Внешний диаметр полого шара равен 16 см. Толщина стенок 2 см Найти

Solve for x<br> -6x+14< -28 or 9x+15≤−12

Free_Kalibri [48]

$- 3 \geqslant x > 6$

Step-by-step explanation:

....

8 0

3 years ago

Write an equation of the line in point-slope form that passes through the given points.

iris [78.8K]

Answer:

Step-by-step explanation:

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Missing: hel ‎| Must include: hel

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4 years ago

Cameron is a high school basketball player. How many points would Cameron score

yuradex [85]

Answer:

30 I THINK

Step-by-step explanation:

7 0

3 years ago

If∠

andreev551 [17]

Answer:

where is tge other half question?

3 0

3 years ago

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