What percent of the second ten natural numbers are prime numbers

Mathematics

2 answers:

Eddi Din [679]3 years ago

8 0

0 based on what I looked up

alexira [117]3 years ago

7 0

40 PERCENT of the second 10 natural prime numbers are prime numbers

You might be interested in

Easy Question pls help. (Algebra II type problems)

Roman55 [17]

Answer:

$\large\boxed{if\ x\neq-3\ \wedge\ y\neq0\ \wedge\ x\neq y\ \wedge\ x\neq -y}$

second

$\large\boxed{y\neq0\ \wedge\ x\neq-y\ (y\neq-x)}$

Step-by-step explanation:

We know that dividing by 0 is impossible.

Therefore, the denominator of the expression must be different from 0.

$\text{For}\\\\\dfrac{(x-y)^2}{2xy+6y}\times\dfrac{4x+12}{x^2-y^2}\\\\\\2xy+6y\neq0\qquad(1)\\\\x^2-y^2\neq0\qquad(2)$

$(1)\\2xy+6y\neq0\qquad\text{distribute}\\\\2y(x+3)\neq0\iff2y\neq0\ \wedge\ x+3\neq0\\\\2y\neq0\qquad\text{divide both sides by 2}\\\boxed{y\neq0}\\\\x+3\neq0\qquad\text{subtract 3 from both sides}\\\boxed{x\neq-3}\\==========================\\(2)\\x^2-y^2\neq0\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\(x-y)(x+y)\neq0\iff x-y\neq0\ \wedge\ x+y\neq0\\\boxed{x\neq y}\ \wedge\ \boxed{x\neq-y}$