<u>Answer:</u>
<u>Step-by-step explanation:</u>
Given dimensions of the box = 20cm × 6cm × 4cm .
Dimension of the cube = 2cm × 2cm × 2cm .
Therefore the number of cubes that can be fitted into the box will be equal to the Volume of box divided by the Volume of the cube. So ,
<h3>
<u>Hence</u><u> the</u><u> </u><u>number</u><u> </u><u>of</u><u> </u><u>cubes</u><u> </u><u>that</u><u> </u><u>can</u><u> </u><u>be</u><u> </u><u>fitted</u><u> </u><u>in</u><u> the</u><u> </u><u>box </u><u>is</u><u> </u><u>6</u><u>0</u><u> </u><u>.</u></h3>
Answer:
54
Step-by-step explanation:
4x + 6 movies
12 movies,
x =12
4x + 6 for x= 12
4(12) + 6 = 54
The answer to your question is b.
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.
Answer:
can you please give me an example?
i will help you but give me your teacher's example
Step-by-step explanation:
let me know if it's done