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Phantasy [73]
2 years ago
9

Help me please help ASAP help

Mathematics
1 answer:
Tomtit [17]2 years ago
4 0

Answer:

98.1 unit²

Step-by-step explanation:

area of semi-circle= ½πr²

½π(3)²

= 9/2π

area of isosceles trapezoid= ½(b1+b2)h

½(15+6)8

=84

area of figure= 9/2π +84

98.1 unit²

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Solve 13+x>11. Enter your answer as an equality
Wittaler [7]

Answer:

x>-2

Step-by-step explanation:

13+x>11

Subtract 13 from each side

13-13+x>11-13

x>-2

5 0
3 years ago
Whats 4.25 increased by 15%
Firlakuza [10]
You would turn the 15% into a decimal which is .15 then multiply .15 by 4.25. After multiplying those together add 4.25 to the answer.
5 0
3 years ago
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Find the distance between the points given. (-3, -4) and (0, 0)
melomori [17]
The distance between the points is 5
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How many solutions exist for the given equation?
zmey [24]
<h3> Answer </h3>

x = 2

<h3>Step by step</h3>

1/2(x + 12) = 4x - 1

1/2x + 12/2 = 4x - 1

1/2x + 6 = 4x - 1

1/2x - 4x = -1 - 6

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8 0
2 years ago
A device produces random 64-bit integers at a rate of one billion per second. After how many years of running is it unavoidable
nikitadnepr [17]

Answer:

3171 × 10^(44) years

Step-by-step explanation:

For each bit, since we are looking how many years of running it is unavoidable that the device produces an output for the second time, the possible integers are from 0 to 9. This is 10 possible integers for each bit.

Thus, total number of possible 64 bit integers = 10^(64) integers

Now, we are told that the device produces random integers at a rate of one billion per second (10^(9) billion per second)

Let's calculate how many it can produce in a year.

1 year = 365 × 24 × 60 × 60 seconds = 31,536,000 seconds

Thus, per year it will produce;

(10^(9) billion per second) × 31,536,000 seconds = 3.1536 × 10^(16)

Thus;

Number of years of running is it unavoidable that the device produces an output for the second time is;

(10^(64))/(3.1536 × 10^(16)) = 3171 × 10^(44) years

6 0
3 years ago
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