D. Plug in the options to both equations and find which makes both true.
Answer:
x = 0
Step-by-step explanation:
Rewrite (6^2)^x =1 as 6^(2x) = 1.
Next, take the logarithm (common or natural) of both sides, obtaining
2x log 6 = log 1
2x log 6 = 0
Then x = 0.
Note that (6^2)^0 = 1, since any positive number raised to the prower 0 is 1.
BG ≅ AG
BG is the perpendicular to the side of the triangle while AG is the angle bisector , So BG cannot equal AG , So BG cannot be congruent to AG. Hence first is false.
DG ≅ FG
DG And FG both are the perpendicular to the sides from the incentre of the circle , Hence DG and FG are congruent , So second statement is true.
DG ≅ BG
Again DG and BG both are the perpendicular to the sides from the incentre of the circle , Hence DG and BG are congruent , So third statement is true.
GE bisects ∠DEF
As said in the question GE is the angle bisector , So yes GE bisects ∠DEF.
This Statement is true.
GA bisects ∠BAF
Again As said in the question GA is the angle bisector , So yes GA bisects ∠BAF.
Hence 2nd, 3rd , 4th , and 5th options are correct.
Answer:
Write the equation y = mx + b, substituting the values for m and b you calculated or determined. The m will be your slope, and the b will be your y-intercept. Leave the y and x variables in the equation as letter variables. Include the sign of the numbers you plug in.
Step-by-step explanation:
