HELP PLZZZZZZZZZZZZZZ

A circle has a circumference of 28 centimeters. If an arc subtends a central angle of 55 degrees, what is the arc length?

Harlamova29_29 [7]

Answer:

s = 4.28 cm (Answer A)

Step-by-step explanation:

1) The formula for the circumference of a circle of radius r is C = 2(pi)r, and this corresponds to a full circle, with central angle 360 degrees.

2) Here the circumference is 28 cm. Therefore, solving for the radius involves the following calculation: 28 cm = 2(pi)r, or

r = (28 cm) / (2 pi), or (28 cm) / 6.28 = 4.46 cm.

3) The arc length s = r(theta), when the central angle (theta) is 55 degrees, is (55/360) times the circumference of the circle: (55/360)(28 cm), or

s = 4.28 cm (Answer A)

3 0

3 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.