is this the whole question because there's nothing to find?
It is impossible as 8 is bigger than 7 and you can’t take it away ur welcome
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.
Are you asking how to solve this equation?
Answer: 32,000
Step-by-step explanation:
the thousands place (1000) is the fourth number to the left of the decimal, so round that one either up or down based on the number just one spot closer to the decimal, in this case, the hundreds place (100).
so we have 32,420
we can ignore the numbers that are going to become zeros as a result of the rounding . .
32,400
let's round 32,XXX either up or down based on the value in the hundreds place. if that value is 5 or greater, than we round up. if it is less than 5, than we round down.
4 is in the hundreds place, so we round down
32,000 is the answer
Hope this helps!!! Good luck!!! ;)
