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2 years ago

6

( URGENT) What are the roots of the polynomial equation x Superscript 4 Baseline + x cubed = 4 x squared + 4 x? Use a graphing c

alculator and a system of equations.
–2, –1, 0, 2
–2, 0, 1, 2
–1, 0
0, 1

2 answers:

Finger [1]2 years ago

4 0

Answer:

A.

Step-by-step explanation:

Feliz [49]2 years ago

3 0

Answer: A

Step-by-step explanation:

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Recall that a 6-bit string is a bit strings of length 6, and a bit string of weight 3, say, is one with exactly three 1's. How m

strojnjashka [21]

Answer:

1.. Total number of 6 bit strings is 64

2. Number of 6-bit strings with weight of 0 is 1

3. Number of 6-bit strings with weight of 1 is 6

4. Number of 6-bit strings with weight of 3 is 20

5. Number of 6-bit strings with weight of 5 is 6

6. Number of 6-bit strings with weight of 6 is 1

7. Number of 6-bit strings with weight of 7 is 0

Step-by-step explanation:

A bit string is a string that contains 0 and 1 only

1. Total number of 6 bit strings is 2^6 = 64

2. Number of 6 bit strings with weight 0 is 1

Explanation

Weight 0 means a string with no occurrence of 1

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 0 i.e. no occurrence of 1

6C0 = 6!/(6-0)!0!

6C0 = 6!/6!0!

6C0 = 1

Hence, the number of string with weight 0 (i.e. no occurrence of 1 ) is 1

3. Number of string with weight 1 is 6

Explanation

Weight 0 means a string with exactly 1 occurrence of '1'

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 1

6C1 = 6!/(6-1)!1!

6C1 = 6!/5!1!

6C1 = 6

Hence, the number of string with weight 6

4. Number of string with weight 3 is 20

Explanation

n = 6 and r = 3

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 20

Hence, the number of string with weight 3 is 20

5. Number of string with weight 5 is 6

Explanation

n = 6 and r = 5

6C5 = 6!/(6-5)!5!

6C5 = 6!/1!5!

6C5 = 6

Hence, the number of string with weight 5 is 6

6. Number of string with weight 6 is 1

Explanation

n = 6 and r = 6

6C6 = 6!/(6-6)!6!

6C6 = 6!/0!6!

6C6 = 1

Hence, the number of string with weight 6 is 1

7. Number of string with weight 7 is 0

Weight of 7 means that a string that has 7 occurrence of 1

The total length of a 6 bit is 6

Since 6 is less than 7, there's no way a bit of weight 7 can occur.

So, the right answer for this is 0.

8 0

3 years ago

Does the number -2.675 belong to the set of integers

tankabanditka [31]

Answer:

yes

Step-by-step explanation:

4 0

3 years ago

Y-intercept of a line that has a slope of 2 and passes<br> through the point (4, 6).

tia_tia [17]

Answer:

<h2> b = -2</h2>

Step-by-step explanation:

The point-slope form of equation is: y - y₀ = m(x - x₀), where (x₀, y₀) is any point the line passes through and m is the slope.

m = 2

(4, 6) ⇒ x₀ = 4, y₀ = 6

So, the point-slope form of equation:

y - 6 = 2(x - 4)

Changing to the slope-intercept form of the equation of the line (y = mx + b, where m is the slope and b is the y-intercept of the line):

y - 6 = 2x - 8 {add 6 to both sides}

y = 2x - 2 ⇒ b = -2

6 0

2 years ago

Quanto fa 2 +2 ,qualcuno capisce l’italiano?

creativ13 [48]

Answer:

i dont know which language is this but the answer is 4

6 0

2 years ago

Read 2 more answers

You roll a 6-sided die.<br>What is P(prime)?<br>Write your answer as a percentage.<br>Submit

Cloud [144]

You roll a 6-sided die. The probability of getting prime is 50%.

<u>Solution:</u>

Given, that we have rolled a 6 – sided die.

We have to find the P(prime) as percentage.

Now, It means that we have to find the probability of getting a prime number on the face.

On rolling a six sided die, the total possible outcomes are 1, 2, 3, 4, 5, 6

Number of possible outcomes = 6

We have to get a prime number on rolling a die

The prime numbers in possible outcomes are 2, 3, 5

So number of favorable outcomes = 3

$\text {Probability of an event as percentage }=\frac{\text { favourable number of outcomes }}{\text { total number of outcomes }} \times 100$

$\text {Probability of getting prime number on face of die }=\frac{3}{6} \times 100$

$\begin{array}{l}{\text { P(prime) }=\frac{1}{2} \times 100} \\\\ {\text { P(prime) }=50 \%}\end{array}$

Hence, probability of getting prime is 50%.

3 0

3 years ago