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bogdanovich [222]
3 years ago
12

PLEASE HELP!!!!!!!!!!! I SUCK AT MATH AND I CAN'T FIGURE THIS OUT!

Mathematics
2 answers:
nikdorinn [45]3 years ago
8 0

Answer:

$25.97

Step-by-step explanation:

2.50+8.75+3+10.25 = 24.50    This is the items be bought and now needs to pay 6% tax.

24.50/100*106 = 25.97

valentinak56 [21]3 years ago
6 0

First find the subtotal - this is the total cost of all items before tax.

$2.50 + $8.75 + $3.00 + $10.25 = $24.50

The 6% tax is applied to the subtotal - this is essentially adding 6% of the subtotal to the subtotal. 6% of the subtotal is

0.06 * $24.50 = $1.47

so the tax is $1.47. (This was done without using the distributive property.)

Computed another way, we can instead apply the 6% tax to all items individually:

0.06 * $2.50 = $0.15

0.06 * $8.75 = $0.525

0.06 * $3.00 = $0.18

0.06 * $10.25 = $0.615

Then we total this to find the total tax:

$0.15 + $0.525 + $0.18 + $0.615 = $1.47

(with the distributive property)

- - -

In both cases, we're computing

0.06 * ($2.50 + $8.75 + $3.00 + $10.25)

Without the distributive property, the idea is to simplify the terms being added first, then multiplying that by the tax rate.

With the distributive property, we distribute the tax rate to each item, then add those products together.

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3 0
3 years ago
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)
zlopas [31]

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}

6 0
3 years ago
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