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xz_007 [3.2K]
3 years ago
12

PLEASE HELP ME ON THIS

Mathematics
1 answer:
Whitepunk [10]3 years ago
4 0
F(t)=262(1.6)^t
Initially the tundra has: f(0)=?
t=0→f(0)=262(1.6)^0
f(0)=262(1)
f(0)=262 caribou

Answer:
Initially, the tundra has 262 caribou, and every 1 year, the number of caribou increases by a factor of 1.6


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Solve for x ....... thanks !
Len [333]
If we can match teh bases we can solve
because if x=x and xᵃ=xᵇ, we can conclude that a=b

16=2⁴
32=2⁵
rememeber that (x^m)^n=x^{mn}

16^{3x+2}=32^{-2x-7}
(2^4)^{3x+2}=(2^5)^{-2x-7}
2^{4(3x+2)}=2^{5(-2x-7)}
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expand/distribute
12x+8=-10x-35
add 10x both sides
22x+8=-35
minus 8 both sides
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8 0
3 years ago
A glass pitcher is filled with water 1/8 of the water is poured equally into two glasses if one fourth of the remaining water is
Lemur [1.5K]
Well we are starting with a full pitcher(1)
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5 0
3 years ago
Comporueba si las siguientes fracciones son equivalentes.
NikAS [45]

Answer:

No sé si este es un problema de opción múltiple, pero tanto (A) como (B) son fracciones equivalentes.

Step-by-step explanation:

4 0
3 years ago
The price of a gallon of unleaded gas was $2.81 yesterday l. Today, the price rose to $2.88. Find the percentage increase
Allisa [31]

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7%

Step-by-step explanation:

4 0
3 years ago
The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%
liq [111]
<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

   Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

   From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by \frac{100+1.8}{100} = \frac{101.8}{100}.

   So, the population in the year t can be given by P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}

   Population in the year 2000 = 3,381,000\textrm{x}(\frac{101.8}{100})^{6}=3,762,979.38

Population in year 2000 = 3,762,979

   Let us assume population doubles by year y.

2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}

log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})

y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537

y≈2033

∴ By 2033, the population doubles.

4 0
3 years ago
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