180 is the answer . Just multiply both numbers.
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>
Answer:
1) x = 14°, y = 5°
2) x = 18.5°, y = 37°
Step-by-step explanation:
1) ∠AOC and ∠BOD are vertical angles, then ∠BOD = 25°
∠MOD = ∠MOB + ∠BOD = 90°
3x + 23° + 25° = 90°
3x = 90° - 23° - 25°
x = 42°/3
x = 14°
∠LOB = ∠LOM + ∠MOB = 90°
5y + 3x + 23° = 90°
5y = 90° - 23° - 3(14°)
y = 25°/5
y = 5°
2) ∠AOC and ∠BOD are vertical angles, then ∠BOD = 16°
∠EOB = ∠EOD + ∠DOB = 90°
2y + 16° = 90°
y = (90° - 16°)/2
y = 37°
∠DOF = ∠BOF + ∠DOB = 90°
4x + 16° = 90°
x = (90° - 16°)/4
x = 18.5°
Answer:
The couch would be $163.85
Step-by-step explanation: