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-BARSIC- [3]
3 years ago
9

Please help!! ASAP!!

Mathematics
1 answer:
posledela3 years ago
4 0

Answer:

26. t<-70 see description below.

Step-by-step explanation:

To solve an inequality in one variable, solve using inverse operations like any other equations. However, if dividing or multiplying by a negative at any points, flip the sign of the inequality.


Example:

\frac{t}{-7}>10\\\\ -7*\frac{t}{-7} >10*-7\\\\t

-7 was multiplied to as an inverse operation to division by -7. Because it was negative, the inequality sign was flipped. To graph, plot -70 on the number line with an open circle. Then draw an arrow from -70 to the left.

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-8(6x +3) <br> pleasegghhjfhnmyufgbdz
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When rabbits were introduced to the continent of Australia they quickly multiplied and spread across the continent since there w
Lady bird [3.3K]

Answer:

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.

-Let's take rabbit population as a function of time to be

y=ab^x

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

#We substitute our values to calculate the value of the constant b:

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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