Explanation:
The function is
To graph the function, let us find the x and y intercepts.
To find x-intercept, let us substitute y=0 in the function
Thus, the x-intercept is
To find the y-intercept, let us substitute x=0, we get,
Thus, the y-intercept is
The graph has no asymptotes.
To plot the points in the graph, we need to substitute the values for x in the function , to find the y-values.
The points are . The image of the graph and table is attached below:
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:
Step-by-step explanation:
Total outcomes possible: 36
A. Divisible by 3
Possible options are:
3, 6, 9 and 12.
Possible outcomes for 3 are: {(1,2), (2,1)} Count 2
Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5
Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4
Possible outcomes for 12 are: {(6,6)} Count 1
Total count = 2 + 5 + 4 + 1 = 12
Probability of an event E can be formulated as:
B. Less than 7:
Possible sum can be 2, 3, 4, 5, 6
Possible cases for sum 2: {(1,1)} Count 1
Possible cases for sum 3: {(1,2), (2,1)} Count 2
Possible cases for sum 4: {(1,3), (3,1), (2,2)} Count 3
Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)} Count 4
Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5
Total count = 1 + 2 + 3 + 4 + 5 = 15
C. Divisible by 3 and less than 7:
Here, common cases are:
Possible outcomes for 3 are: {(1,2), (2,1)} Count 2
Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5