A B C or D you can only choose one

Mathematics

2 answers:

Vlada [557]3 years ago

4 0

Answer: first one

Step-by-step explanation:

Aneli [31]3 years ago

4 0

Answer:

Step-by-step explanation:

I used my calculator to convert the percent to a mixed number

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Pls help and show your work/explain if you can

Mumz [18]

B: 2? because there are only 2 yellow marbles

3 0

3 years ago

Read 2 more answers

Solve this rational equation

tino4ka555 [31]

Answer:

Step-by-step explanation:

$\text{Domain:}\\\\x-4\neq0\ \wedge\ x-2\neq0\ \wedge\ x^2-6x+8\neq0\\\\x\neq4\ \wedge\ x\neq2\ \wedge\ x^2-4x-2x+8\neq0\\\\x\neq4\ \wedge\ x\neq2\ \wedge\ x(x-4)-2(x-4)\neq0\\\\x\neq4\ \wedge\ x\neq2\ \wedge\ (x-4)(x-2)\neq0\\\\x\neq4\ \wedge\ x\neq2\ \wedge\ x-4\neq0\ \wedge\ x-2\neq0\\\\\boxed{D:x\neq4\ \wedge\ x\neq2}$

$\dfrac{1}{x-4}+\dfrac{x}{x-2}=\dfrac{2}{x^2-6x+8}\\\\\dfrac{1(x-2)}{(x-4)(x-2)}+\dfrac{x(x-4)}{(x-4)(x-2)}=\dfrac{2}{(x-4)(x-2)}\\\\\dfrac{x-2+x(x-4)}{(x-4)(x-2)}=\dfrac{2}{(x-4)(x-2)}\iff x-2+x(x-4)=2\\\\x-2+(x)(x)+(x)(-4)=2\\\\x-2+x^2-4x=2\qquad\text{subtract 2 from both sides}\\\\x^2-3x-4=0\\\\x^2-4x+x-4=0\\\\x(x-4)+1(x-4)=0\\\\(x-4)(x+1)=0\iff x-4=0\ \wedge\ x+1=0\\\\x-4=0\qquad\text{add 4 to both sides}\\x=4\notin D\\\\x+1=0\qquad\text{subtract 1 from both sides}\\x=-1\in D$