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Aleonysh [2.5K]
3 years ago
8

27

Mathematics
1 answer:
asambeis [7]3 years ago
3 0

Answer:

  The sum of 2 and the quotient of 3 and y

Step-by-step explanation:

The expression 2 + 3/y without any parentheses is evaluated as ...

  2 + (3/y)

which is a sum. The first contributor to the sum is 2, and the second contributor is the quotient of 3 and y. (Usually, "the quotient of "a" and "b" means "a/b".)

Hence the answer shown above is a good description of the expression.

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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
3 years ago
24 fl oz =___ pt ___c. i hope you can understand wat i have nean pliz
tiny-mole [99]
24 fl oz= 1 pint and 1 cup
6 0
4 years ago
Read 2 more answers
6. Given the function
fredd [130]

Answer:

See below.

Step-by-step explanation:

The  x - 1 will  move whole graph 1 unit to the right.

The 1/2 will stretch it horizontally by a factor 2.

The 2 will  stretch it vertically by a factor 2.

The -7 translates it 7 units down.

7 0
4 years ago
One box of crackers costs $1.75. The crackers are advertised as “3 boxes for $5.25.” Which proportion can be used to represent t
Daniel [21]
A.  1/1.75 = 3/5.25 


Ignore this part I'm just trying to get at least 20 characters.
5 0
4 years ago
Read 2 more answers
Joely's Tea Shop, a store that specializes in tea blends, has available 45 pounds of A grade tea and 70 pounds of B grade tea. T
prisoha [69]

Answer:

Hence, (75,40) is the maximum point and maximum profit is $192.5

Step-by-step explanation:

Let the breakfast be x

And  let the lunch or afternoon be y

Inequalities will become

\frac{1}{3}x+\frac{1}{2}y\leq 45

\frac{2}{3}x+\frac{1}{2}y\leq 70

And objective function will be

Z=1.5x+2.0y  

You can see the graph in the attachement

ABCD is the feasible region

Points of feasible region is (0,90) ,(0,0) ,(75,40) and(105,0)

We have to find Z by substituting the points of feasible region

At (0,90) we get

Z=1.5(0)+2(90)

Z=180

At (0,0)

Z=1.5(0)+2(0)

Z=0

At (75,40)

Z=1.5(75)+2(40)

Z=192.5

At (105,0)

Z=1.5(105)+2(0)

Z=157.5

The maximum number we are getting is 192.5 which is at (75,40)

Hence, (75,40) is the point of maximum

4 0
4 years ago
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