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Mademuasel [1]
3 years ago
15

IM REALLY STUCK WITH GEOMETRY CAN SOMEONE PLEASE HELP? ON MY PAGE THERE IS 2 QUESTIONS! P L E A S E HELP

Mathematics
1 answer:
statuscvo [17]3 years ago
4 0

hey friend i have answered your most of the question so can you please mark my all answers as brainlist

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Find the area of the sector.<br> 1350<br> 18 km
harina [27]

Answer:

21870 km^2

Step-by-step explanation:

Area of sector=1/2r^2(theta)

Where r=18 km and theta=135

Area of sector=1/2(18)^2(135)

=1/2(324)(135)

=1/2(43740)

=21870 km^2

7 0
3 years ago
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timofeeve [1]

Answer:

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3 years ago
"Gina wrote the following paragraph to prove that the segment joining the midpoints of two sides of a triangle is parallel to th
Aleonysh [2.5K]
What is the flaw in Gina’s proof?
A) Points D and E must be constructed, not simply labeled, as midpoints.

B) Segments DE and AC are parallel by construction.  THIS IS THE FLAW. THE SEGMENTS WERE NOT DRAWN NOR PROPERLY IDENTIFIED.

C) The slope of segments DE and AC is not 0.

D) The coordinates of D and E were found using the Distance between Two Points Postulate 
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3 years ago
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Please find the square roots of the following complex number.
Nana76 [90]

Hey there,

First, find the square root of the given equation.

\sqrt{z}=\sqrt{81(\text{cos}(\frac{4\pi}{9})+i\text{sin}(\frac{4\pi}{9})}\\=9\text{cos}\frac{1}{2}(2k\pi+(\frac{4\pi}{9}))+i\text{sin}\frac{1}{2} (2k\pi+(\frac{4\pi}{9}))

Keep in mind that this is solved for k = 0 and 1.

Second, simplify.

=9\text{cos}(\frac{2\pi}{9})+i\text{sin}(\frac{2\pi}{9})~\text{or}~9\text{cos}(\frac{11\pi}{9})+i\text{sin}(\frac{11\pi}{9})

Best of Luck!

6 0
3 years ago
I really need help with this and how you got it please 10 points!!
Kamila [148]

Answer:

1 and 7  are congruent

2 and 5 are supplementary

4 and 6 are congruent

3 and 8 are supplementary

8 0
3 years ago
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