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Jlenok [28]
2 years ago
6

Which expression is equivalent to i 233

Mathematics
2 answers:
Fiesta28 [93]2 years ago
5 0
Start with second, third and fourth  degree of imaginary unit i:

i^2=-1, \\ i^3=i^2\cdot i=-1\cdot i=-i, \\ i^4=i^2\cdot i^2=-1\cdot (-1)=1.

Since 233=232+1=4·58+1, then i^{233}=i^{4\cdot 58+1}=(i^4)^{58}\cdot i^1=1^{58}\cdot i=i.

Answer: i^{233}=i
natta225 [31]2 years ago
5 0

Answer:C

Step-by-step explanation:

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Step-by-step explanation:

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g A car company is testing a new type of tire. They want to determine the stopping distance if the car has anti-lock brakes or n
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2 years ago
Joanne has a goal of buying a townhome and saved for 5 years. She expected to have the down payment of $15,000 saved by now but
serg [7]

Answer:


Step-by-step explanation:

Average amount saved per year = \frac{15,000}{5}

                                                      =$3000

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\frac{12}{3000} \times2000

= 8 Months.

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2 years ago
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Imaginá que tenés 125 dados cúbicos del mismo tamaño ¿Cuantos dados de altura tiene el cubo de mayor tamaño que podés armar apil
kumpel [21]

Answer:

(i) Debemos apilar 5 dados para construir el cubo de mayor tamaño.

(ii) Se necesita 121 dados cuadrados para formar el cuadrado con la mayor cantidad de dados posibles, quedando 4 dados sobrantes.

Step-by-step explanation:

(i) Sabemos por la Geometría Euclídea del Espacio que un cubo es un sólido regular con 6 caras cuadradas y longitudes iguales. Cada dado tiene un volumen de 1 dado cúbico y 125 dados dan un volumen total de 125 dados cúbicos.

El volumen de un cubo está dado por la siguiente fórmula:

V = L^{3}

Donde:

L - Longitud de la arista, medida en dados.

V - Volumen del cubo, medido en dados cúbicos.

Ahora, necesitamos despejar la longitud de la arista para calcular la altura máxima posible:

L = \sqrt[3]{V}

Dado que V = 125\,dados^{3}, encontramos que la altura del cubo de mayor tamaño sería:

L =\sqrt[3]{125\,dados^{3}}

L = 5\,dados

Debemos apilar 5 dados para construir el cubo de mayor tamaño.

(ii) El área cuadrada formada por cubos está determinada por la siguiente fórmula:

A = L^{2}

Donde:

L - Longitud de arista, medida en dados.

A - Área, medida en dados cuadrados.

Puesto que la longitud de arista se basa en un conjunto discreto, esto es, el número de dados disponibles, debemos encontrar el valor máximo de L tal que no supere 125 y de un área entera. Es decir:

L \leq 125\,dados

Si cada cubo tiene un área de 1 dado cuadrado, entonces un cuadrado conformado por 125 dados tiene un área total de 125 dados cuadrados. Entonces:

L^{2}< 125\,dados^{2}

Esto nos lleva a decir que:

L < 11.180\,dados

Entonces, la longitud máxima del cuadrado con la mayor cantidad de cubos posible es de 11 dados. El número total requerido de cubos es el cuadrado de esa cifra, es decir:

n = (11\,dados)^{2}

n = 121\,dados

Se necesita 121 dados cuadrados para formar el cuadrado con la mayor cantidad de dados posibles, quedando 4 dados sobrantes.

4 0
2 years ago
James folds a piece of paper in half several times,each time unfolding the paper to count how many equal parts he sees. After fo
Snezhnost [94]

Answer:

There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

The needed function is y = 2 ^n

Step-by-step explanation:

Let us assume the piece of paper James decides to fold is a SQUARE.

Now, let us assume:

n : the number of times the paper is folded.

y : The number of parts obtained after folds.

Now, if the paper if folded ONCE ⇒  n = 1

Also, when the pap er is folded once, the parts obtained are TWO equal parts.

⇒  for n = 1 , y = 2       ..... (1)

Similarly, if the paper if folded TWICE  ⇒  n = 2

Also, when the paper is folded twice, the parts obtained are FOUR equal parts.

⇒  for n = 2 , y = 4       ..... (2)

⇒y  = 2^2  =  2^n

Continuing the same way, if the paper is folded SEVEN times  ⇒  n = 7

So, y = 2^ n = 2^7 = 128

⇒  There are total 128 equal parts.

Lastly,  if the paper is folded ELEVEN  times  ⇒  n = 11

So, y = 2^ n = 2^{11} = 2048

⇒  There are total 2048 equal parts.

Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

And the needed function is y = 2 ^n

8 0
3 years ago
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