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kkurt [141]
3 years ago
5

Pls help with 2 and 3 with work SHOWN Please THANKS!

Mathematics
1 answer:
madreJ [45]3 years ago
3 0
22222222222 is b
33333 is c
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Pls help me :(((( Thank you
Alina [70]

Step-by-step explanation:

\frac{2}{ \sqrt{9} }

\frac{2 \times  \sqrt{9} }{ \sqrt{9}  \times  \sqrt{9} }

\frac{2 \sqrt{9} }{9}

3 0
4 years ago
Read 2 more answers
What is the circumference of this circle
exis [7]

Answer:<em> 12.56 km</em>

Step-by-step explanation:

C=2(3.14)r

C=2(3.14)2

C=12.56

6 0
3 years ago
Which of the follow box-and-whisker plots correctly displays this data set?
sdas [7]

Answer:

d

Step-by-step explanation:

4 0
3 years ago
Find the value of the following expression: (2^8 • 5^-5 • 19^0)^-2 • (5^-2 over 2^3)^4 • 2^28. Write your answer in simplified f
DaniilM [7]
Recognize that "( ) over 2^3" means ( ) • 2^-3. Use the rule of exponents
.. (a^b)^c = a^(b•c)

= 2^-16•5^10•19^-2 • 5^-8*2^-12 • 2^28

Now, you can use the rule of exponents
.. (a^b)*(a^c) = a^(b+c)

= 2^(-16 -12 +28) • 5^(10 -8) • 19^-2
= 5^2 • 19^-2
= 5^2 / 19^2
= 25/361
3 0
4 years ago
A radioactive substance decays exponentially. A scientist begins with 190 milligrams of a radioactive substance. After 19 hours,
Lelu [443]

Answer:

  51 milligrams

Step-by-step explanation:

Exponential growth or decay can be modeled by the equation ...

  y = a·b^(x/c)

where 'a' is the initial value, 'b' is the "growth factor", and 'c' is the time period over which that growth factor applies. The time period units for 'c' and x need to be the same.

In this problem, we're told the initial value is a = 190 mg, and the value decays to 95 mg in 19 hours. This tells us the "growth factor" is ...

  b = 95/190 = 1/2

  c = 19 hours

Then, for x in hours the remaining amount can be modeled by ...

  y = 190·(1/2)^(x/19)

__

After 36 hours, we have x=36, so the remaining amount is ...

  y = 190·(1/2)^(36/19) ≈ 51.095 . . . . milligrams

About 51 mg will remain after 36 hours.

8 0
2 years ago
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