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lisov135 [29]
3 years ago
15

A certain calling card gives the following rate to call Malaysia. There is a connection fee of $0.39 which allows you to connect

for up to (but not including) one minute. For each additional minute or part of a minute, there is a $0.24 cent charge. (For example, 1 minute up to but not including 2 minutes costs $0.63.) This fare is only valid for up to (but not including) 40 minutes.
15. Express this information as a function, where f(x) represents the amount of money charged and x represents the number of minutes which have elapsed.
16. What is the domain for this function?
17. What is the range for this function?
18. What is f(2.5)? What does this value represent in the context of the problem?
19. Is this function one-to-one? How do you know?
Mathematics
1 answer:
Len [333]3 years ago
8 0

f(x)=.24x+.39

f(x)=.24(19)+.39

f(x)=4.95

You would be spending $4.95.

Hope this helps!

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Answer:

An electronics company can be produce 350 transistors and 340 computer chips, they can´t produce resistors.

Step-by-step explanation:

1. We will name the variables for transistors, resistors and the computer chips.

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\left \{ {{3a+3b+2c=1730} \atop {a+2b+c=690}}\atop {2a+b+2c=1380}} \right.

3. We write the matrix form as Ax=d

A=\left(\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right)

x=\left(\begin{array}{ccc}a\\b\\c\end{array}\right)

A=\left(\begin{array}{ccc}1730\\690\\1380\end{array}\right)

With this formula the solution of x is x=\frac{d}{A} or x=A^{-1}d

4. We will find the inverse matrix A^{-1} using the formula:

A^{-1} = \frac{1}{detA} (C_{A})^{T}

a. det A

det A=\left[\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right] =3*(4-1)-3*(2-2)+2*(1-4)=9-0-6=3

b. (C_{A})^{T}

C_{A}=\left(\begin{array}{ccc}4-1&.(2-2)&1-4\\-(6-2)&6-4&-(3-6)\\3-4&-(3-2)&6-3\end{array}\right)

C_{A}=\left(\begin{array}{ccc}3&.0&-3\\-4&2&3\\-1&-1&3\end{array}\right)

(C_{A}) ^T=\left(\begin{array}{ccc}3&0&-3\\-4&2&3\\-1&-1&3\end{array}\right)^T

(C_{A}) ^T=\left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)

c.A^{-1}

A^{-1}=\frac{1}{3} \left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)

5. As x=\frac{d}{A} or x=A^{-1}d, the solution of x is:

x=\frac{1}{3}\left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)\left(\begin{array}{ccc}1730\\690\\1380\end{array}\right)

x=\frac{1}{3}\left(\begin{array}{ccc}(3*1730)+(-4*690)+(-1*1380)\\(0*1730)+(2*690)+(-1*1380)\\(-3*1730)+(3*690)+(3*1380)\end{array}\right)

x=\frac{1}{3}\left(\begin{array}{ccc}1050\\0\\1020)\end{array}\right)

X=\left[\begin{array}{ccc}350\\0\\340\end{array}\right]

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<u><em>a= 350 Transistors</em></u>

<u><em>b=0 Resistors</em></u>

<u><em>c=340 Computer chips</em></u>

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