Answer:
The first four terms of the sequence are-6, -2, 2 and 6
Step-by-step explanation:
The given sequence is 
In order to find the first four term of the sequence we put n =0, 1, 2 and 3.
For n =0

For n =1

For n =2

For n =3

Therefore, the first four terms of the sequence are
-6, -2, 2 and 6
Answer:
t ≈ -2.014 or 3.647
Step-by-step explanation:
Add the opposite of the expression on the right side of the equal sign to put the equation into standard form.
4.9t² -8t -36 = 0
You can divide by 4.9 to make this a little easier to solve.
t² -(8/4.9)t -36/4.9 = 0
Now, add and subtract the square of half the x-coefficient to "complete the square."
t² -(8/4.9)t +(4/4.9)² -36/4.9 -(4/4.9)² = 0
(t -4/4.9)² -192.4/4.9² = 0 . . . . simplify
Add the constant term, then take the square root.
(t -4/4.9)² = 192.4/4.9²
t -4/4.9 = ±(√192.4)/4.9
t = (4 ± √192.4)/4.9
t ≈ {-2.014, 3.647}
first convert 0.88 to degrees
so arcsin(0.88) = 61.642 degrees
Arcsin on a calculator is sin^-1
then to convert to radians divide degrees by 180 and multiply that by PI
so (61.642/180) x PI = 1.076 radians
Step-by-step explanation:
Let x be the miles Kenyon ran on tuesday
Let y be the miles Kenyon ran on thursday
Given,
Mon + Tues + Wed + Thu = 8mi

also given,

Now we have x + y = 5.2 as equation 1 and -x + y = 0.4 as equation 2
Using equation 1,

Substitute this equation into equation 2.

Substitute y = 2.8 into equation 1.

Therefore Kenyon ran 2.4 miles on tuesday, and ran 2.8 miles on thursday.
Answer:
n times 5
Step-by-step explanation:
A matrix Anxn of this way is called an upper triangular matrix. It can be proved that the determinant of this kind of matrix is

In this case, it would be 5+5+...+5 (n times) = n times 5
We are going to develop each determinant by the first column taking as pivot points the elements of the diagonal
![det\left[\begin{array}{cccc}5&a_{12}&a_{13}...&a_{1n}\\0&5&a_{23}...&a_{2n}\\...&...&...&...\\0&0&0&5\end{array}\right] =5+det\left[\begin{array}{ccc}5&a_{23}...&a_{2n}\\0&5&a_{3n}\\...&...&...\\0&0&5\end{array}\right]=5+5+...+det\left[\begin{array}{cc}5&a_{n-1,n}\\0&5\end{array}\right]=5+5+...+5+5\;(n\;times)](https://tex.z-dn.net/?f=det%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D5%26a_%7B12%7D%26a_%7B13%7D...%26a_%7B1n%7D%5C%5C0%265%26a_%7B23%7D...%26a_%7B2n%7D%5C%5C...%26...%26...%26...%5C%5C0%260%260%265%5Cend%7Barray%7D%5Cright%5D%20%3D5%2Bdet%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26a_%7B23%7D...%26a_%7B2n%7D%5C%5C0%265%26a_%7B3n%7D%5C%5C...%26...%26...%5C%5C0%260%265%5Cend%7Barray%7D%5Cright%5D%3D5%2B5%2B...%2Bdet%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D5%26a_%7Bn-1%2Cn%7D%5C%5C0%265%5Cend%7Barray%7D%5Cright%5D%3D5%2B5%2B...%2B5%2B5%5C%3B%28n%5C%3Btimes%29)