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Sonja [21]
3 years ago
5

What is 6/12 in simplist form?

Mathematics
1 answer:
patriot [66]3 years ago
3 0
It’s 1/2 there you go I hope this was helpful
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Step-by-step explanation:

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2 years ago
The student government association at a certain college assigns a number to each student enrolled, puts the numbers in order, an
snow_tiger [21]

Answer: Systematic

Step-by-step explanation:

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3 years ago
What is the thirty-second term of the arithmetic sequence -12, -7, -2, 3, ... ? Show your work.
anzhelika [568]

The thirty-second term is 143

Further explanation:

As it is already given that the given sequence is an arithmetic sequence

We have to find the common difference first

So,

a_1=-12\\a_2=-7\\a_3=-2\\a_4=3

d=a_2-a_1=-7-(-12)=-7+12=5\\a_3-a_2=-2-(-7)=-2+7=5

The common difference is 5.

And first term is -12

The explicit formula for arithmetic sequence is:

a_n=a_1+(n-1)d\\Here,\\a_n\ is\ nth\ term\\a_1\ is\ first\ term\\d\ is\ common\ difference

Putting the values in the formula

a_n=-12+(n-1)(5)\\a_n=-12+5n-5\\a_n=-17+5n

Putting n=32 in explicit formula

a_{32}=-17+5(32)\\=-17+160\\=143

The thirty-second term is 143

Keywords: Common difference, Arithmetic Sequence

Learn more about arithmetic sequence at:

  • brainly.com/question/10879401
  • brainly.com/question/10940255

#LearnwithBrainly

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2 years ago
PLEASE HELP ME WITH EMPIRICAL RULE!! ASAP!
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Answer:

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8 0
3 years ago
The number of bacteria in a culture is given by the function n(t) =950e^0.45t Where t is measured in hours A) what is the contin
tatiyna

Answer:

a) The continuous rate of growth of this bacterium population is 45%

b) Initial population of culture at t = 0 is 950 bacteria

c) number of bacterial culture contain at t = 5 is 9013 bacteria

Step-by-step explanation:

The number of bacteria in a culture is given by

n(t) = 950e^{0.45t}

a) rate of growth is

Bacteria growth model N(t) = noe^{rt}

Where r is the growth rate

hence r = 0.45

              = 45%

b) Initial population of culture at t = 0

n(0) = 950e^{0.45(0)}

       = 950e^{0}

       = 950 bacteria

c) number of bacterial culture contain at t = 5

n(5) =  950e^{0.45(5)}

      = 950e^{2.25}

      = 9013.35

      = 9013 bacteria

6 0
3 years ago
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