Option A. is a function...
here For different values of x in Domain of(x,y) ;There is a unique image y in codomain.
All others have more than one image for a single value x.
Hope it helps...
Regards;
Leukoniv/Olegion.
The problem can be translated into an equation that is something like 4/5 + 3/x = 1/2
we cannot have x equal to zero because the number can be infinite.
So the LCD here is 10x, so multiply both sides by that to get:
8x + 30 = 5x
Subtract 5x and 30 from both sides:
3x = -30
divide:
x = -10
The solution isn't zero so there is a solution.
Consider a homogeneous machine of four linear equations in five unknowns are all multiples of 1 non-0 solution. Objective is to give an explanation for the gadget have an answer for each viable preference of constants on the proper facets of the equations.
Yes, it's miles true.
Consider the machine as Ax = 0. in which A is 4x5 matrix.
From given dim Nul A=1. Since, the rank theorem states that
The dimensions of the column space and the row space of a mxn matrix A are equal. This not unusual size, the rank of matrix A, additionally equals the number of pivot positions in A and satisfies the equation
rank A+ dim NulA = n
dim NulA =n- rank A
Rank A = 5 - dim Nul A
Rank A = 4
Thus, the measurement of dim Col A = rank A = five
And since Col A is a subspace of R^4, Col A = R^4.
So, every vector b in R^4 also in Col A, and Ax = b, has an answer for all b. Hence, the structures have an answer for every viable preference of constants on the right aspects of the equations.
