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2 years ago

The lines shown below are perpendicular. If the green line has a slope of 2/5 what is the slope of the red line

1 answer:

6 0

In this item, we are not given with the figure but knowing that these lines ought to be perpendicular then we will be able to derive the relationship between the slopes of the line.

The slopes of the perpendicular line are the negative reciprocals of one another. If we represent the slopes of the lines as m₁ and m₂, the relationship can be written in the form,
(m₁)(m₂) = -1

We are given with one of the slopes. To determine the value of the second slope then,
m₂ = -1/m₁
m₂ = -1/(2/5)
m₂ = -5/2

<em>ANSWER: m₂ = -5/2</em>

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What term is 1/1024 in the geometric sequence,-1,1/4,-1/6..?

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Answer:

$\large\boxed{\text{sixth term is equal to}\ \dfrac{1}{1024}}$

Step-by-step explanation:

The explicit formula for a geometric sequence:

$a_n=a_1r^{n-1}$

$a_n$ - n-th term

$a_1$ - first term

$r$ - common ratio

$r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=...=\dfrac{a_n}{a_{n-1}}$

We have

$a_1=-1,\ a_2=\dfrac{1}{4},\ a_3=-\dfrac{1}{6},\ ...$

The common ratio:

$r=\dfrac{\frac{1}{4}}{-1}=-\dfrac{1}{4}\\\\r=\dfrac{-\frac{1}{6}}{\frac{1}{4}}=-\dfrac{1}{6}\cdot\dfrac{4}{1}=-\dfrac{2}{3}\neq-\dfrac{1}{4}$

<h2>It's not a geometric sequence.</h2>

If $a_3=-\dfrac{1}{16}$ then the common ratio is $r=\dfrac{-\frac{1}{16}}{\frac{1}{4}}=-\dfrac{1}{16}\cdot\dfrac{4}{1}=-\dfrac{1}{4}$

Put to the explicit formula:

$a_n=-1\left(-\dfrac{1}{4}\right)^{n-1}$

Put $a_n=\dfrac{1}{1024}$ and solve for <em>n </em>:

$-1\left(-\dfrac{1}{4}\right)^{n-1}=\dfrac{1}{1024}\qquad\text{use}\ a^n:a^m=a^{n-m}\\\\-\left(-\dfrac{1}{4}\right)^n:\left(-\dfrac{1}{4}\right)^1=\dfrac{1}{1024}\\\\-\left(-\dfrac{1}{4}\right)^n\cdot(-4)=\dfrac{1}{1024}\\\\(4)\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{1024}\qquad\text{divide both sides by 4}\ \text{/multiply both sides by}\ \dfrac{1}{4}/\\\\\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{4096}\\\\\dfrac{(-1)^n}{4^n}=\dfrac{1}{4^6}\qquad n\ \text{must be even number. Therefore}\ (-1)^n=1$

$\dfrac{1}{4^n}=\dfrac{1}{4^6}\iff n=6$

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