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SCORPION-xisa [38]
3 years ago
11

Milli plans to rent a room for her birthday party for $40. The catering charge is $12 per person but there is no charge for Mill

i's meal since she is the guest of honor. If Milli has $100 to spend on her birthday party, what is the maximum number of friends Milli can invite?
Mathematics
1 answer:
Mrac [35]3 years ago
8 0
This is a 2 part question. 

Milli has $100 to spend and pays for the room first. $100 - $40 = $60 to pay for meals.

Each meal is $12 - how many meals can she buy with $60? 

60 / 12 = 5 meals 

So Milli and 5 friends can be at the party. 

To check: 5($12) + $40 = $100
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Answer:

Option D. y=-4x^{2} -16x-12

Step-by-step explanation:

using a graphing tool

Graph and determine the vertex in each case

we know that

If the equation has a maximum value at x=-2, then the x-coordinate of the vertex must be  equal to -2 and the parabola open downward

case A) y=-x^{2} -20x-16

The vertex is the point (-10,84)

case B) y=-x^{2} -16x-12

The vertex is the point (-8,52)

case C) y=-4x^{2} -20x-16

The vertex is the point (-2.5,9)

case D) y=-4x^{2} -16x-12

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Let an = –3an-1 + 10an-2 with initial conditions a1 = 29 and a2 = –47. a) Write the first 5 terms of the recurrence relation. b)
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We can express the recurrence,

\begin{cases}a_1=29\\a_2=-47\\a_n=-3a_{n-1}+10a_{n-2}7\text{for }n\ge3\end{cases}

in matrix form as

\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}

By substitution,

\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}\implies\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^2\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}

and continuing in this way we would find that

\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}\begin{bmatrix}a_2\\a_1\end{bmatrix}

Diagonalizing the coefficient matrix gives us

\begin{bmatrix}-3&10\\1&0\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}

which makes taking the (n-2)-th power trivial:

\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}^{n-2}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}

\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}

So we have

\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}

and in particular,

a_n=\dfrac{29\left(2(-5)^{n-1}+5\cdot2^{n-1}\right)-47\left(-(-5)^{n-1}+2^{n-1}\right)}7

a_n=\dfrac{105(-5)^{n-1}+98\cdot2^{n-1}}7

a_n=15(-5)^{n-1}+14\cdot2^{n-1}

\boxed{a_n=-3(-5)^n+7\cdot2^n}

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