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egoroff_w [7]
2 years ago
15

On the first Saturday of every month, a local amusement park allows children under the age of 12 to ride for free. Which of thes

e children can ride for free: Jamie (12 years), Su (11 years, 9 months), Katalin (12 years, 3 months), Devon (13 years, 2 months)?
Mathematics
1 answer:
DochEvi [55]2 years ago
5 0
Su is the only child under 12 years of age, therefore, she is only child entering for free.
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Answer:

(x+12)(x+12)-x^2

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=24x+144

Step-by-step explanation:

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4. Andrés desea embaldosar el piso de su casa que tiene 375 cm de ancho y 435 cm de largo. Calcula la longitud del lado que tend
svetlana [45]

Answer:

Sabemos que el piso es un rectángulo de 435 cm de largo y 375 cm de ancho.

Recordar que para un rectángulo de largo L, y ancho W, el área es:

A = L*W

Entonces el área del piso, será:

A = 435cm*375cm = 163,125 cm^2

Primero, sabemos que se utilizaran baldosas (las cuales son cuadradas) y queremos saber la longitud de lado que tendrían las baldosas.

No tenemos ningún criterio para encontrar este lado, solo que (si queremos usar un número entero de baldosas) el largo L del lado de la baldosa deberá ser un divisor de tanto el ancho como el largo del suelo.

Dicho de otra forma

el largo, 435cm, tiene que ser múltiplo de L

el ancho, 375cm, tiene que ser múltiplo de L.

Por ejemplo, ambos números son múltiplos de 5, entonces podríamos tomar L = 5cm

En este caso, el área de cada baldosa es:

a = L^2 = 5cm*5cm = 25cm^2

Y el número total de baldosas que necesitaría usar esta dado por el cociente entre el área del suelo y el area de cada baldosa.

N = ( 163,125 cm^2)/(25cm^2) = 6,525 baldosas.

También sabemos que ambos números (435cm y 375cm) son múltiplos de 15cm

Entonces las baldosas podrían tener 15cm de lado.

En este caso, el área de cada baldosa es:

A = (15cm)^2 = 225cm

En este caso el número total de baldosas necesarias será:

N =  ( 163,125 cm^2)/(225cm^2) = 725 baldosas.

5 0
3 years ago
BRAINIEST ANSWER HERRREEE!!! :D
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3 years ago
Mr. Toshio lent out 11 rulers at the beginning of class, collected 4 rulers in the middle of class, and gave out 7 at the end of
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Mr. Toshio had 32 rulers. 32-11 = 21 rulers. 21+4 = 25 rulers. and 25-7 = 18 rulers
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3 years ago
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