What is 34÷18? Explain or show your reasoning.

Where r is the radius of the cylinder and h is the height of the cylinder.

AnnyKZ [126]

<h3>Given:-</h3>

radius = 7 inches
height = 9 inches

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Surface area of cylinder

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we know:-

$\bigstar \boxed{ \rm Surface \: Area \: of \: Cylinder=2\pi rh+2\pi {r}^{2} }$

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So:-

⇢SA of cylinder=2πrh+2πr²

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⇢SA of cylinder = 2πr(h+r)

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⇢SA of cylinder = 2 × π × 7(7 + 9)

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⇢SA of cylinder = 2 × π × 7 (16)

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⇢SA of cylinder = 2 × π × 112

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⇢SA of cylinder = 2 × 112 × π

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⇢SA of cylinder = 224π inches ²

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.°. Surface Area of cylinder is 224π inches².

3 0

2 years ago

Determine the exact formula for the following discrete models:

marshall27 [118]

I'm partial to solving with generating functions. Let

$T(x)=\displaystyle\sum_{n\ge0}t_nx^n$

Multiply both sides of the recurrence by $x^{n+2}$ and sum over all $n\ge0$ .

$\displaystyle\sum_{n\ge0}2t_{n+2}x^{n+2}=\sum_{n\ge0}3t_{n+1}x^{n+2}+\sum_{n\ge0}2t_nx^{n+2}$

Shift the indices and factor out powers of $x$ as needed so that each series starts at the same index and power of $x$ .

$\displaystyle2\sum_{n\ge2}2t_nx^n=3x\sum_{n\ge1}t_nx^n+2x^2\sum_{n\ge0}t_nx^n$

Now we can write each series in terms of the generating function $T(x)$ . Pull out the first few terms so that each series starts at the same index $n=0$ .

$2(T(x)-t_0-t_1x)=3x(T(x)-t_0)+2x^2T(x)$

Solve for $T(x)$ :

$T(x)=\dfrac{2-3x}{2-3x-2x^2}=\dfrac{2-3x}{(2+x)(1-2x)}$

Splitting into partial fractions gives

$T(x)=\dfrac85\dfrac1{2+x}+\dfrac15\dfrac1{1-2x}$

which we can write as geometric series,

$T(x)=\displaystyle\frac8{10}\sum_{n\ge0}\left(-\frac x2\right)^n+\frac15\sum_{n\ge0}(2x)^n$

$T(x)=\displaystyle\sum_{n\ge0}\left(\frac45\left(-\frac12\right)^n+\frac{2^n}5\right)x^n$

which tells us

$\boxed{t_n=\dfrac45\left(-\dfrac12\right)^n+\dfrac{2^n}5}$

# # #

Just to illustrate another method you could consider, you can write the second recurrence in matrix form as

$49y_{n+2}=-16y_n\implies y_{n+2}=-\dfrac{16}{49}y_n\implies\begin{bmatrix}y_{n+2}\\y_{n+1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}\begin{bmatrix}y_{n+1}\\y_n\end{bmatrix}$

By substitution, you can show that

$\begin{bmatrix}y_{n+2}\\y_{n+1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}^{n+1}\begin{bmatrix}y_1\\y_0\end{bmatrix}$

or

$\begin{bmatrix}y_n\\y_{n-1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}^{n-1}\begin{bmatrix}y_1\\y_0\end{bmatrix}$

Then solving the recurrence is a matter of diagonalizing the coefficient matrix, raising to the power of $n-1$ , then multiplying by the column vector containing the initial values. The solution itself would be the entry in the first row of the resulting matrix.

5 0

3 years ago

I need help with square roots. I would really appreciate it if you help me understand them. Thank you! I'll give you 10 points a

Juliette [100K]

Answer: 9/8

Step-by-step explanation:

The square root of this is 9/8 because when you square 9/8, you get 81/64, and when you square root it, you get 9/8.

The easiest way tho is knowing that when you square a number and square root the square, you get the same number, cuz the square root cancels the square. Does that make sense?

Anyways hope this helped!

8 0

3 years ago

How do you find a reasonable domain and range for number 13?

kogti [31]

They have 3 packages so p=3
so N(p) = 90p
therefore N(3) = 270
they cannot feed more than 270 people so the range must be N(p)>270

7 0

3 years ago

PLEASE HELP

padilas [110]

Answer:

Q1 B