Answer:
C
Step-by-step explanation:
I did this question today, pls lmk if u get it right
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>
Let
x = wristbands
y = headbands
We then have the following inequations:
2x + 3y> = 50 x> = 5 The graph that represents the solution for this system of inequations is shown in the attached image.
The set of solutions is the shaded region.
Answer:
The maximum number of quarters that he could have is 13
Step-by-step explanation:
Let x represent the number of quarters he could have. If he has 10 dimes and has at least 20 coins, then
→ x + 10 >= 20
→ x >= 10 (1)
His coins worth at most $4.25. Also, it is known that 1 dollar is equal to 100 cents, 1 dime is equal to 10 cents and 1 quarter is equal to 25 cents.
→ (x * 25) + (10 * 10) <= (4.25 * 100)
→ 25x + 100 <= 425
→ 25x <= 325
→ x <= 13 (2)
If we combine the equation 1 and 2:
10 <= x <= 13
The maximum value of x is 13