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Paul [167]
3 years ago
15

HELP!!!ITS DUE IN AN HOUR

Mathematics
2 answers:
sladkih [1.3K]3 years ago
5 0

Answer:

26.5

Step-by-step explanation:

dont worry I'm actually the salesperson at the  jewlery store

andreyandreev [35.5K]3 years ago
4 0

Step-by-step explanation:

it is not confrmed answer

commission:106%

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Fill in the missing reason in proof.
KiRa [710]

Answer:

  • SAS

Step-by-step explanation:

<u>As per steps given the congruent pairs are:</u>

  • AB and AC
  • BD and CD
  • Angle B and angle C

So two sides and included angle - it represents side-angle-side or SAS congruency theorem

8 0
3 years ago
Write the equation of the line that passes through the points (−7,5) and (-7,-8).
Viefleur [7K]

Answer:

x = -7

Step-by-step explanation:

First we find the slope using

m = ( y2-y1)/(x2-x1)

   = ( -8 - 5)/( -7 - -7)

   = (-8-5)/(-7+7)

   = -13/0

This means the slope is undefined and the line is vertical

Vertical lines are in the form

x= constant and the constant is the x value of the points

x = -7

3 0
2 years ago
Suppose that a dart board is given by a circle of radius 1 in R², centered at (0,0). You throw a dart at the dart board, and the
Gnesinka [82]

Answer:

Answer B

Step-by-step explanation:

7 0
2 years ago
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:

<u>When k=1</u>

<u />\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

<u>When k=0</u>

<u />\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
1 year ago
<img src="https://tex.z-dn.net/?f=2x%20%3D%2044" id="TexFormula1" title="2x = 44" alt="2x = 44" align="absmiddle" class="latex-f
FrozenT [24]

Answer:

<h2>4</h2>

Step-by-step explanation:

2x = 44

=  > x =  \frac{44}{2}

=  > x = 22

Now if the 2 at the end is an power then it will turn to

22 =   {2}^{2}

= 2 \times 2

= 4

So the answer will be 4.

5 0
2 years ago
Read 2 more answers
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