$\bf x^{-1}-1\div x-1\implies \implies \cfrac{1}{x}-1\div x-1\implies \cfrac{\frac{1}{x}-1}{~~x-1~~}\implies \cfrac{~~\frac{1-x}{x}~~}{\frac{x-1}{1}} \\\\\\ \cfrac{1-x}{x}\cdot \cfrac{1}{x-1}\implies \cfrac{-(\begin{matrix} x-1 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix})}{x}\cdot \cfrac{1}{\begin{matrix} x-1 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}\implies -\cfrac{1}{x}\implies -x^{-1}$

3 0

3 years ago

There are 20 players in a table tennis competition. each player must play exactly one match against every other player. how many

Flauer [41]

There are 10 matches in the competition if you count every other player you'll end up with 10

7 0

3 years ago

Given a term and the common difference, find the explicit formula.<br> 8. a(19) = 135<br> d = 15

Rasek [7]

The explicit formula is a(n) = 15(n – 10)

<u>Solution:</u>

Given, a term a(19) = 135 and common difference d = 15

We have to find the explicit formula.

Now, we know that, a(n) = a + (n – 1)d where a(n) is nth term, a is first term, d is common difference,

So, for a(19)

$\begin{array}{l}{\rightarrow a(19)=a+(19-1) 15} \\\\ {\rightarrow 135=a+18 \times 15} \\\\ {\rightarrow a=135-270} \\\\ {\rightarrow a=-135}\end{array}$

Now, we know that, an explicit formula is an expression for finding the nth term,

So, in our problem, expression for finding nth term is a + (n – 1)d

$\begin{array}{l}{\rightarrow-135+(n-1) 15} \\\\ {\rightarrow-135+15 n-15} \\\\ {\rightarrow 15 n-150} \\\\ {\rightarrow 15(n-10)}\end{array}$

Hence, the explicit formula is a(n) = 15(n – 10).

7 0

3 years ago