Answer:
30000+0+400+0+5
6000000+800000+0+4000+100+50+2
Step-by-step explanation:
hope it's right plz mark me brainlists
Current amount in account
P=36948.61
Future value of this amount after n years at i=11% annual interest
F1=P(1+i)^n
=36948.61(1.11)^n
Future value of $3000 annual deposits after n years at i=11%
F2=A((1+i)^n-1)/i
=3000(1.11^n-1)/0.11
We'd like to have F1+F2=280000, so forming following equation:
F1+F2=280000
=>
36948.61(1.11)^n+3000(1.11^n-1)/0.11=280000
We can solve this by trial and error.
The rule of 72 tells us that money at 11% deposited will double in 72/11=6.5 years, approximately.
The initial amount of 36948.61 will become 4 times as much in 13 years, equal to approximately 147800 by then.
Meanwhile the 3000 a year for 13 years has a total of 39000. It will only grow about half as fast, namely doubling in about 13 years, or worth 78000.
Future value at 13 years = 147800+78000=225800.
That will take approximately 2 more years, or 225800*1.11^2=278000.
So our first guess is 15 years, and calculate the target amount
=36948.61(1.11)^15+3000(1.11^15-1)/0.11
=280000.01, right on.
So it takes 15.00 years to reach the goal of 280000 years.
3(x - 1) = 2x + 9
3x - 3 = 2x + 9
3x (-2x) -3 (+3) = 2x (-2x) + 9 (+9)
(3x -2x) = (9 + 9)
(x) = (18)
x = 18
x = 18 is your answer
hope this helps
Answer:
i think it is y=2x
Step-by-step explanation:
I just can help u with
D: all real numbers
R: [3,infinity)
y -intercept :(0,4)
max.: none
min.: (1,3)
