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damaskus [11]
3 years ago
9

At a local gardening store, there is a small group of plants located near the gardening

Mathematics
1 answer:
Veronika [31]3 years ago
8 0

Answer:

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Step-by-step explauuuuuuuuuuu

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You are playing a card game with friends. at one point. you double your score. in the next hand you lose 75 points. you end up w
Gnoma [55]

Answer:

i) x points; ii) 2x - 75 = -25; iii) x = 25

Step-by-step explanation:

i) Starting points

Let's say you had x points before doubling your score.

(ii) The equation

 After doubling score: 2x points

After losing 75 points: (2x - 75) points

         At end of game: -25 points

The equation is

2x - 75 = -25

iii) Solution to equation

2x - 75 = -25     Add 75 to each side

      2x =  50     Divide each side by 2

        x =  25

3 0
4 years ago
Darrel divided 575 by 14 by using partial quotients, what is the quotient?
Usimov [2.4K]
143r3 (143 remainder 3)
3 0
3 years ago
T-5-2t+5 simplify expression
pav-90 [236]

Answer:

-t

Step-by-step explanation:

t - 5 - 2t + 5 =

t - 5 + 5 - 2t =

t - 2t =

-t

3 0
3 years ago
Read 2 more answers
Choose the three shapes that have an area of 36 square centimeters
Artemon [7]

Choose the three shapes that have and Area of 36 square centimeters

Step-by-step explanation:

3 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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