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3 years ago

A ratio is a comparison of two numbers by addition. always sometimes never

Mathematics

2 answers:

vesna_86 [32]3 years ago

6 0

<span>A ratio is never a comparison of two numbers by addition. [never]</span>

Viktor [21]3 years ago

3 0

Answer:

C) never

Step-by-step explanation:

A ratio is a comparison of two things with their numbers by division.

The ratio can be written as a:b, a/b and a to b.

The ration cannot be compared by addition.

Therefore, answer is C ) never

Thank you.

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Which equation does not have the same solution as the others?

dem82 [27]

<span>The answer is B. X-9=17
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3 years ago

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Find x such that x - 4, x, and 3x - 8 are three consecutive terms in a geometric sequence

Nitella [24]

In a geometric sequence dividing any term by the term before it yields the common multiplier.
Then x/(x+8) = (3x-8)/x
Solving for x: reduces to x^2-10x +16 factor into (x-2)(x-8)
Then x = 2 or 8
2 produces a sequence of -2, 2, -2 ...
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3 years ago

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

Trava [24]

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>:

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