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3241004551 [841]

3 years ago

15

Solve the inequality and describe the solution set: y-6 ≥ 12

2 answers:

LUCKY_DIMON [66]3 years ago

8 0

Answer: The solution set is y≥18.

Step-by-step explanation:

Since we have given that

$y-6\geq 12$

We need to first find the inequality and find the solution set.

$y-6\geq 12\\\\y\geq 12+6\\\\y\geq 18$

So, the solution set for this would be all the values of y which is greater than 18.

i.e. y≥18.

Finger [1]3 years ago

3 0

Y<span>≥18

You want to get the y only on one side of the equation. To get rid of the -6 you have to add 6 to both sides to even out the equation.

y-6</span><span>≥</span><span>12
+6 +6
y </span>≥18

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Given a right triangle with a hypotenuse of 6 cm and a leg of 4cm, what is the measure of the other leg of the triangle rounded

slega [8]

Answer:

$4.5 cm=L2$

Step-by-step explanation:

This problem can be solved using the pythagorian theoreme:

$H=\sqrt{L1^{2}+L2^{2} }$

Where:

$H=6 cm$

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5 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