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lisov135 [29]
2 years ago
11

What is next number in pattern 125,-25,5,-1,

Mathematics
1 answer:
MatroZZZ [7]2 years ago
7 0
1/5 or you may also say 2
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It's a lovely day. It's about 75 degrees - finally shorts weather! - and I can't wait to take a long walk. I leave from my front
maw [93]

Answer:

You'll be back at your friends house

Step-by-step explanation:

North goes up

East goes right

South goes down

West goes left

And he traveled 3 blocks each time. its sort if like a square.

7 0
2 years ago
Two step equations x−17/3=−3
vovikov84 [41]

Answer:

the answer would be x= 8/3 happy to help ya:) and pls mark me as brainliest!!!!!!!!!

Step-by-step explanation:

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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
Zina [86]

Answer:

Step-by-step explanation:

The given differential equation is:

x^3y'' + 2x^2y' + 4y

the main task here is to determine the singular points of the given differential equation and Classify each singular point as regular or irregular.

So, for a regular singular point ;  x=x_o is  located at the first power in the denominator of P(x) likewise at the Q(x) in the second power of the denominator. If that is not the case, then it is termed as an irregular singular point.

Let first convert it to standard form by dividing through with x³

y'' + \dfrac{2x^2y'}{x^3} + \dfrac{4y}{x^3} =0

y'' + \dfrac{2y'}{x} + \dfrac{4y}{x^3} =0

The standard form of the differential equation is :

\dfrac{d^2y}{dy} + P(x) \dfrac{dy}{dx}+Q(x)y =0

Thus;

P(x) = \dfrac{2}{x}

Q(x) = \dfrac{4}{x^3}

The zeros of x,x^3  is 0

Therefore , the singular points of above given differential equation is 0

Classify each singular point as regular or irregular.

Let p(x) = xP(x)    and q(x) = x²Q(x)

p(x) = xP(x)

p(x) = x*\dfrac{2}{x}

p(x) = 2

q(x) = x²Q(x)

q(x) = x^2 * \dfrac{4}{x^3}

q(x) =\dfrac{4}{x}

The function (f) is analytic if at a given point a it is represented by power series in x-a either with a positive or infinite radius of convergence.

Thus ; from above; we can say that q(x) is not analytic  at x = 0

Q(x) = \dfrac{4}{x^3}  do not satisfy the condition,at most to the second power in the denominator of Q(x).

Thus, the point x =0 is an irregular singular point

6 0
3 years ago
Please help me. Brainliest xD
masya89 [10]

\pi \alpha \alpha \beta \\

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3 years ago
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What’s the correct answer for this?
Maurinko [17]

Answer:

c

Step-by-step explanation:

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