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Lynna [10]
3 years ago
15

What is 1/6 + 1/6 + 1/6

Mathematics
2 answers:
tatyana61 [14]3 years ago
6 0

Answer:

1/6 + 1/6 + 1/6 =

1/2 or 0.5

ankoles [38]3 years ago
5 0
1/6+1/6+1/6+=2/6+1/6 2/6+1/6=3/6
3/6=1/2
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please answer will mark brainalist .the area of a triangle of altitude 15 m is 225 sq.m.whatbis the length of its base please sh
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Step-by-step explanation:

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7 0
3 years ago
6j+8j+(5+5) Use the distributive property
vova2212 [387]

Hello!

Answer:

\boxed{14j+10}

Step-by-step explanation:

Distributive property: →→ a(b+c)=ab+ac

Order of operations stands for:

↓

Parenthesis

Exponent

Multiply

Divide

Add

Subtract

*Left to right*

Please Excuse My Dear Aunt Sally!

First, you remove parenthesis.

6j+8j+5+5

Then, you add by the numbers from left to right.

6+8=14

14j+5+5

5+5=10

<u><em>14j+10 is the final answer.</em></u>

Hope this helps!

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5 0
3 years ago
The robotics team needs to purchase $350 of new equipment. Each of the x students on the team plans to fundraise and contribute
Thepotemich [5.8K]
How many people are there?
3 0
3 years ago
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Neporo4naja [7]

Answer:

b

Step-by-step explanation:

6 0
3 years ago
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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