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ella [17]
3 years ago
6

The number of bacteria in a culture is given by the function n(t) =950e^0.45t Where t is measured in hours A) what is the contin

uous rate of growth of this bacterium population? Your answer is __ percent B) what is the initial population of the culture (at t=0) your answer is __ C) how many bacteria will the culture contain at time t=5 ? Your answer is __ Round to the nearest bacteria
Mathematics
1 answer:
tatiyna3 years ago
6 0

Answer:

a) The continuous rate of growth of this bacterium population is 45%

b) Initial population of culture at t = 0 is 950 bacteria

c) number of bacterial culture contain at t = 5 is 9013 bacteria

Step-by-step explanation:

The number of bacteria in a culture is given by

n(t) = 950e^{0.45t}

a) rate of growth is

Bacteria growth model N(t) = noe^{rt}

Where r is the growth rate

hence r = 0.45

              = 45%

b) Initial population of culture at t = 0

n(0) = 950e^{0.45(0)}

       = 950e^{0}

       = 950 bacteria

c) number of bacterial culture contain at t = 5

n(5) =  950e^{0.45(5)}

      = 950e^{2.25}

      = 9013.35

      = 9013 bacteria

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3 years ago
Anyone taking geometry at connexus ?
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2 years ago
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Solve the equation in the interval [0,2π]. If there is more than one solution write them separated by commas.
Sedaia [141]
\large\begin{array}{l} \textsf{Solve the equation for x:}\\\\ 
\mathsf{(tan\,x)^2+2\,tan\,x-4.76=0}\\\\\\ \textsf{Substitute}\\\\ 
\mathsf{tan\,x=t\qquad(t\in \mathbb{R})}\\\\\\ \textsf{so the equation 
becomes}\\\\ \mathsf{t^2+2t-4.76=0}\quad\Rightarrow\quad\begin{cases} 
\mathsf{a=1}\\\mathsf{b=2}\\\mathsf{c=-4.76} \end{cases} 
\end{array}


\large\begin{array}{l} \textsf{Using 
the quadratic formula:}\\\\ \mathsf{\Delta=b^2-4ac}\\\\ 
\mathsf{\Delta=2^2-4\cdot 1\cdot (-4.76)}\\\\ 
\mathsf{\Delta=4+19.04}\\\\ \mathsf{\Delta=23.04}\\\\ 
\mathsf{\Delta=\dfrac{2\,304}{100}}\\\\ 
\mathsf{\Delta=\dfrac{\diagup\!\!\!\! 4\cdot 576}{\diagup\!\!\!\! 4\cdot
 25}}\\\\ \mathsf{\Delta=\dfrac{24^2}{5^2}} \end{array}

\large\begin{array}{l}
 \mathsf{\Delta=\left(\dfrac{24}{5}\right)^{\!2}}\\\\ 
\mathsf{\Delta=(4.8)^2}\\\\\\ 
\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\ 
\mathsf{t=\dfrac{-2\pm\sqrt{(4.8)^2}}{2\cdot 1}}\\\\ 
\mathsf{t=\dfrac{-2\pm 4.8}{2}}\\\\ \mathsf{t=\dfrac{\diagup\!\!\!\! 
2\cdot (-1\pm 2.4)}{\diagup\!\!\!\! 2}}\\\\\mathsf{t=-1\pm 2.4} 
\end{array}

\large\begin{array}{l} \begin{array}{rcl} 
\mathsf{t=-1-2.4}&~\textsf{ or }~&\mathsf{t=-1+2.4}\\\\ 
\mathsf{t=-3.4}&~\textsf{ or }~&\mathsf{t=1.4} \end{array} 
\end{array}


\large\begin{array}{l} \textsf{Both 
are valid values for t. Substitute back for }\mathsf{t=tan\,x:}\\\\ 
\begin{array}{rcl} \mathsf{tan\,x=-3.4}&~\textsf{ or 
}~&\mathsf{tan\,x=1.4} \end{array}\\\\\\ \textsf{Take the inverse 
tangent function:}\\\\ \begin{array}{rcl} 
\mathsf{x=tan^{-1}(-3.4)+k\cdot \pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi}\\\\ 
\mathsf{x=-tan^{-1}(3.4)+k\cdot \pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi} \end{array}\\\\\\ 
\textsf{where k in an integer.} \end{array}

__________


\large\begin{array}{l}
 \textsf{Now, restrict x values to the interval 
}\mathsf{[0,\,2\pi]:}\\\\ \bullet~~\textsf{For }\mathsf{k=0:}\\\\ 
\begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)


\large\begin{array}{l}
 \bullet~~\textsf{For }\mathsf{k=1:}\\\\ \begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)+\pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}\\\\\\ 
\boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+\pi} 
\end{array}}\textsf{ is in the 2}^{\mathsf{nd}}\textsf{ quadrant.}\\\\ 
\mathsf{x\approx 1.86~rad~~(106.39^\circ)}\\\\\\ 
\boxed{\begin{array}{c}\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}}\textsf{
 is in the 3}^{\mathsf{rd}}\textsf{ quadrant.}\\\\ \mathsf{x\approx 
4.09~rad~~(234.46^\circ)}\\\\\\ \end{array}


\large\begin{array}{l}
 \bullet~~\textsf{For }\mathsf{k=2:}\\\\ \begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)+2\pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+2\pi>2\pi~~\textsf{(discard)}} 
\end{array}\\\\\\ \boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+2\pi} 
\end{array}}\textsf{ is in the 4}^{\mathsf{th}}\textsf{ quadrant.}\\\\ 
\mathsf{x\approx 5.00~rad~~(286.39^\circ)} \end{array}


\large\begin{array}{l}
 \textsf{Solution set:}\\\\ 
\mathsf{S=\left\{tan^{-1}(1.4);\,-tan^{-1}(3.4)+\pi;\,tan^{-1}(1.4)+\pi;\,-tan^{-1}(3.4)+2\pi\right\}}
 \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2071152</span>


\large\textsf{I hope it helps.}


Tags: <em>trigonometric trig quadratic equation tangent tan solve inverse symmetry parity odd function</em>

6 0
3 years ago
7HELP________________
tatuchka [14]

Answer:

The statement If ∠A ≅ ∠C not prove that Δ ABD ≅ Δ CBD by SAS ⇒ C

Step-by-step explanation:

* Lets revise the cases of congruence  

- SSS  ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ  

- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and  

 including angle in the 2nd Δ

- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ

 ≅ 2 angles and the side whose joining them in the 2nd Δ

- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles

 and one side in the 2ndΔ

- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse  

 leg of the 2nd right angle Δ

* Lets solve the problem

- In the 2 triangles ABD , CBD

∵ AB = CB

∵ BD is a common side in the two triangles

- If AD = CD

∴ Δ ABD ≅ Δ CBD ⇒ SSS

- If BD bisects ∠ABC

∴ m∠ABD = m∠CBD

∴ Δ ABD ≅ Δ CBD ⇒ SAS

- If ∠A = ∠C

∴ Δ ABD not congruent to Δ CBD by SAS because ∠A and ∠C

  not included between the congruent sides

* The statement If ∠A ≅ ∠C not prove that Δ ABD ≅ Δ CBD by SAS

5 0
2 years ago
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