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Liula [17]
3 years ago
13

Help me please ..the Oakland football stadium, home of the oakland raiders , is capable of seating 63,026 fans the oakland raide

rs sell fuel for $30.00 functions f(x)=30x represents the situation where x is the number of people in attendance and f(x) is the amount of money that the what is the domain of the situation where x is the number of people in attendance and f(x)is the amount of money that the roadding part A) what is the domain of the situation part B) what is the range of the situation?
Mathematics
1 answer:
guapka [62]3 years ago
3 0

Answer:

A) Domain = 0 \leq x \leq 63,026

B) Range = 0 \leq y \leq 1,890,780

Step-by-step explanation:

Domain is the number of people and Range is the amount of money they make.

A)

Since seating capacity is at max 63,026, and minimum people to attend would be 0. Hence the domain is between 0 and 63,026

We can say:

Domain = 0 \leq x \leq 63,026

B)

The tickets cost 30 each and if there are no people in attendance, they raise $0. And if the stadium is full, they can raise:

63,026 * 30 = 1,890,780

Hence, the range would be minimum $0 and maximum $1,890.780

We can say:

Range = 0 \leq y \leq 1,890,780

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\huge \boxed{\mathfrak{Question} \downarrow}

\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right.

First, write both the equations in its standard form.

8x+2y=46\\ 7x+3y=47

Now, write the equations in form of matrix.

\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}46\\47\end{matrix}\right)

Then, multiply the equation towards the left by using the inverse of matrix \left(\begin{matrix}8&2\\7&3\end{matrix}\right)

\sf \: inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)

The product of the matrix & its inverse will be the identity matrix.

\sf\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)

Now, multiply the matrices that lie on the left-hand side of the equal sign.

\sf\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)

For the 2 × 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is ⇨ \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right).

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{8\times 3-2\times 7}&-\frac{2}{8\times 3-2\times 7}\\-\frac{7}{8\times 3-2\times 7}&\frac{8}{8\times 3-2\times 7}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)

Do the calculations.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}&-\frac{1}{5}\\-\frac{7}{10}&\frac{4}{5}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)

Multiply the matrices.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}\times 46-\frac{1}{5}\times 47\\-\frac{7}{10}\times 46+\frac{4}{5}\times 47\end{matrix}\right)

Do the arithmetics again.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{22}{5}  \\\frac{27}{5}\end{matrix}\right)

Finally, extract the matrix elements x & y & write them separately.

\large \boxed{ \boxed{ \bf \: x=\frac{22}{5},y=\frac{27}{5} }}

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