if the original size of an image is 540 x 680 which of the following image sizes cannot be achieved without disturbing the ratio​

Answer:

ummm what ?

Explanation:

Answer:

2520

Step-by-step explanation:

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

To find: The smallest number that can be divided by each of the numbers from 1 to 20 without any remainder.

• We just need to take L.C.M. the smallest number that can be divided by each of the numbers from 1 to 20 without any remainder.

• The Least Common Multiple (LCM) is also referred to as the Lowest Common Multiple (LCM) and Least Common Divisor (LCD). For two integers a and b, denoted LCM(a,b), the LCM is the smallest positive integer that is evenly divisible by both a and b. For example, LCM (2,3) = 6 and LCM (6,10) = 30.

• The LCM of two or more numbers is the smallest number that is evenly divisible by all numbers in the set.

Let's do this in listing factors method.

Write down factoriaztion of each number:

1 = 1

2 = 2

3 = 3

4 = 2 * 2

5 = 5

6 = 2 * 3

7 = 7

8 = 2 * 2 * 2

9 = 3 * 3

10 = 2 * 5

11 = 11

12 = 3 * 2 * 2

13 = 13

14 = 2 * 7

15 = 3 * 5

16 = 2 * 2 * 2 * 2

17 = 17

18 = 2 * 3 * 3

19 = 19

20 = 2 * 2 * 5

Write the primes vertically. (As in attachment 1)

Bring down numbers in each column and multiply to get LCM. (As in attachment 2)

Answer:

Complete step-by-step answer:

Arithmetic mean of numbers 2, 4, 6, 8 and 10 is 6.

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The arithmetic mean of 2 , 4 , 6 , 8 , 10 is 6

Given :

The observations 2 , 4 , 6 , 8 , 10

To find :

The arithmetic mean of 2 , 4 , 6 , 8 , 10

Solution :

Step 1 of 2 :

Calculate sum of the observations

Here the given observations are 2 , 4 , 6 , 8 , 10

Number of observations = 5

Sum of the observations

= 2 + 4 + 6 + 8 + 10

= 30

Step 2 of 2 :

Find the arithmetic mean

The required arithmetic mean

[tex]\displaystyle \sf{ = \frac{Sum \: of \: the \: observations}{Number \: of \: observations} }[/tex]

[tex]\displaystyle \sf{ = \frac{30}{5} }[/tex]

[tex]\displaystyle \sf{ = 6}[/tex]

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Explanation:

initial velocity of particle, u = 10j m/s

acceleration of particle , a = (8i + 2j) m/s²

displacement of particle in x direction , x = 16m

(a) so, use formula x=u_xt+\frac{1}{2}a_xt^2x=u

x

t+

2

1

a

x

t

2

here, u_x=0,a_x=8m/s^2u

x

=0,a

x

=8m/s

2

so, 16 = 0 + 1/2 × 8 × t²

or, 16 = 4t² => t = 2

hence, after 2 sec particle moves 16m in x direction.

now we have to find y - co-ordinate of particle at 2 sec.

so, use formula, y=u_yt+\frac{1}{2}a_yt^2y=u

y

t+

2

1

a

y

t

2

here, u_y=10m/s^2u

y

=10m/s

2

, a_y=2m/s^2a

y

=2m/s

2

, t = 2sec

so, y = 10 × 2 + 1/2 × 2 × 2²

y = 20 + 4 = 24m

(b) velocity of particle in x direction after 2 sec, v_x=u_x+a_xtv

x

=u

x

+a

x

t

= 0 + 8i × 2 = 16i m/s

velocity of particle in y direction after 2 sec ,

v_y=u_y+a_ytv

y

=u

y

+a

y

t

= 10 j + 2j × 2= 14j m/s

hence, speed , |v| = \sqrt{v_x^2+v_y^2}

v

x

2

+v

y

2

= \sqrt{16^2+14^2}

16

2

+14

2

= 21.26m/s