A) There are a number of ways to compute the determinant of a 3x3 matrix. Since k is on the bottom row, it is convenient to compute the cofactors of the numbers on the bottom row. Then the determinant is ...
1×(2×-1 -3×1) -k×(3×-1 -2×1) +2×(3×3 -2×2) = 5 -5k
bi) Π₁ can be written using r = (x, y, z).
Π₁ ⇒ 3x +2y +z = 4
bii) The cross product of the coefficients of λ and μ will give the normal to the plane. The dot-product of that with the constant vector will give the desired constant.
Π₂ ⇒ ((1, 0, 2)×(1, -1, -1))•(x, y, z) = ((1, 0, 2)×(1, -1, -1))•(1, 2, 3)
Π₂ ⇒ 2x +3y -z = 5
c) If the three planes form a sheath, the ranks of their coefficient matrix and that of the augmented matrix must be 2. That is, the determinant must be zero. The value of k that makes the determinant zero is found in part (a) to be -1.
A common approach to determining the rank of a matrix is to reduce it to row echelon form. Then the number of independent rows becomes obvious. (It is the number of non-zero rows.) This form for k=-1 is shown in the picture.
Answer:
Step-by-step explanation:
By applying tangent rule in the given right triangle AOB,
tan(30°) = 


By applying tangent rule in the given right triangle BOC,
tan(60°) = 
OC = BO(√3)
OA + OC = AC

2√3(BO) = 60
BO = 10√3
OC = BO(√3)
OC = (10√3)(√3)
OC = 30
By applying tangent rule in right triangle DOC,
tan(60°) = 
OD = OC(√3)
OD = 30√3
Since, BD = BO + OD
BD = 10√3 + 30√3
BD = 40√3
≈ 69.3
The two graphs in D are not inverses of each other.
Hey! You just have to square root the 2 and 5 then multiply by 11 so the answer I came up with is 34.78505426185218
I hope I helped!
Answer: 49
Step-by-step explanation:
50−16/4+3
=50−4+3
=46+3
=49