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Dominik [7]
3 years ago
10

Is 3/6 divided by 5 greater than 1?

Mathematics
2 answers:
Naily [24]3 years ago
6 0

It is not greater than one.

When you divide 3/6 by 5, your quotient is 0.1, which has a value that is less than 1.

Hope this helps!

- sahira :>

Kisachek [45]3 years ago
5 0

Step-by-step explanation:

3/6 divided by 5

The numbers in 3/6 divided by 5 are labeled below:

3 = numerator

6 = denominator

5 = whole number

To make it a fraction form answer, you keep the numerator and multiply the denominator by the whole number to make a new denominator:

3/6 times 5 = 3/30

Thus, the answer to 3/6 divided by 5 in fraction form is:

3/30

to make the answer to 3/6 divided by 5 in decimal form, you simply divide the numerator by the denominator from the fraction answer above:

3/30 = 0.1

The answer is rounded to the nearest four decimal points if necessary.

3/30 can be simplified to 1/10

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Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y
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Step-by-step explanation:

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The differential equation; (x^2-4)^2y'' + (x + 2)y' + 7y = 0

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The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

p(x) = \dfrac{(x+2)}{(x^2-4)^2} \

p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \

p(x) = \dfrac{1}{(x+2)(x-2)^2}

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q(x) = \dfrac{7}{(x^2-4)^2}

q(x) = \dfrac{7}{(x+2)^2(x-2)^2}

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

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\lim \limits_{x \to-2} (x+ 2) p(x) =  \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{1}{(x-2)^2}

\implies \dfrac{1}{16}

\lim \limits_{x \to-2} (x+ 2)^2 q(x) =  \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{7}{(x-2)^2}

\implies \dfrac{7}{16}

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

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3 years ago
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