In 1993, Moose Population: 3280
In 1999, population became: 4960
P (Population) , t (years)
t = 6 —> 4960 - 3280 = 1680
Average Change —> 1680/6 = 280 moose/year
• In terms of 1990:
t = 3 —> 3280-3 (280)
P(1990) = 2440
P(t) = 2440 + 280t
• In 2003; t = 13
P(13) = 2440 + 280 (13)
P(13) = 2440 + 3640
P(13) = 6080
• Moose population in 2003
= 6080
Answer:
ENLARGEMENT
Step-by-step explanation:
PLEASE include a picture so we can answer!!!
To find all the positive integers less than 2018 that are divisible by 3, 11, and 61, you will use what you know about factors.
3, 11, and 61 are all answers. So are 33, 183, 671, and 2013.
If you put these in factors, the product will be divisible by them!
3 x 11 = 33
3 x 61 = 183
11 x 61 = 671
3 x 11 x 61 = 2013
Take each number and square it, cube it, etc...
9, 27, 81, 243, 729
121, 1331
9 x 11 = 99
27 x 11 = 297
81 x 11 = 891
121 x 9 =1089
121 x 3 = 363
61 x 9 = 549
61 x 27 = 1647
Everything in bold is a correct answer.
