics

2 answers:

Diano4ka-milaya [45]3 years ago

6 0

I’m pretty sure it’s b because
You find your midpoint then you would divide after that I’m pretty sure you round

dybincka [34]3 years ago

4 0

Answer:

Step-by-step explanation:

Use the midpoint calculator to find out the midpoint of a line segment, ... A graph showing how to find the midpoint of a segment on the Cartesian plane ... As a supplement to this calculator, we have written an article below that ... Divide 10 by 2, the result of which is 5, this is the y coordinate of the midpoint.

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I need help with geometry

Soloha48 [4]

Answer:

It 12

Step-by-step explanation:

This is because the scale factor between the triangles is 1.75, so I divide 21 by 1.75 and got 12

Hope this helped

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3 years ago

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If (x/3)-1 is twice as large as (x/4)-3, what is the value of x?

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Answer:

Step-by-step explanation:

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Which of the following is not one of the 8th roots of unity?

Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1. Since $z=1=1+0\cdot i,$ then

$Rez=1,\\ \\Im z=0$

and

$|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.$

This gives you $\varphi=0.$

Thus,

$z=1\cdot(\cos 0+i\sin 0).$

2. The 8th roots can be calculated using following formula:

$\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.$

Now

at k=0, $z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;$

at k=1, $z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=2, $z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;$

at k=3, $z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=4, $z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;$

at k=5, $z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

at k=6, $z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;$

at k=7, $z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

The 8th roots are

$\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.$

Option C is icncorrect.

5 0

3 years ago

If the discount is 27%, then what percentage of the original price do you pay? Please give an explanation as well if you can! Th

lina2011 [118]

100-27=73
Therefore you paid 73% of the original price. That’s the remaining percentage

7 0

3 years ago

Consider the system of equations:

vodka [1.7K]

Answer:

(x, y) = (5, 1)

Step-by-step explanation:

To <em>eliminate</em> x, you can double the second equation and subtract the first.

... 2(x +4y) -(2x -3y) = 2(9) -(7)

...11y = 11 . . . . . simplify

... y = 1 . . . . . . divide by 11

Using the second equation to find x, we have ...

... x + 4·1 = 9

... x = 5 . . . . . subtract 4

_____

<u>Check</u>

2·5 -3·1 = 10 -3 = 7 . . . . agrees with the first equation

(Since we used the second equation to find x, we know it will check.)

5 0

3 years ago

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