Answer:
A. Yes, the triangles are congruent by SAS.
Step-by-step explanation:
EF = FG and DF = FH-> Given
angle EFD = angle HFG -> Vertical angles are congruent
DE F = HGF -> SAS Triangle Congruence Theorem
One is the amount you have, also its the numerator
Explanation:
Basically, you can do it in many ways. But just, in my opinion, exactly linear algebra was made for such cases.
the optimal way is to do it with Cramer's rule.
First, find the determinant and then find the determinant x, y, v, u.
Afterward, simply divide the determinant of variables by the usual determinant.
eg. 
and etc.
I think that is the best way to solve it without a hustle of myriad of calculations reducing it to row echelon form and solving with Gaussian elimination.
Step-by-step explanation:
that is only possible, if we don't need to use a 2 and a 5 (but especially the 2) to create the 1 in the middle (log(10) = log(2×5) or log (5×2) = 1).
in other words, if the 1 in the middle is just a given, and does not use any of our digits for is creation, then a solution is possible.
first of all, the upper left corner has to be log(0⁰). we cannot use the digit 0 for anything else. and 0^n (except for n = 0) is not a valid argument for a logarithm.
log(0⁰) = 0 log(2×1) = 0.3... log(9/4) = 0.35...
log(1×2) = 0.3... 1 log(3×6) = 1.26...
log(7/3) = 0.37... log(4×5) = 1.3... log(6⁹) = 7.0...
but as soon as we need to build the center 1 by log(10), which takes then away one of the 2s (as the only way to build 10 by multiplication is 2×5 or 5×2), there is no possible solution.
