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:
first cross multiply
2x×4=3×8
8x=24
then, divide both sides by "8" so x could be alone
x=24/8
x=3
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Step-by-step explanation:
Well if you were to solve for y it would be “y=300•1.005^12t
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If you were to solve for t it’d be “t=In(y/300)/12In(1.005)
Answer:
x=
/2-2
Step-by-step explanation:
Hector is h. Sister is s.
h-22=s
h+5= 3(s+5)
Plug in s into the second equation
h+5=3(h-22+5)
h+5= 3(h-17)
h+5=3h-51
h+56=3h
56=2h
28=h
Final answer: Hector-28, Sister-6
