All categories

3 years ago

Please I really need help on this please

Mathematics

1 answer:

AlexFokin [52]3 years ago

5 0

Answer:

Option D: $\frac{log \hspace{1mm} 3mn}{p}$

Step-by-step explanation:

We have the following rules:

$log \hspace{1mm} a + log \hspace{1mm} b = log \hspace{1mm}(a + b)$

$log \hspace{1mm} a - log \hspace{1mm} b = log\hspace{1mm} \bigg( \frac{a}{b} \bigg )$

Here, we have: $log \hspace{1mm} 3 + log \hspace{1mm} m + log \hspace{1mm} n - log \hspace{1mm} p$

$= \bigg ( log \hspace{1mm} 3 + log \hspace{1mm} m + log \hspace{1mm} n \bigg ) - log \hspace{1mm} p$

$= log \hspace{1mm} 3mn - log \hspace{1mm} p$

$= log \bigg( \frac{3mn}{p} \bigg )$

Hence, the answer.

You might be interested in

Raju has 2020 one-centimetre square tiles. He

Step2247 [10]

84 the process is shown in the following picture

6 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

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

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

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.