HELP WILL MARL BRAINLIEST JUST HELP ME DO B

Mathematics

1 answer:

Svetllana [295]3 years ago

3 0

Answer: Stephen did not win four games because when you read the problem you see that they put he won first so the number of games he won goes before the colon and the number of games he lost goes after the colon.. Hope this answer helps... Have a great weekend and stay safe... :D

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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.

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

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