His average is 80%. You take the two scores add them up and divide by the number of scores.
Answer:
The signed number that indicates how much of the levee will erode during the next decade is:
-30 feet
Step-by-step explanation:
a) Data and Calculations:
Erosion rate of the banks of the levee = 3 feet per year
For 10 years, if nothing is done to correct the problem, the eroded banks will be equal to 10 * 3 = 30 feet
Therefore, the signed erosion in the next decade is -30 feet.
b) The quantity of erosion can be given as the number of years involved without correction, multiplied by the rate of erosion per year. This implies that the quantity of erosion depends on the number of years and the rate of erosion per year.
A(n)=ar^(n-1) so
100=4r^9
25=r^9
ln25=9lnr
ln25/9=lnr
r=e^(ln25/9)
a(n)=4e^((n-1)(ln25/9) if you wanted an approximation...
a(n)=4(1.43^(n-1))
It’s best if you download Photomath
Answer:
Step-by-step explanation:
a).
= 
= 
=
[Since, i =
]
b).
= 
= 
= 5 ± 2i [Since, i =
]
c).
= 
= 
= 
=
[Since, i =
]