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3 years ago

15

Help me please ^^ much appreciated

1 answer:

Reil [10]3 years ago

8 0

Answer:

C

Step-by-step explanation:

Using the sine ratio in the right triangle.

sinC = $\frac{opposite}{hypotenuse}$ = $\frac{AB}{BC}$ = $\frac{16}{28}$ , thus

∠ C = $sin^{-1}$ ( $\frac{16}{28}$ ) ≈ 34.85° ( to 2 dec. places )

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3 years ago

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3 years ago

When rabbits were introduced to the continent of Australia they quickly multiplied and spread across the continent since there w

Lady bird [3.3K]

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a. $y=6(1.7472)^x$

b. $y=6e^{0.558t}$

c.13.3 months

Step-by-step explanation:

a.-Given the first term at $t_0$ is 6 and the second term at $t_3$ is 32.

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$y=ab^x$

where y is the population at time x and a the initial population at $t_0\\$

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$y_x=ab^x\\\\y_3=ab^3\\\\32=6b^3\\\\b=1.472$

#Replace b in the population function:

$y=ab^x, b=1.7472,a=6\\\\\therefore y=6(1.7472)^x$

Hence, the regression for the rabbit population as a function of time x is $y=6(1.7472)^x$

b. The exponential function in terms of base $e$ is usually expressed as:

$A=A_0e^{kt}$

Where:

$A_0$ -is the initial population at $t_o$

$A$ -is the population at time t.

$k-$ is the exponential growth constant.

$e-$ the exponent

Our function in terms of base exponent is rewritten as:

$y=A_0e^{kt}$

#Substitute with actual figures to solve for t:

$y=A_0e^{kt}, y=32, xt=3, A_0=6\\\\32=6e^{3k}\\\\3k=In (32/6)\\\\k=0.5580$

Hence, the regression equation in terms of base e is $y=6e^{0.558t}$

c. We substitute y with any number higher than 10,000 to estimate the time for the rabbits to exceed 10,000.

-We know that $y=6e^{0.558t}.$

Therefore we calculate t as(take y=10001):

$y=6e^{0.558t}, y=10001\\\\10001=6e^{0.558t}\\\\1666.8333=e^{0.558t}\\\\0.558t= In 1666.8333\\\\t=13.2951$

Hence, it takes approximately 13.3 months for the population to exceed 10000

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4 years ago