Using the law of logarithm. LogA - LogB = Log(A/B)
log(x²-y²) - log(x-y) = log((x²-y²)/(x-y))
Note by difference of two squares, (x²-y²) = (x-y)(x+y)
Simplifying (x²-y²)/(x-y) = (x-y)(x+y)/(x-y) = (x+y)
Therefore log(x²-y²) - log(x-y) = log((x²-y²)/(x-y)) = log((x-y)(x+y)/(x-y)) = log(x+y)
-11= -5+6n+6
-11+5-6=6n
-12=6n
n= -2
(hope i helped :D!)
To answer this you will need to write the ratio of difference in the tides over the amount of time that goes by.
1. 3.21 ft - 0.20 ft= 3.01 ft change.
2. 3.01 change over 6 5/12 hr.
3. Divide these to find the lapse rate per hour. This is about 0.47 ft per hour.
Answer:
x=151/64 y=9/8 z=-51/32
Step-by-step explanation:
2x+2y+5z- (2x-y+z)=-1-2
3y+4z=-3
2x+2y+5z- (2x+4y-3z)=-1-14
-2y+8z=-15
We have two equations 3y+4z=-3 and -2y+8z=-15 (We get it due to elimination method)
3y+4z=-3 (2)
-2y+8z=-15 (-3)
6y+8z=-6
6y-24z=45
(6y+8z)- (6y-24z)= -6-45
32z=-51
z=-51/32
3y-204/32=-3
3y= -96/32+204/32
3y=108/32
y=36/32=9/8
2x+2*9/8+5*(-51/32)=-1
2x+9/4-255/32=-1
2x+72/32-255/32=-1
2x-183/32=-32/32
2x=151/32
x=151/64
I cannot see the mass of the moon, can you comment the two masses so I can explain?
