Answer:
We have the equation
![c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]](https://tex.z-dn.net/?f=c_1%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_2%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_3%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_4%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Then, the augmented matrix of the system is
![\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D0%260%260%268%5C%5C0%260%264%264%5C%5C0%263%263%263%5C%5C1%261%261%261%5Cend%7Barray%7D%5Cright%5D)
We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:
![\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%261%261%5C%5C0%263%263%263%5C%5C0%260%264%264%5C%5C0%260%260%268%5Cend%7Barray%7D%5Cright%5D)
This matrix is in echelon form. Then, now we apply backward substitution:
1.
2.
3.
4.
Then the system has unique solution that is 
and this imply that the vectors 
are linear independent.
Answer: Hello!
Let's start with the word TRISKAIDEKAPHOBIA wich has 17 letters (some of them repeat, but it does not matter in this problem)
We want to know how many permutations we can do with 17 letters: then think this way, Lets compose a word. The first letter of this word has 17 options, the second letter of the word has 16 options (you already took one of the set) the third letter of the word has 15 options, and so on.
The total number of permutations is the product of the number of options that you have for each letter, this is:
17*16*15*14*....*3*2*1 = 17! = 3.6e+14
(b) FLOCCINAUCINIHILIPILIFICATION now we have 30 letters in total, using the same reasoning as before, here we have 30! permutations; this is
30! = 2.65e+32
(c) PNEUMONOULTRAMICROSCOPICSILICOVOLCANOCONIOSIS: now there are 47 letters.
then P = 47! = 2.59e+59
(d) DERMATOGLYPHICS: here are 18 letters, then:
p = 18! = 6.4e+15
- Quadratic Formula:
, with a = x^2 coefficient, b = x coefficient, and c = constant
With our equation, plug in the values:
Next, solve the exponent and multiplications:
Next, solve the subtraction:
Next, factor out i (i = √-1):
Next, solve the square root:
Lastly, divide and <u>your answer is:</u>
Answer:
= (3t+2)(3t-2)(3t-4)
Step-by-step explanation:
Given the expression 27t^3 - 36t^2 - 12t + 16
On factoring:
(27t^3 - 36t^2) - (12t + 16)
= 9t²(3t-4)-4(3t-4)
= (9t²-4)(3t-4)
factoring 9t²-4
9t²-4 = (3t)² - 2²
From different of two square, a²-b² = (a+b)(a-b)
Hence (3t)² - 2² = (3t+2)(3t-2)
= (9t²-4)(3t-4)
= (3t+2)(3t-2)(3t-4)
Hence the factored form of the expression is (3t+2)(3t-2)(3t-4)
Answer: Y= -10
Step-by-step explanation:
