The differential equation has as a solution. x^2 d^2y/dx^2- 7x dy/dx + 16y = 0
Applying reduction order we set y^2 = ux^4. Then (using the prime notation for the derivatives). So, plugging y^2 into the left side of the differential equation, and reducing, we get
x^2y"^2 - 7xy'^2 + 16y^2 =
The reduced form has a common factor of x^5 which we can divide out of the equation so that we have xu" + u' = 0. Since this equation does not have any u terms in it we can make the substitution w = u' giving us the first order linear equation xw' + w = 0. This equation has integrating factor for x > 0. If we use a as the constant of integration, the solution to this equation is w =. Integrating to get u, and using b as our second constant of integration we have u = 0. Finally y^2 =and the general solution is.

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .

well we know there are 4 clubs in 52 cards so it’s replaced so it should be 3/51

step-by-step explanation:

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation:

a rotation about point l

step-by-step explanation:

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