5. a carpenter is building a bookcase. how can the carpenter use a tape measure to check if the bookshelf is rectangular? a. she can measure the sides. if all of the sides are the same length, then the bookshelf is rectangular. b. she can measure the opposite sides. if they are congruent, then the bookshelf is in the shape of a parallelogram. then she can measure the diagonals. if they are congruent, then the bookshelf is rectangular. c. she can draw the diagonals, find the point where they intersect, and then measure to see whether they bisect each other. if they do, then the bookshelf is rectangular. d. she could use any of the three methods. 6. suppose the diagonals of a parallelogram are both perpendicular and congruent. which type of special quadrilateral it is? use the most descriptive name possible. a. rhombus b. rectangle c. square d. parallelogram

5.

b. She can measure the opposite sides. If they are congruent, then the bookshelf is in the shape of a parallelogram. Then she can measure the diagonals. If they are congruent, then the bookshelf is rectangular.

6.

c. Square

Step-by-step explanation:

5. Let us consider the properties of a parallelogram,

1. Opposite sides are congruent and parallel to each other

2. Diagonals of parallelogram bisect each other

3. If one  angle of a parallelogram is 90 degrees then all the angles are 90 degrees.

In case of a rectangle, it obeys all the properties of parallelogram except, one thing, the diagonals of a rectangle are equal in measure.

So if the carpenter finds out that the diagonals of the bookshelf are congruent and the opposite sides are congruent then it is a rectangle.

6. If the diagonals of a parallelogram are perpendicular and congruent then it is a square because in a square, both the diagonals are congruent to each other and they bisect each other at right angles. Therefore, the most descriptive name possible is a square.

5: B

6: C

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:

p

step-by-step explanation: