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Pavel [41]
3 years ago
6

A certain species of bird migrates 14,000 miles in 90 days. It rests 8 hours each day and

Mathematics
1 answer:
LuckyWell [14K]3 years ago
4 0

Answer:

The bird moves faster

Step-by-step explanation:

Here, we want to get which of the two moves faster

This is a matter of speed

14,000 to km would be;

14,000/0.62 = 22,581 km

There is a rest of 8 hours per day

So 16 hours is used in moving

for 90 days, the total

time spent moving is 16(90) = 1,440 hours

So, the speed is generally

distance/time = 22.581/1,440 = 15.68215 km/h

For the person

100 minutes is same as 60 + 40 minutes

= 1 hour + 40/60

40/60 = 2/3 = 0.667

So the total

time is 1.667 hours

The speed is thus;

20/1.667 = 12 km/h

Comparing the two speeds, we can see that the bird moves faster

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Answer:

\huge\boxed{\sf 87.5 \ \%}

Step-by-step explanation:

35 marks out of 40<u> as a fraction:</u>

35 / 40

<u>As a percentage:</u>

<u></u>\sf \frac{35}{40} * 100 \ \%\\\\0.875 * 100 \ \%\\\\87.5 \ \%\\\\\rule[225]{225}{2}

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En una caja de cartón se empacan 400 latas de atún al acomodarlas resultan que caben 5 latas más a lo largo que a lo ancho y a l
Anvisha [2.4K]

Answer:

Hay 50 latas en total que tocan el fondo de la caja.

Step-by-step explanation:

Una caja de cartón es representada por un cuadrilátero, cuyas caras son rectángulos. Si la caja esta ocupada por completo, entonces la cantidad total de latas es representada por la siguiente expresión:

n_{V} = n_{w}\cdot n_{h}\cdot n_{l} (1)

Donde:

n_{V} - Total de latas de atún, adimensional.

n_{w} - Cantidad de latas de atún a lo ancho de la caja, adimensional.

n_{h} - Cantidad de latas de atún a lo alto de la caja, adimensional.

n_{l} - Cantidad de latas de atún a lo largo de la caja, adimensional.

De acuerdo con el enunciado, tenemos las siguientes relaciones:

n_{V} = 400 (2)

n_{l} = n_{w} +5 (3)

n_{h} = n_{w}+3 (4)

Si aplicamos estas fórmulas a (1), tenemos el siguiente polinomio de tercer orden:

n_{w} \cdot (n_{w}+5)\cdot (n_{w}+3) = 400

n_{w}\cdot (n_{w}^{2}+8\cdot n_{w}+15) = 400

n_{w}^{3}+8\cdot n_{w}^{2}+15\cdot n_{w} -400 = 0

Este polinomio se puede resolver por vía analítica por el Método de Cardano o por vía numérica, sus raíces son:

n_{w,1} = 5, n_{w,2} = -6.5+i\,6.144, n_{w,3} = -6.5-i\,6.144

En consecuencia, la única solución válida es n_{w} = 5 y las variables restantes son por (3) y (4):

n_{l} = 10 y n_{h} = 8

La cantidad de latas que tocan el fondo de la caja es igual al producto de la cantidad de latas a lo largo de la caja y la cantidad de latas a lo ancho de la caja, es decir:

n_{A} = n_{w}\cdot n_{l} (5)

n_{A} = (5)\cdot (10)

n_{A} = 50

Hay 50 latas en total que tocan el fondo de la caja.

4 0
2 years ago
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