okay. the point has an x and y value. place them into the equation.
1=m(1)+b
m=slope, and theequation tells you that slope is 7.
1=7(1)+b
now you need to figure out what b is.
1=7(1)+b
^
1= 7 +b
-7 -7
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-6=B
b is 6. now place it into the equation, replacing the x and y values back.
y=7x-6.
write 7 and 6 in the boxes (the negative for the six has already been provided)
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.
The line segment shown in the figure is C. XZ. A line segment is a line that is bounded by 2 endpoints, hence the line is not infinite. As shown in the figure, only points X and Z are visibly connected. There are no visible lines connecting the other points together.
B, the first differences are not the same , that proves it is not linear. The second set of differences are the same this shows it is a quadratic function.