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cestrela7 [59]
3 years ago
8

UN PASTEL SE CORTA QUITANDO CADA VEZ LA TERCERA PARTE DEL PASTEL QUE HAY EN EL MOMENTO DE CORTAR ¿QUE FRACCION DEL PASTEL ORIGIN

AL QUEDO DESPUES DE CORTAR TRES VECES?
Mathematics
1 answer:
oee [108]3 years ago
7 0

Answer:

No queda nada del pastel.

Step-by-step explanation:

El pastel representaria un entero, si al entero lo dividimos en tercios, son tres tercios los que los conforman, por ende si lo cortamos tres veces, en el primer corte quedarian dos tercios de pastel, en el segundo corte un tercio, y en el tercer corte el ultimo tercio , generando asi que no quede nada de este.

Los enteros se dividen en cuartos, tercios y hasta octavos, o muchas mas fracciones.

Un entero es equivalente a 4 porciones de 1/4 y a 3 porciones de 1/3

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3. At noon Lan is in his blue mini-van 300 km north of Makenna in her Bugatti. Lan heads south
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Answer:

The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h

Step-by-step explanation:

At noon the location of Lan = 300 km north of Makenna

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Lan's speed = 60 km/h

Makenna's direction and speed = West at 75 km/h

The distance  Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km

The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km

The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km

Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.

By Pythagoras' theorem, we have;

s² = x² + y²

The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61

ds²/dt = dx²/dt + dy²/dt

2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt

2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900

ds/dt = 900/(2×30·√61) ≈ 1.92

The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h

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