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AlekseyPX
3 years ago
12

Adventure play sells an mp3 for 18.00 and charges $3.25 per song. King music sell a player for $23.00 and charges $2.00. For how

many songs will the cost be the same
Mathematics
1 answer:
Sveta_85 [38]3 years ago
4 0

Answer:

4 songs

Step-by-step explanation:

Given data,

Adventure play sells mp3 = $18

Charges per song = $3.25

King music sell Mp3 = $23

Charges per song = $2

Let number of songs be Y, then we can calculate by using following method,

Adventure play MP3 cost + charges per song = Kings Music MP3 cost + charges per song

By putting the value, we get

$18 + $3.25 Y = $23 + $2 Y

By solving the equation,

$3.25 Y - $2Y = $23 - $18

$1.25 Y = $5

Y = $5 ÷ $1.25

Y = 4 songs

Hence, for 4 songs the cost be same.

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Fit a trigonometric function of the form f(t)=c0+c1sin(t)+c2cos(t)f(t)=c0+c1sin⁡(t)+c2cos⁡(t) to the data points (0,5.5)(0,5.5),
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Answer

f(t)=-0.2+4.1sin(t)+4cos(t)

Step-By-Step Explanation

Given the function f(t)=c_0+c_1sin(t)+c_2cos(t).

For each pair (t, f(t)) in the data points (0,5.5), (π/2,0.5), (π,−2.5), (3π/2,−7.5)

f(0)=c_0+0c_1+c_2=5.5.

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f(\pi)=c_0+0sin(t)-c_2=-2.5.

f(3\pi /2)=c_0-c_1+0c_2=-7.5.

Expressing this as a system of linear equations in matrix form AX=B

\left(\begin{array}{ccc}   1 & 0 & 1 \\   1 & 1 & 0 \\   1 & 0 & -1 \\   0 & -1 & 0    \end{array}   \right)\left(   \begin{array}{c}   c_{0} \\   c_{1} \\   c_{2}\\   \end{array}   \right)=\left(\begin{array}{c}   5.5 \\   0.5 \\   -2.5 \\   -7.5    \end{array}   \right)      

Where    

A=\left(\begin{array}{ccc}   1 & 0 & 1 \\   1 & 1 & 0 \\   1 & 0 & -1 \\   0 & -1 & 0    \end{array}   \right),      

B=\left(\begin{array}{c}5.5\\0.5\\-2.5\\-7.5\end{array} \right)

X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)     

To determine the values of X, we use the expression  

X=(A^{T}A)^{-1}A^{T}B      

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(A^{T}A)^{-1}= \left(\begin{array}{ccc}   0.4 & -0.2 & 0 \\   -0.2 & 0.6 & 0 \\   0 & 0 & 0.5    \end{array}   \right)      

A^{T}B=\left(\begin{array}{c}   3.5 \\   8 \\   8    \end{array}   \right)      

Therefore:    

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