So what this is is
many words
assuming year 0 is 2017
so compound first thing till 2020, take out 30000
the remaining is copmpounded til 2022, take out 50000
remaining is compounded for 1 more year and that is equal to 80000
so from 2017 to 2020, that is 5 years
from 2020 to 2022 is 2 years
from 2022 to 2023 is 1 year
work backwards
A=P(r+1)^t
last one
A=80000
P=?
r=0.08
t=1 year
80000=P(1.08)^1
divide both sides by 1.08
I would leave in fraction
20000000/27=P
now that is the remaining after paying 50000, after 2 years of compounding
so
50000+(2000000/27)=P(1.08)^2
solve using math
about
106374=P
now reverse back
5 years
paid 30000
30000+106374=P(1.08)^5
solve using math
92813.526=P
round
$92813.53
put $92813.53 in the fund
For x you insert zero, and
0 = 2 - 8 * 0
0 = 2
I think A because 560×3=1,680
Number of zids=x
number of zods=y
number of zid legs=5x
number of zod legs=7y
so
5x+7y=140
try to get it into slope intercept from so you can graph is (y=mx+b)
5x+7y=140
subtract 5x from both sides
7y=140-5x
divide both sides by 7
y=20-5/7x
y=-5/7x+20
plug in numbers for x and get numbers for y (you can only plug in multiples of 7 for x so that there are a whole number of zids and since you are counting, x and y must never be negative)
lets try 0 for x
y=-5/7(0)+20
y=20
so x=0 and y=20 is an answer (if you can have only one of that species)
lets try 7 for x
y=-5/7(7)+20
y=-5+20
y=15
so x=7 and y=15 is an answer
lets try 14 for x
y=-5/7(14)+20
y=-10+20
y=10
so x=14 and y=10 is another answer
lets try 21 for x
y=-5/7(21)+20
y=-15+20
y=5
so x=21 and y=5 is another answer
lets try 28 for x
y=-5/7(28)-20
y=-20+20
y=0
so x=28 and y=0 is an answer (if there can be only one of a species)
if we go further, then y will be negative so the answers are
(x,y)
(0,20)
(7,15)
(14,10)
(21,5)
(28,0)
if it is allowed that only one species exists then there are 5 possible answers
if both must exist simultaneously, then there are only 3 answers
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