The picture below shows a right-triangle-shaped charging stand for a gaming system:
2 answers:
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Answer:
Step-by-step explanation:
Our expression is: .
Let's focus on the cube root of 81 first. What's the prime factorisation of 81? It's simply: 3 * 3 * 3 * 3, or . Put this in for 81:
We know that the cube root of 3 cubed will cancel out to become 3, but the cube root of 3 cannot be further simplified, so we keep that. Our outcome is then:
Now, let's multiply this by 1/3, as shown in the original problem:
Thus, the answer is .
<em>~ an aesthetics lover</em>
Answer:
Only d) is false.
Step-by-step explanation:
Let be the characteristic polynomial of B.
a) We use the rank-nullity theorem. First, note that 0 is an eigenvalue of algebraic multiplicity 1. The null space of B is equal to the eigenspace generated by 0. The dimension of this space is the geometric multiplicity of 0, which can't exceed the algebraic multiplicity. Then Nul(B)≤1. It can't happen that Nul(B)=0, because eigenspaces have positive dimension, therfore Nul(B)=1 and by the rank-nullity theorem, rank(B)=7-nul(B)=6 (B has size 7, see part e)
b) Remember that . 0 is a root of p, so we have that
.
c) The matrix T must be a nxn matrix so that the product BTB is well defined. Therefore det(T) is defined and by part c) we have that det(BTB)=det(B)det(T)det(B)=0.
d) det(B)=0 by part c) so B is not invertible.
e) The degree of the characteristic polynomial p is equal to the size of the matrix B. Summing the multiplicities of each root, p has degree 7, therefore the size of B is n=7.