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victus00 [196]
2 years ago
9

What the meaning of the acronym lcm​

Mathematics
1 answer:
Korvikt [17]2 years ago
5 0

Answer:

Least Common Multiple

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The cattle at the Clinton Farm are fed 1/4 of a bale of hay each day. The horses are fed 2 times as much hay as the cattle. How
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1/2 of a bale

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$453 with a sales tax of 6%
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A star is a swirling mass of dense gases powered by thermonuclear reactions. The great solar furnace transforms 7 million tons o
marta [7]

Answer:

2.20752 x 10^{24}

Step-by-step explanation:

The first thing you have to do is to look for the number of seconds in one year. You have to multiply 365 days by 86,400 seconds.

  • 365 days refer to the total number of days per year
  • 86,400 seconds refer to the number of seconds per day

Let's solve.

  • 365 x 86,400 = 31,536,000 or 3.1536 x 10^{7}

<em>Therefore, one year has 3.1536 x </em>10^{7}<em> seconds.</em>

Next, you have to know the number of seconds in 10 billion years.

  • (3.1536 x 10^{7}) x  (1.0 x 10^{10}) = 3.1536 x 10^{17}

The last step is to multiply the number of seconds in 10 billion years to 7 million tons of mass per second.

  • (3.1536 x 10^{17}) x 7,000,000 = 2.20752 x 10^{24}
6 0
2 years ago
Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y
ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

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

The above equation can be better expressed as:

y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0

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}

Also;

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

When x = - 2

\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.

5 0
3 years ago
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