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Oksana_A [137]
2 years ago
11

Problema 6 Descompongan los siguientes números en factores primos. ¿Es posible encontrar, para cada número, más de una descompos

ición en factores primos? a. 42 b. 31 c. 36 d. 45
Mathematics
1 answer:
Bogdan [553]2 years ago
8 0

Answer:

a: 42 =  2*3*7

b: 31 = 31*1

c: 36 = 2*2*3*3

d: 45 = 3*3*5

Cada descomposición es unica.

Step-by-step explanation:

Sabemos que todo número entero puede ser reescrito como un producto de números primos.

Esta descomposición es única, dado que una vez tenemos un número escrito como producto de primos, esos números primos no pueden descomponerse en otra cosa, por lo que se concluye que una descomposición en primos es única.

a: 42

Para obtener la descomposición, podemos comenzar dividiendo por primos, comenzando por los más bajos.

En este caso, podemos comenzar por 2:

42/2 = 21

asi, podemos reescribir:

42 = 2*21

Ahora ya tenemos un factor que es primo, el 2, y un factor que no lo es, el 21.

Asi que debemos reescribir el 21 como producto de primos.

Y sabemos que 3*7 = 21, donde ambos 3 y 7 son primos, entonces:

42 = 2*21 = 2*3*7

42 = 2*3*7

así hemos reescrito 42 como un producto entre números primos.

b. 31

31 es un número primo, por lo que su descomposición es:

31 = 31*1

c. 36

Comenzamos dividiendo por 2.

36/2 = 18

36 = 2*18

Volvemos a dividir por 2 el 18.

18/2 = 9

18 = 2*9

reemplazando eso en nuestra descomposición obtenemos:

36 = 2*18 = 2*2*9

9 es impar así que no podemos dividir por 2, pasamos al próximo número primo, el 3.

9/3 = 3

9 = 3*3

Reemplazando eso, obtenemos:

36 = 2*2*3*3

d. 45

no podemos dividir por 2, puesto que es un número impar, así que pasamos al próximo primo, el 3.

45/3 = 15

45 = 3*15

Ahora descompongamos el 15.

15/3 = 5

15 = 3*5  (notar que 3 y 5 son primos)

Reemplazando eso, obtenemos:

45 = 3*15 = 3*3*5

45 = 3*3*5

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rodikova [14]

Answer:

1022 of $10 tickets, 1326 of $20 tickets, 948 of VIP $30 tickets

Step-by-step explanation:

We can set-up a system of equations to find the number of each kind of ticket. We know there are the cheap tickets c, the medium tickets m and the VIP tickets v. Since 3296 tickets were purchased, then c+m+v=3296.

We also know that 304 more medium priced tickets have been sold then cheap tickets. We write 304+c=m.

Lastly, we know that a total of $65,180 was sold. We write 10c+20m+30v=65,180.

We will solve by substituting one equation into the other. We substitute m=304+c into c+m+v=3296. Simplify and isolate the variable v.

c+m+v=3296

c+(304+c)+v=3296

c+304+c+v=3296

2c+304-304+v=3296-304

2c-2c+v=2992-2c

v=2992-2c

We will now substitute this into the remaining equation with our first substitution.

10c+20m+30v=65,180

10c+20(304+c)+30(2992-2c)=65,180

10c+6080+20c+89760-60c=65180

-30c+95840=65180

-30c+95840-95840=65180-95840

-30c=-30660

c=1022

This means 1,022 tickets purchased were $10 tickets. We substitute this equation back into m=304+c to find m.

m=304+1022

m=1326

This means 1,326 were $20 tickets.

Lastly we know 948 VIP tickets were bought because 1022+1326+948=3296


7 0
3 years ago
The distance time graph shows information about part of a bicycle journey. Use the graph to estimate the speed in m/s at time 6
svetlana [45]

Answer:

2.16

Step-by-step explanation:

if x is seconds on the graph and y is the distance then you just look for what is y in the coordinate (6, _ ), in which the line on the graph shows it is 13

hope this helps!!

edit:

it seems i was incorrect, ahh im sorry

so you do have to divide the answer in the end, so 13/6 which is equaled to 2.16

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Sphinxa [80]

Answer:

It would be a decimal not a remainder :)

Step-by-step explanation:

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Lee can row ⅞ of a mile in ½ of an hour. What is the unit rate that describes the miles per hour?
MatroZZZ [7]

Answer:

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Step-by-step explanation:

To find miles per hour, divide miles by hours:

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28 Jess bought a new car for £12 000.
devlian [24]

The value of the car after three years is £6936

<h3>How to determine the value of the car?</h3>

The initial value of the car is given as:

Value =  £12000

When the value depreciates by 20% after the first year, the new value of the car is:

New =  £12000 * (1 - 20%)

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When the value depreciates by 15% after the second year, the new value of the car is:

New =  £9600* (1 - 15%)

Evaluate

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When the value depreciates by 15% after the third year, the new value of the car is:

New =  £8160 * (1 - 15%)

Evaluate

New =  £6936

Hence, the value of the car after three years is £6936

Read more about depreciation at:

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