PLEASE HELP MEE i dont really understand this<br><br> 4(p+1.25)=7 2/5

Find the complex zeros of the following polynomial tunction and use the complex zeros to factor f

kramer

Answer:

x = 5 or x = 5 + 2 i or x = 5 - 2 i

Step-by-step explanation:

Solve for x:

x^3 - 15 x^2 + 79 x - 145 = 0

Hint: | Factor the left-hand side.

The left hand side factors into a product with two terms:

(x - 5) (x^2 - 10 x + 29) = 0

Hint: | Find the roots of each term in the product separately.

Split into two equations:

x - 5 = 0 or x^2 - 10 x + 29 = 0

Hint: | Look at the first equation: Solve for x.

Add 5 to both sides:

x = 5 or x^2 - 10 x + 29 = 0

Hint: | Look at the second equation: Solve the quadratic equation by completing the square.

Subtract 29 from both sides:

x = 5 or x^2 - 10 x = -29

Hint: | Take one half of the coefficient of x and square it, then add it to both sides.

Add 25 to both sides:

x = 5 or x^2 - 10 x + 25 = -4

Hint: | Factor the left-hand side.

Write the left hand side as a square:

x = 5 or (x - 5)^2 = -4

Hint: | Eliminate the exponent on the left-hand side.

Take the square root of both sides:

x = 5 or x - 5 = 2 i or x - 5 = -2 i

Hint: | Look at the second equation: Solve for x.

Add 5 to both sides:

x = 5 or x = 5 + 2 i or x - 5 = -2 i

Hint: | Look at the third equation: Solve for x.

Add 5 to both sides:

Answer: x = 5 or x = 5 + 2 i or x = 5 - 2 i

7 0

4 years ago

Using the BOX and WHISKER above, what is the LOWER QUARTILE of the information?

kifflom [539]

Answer:

Lower Quartile is 8 according to the diagram.

6 0

2 years ago

In this exercise, we consider strings made from uppercase letters in the English alphabet and decimal digits.

Yuri [45]

Answer:

a) $26^2 36^8$

b) $21\cdot10\cdot36^7$

c) $5^3 31^7$

d) $10\cdot 9\cdot 8 \cdot 7 \cdot 26^6$

Step-by-step explanation:

We will use the product rule from combinatorics.

a) There are 26 letters in the English alphabet, so there are 26 possible choices for the first character and 26 possible choices for the last one. Each one of the remaining eight characters of the string has 36 choices (letters or digits). By the product rule, there are $26\cdot36\cdot 36\cdots 36\cdot 26=26^2 36^8$ strings.

b) We have 5 possible choices for the first character, it must be some vowel a,e,i,o,u. The second character can be chosen in 21 ways, selecting some consonant. There are 10 possibilities for the last character because only of the digits are allowed. The other seven characters have no restrictions, so each one can be chosen in 36 ways. By the product rule there are $21\cdot 10\cdot 36^7$ strings.

c) The third character has 5 possibilities. Repetition of vowels is allowed, so the sixth and eighth characters have each one 5 possible choices. There are seven characters left. None of them are a vowel, but they are allowed to take any other letter or digit, so each one of them can be chosen in 36-5=31 ways. Therefore there are $5^3 31^7$ strings.

d) Remember that the binomial coefficient $\binom{n}{k}$ is the number of ways of choosing k elements from a set of n elements. In this case, to count all the possible strings, we first need to count in how many ways we can select the four positions that will have the digits. This can be done in $\binom{10}{4}$ ways, since we are choosing four elements from the set of the ten positions of the string. Now, for the first position, we can choose any digit so it has 10 possibilities. The second position has 9 possibilities, because we can't repeat the digit used on the first position. Similarly, there are 8 choices for the third position and there are 7 choices for the fourth. Now, these are the only digits on the string, so the remaining 6 characters must be letters, then each one of them has 26 possibilities. By the product rule, there are $10\cdot 9\cdot 8 \cdot 7 \cdot 26^6$ strings.