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3 years ago

I require assistance once again.

Mathematics

2 answers:

stealth61 [152]3 years ago

4 0

Answer:

I can help you with 1, 2 and 3.

Step-by-step explanation:

1. 2x + 3y = 1470

Subtract 2x from both sides: 3y= -2x+1470

Divide three from both sides: y= \frac{-2}{3}x +490

2. To graph it, you would first go to (0,490) because that is where the y-intercept is. Since slope is ride over run, you can either go up 2 left3 or down 2 right 3. You can proceed doing that until you have enough points to draw the line.

3. f(x)= -2/3x +490. This function means represents how many wraps are sold based on how many rolls are sandwich.

Lera25 [3.4K]3 years ago

3 0

Answer:

Step-by-step explanation:

1. Equation changed to slope-intercept as thus

y = 490 - 0.667x

with slope = -0.667

intercept on y axis = 490

intercept on x axis = 735

2. Description about graphing the line using slope-intercept method

On the y (vertical axis): maximum value of y = 490 and that is when x = 0

On the x (horizontalal axis): maximum value of x = 735 and that is when y = 0

Incrementing on x axis cause a decline of y-axis (as shown in the graph as attached)

3. Equation of the function notation

F(x) = 490 - 0.667x

where wrap, y= F(x) and sandwich, x

4. See attached graph but try and continue the projection to intercept x at 735

5. Equation for next month is 2x + 3y = 1593

For the graph;

Slope remains the same as previous month = -0.667

intercept on y axis = 531

intercept on x axis = 796.5

The graphs look a little similar just that the intercepts are different.

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4 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

I require assistance once again.​

I require assistance once again.