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MariettaO [177]
2 years ago
8

Suppose f(x) is a function which satisfies f'(3) = 0,f'(5) = 0, f"(3) = -4, and f"(5) = 5.

Mathematics
1 answer:
Dennis_Churaev [7]2 years ago
8 0
If you find the answer let me know please e
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Bad White [126]

Answer:

ASA

Step-by-step explanation:

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Dafna1 [17]

Answer: The first number that appears in both sequences is 28.

Step-by-step explanation:

Let's write down numbers from each of the sequences

Sequence 1) We need to start from 7 and multiply 4

7x4=28, 28x4=112

The sequence is 7,28,112...

Sequence 2) We need to start from 8 and add 5

5+8=13, 13+5=18, 18+5=23, 23+5=28

The sequence is 8,13,18,23,28...

They both have 28

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2 years ago
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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.

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3 years ago
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Ludmilka [50]

Answer:

<em>Hi</em><em> </em><em>there</em><em>!</em><em>!</em>

<em>The</em><em> </em><em>answer</em><em> </em><em>would be</em><em> </em><em>x</em><em>=</em><em>2</em><em>0</em><em> </em><em>and</em><em> </em><em>y</em><em>=</em><em> </em><em>1</em><em>0</em><em> </em><em>root</em><em> </em><em> </em><em>3</em><em> </em><em>or</em><em> </em><em>on</em><em> </em><em>decimal</em><em> </em><em>it's</em><em> </em><em>1</em><em>7</em><em>.</em><em>3</em><em>2</em><em>.</em>

<em>explanation</em><em> </em><em>look</em><em> </em><em>in</em><em> </em><em>picture</em><em>,</em><em> </em><em>alright</em><em>. </em>

<em><u>I</u></em><em><u> </u></em><em><u>hope</u></em><em><u> </u></em><em><u>it will</u></em><em><u> </u></em><em><u>help u</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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