Answer:
Step-by-step explanation:
<h2><u>Part A</u></h2>
in interval ( 0 ; 2)
<h2><u>Part B</u></h2>
in interval (2; 4)
<h2><u>Part C</u></h2>
in interval (4 ; 6 )
<h2><u>Part D</u></h2>
The graph shows that at first the ball rises up ; and then it is seen that it goes down and loses height to zero , from which it can be concluded that the height after 10 seconds remains unchanged and therefore the height of the ball after 16 seconds will be zero
7 ^ 1/5 ^ 5
power to a power is multiplied
7 ^ (1/5 *5)
7^1
7
Choice A
Answer: which problem is it ?
Step-by-step explanation:
Answer:
(-3,0) = x and (0,4) = y
Step-by-step explanation:
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>