The <span>Hendecagon comes after.</span>
Only the first two options are true.
Step-by-step explanation:
- Step 1: The first two statements are true as they are basic rules pertaining to logarithms. But the third and fourth statements are not true.
I:2x – y + z = 7
II:x + 2y – 5z = -1
III:x – y = 6
you can first use III and substitute x or y to eliminate it in I and II (in this case x):
III: x=6+y
-> substitute x in I and II:
I': 2*(6+y)-y+z=7
12+2y-y+z=7
y+z=-5
II':(6+y)+2y-5z=-1
3y+6-5z=-1
3y-5z=-7
then you can subtract II' from 3*I' to eliminate y:
3*I'=3y+3z=-15
3*I'-II':
3y+3z-(3y-5z)=-15-(-7)
8z=-8
z=-1
insert z in II' to calculate y:
3y-5z=-7
3y+5=-7
3y=-12
y=-4
insert y into III to calculate x:
x-(-4)=6
x+4=6
x=2
so the solution is
x=2
y=-4
z=-1
Answer:
-1
Step-by-step explanation:
Square root 1 is 1. Add that to the 12 to get 13. Subtract 13 from both sides to get: -5x = 5. Divide both sides by -5 to get your answer: -1