In the set of consecutive integers from 15 to 30 inclusive, two integers are multiples of both 3 and 5. How many integers in this set are multiples of neither 3 nor 5?

Number of integers ,multiples of neither 3 nor 5 = 8

Step-by-step explanation:

Given:

Integers

[15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30]

Total number of integer = 16

Find:

Number of integers ,multiples of neither 3 nor 5?

Computation:

Common multiples of both 3 and 5 = 15 ,30 = 2 integers

Multiply of 3 = 18,21,24,27 = 4 integers

Multiply of 5 = 20,25 = 2 integers

Number of integers ,multiples of neither 3 nor 5 = Total number of integer - Common multiples of both 3 and 5 - Multiply of 3 - Multiply of 5

Number of integers ,multiples of neither 3 nor 5 = 16 - 2 - 4 - 2

Number of integers ,multiples of neither 3 nor 5 = 8

christopher is 272,195 year old

step-by-step explanation:

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.