Part of the value of sin(u) is cut off; I suspect it should be either sin(u) = -5/13 or sin(u) = -12/13, since (5, 12, 13) is a Pythagorean triple. I'll assume -5/13.
Expand the tan expression using the angle sum identities for sin and cos :
tan(u + v) = sin(u + v) / cos(u + v)
tan(u + v) = [sin(u) cos(v) + cos(u) sin(v)] / [cos(u) cos(v) - sin(u) sin(v)]
Since both u and v are in Quadrant III, we know that each of sin(u), cos(u), sin(v), and cos(v) are negative.
Recall that for all x,
cos²(x) + sin²(x) = 1
and it follows that
cos(u) = - √(1 - sin²(u)) = -12/13
sin(v) = - √(1 - cos²(v)) = -3/5
Then putting everything together, we have
tan(u + v)
= [(-5/13) • (-4/5) + (-12/13) • (-3/5)] / [(-12/13) • (-4/5) - (-5/13) • (-3/5)]
= 56/33
(or, if sin(u) = -12/13, then tan(u + v) = -63/16)
30 students.
5 + 2 + 4 + 7 + 5 + 9 = 32 students
<span>192 students. 32 students</span> = 6
so,
7 − 2 = 5
5 students × 6 = 30 students
I don’t see the table necessary in order to solve this question
Answer:
8
Step-by-step explanation:
Preety Sure is 8
Answer:
k = 13
Step-by-step explanation:
Matrix is;
A =
[2 -5 9]
[-1 3 -5]
[ 7 1 k ]
Number of unknowns is less than the rank of 2. Thus, the determinant has to be zero for us to find k.
Thus, we have;
2((3 × k) - (-5 × 1)) -(-5)((-1 × k) - (-5 × 7)) + 9((-1 × 1) - (7 × 3)) = 0
Simplifying further gives;
2(3k +5) + 5(-k + 35) + 9(-22) = 0
6k + 10 - 5k + 175 - 198 = 0
k - 198 + 185 = 0
k - 13 = 0
k = 13
