Please give me an answer for that equations
1 answer:
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Answer:
See explanation
Step-by-step explanation:
If then triangle PXY is isosceles triangle. Angles adjacent to the base XY of an isosceles triangle PXY are congruent, so
and
Angles 1 and 3 are supplementary, so
Angles 2 and 4 are supplementary, so
By substitution property,
Hence,
Consider triangles APX and BPY. In these triangles:
- given;
- given;
- proven,
so by ASA postulate.
Congruent triangles have congruent corresponding sides, then
Therefore, triangle APB is isosceles triangle (by definition).
Complete Question:
Attached below as picture.
Answer:
From first graph there is no linear pattern so here linearity assumption violated.
From second graph there is observation is in some pattern like funnel or v shape so there is no constant variance occur that is there is no constant variance for error.
Constant variance for error occur when in residual plot all observation are in scatter everywhere.
From third graph we can say there is positive distribution but for regression analysis we need symmetric that is normal distribution.
Step-by-step explanation:
See graphs attached below.
Answer:The system could have no solution or n number of solution where n is the number of unknown in the n linear equations.
Step-by-step explanation:
To determine if solution exist or not, you test the equation for consistency.
A system is said to be consistent if the rank of a matrix (say B ) is equal to the rank of the matrix formed by adding the constant terms(in this case the zeros) as a third column to the matrix B.
Consider the following scenarios:
(1) For example:Given the matrix A=, to transpose A, exchange rows with columns i.e take first column as first row and second column as second row as follows:
Let A transpose be B.
∵B=
the system Bx=0 can be represented in matrix form as:
=
................................eq(1)
Now, to determine the rank of B, we work the determinant of the maximum sub-square matrix of B. In this case, B is a 2 x 2 matrix, therefore, the maximum sub-square matrix of B is itself B. Hence,
|B|=(1*4)-(3*2)= 4-6 = -2 i.e, B is a non-singular matrix with rank of order (-2).
Again, adding the constant terms of equation 1(in this case zeros) as a third column to B, we have :
=
. The rank of
can be found by using the second column and third column pair as follows:
||=(3*0)-(0*2)=0 i.e,
is a singular matrix with rank of order 1.
Note: a matrix is singular if its determinant is = 0 and non-singular if it is 0.
Comparing the rank of both B and , it is obvious that
Rank of BRank of
since (-2)<1.
Therefore, we can conclude that equation(1) is <em>inconsistent and thus has no solution. </em>
(2) If B= is the transpose of matrix A=
, then
Then the equation Bx=0 is represented as:
=
..................................eq(2)
|B|= (-4*10)-(5*(-8))= -40+40 = 0 i.e B has a rank of order 1.
=
,
||=(5*0)-(0*10)=0-0=0 i.e
has a rank of order 1.
we can therefor conclude that since
rank B=rank =1, equation(2) is <em>consistent</em> and has 2 solutions for the 2 unknown (
and
).
<u>Summary:</u>
- Given an equation Bx=0, transform the set of linear equations into matrix form as shown in equations(1 and 2).
- Determine the rank of both the coefficients matrix B and
which is formed by adding a column with the constant elements of the equation to the coefficient matrix.
- If the rank of both matrix is same, then the equation is consistent and there exists n number of solutions(n is based on the number of unknown) but if they are not equal, then the equation is not consistent and there is no number of solution.