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Temka [501]
2 years ago
7

En ampliar o reducir una fotografía, obtenemos copias de la fotografía original, pero de medida diferente. Si la anchura del ori

ginal de una fotografía mide 15 cm, cuál será la anchura de una copia reducida un 60%?
Mathematics
1 answer:
antoniya [11.8K]2 years ago
7 0

Answer:

La anchura será de:

6 cm

Step-by-step explanation:

El formato original corresponde al 100%

Si se reduce un 60% quiere decir que:

100 - 60 = 40%

la reducción queda al 40%

entonces, por regla de tres:

15cm es al 100%

A cm es al 40%

A = 15*40/100

A = 6cm

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Find the perimeter of the figure
77julia77 [94]

Answer:

51

Step-by-step explanation:

Hey There!

To find the perimeter of any figure all you have to do is add up all of the side lengths

5+5+7+11+7+16=53

so the perimeter is equal to 51

Note: the 18 is from the bottom length of the figure

you would have to add the 5 and 11

the extra seven is from the very right side length

8 0
2 years ago
A group of 4 friends paid a total of $50.24 for tickets to a museum. Each friend paid the same amount for a ticket. Solve this a
Dmitry_Shevchenko [17]

Answer:

$12.56

Step-by-step explanation:

$50.24 divided by the 4 friends = $12.56

I divided the total cost by the amount of friends who went to the museum and so I did $50.24 divided by the 4 friends = $12.56.

6 0
2 years ago
Si elegimos un día de la semana al azar, ¿qué probabilidad hay de que sea fin de semana (ya sabemos que hay dos días…)?
irga5000 [103]

Answer:

La probabilidad que hay de que sea fin de semana es \frac{2}{7}

Step-by-step explanation:

La Probabilidad es la mayor o menor posibilidad de que ocurra un determinado suceso. En otras palabras, la probabilidad establece una relación entre el número de sucesos favorables y el número total de sucesos posibles. Entonces, la probabilidad de un suceso cualquiera A se define como el cociente entre el número de casos favorables (número de casos en los que puede ocurrir o no el suceso A) y el número total de casos posibles. Esta se denomina Ley de Laplace.

P(A)=\frac{numero de casos favorables}{numero de casos posibles}

En este caso:

  • número de casos favorables: 2 (cantidad de días del fin de semana)
  • número de casos posibles: 7 (cantidad de días totales en la semana)

Reemplazando:

P(A)= \frac{2}{7}

<u><em>La probabilidad que hay de que sea fin de semana es </em></u>\frac{2}{7}<u><em></em></u>

3 0
3 years ago
1-What is the sum of the series? ​∑j=152j​ Enter your answer in the box.
tangare [24]

Answer:

Please see the Step-by-step explanation for the answers

Step-by-step explanation:

1)

∑\left \ {{5} \atop {j=1}} \right. 2j

The sum of series from j=1 to j=5 is:

∑ = 2(1) + 2(2) + 2(3) + 2(4) + 2(5)

  =  2 + 4 + 6 + 8 + 10

∑ = 30

2)

This question is not given clearly so i assume the following series that will give you an idea how to solve this:

∑\left \ {{4} \atop {k=1}} \right. 2k²

The sum of series from k=1 to j=4 is:

∑ = 2(1)² + 2(2)² + 2(3)² + 2(4)²

  = 2(1) + 2(4) + 2(9) + 2(16)

  =  2 + 8 + 18 + 32

∑ = 60

∑\left \ {{4} \atop {k=1}} \right. (2k)²

∑ = (2*1)² + (2*2)² + (2*3)² + (2*4)²

  = (2)² + (4)² + (6)² + (8)²

  = 4 + 16 + 36 + 64

∑ = 120

∑\left \ {{4} \atop {k=1}} \right. (2k)²- 4

∑ = (2*1)²-4 + (2*2)²-4 + (2*3)²-4 + (2*4)²-4

  = (2)²-4 + (4)²-4 + (6)²-4 + (8)²-4

  = (4-4) + (16-4) + (36-4) + (64-4)

  = 0 + 12 + 32 + 60

∑ = 104

∑\left \ {{4} \atop {k=1}} \right. 2k²- 4

∑ = 2(1)²-4 + 2(2)²-4 + 2(3)²-4 + 2(4)²-4

  = 2(1)-4 + 2(4)-4 + 2(9)-4 + 2(16)-4

  = (2-4) + (8-4) + (18-4) + (32-4)

  = -2 + 4 + 14 + 28

∑ = 44

3)

∑\left \ {{6} \atop {k=3}} \right. (2k-10)

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)  

  = (6-10) + (8-10) + (10-10) + (12-10)

  = -4 + -2 + 0 + 2  

∑ = -4

4)

1+1/2+1/4+1/8+1/16+1/32+1/64

This is a geometric sequence where first term is 1 and the common ratio is 1/2 So

a = 1

This can be derived as

1/2/1 = 1/2 * 1 = 1/2

1/4/1/2 = 1/4 * 2/1 = 1/2

1/8/1/4 = 1/8 * 4/1  = 1/2

1/16/1/8 = 1/16 * 8/1  = 1/2

1/32/1/16 = 1/32 * 16/1  = 1/2

1/64/1/32 = 1/64 * 32/1  = 1/2

Hence the common ratio is r = 1/2

So n-th term is:

ar^{n-1} = 1(\frac{1}{2})^{n-1}

So the answer that represents the series in sigma notation is:

∑\left \ {{7} \atop {j=1}} \right. (\frac{1}{2})^{j-1}

5)

−3+(−1)+1+3+5

This is an arithmetic sequence where the first term is -3 and the common difference is 2. So  

a = 1

This can be derived as

-1 - (-3) = -1 + 3 = 2

1 - (-1) = 1 + 1 = 2

3 - 1 = 2

5 - 3 = 2

Hence the common difference d = 2

The nth term is:

a + (n - 1) d

= -3 + (n−1)2

= -3 + 2(n−1)

= -3 + 2n - 2

= 2n - 5

So the answer that represents the series in sigma notation is:

∑\left \ {{5} \atop {j=1}} \right. (2j−5)

6 0
3 years ago
Find 2 numbers a and b whose sum is 4 and whose difference is -2​
umka21 [38]

Answer:

a = 1, b = 3.

Step-by-step explanation:

a + b = 4

a - b = -2     Adding to eliminate b:

2a = 2

a = 1

So 1 + b = 4

giving b = 3.

7 0
3 years ago
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