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IgorC [24]
3 years ago
9

The longest side of an obtuse triangle measures 20 cm. The two shorter sides measure x cm and 3x cm. Rounded to the nearest tent

h, what is the greatest possible value of x? 6.3 6.4 7.0 7.1
Mathematics
2 answers:
Flura [38]3 years ago
8 0

Answer:

7.1

Step-by-step explanation:

because pythagorean theorem

c² > a² + b²

20² > x² + 3x²

400 > 4x²

÷4      ÷4

100 > x²

  10 > x           400 > 4(7.1)²

                      400 > 4(50.41)

                      400 > 201.64

tankabanditka [31]3 years ago
6 0
Obtuse triangle Pythagorean Theorem

c² > a² + b²

20² > x² + 3x²
400 > 4x²
<u>÷4      ÷4</u>
<u> 100 > x²</u>
   10 > x
  
400 > 4(7.1)²
400 > 4(50.41)
400 > 201.64

The greatest possible value of x is 7.10
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B

Step-by-step explanation:

Domain and ranges of inverse functions are swapped.

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Photon
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3 years ago
A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 2
Shalnov [3]

Answer:

The expected winnings for a person buying 1 ticket is -0.2.                  

Step-by-step explanation:

Given : A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 25 cents each, find the expected winnings for a person buying 1 ticket.

To find : What are the expected winnings?    

Solution :

There are one first prize, 2 second prize and 20 third prizes.

Probability of getting first prize is \frac{1}{20000}

Probability of getting second prize is \frac{2}{20000}

Probability of getting third prize is \frac{20}{20000}

A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each.

So, The value of prizes is

\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10

If 20000 tickets are sold at 25 cents each i.e. $0.25.

Remaining tickets = 20000-1-2-20=19977

Probability of getting remaining tickets is \frac{19977}{20000}

The expected value is

E=\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10-\frac{19977}{20000}\times 0.25

E=\frac{1000+600+200-4994.25}{20000}

E=\frac{-3194.25}{20000}

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Therefore, The expected winnings for a person buying 1 ticket is -0.2.

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3 years ago
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siniylev [52]

Answer:

Garrett can order stickers from two companies. Company A charges a $30 design fee plus $0.80 per sticker. Company B charges a $14 design fee plus $1.20 per sticker. How many stickers would Garrett have to order for the cost at both companies to be the same?

Step-by-step explanation:

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