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3 years ago

Separate triangular plot lines for two characters showing their stories progressing side by side

Mathematics

2 answers:

shusha [124]3 years ago

8 0

Answer:

A: parallel

Step-by-step explanation:

its A the lines are side by side hence the answer parallel.. also i took the quiz

Savatey [412]3 years ago

6 0

Answer:

Episodic

Step-by-step explanation:

There are different types of plot ,But the type of plot that was illustrated here is reffered to as episodic.

Episodic plot is the plot that can be described as one that is composed of series or varieties of different events, one of the unique features of the Episodic plot is that they are usually one chapter long, and they are usually related to one another.Each chapter of the plot posses its rising action, and this has relationship with somehow with the next and previous chapters.

The total events are joined together with just a common theme as well as characters.

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Ivenika [448]

Answer:

$\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}} = -\dfrac{5}{3}$

Step-by-step explanation:

We are given the expression to be simplified:

$\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}}$

Let us take common a term with a power of 5 from the numerator and the denominator of the given expression.

We know that:

$a^{p+q} = a^p \times a^q$

Let us use it to solve the powers of 5 in the given expression.

$\therefore$ we can write:

$5^{n+2} = 5^{n+1}\times 5= 5^n\times 5^{2}$

$5^{n+1} = 5^n\times 5$

The given expression becomes:

$\dfrac{5^{n+1} \times 5-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n}\times 5}$

Taking common $5^{n+1}$ from the numerator and

Taking common $5^{n}$ from the denominator

$\Rightarrow \dfrac{5^{n+1} (5-6)} {5^{n}(13-2\times5)}\\\Rightarrow \dfrac{5^{n+1} (-1)} {5^{n}(13-10)}\\\Rightarrow -\dfrac{5^{n+1}} {5^{n}\times3}\\\Rightarrow -\dfrac{5^{n}\times 5} {5^{n}\times3}\\\Rightarrow -\dfrac{5}{3}$