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Sladkaya [172]
2 years ago
13

Think about all of the ways in which a circle and a parabola can intersect. Select all of the number of ways in which a circle a

nd a parabola can intersect.

Mathematics
2 answers:
tangare [24]2 years ago
8 0

Answer:

0

1

2

just sacrificed myself on edge

forsale [732]2 years ago
4 0

Answer:

A circle can intersect a parabola in

1. One point [ when circle just touches the parabola]

2. Two points [ When circle cuts the parabola in two distinct points. ]

3. Three points [Circle just touches at one point and cuts the parabola in two distinct points]

4. Four points [ Either parabola or circle meeting each other or crossing at four distinct points]




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The Parsons have a balance of $3,664.47 in their check register. On their bank statement, the balance on the account is $2,346.8
Lapatulllka [165]
$2,346.89 + $500 + $1,456.81 = $4,303.70

$4,303.70 - $324.56 - $25.50 - $78.92 - $30.25 - $60 - $60 - $60 = $3,664.47

$3,664.47 is exactly what they have in their checkbook.
5 0
2 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
2 years ago
You and your friend are rolling number cubes. Each cube has 6 sides with the numbers 1 to 6. If a sum of 7 is rolled, your frien
Ad libitum [116K]
You don't list the "sums listed below" but I can still help you.
There are 6 ways to get a sum of 7 using a pair of dice:
1+6
6+1    You might be asking why 2+5 isn't the same as 5+2
2+5      Imagine the dice are different colors.  Is a red 2 and green 5 the
5+2          same as a green 2 and red 5?  (No, it's not)
3+4
4+3
You need two numbers that also provide 6 ways to win.  One pair that would work is 11 and 9   There are two ways to make 11 (5+6) and (6+5)  and there are four ways to make 9 (3+6), (6+3) (4+5) and (5+4) - that's a total of six ways.  So to balance your friend's 7 - you could go for sums of 11's and 9's.  Check the choices you were given and find the numbers (it could be three numbers like 2,3, and 12) that give you 6 ways to win.
Here are the numbers and their ways
Sum of the dice                          Ways to get it
2                                                      1
3                                                      2
4                                                      3
5                                                      4
6                                                      5
7                                                      6
8                                                      5
9                                                      4
10                                                    3
11                                                    2
12                                                    1


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The amount of change totaled -$100.

7 0
1 year ago
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