Answer:
Let the vectors be
a = [0, 1, 2] and
b = [1, -2, 3]
( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.
Let the cross product be another vector c.
To find the cross product (c) of a and b, we have
![\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0%261%262%5C%5C1%26-2%263%5Cend%7Barray%7D%5Cright%5D)
c = i(3 + 4) - j(0 - 2) + k(0 - 1)
c = 7i + 2j - k
c = [7, 2, -1]
( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:
c / | c |
Where | c | = √ (7)² + (2)² + (-1)² = 3√6
Therefore, the unit vector is
or
[
,
,
]
The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:
[
,
,
]
In conclusion, the two unit vectors are;
[
,
,
]
and
[
,
,
]
<em>Hope this helps!</em>
A sequence is a set of numbers, called terms, arranged in some particular order. An arithmetic sequence is a sequence with the difference between two consecutive terms constant. The difference is called the common difference. A geometric sequence is a sequence with the ratio between two consecutive terms constant.
In this problem, it is important to take note that the number of numbers to be utilized isn't specified so it can be up to a thousand numbers. It wasn't also specified if repeating of numbers is allowed or not. So with those taken into consideration and the condition presented in mind, the numbers that can give you 8 when added and 30 when multiplied are 2, 3, 5, -1, and another -1. The derivation from this is mainly from factorization and a little bit of logic.
here is the solution.
2 x 3 x 5 x -1 x -1 = 30
6 x 5 x -1 x -1 = 30
30 x -1 x -1 = 30
-30 x -1 = 30
30 = 30
2 + 3 + 5 + -1 + -1 = 8
5 + 5 + -1 + -1 = 8
10 + -1 + -1 = 8
9 + -1 = 8
8 = 8
Answer:
B. 100
Step-by-step explanation:

143-13=130
10x=130 divide both sides by 10 130/10= 13 x=13
MN=7(13)=91+9=100
LN=3(13)=39+4=43
100+43
16)
s^1/4 * s^1/4 * s^1/4 * s^1/4 = s^(1/4 + 1/4+1/4+1/4) = s^4/4 = s
Answer is s
15) Answer is 1. (because any number to the 0 exponent is equal 1 , except 0)
17)
= (t^3/6)^2 = t^6/6 = t
Answer is t