Which statement is true of a rectangle that has an area of 4x2 + 39x – 10 square units and a width of (x + 10) units? a) the rectangle is a square. b) the rectangle has a length of (2x – 5) units. c) the perimeter of the rectangle is (10x + 18) units. d) the area of the rectangle can be represented by (4x2 + 20x – 2x – 10) square units.

its C

Step-by-step explanation:

its c

C) The perimeter of the rectangle is (10x + 18) units.  

Step-by-step explanation:

First we will find the length by dividing the polynomials.

We first put the dividend, 4x²+39x-10, under the box.  The divisor, x+10, goes on the outside of the box (to the left).

We divide 4x² by x; this is 4x.  We put 4x above 39x in the quotient.

Next we multiply back through by 4x:

4x(x+10) = 4x²+40x.  This goes under 4x²+39x in the dividend.  We subtract, giving us -1x.  We then bring down the -10, giving us the expression -1x-10.

We divide -1x by x.  This is -1.  This goes beside the 4x in the quotient.  We multiply back through by -1:

-1(x+10) = -1x-10.  This goes under the -1x-10 in the work.  Subtracting cancels both of these, leaving us with the quotient 4x-1.

This is not the same as the width, so the rectangle is not a square.  The length was not 2x-5.

The perimeter is found by adding

(4x-1)+(4x-1)+(x+10)+(x+10)

Combining like terms, we have

4x+4x+x+x-1-1+10+10

10x+18 is the perimeter.

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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: