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Zarrin [17]
3 years ago
9

Hugo plays basketball.

Mathematics
1 answer:
djyliett [7]3 years ago
4 0

Answer:

170

Step-by-step explanation:

multiply 12×15=170

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Two solids cones geometrically similar and the height of one cone is 1 1/2 times that of the other. Given that the height of the
Nitella [24]
I didnt bring my calculator
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Please help.<br> Algebra.
mr Goodwill [35]

Answer:  m= -1/2

Step-by-step explanation:

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4 0
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plzzz help smart people, its math will mark branilyest (DONT ANSWER IF YOU DONT KNOW THIS TYPE OF MATH PLZZ SHOW WORK)
sveta [45]

Answer:

6) 15

7)5

8)120 degrees

9)60 degrees

10)9

Step-by-step explanation:

GHIJ is a parallelogram.

Opposite sides of a parallelogram are congruent.

3y - 1 = 2y + 1

3y - 2y = 1 + 1

y = 2

Opposite sides of a parallelogram are congruent.

4x + 3 = x + 12

4x - x = 12 - 3

3x = 9

x = 9/3

x = 3

6)GH = ?

GH = 4x + 3

GH = 4(3) + 3

= 12 + 3

= 15

therefore, GH = 15

7) HI = ?

HI = 2y + 1

= 2(2) + 1

= 4 +1

= 5

therefore, HI = 5

Opposite angles of a parallelogram are equal.

8) m(angle I) = 120 degrees...... (given)

therefore, measure of angle G = measure of angle I

therefore, m(angle G) = 120 degrees

Consecutive angles of a parallelogram are supplementary.

9) m(angle I) + m(angle J) = 180 degrees

120 + m (angle J) = 180

m(angle J) = 180 - 120

= 60 degrees.

The diagonals of a parallelogram bisect each other.

10) JK = 9 .........( given)

JK = HK

therefore, HK = 9

5 0
3 years ago
Write <br> 2/9 as a repeating decimal.
grigory [225]

Answer:

.2 repeating

Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
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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