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e-lub [12.9K]
3 years ago
14

A store had 235 MP3 players in the month of January. Every month, 30% of the MP3 players were sold and 50 new MP3 players were s

tocked in the store.
Which recursive function best represents the number of MP3 players in the store f(n) after n months?
a f(n) = 0.7 × f(n − 1) + 50, f(0) = 235, n > 0
b f(n) = 237 − 0.7 × f(n − 1) + 50, f(0) = 235, n > 0
c f(n) = 0.3 × f(n − 1) + 50, f(0) = 235, n > 0
d f(n) = 237 + 0.7 × f(n − 1) + 50, f(0) = 235, n > 0
Mathematics
1 answer:
neonofarm [45]3 years ago
7 0
The answer is A cause 30% is 0.7 and f(n-1) equals the amount being sold the +50 is for every 50 new one and the the 235 is for what they started with                                         
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List all the perfect cubes up to 1000 and their cube roots.
DiKsa [7]

Hello from MrBillDoesMath!

Answer:

1^3    =     1

2^3   =     8

3^3   =     27

4^3   =     64

5^3   =    125

6^3   =    216

7^3   =    343

8^3  =     512

9^3   =    729

10^3   =    1000

Thank you,

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4 0
3 years ago
140/96 into a mixed number
FinnZ [79.3K]
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3 0
3 years ago
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Choose the value for x that makes the proportion true.A.3B.4C.5D.6​
kondor19780726 [428]

Answer:

B) 4

Step-by-step explanation:

It would be 4 because when you cross multiply 16*x and 1*64, you get 4/1.

7 0
3 years ago
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
How do you work out the equation 1-x=9.
mafiozo [28]

Answer:

x=8

Step-by-step explanation:

x + 1 = 9

Simplifying

x + 1 = 9

Reorder the terms:

1 + x = 9

Solving

1 + x = 9

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-1' to each side of the equation.

1 + -1 + x = 9 + -1

Combine like terms: 1 + -1 = 0

0 + x = 9 + -1

x = 9 + -1

Combine like terms: 9 + -1 = 8

x = 8

Simplifying

x = 8

7 0
3 years ago
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