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s2008m [1.1K]
3 years ago
12

A waitress sold 15 ribeye steak dinners and 14 grilled salmon​ dinners, totaling ​$583.59583.59 on a particular day. another day

she sold 23 ribeye steak dinners and 7 grilled salmon​ dinners, totaling ​$580.74. how much did each type of dinner​ cost?
Mathematics
1 answer:
Ratling [72]3 years ago
7 0

Answer:


Step-by-step explanation:

This is a systems of equations question.  Let's set ribeye steak to the variable r, and grilled salmon to s.

We get these two equations:

13r + 18s = 550.25

22r + 6s = 582.08

First, we'll isolate one variable in the first equation.  Let's choose r:

13r = 550.25 - 18s

r = 42.33 - 1.38s

Now, we'll take this value for r and plug it into the second equation:

22r + 6s = 582.08

22 (42.33 - 1.38s) + 6s = 582.08

(931.19 - 30.46s) + 6s = 582.08          | multiply values by 22

931.19 - 24.46s = 582.08                   | combine s values

349.11 = 24.46s                               | move 582.08 to left side, and combine;  move -24.46s to right side

14.27 = s                                        | solve for s

Now, plug this value for s back into the first equation:

13r + 18s = 550.25

13r + 18 (14.27) = 550.25

13r = 256.91 = 550.25

13r = 293.34

r = 22.56

So r = 22.56 and s = 14.27.

The ribeye steak dinners cost $22.56 each and the grilled salmon dinners cost $14.27 each.

Note, this solution does not factor in any tips the waitress makes on each dinner!

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kari74 [83]

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<h3>What is the equation which represents the data in the table as attached?</h3>

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Hence, the equation which represents the function is;

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2 years ago
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iren2701 [21]

Answer:

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Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

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Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

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From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

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faust18 [17]

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<u>Step-by-step explanation:</u>

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Since the base on both sides as ‘12’ are the same, we can write it as

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Often, the value of x is easiest to solve by a x^{2}+b x+c=0 by factoring a square factor, setting each factor to zero, and then isolating each factor. Whereas sometimes the equation is too awkward or doesn't matter at all, or you just don't feel like factoring.

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                       x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Where, a = 1, b = 3 and c = -10

                       x=\frac{-3 \pm \sqrt{(-3)^{2}-4(1)(-10)}}{2(1)}

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