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yulyashka [42]
3 years ago
6

Least common denominator?

Mathematics
1 answer:
Over [174]3 years ago
4 0

Answer:

\large\boxed{(s-2)(s+2)(s-3)=(s^2-4)(s-3)=s^3-3s^2-4s+12}

Step-by-step explanation:

s^2-4=s^2-2^2=(s-2)(s+2)\\\\\text{used}\ a^2-b^2=(a-b)(a+b)\\\\s^2-5s+6=s^2-2s-3s+6=s(s-2)-3(s-2)=(s-2)(s-3)\\\\\text{Therefore}\ LCD(s^2-4,\ s^2-5s+6)=(s-2)(s+2)(s-3)

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What is 1 2/3 as an inproper fraction?​
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Answer:

5/3

Step-by-step explanation

1 2/3 = 5/3

3/3 + 2/3 = 5/3

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G(5)= please help asap
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Answer:

The answer is 5G

Step-by-step explanation:

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Select all the expressions that are equivalent to −5/6 / -1/3
Sindrei [870]

Step-by-step explanation:

We need to find an expression for \dfrac{\dfrac{-5}{6}}{\dfrac{-1}{3}}.

We can solve it as follows.

We know that,

\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}

So,

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7 0
3 years ago
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A) Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s.
Fed [463]

Answer:

A) a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

B) a_{0} = a_{1} = 0

C)   for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

Step-by-step explanation:

A) A recurrence relation for the number of bit strings of length n that contain a  pair of consecutive Os can be represented below

if a string (n ) ends with 00 for n-2 positions there are a pair of  consecutive Os therefore there will be : 2^{n-2} strings

therefore for n ≥ 2

The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

b ) The initial conditions

The initial conditions are : a_{0} = a_{1} = 0

C) The number of bit strings of length seven containing two consecutive 0s

here we apply the re occurrence relation and the initial conditions

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

7 0
3 years ago
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