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Andrei [34K]
3 years ago
9

Which type of triangle has special proportions

Mathematics
1 answer:
MrRissso [65]3 years ago
5 0

Answer: right triangle

Step-by-step explanation:

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2. The electrical resistance R of a wire varies inversely with the square of its diameter d. If a wire with a
Mariulka [41]

Answer:

0.225 \Omega =225 m\Omega

Step-by-step explanation:

We know (Ohm second law) that R= \frac{k}{d^2} where k inlcudes the rest of the parameters (material, lenght). In our situation we have k= 0.4\cdot9 \Omega mm^2. The moment the diameter becomes 8mm R becomes

R= {{0.4\cdot 9}\over{16}} \Omega = 0.225 \Omega

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3 years ago
Find the area. Simplify your answer​
Wittaler [7]

Answer:

14

Step-by-step explanation:

6 0
3 years ago
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8. There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random without replacement.
PIT_PIT [208]

Answer:

a. 720

b. 0.3

Step-by-step explanation:

a.

Three balls are selected at random without replacement.

First ball can be selected in 10 ways

Second ball can be selected in 9 ways

Third ball can be selected in 8 ways

Thus, there are 10×9×8 =720 different ways of selecting three balls.

b.

There are three cases that satisfy this condition

  1. first ball picked is 5, the others are not
  2. first ball is not 5, second is 5, third is not
  3. first and second ball is not five, third ball is 5

Each probabilities are:

  1. \frac{1}{10} * \frac{9}{9} *\frac{8}{8} =\frac{1}{10}
  2. \frac{9}{10} *\frac{1}{9} *\frac{8}{8} =\frac{1}{10}
  3. \frac{9}{10} *\frac{8}{9} *\frac{1}{8} =\frac{1}{10}

Thus, one of the balls selected is the number 5 is then \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}

5 0
3 years ago
How do I simplify the square root: 7 times the square root of 81
DENIUS [597]
7\cdot\sqrt{81}=7\cdot\sqrt{9\cdot9}=7\cdot9=\boxed{63}
7 0
3 years ago
Can someone please help ? Thank you
goldfiish [28.3K]

Answer: 25 times 50 = 1,250 inches every inch is 50 miles on the second map

Step-by-step explanation: hope i could help you

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3 years ago
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