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anzhelika [568]
3 years ago
9

cole spent $55.00 buying songs and movies at an online store that charges $1.25 for each song and $2.75 for each movie. He purch

ased a total of 26 song and movies combined. Write and solve a system of equations that represent this situation
Mathematics
1 answer:
Sergeu [11.5K]3 years ago
6 0
If you would like to write and solve a system of equations that represent the situation above, you can do this using the following steps:

s ... number of songs
m ... number of movies

$55.00 = $1.25 * s + $2.75 * m
55 = 1.25 * s + 2.75 * m

26 songs and movies = s + m
26 = s + m
s = 26 - m

55 = 1.25 * s + 2.75 * m
55 = 1.25 * (26 - m) + 2.75 * m
55 = 1.25 * 26 - 1.25 * m + 2.75 * m
55 - 32.5 = 1.5 * m
22.5 = 1.5 * m
m = 22.5 / 1.5 = 15 movies

s = 26 - m = 26 - 15 = 11 songs

The correct result would be 15 movies and 11 songs.
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For the cost function c equals 0.1 q squared plus 2.1 q plus 8​, how fast does c change with respect to q when q equals 11​? Det
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Answer:

Rate of change of c with respect to q is 4.3

Percentage rate of change c with respect to q is  9.95%

Step-by-step explanation:

Cost function is given as,  c=0.1\:q^{2}+2.1\:q+8

Given that c changes with respect to q that is, \dfrac{dc}{dq}. So differentiating given function,  

\dfrac{dc}{dq}=\dfrac{d}{dq}\left (0.1\:q^{2}+2.1\:q+8 \right)

Applying sum rule of derivative,

\dfrac{dc}{dq}=\dfrac{d}{dq}\left(0.1\:q^{2}\right)+\dfrac{d}{dq}\left(2.1\:q\right)+\dfrac{d}{dq}\left(8\right)

Applying power rule and constant rule of derivative,

\dfrac{dc}{dt}=0.1\left(2\:q^{2-1}\right)+2.1\left(1\right)+0

\dfrac{dc}{dt}=0.1\left(2\:q\right)+2.1

\dfrac{dc}{dt}=0.2\left(q\right)+2.1

Substituting the value of q=11,

\dfrac{dc}{dt}=0.2\left(11\right)+21.

\dfrac{dc}{dt}=2.2+2.1

\dfrac{dc}{dt}=4.3

Rate of change of c with respect to q is 4.3

Formula for percentage rate of change is given as,  

Percentage\:rate\:of\:change=\dfrac{Q'\left(x\right)}{Q\left(x\right)}\times 100

Rewriting in terms of cost C,

Percentage\:rate\:of\:change=\dfrac{C'\left(q\right)}{C\left(q\right)}\times 100

Calculating value of C\left(q \right)

C\left(q\right)=0.1\:q^{2}+2.1\:q+8

Substituting the value of q=11,

C\left(q\right)=0.1\left(11\right)^{2}+2.1\left(11\right)+8

C\left(q\right)=0.1\left(121\right)+23.1+8

C\left(q\right)=12.1+23.1+8

C\left(q\right)=43.2

Now using the formula for percentage,  

Percentage\:rate\:of\:change=\dfrac{4.3}{43.2}\times 100

Percentage\:rate\:of\:change=0.0995\times 100

Percentage\:rate\:of\:change=9.95%

Percentage rate of change of c with respect to q is 9.95%

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