Kaitlin will donate up to $460 to charity. The money will be divided between two charities: the City Youth Fund and the Educational Growth Foundation. Kaitlin would like the amount donated to the Educational Growth Foundation to be at least four times the amount donated to the City Youth Fund. Let x denote the amount of money (in dollars) donated to the City Youth Fund. Let y denote the amount of money (in dollars) donated to the Educational Growth Foundation. Shade the region corresponding to all values of x and y that satisfy these requirements. Note: Graph

The question states "up to $460" in charity, which means it can be less than or equal to 460, but not any higher. Then,

$460 to charity (max)

$160+ to Educational (160 being the minimum)

Education is 3x (or higher) than City.

If we denote x to City, and y to Educational, then,

x + y = 460 . And we know that y >= 3x

x + 3x = 4x = 460

x = 115. This means when she donates $115 to City, she will donate 3x (or $345) to Educational. These are the max values; she will not donate higher than $460.

A minimum requirement was placed on Educational earlier. When y = 160, x will be y/3, or $53.34. (This isn't very meaningful, but: This means she donated 213.34, which is still less than $460)

Therefore, x (which is City) can be the range of $53-ish dollars (depending on your region/graph) but no higher than $115.

y (which is Educational) will be as low as $160, but no higher than $345.

Hope this helps

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: