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Vinil7 [7]
3 years ago
9

Please help with geometry! #9!!

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
4 0

Answer:

           6, 7, 8 are correct

<h3>           9.  2) TU = 58</h3>

Step-by-step explanation:

XV=\dfrac{RS+TU}2\\\\\\44=\dfrac{30+TU}2\\\\44\cdot2=30+TU\\\\88-30=TU\\\\TU=58

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Find the sum \[\sum_{k = 1}^{2004} \frac{1}{1 + \tan^2 (\frac{k \pi}{2 \cdot 2005})}.\]
vampirchik [111]

Answer:1002

Step-by-step explanation:

\Rightarrow \sum_{k=1}^{2004}\left ( \frac{1}{1+\tan ^2\left ( \frac{k\pi }{2\cdot 2005}\right )}\right )

and 1+\tan ^2\theta =\sec^2\theta

and \cos \theta =\frac{1}{\sec \theta }  

\Rightarrow \sum_{k=1}^{2004}\left ( \cos^2\frac{k\pi }{2\cdot 2005}\right )

as \cos ^2\theta =\cos ^2(\pi -\theta )

Applying this we get

\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]

every \thetathere exist \pi -\theta

such that \sin^2\theta +\cos^2\theta =1

therefore

\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]=1002                          

6 0
3 years ago
Solve for x 3+5x=18
Nikitich [7]

Answer:

x=3

Step-by-step explanation:

7 0
3 years ago
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Eric and mark went t a ball game and visited the concession stand.They each got a hot dog,and mark got a large soda and a bag of
LenKa [72]
The soda was $1.00.


Hot Dog- $3.50
Peanuts- $4.00

3.50+4.00= 7.50

Total paid is $11.50-$7.50= $4.00-$3.00(coupon)= $1.00
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IgorLugansk [536]

Step-by-step explanation: The price of3.8 pounds of almonds=$39.90 thus we can write Hence the price of almonds is $10.50 per pounds.

AWNSERS:$10.50

3 0
3 years ago
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Select all the correct answers.
harkovskaia [24]

Answer:Which four inequalities can be used to find the solution to this absolute value inequality?3+2|x-11 <9

2(x - 1) <-3

x-1 <9

X-1 >-9

-2(x - 1) > 3

x-1-3

X-1 <3

Step-by-step explanation:

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3 years ago
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