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Minchanka [31]
3 years ago
6

You wish to take flying lessons. The introductory course cost $125 per session and will include 12 lessons. Smith Flying Academy

also charges a $50 registration fee. You currently have $700 saved for your flying lessons. How much more money do you need to pay for the classes in full?
Mathematics
1 answer:
allsm [11]3 years ago
3 0
125 x 12 = 1500
1500 + 50 = 1550
1550 - 700 = 850
You would need $850
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Given the following functions f(x) and g(x), solve (f/g)(-4) and select the correct answer below. F(x)=4x-4 G(x)=x-1
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To start set up a fraction with the f(x) on top and g(x) on bottom
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An electronics company produces​ transistors, resistors, and computer chips. Each transistor requires 3 units of​ copper, 1 unit
padilas [110]

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.

a = Transistors

b= Resistors

c = Computer chips

2. We propose three linear equations, one for the copper, one for the zinc and one for the glass.

\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]

<u><em>Therefore:</em></u>

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

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

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

4 0
2 years ago
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