For this case we have:
By definition, two points are needed to find the equation of a line. We have a series of points given by:

We chose two points:

By definition, the formula of the slope is given by:

Substituting:


Being a line of the form
, we substitute a point and the slope found to find the cut point b.

Thus, the equation of the line is given by:

Answer:
Option C
Answer:
a) False
b) False
c) True
d) False
e) False
Step-by-step explanation:
a. A single vector by itself is linearly dependent. False
If v = 0 then the only scalar c such that cv = 0 is c = 0. Hence, 1vl is linearly independent. A set consisting of a single vector v is linearly dependent if and only if v = 0. Therefore, only a single zero vector is linearly dependent, while any set consisting of a single nonzero vector is linearly independent.
b. If H= Span{b1,....bp}, then {b1,...bp} is a basis for H. False
A sets forms a basis for vector space, only if it is linearly independent and spans the space. The fact that it is a spanning set alone is not sufficient enough to form a basis.
c. The columns of an invertible n × n matrix form a basis for Rⁿ. True
If a matrix is invertible, then its columns are linearly independent and every row has a pivot element. The columns, can therefore, form a basis for Rⁿ.
d. In some cases, the linear dependence relations among the columns of a matrix can be affected by certain elementary row operations on the matrix. False
Row operations can not affect linear dependence among the columns of a matrix.
e. A basis is a spanning set that is as large as possible. False
A basis is not a large spanning set. A basis is the smallest spanning set.
Answer:
1/4
Step-by-step explanation:
y2-y1/x2-x1=m
6-5/9-5=1/4
Answer:
<em>6, 10 and 8</em>
Step-by-step explanation:
Given the recursive function an = an-1-(an-2 - 4), when a5 =-2 and a6 = 0?
a6 = a5 - (a4 - 4)
0 = -2 - (14 - 4)
2 = - (a4 - 4)
-2 = a4 - 4
a4 = -2 + 4
a4 = 2
a5 = a4 - (a3 - 4)
-2 = 2 - (a3 - 4)
-2-2 = - (a3 - 4)
-4 = -(a3 - 4)
4 = a3 - 4
a3 = 4+4
a3 = 8
Also
a4 = a3 - (a2 - 4)
2 = 8 - (a2 - 4)
2-8 = - (a2 - 4)
-6 = -(a2 - 4)
6 = a2 - 4
a2 = 6+4
a2 = 10
a3 = a2 - (a1 - 4)
8 = 10 - (a1 - 4)
8-10 = - (a1 - 4)
-2 = -(a1 - 4)
2 = a1 - 4
a1 = 2+4
a1 = 6
<em>Hence the first 3 terms are 6, 10 and 8</em>
Ones because it is the last place value