What’s the answer to this algebra problem? 12x +4 = 85

Mathematics

2 answers:

Elena-2011 [213]3 years ago

7 0

Answer:

Step-by-step explanation:

12x +4 = 85

1) Subtract 4 from both sides

12x +4 = 85

-4 = -4

12x = 81

2) Divide both sides by 12

12x/12 = 81/12

It gives you: x = 6.75

Pavel [41]3 years ago

5 0

Answer: 6 3/4

Step-by-step explanation: you subtract 4 from both sides and then you will divide 12 from both sides

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$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\\\\\tan3x=\tan(2x+x)=\dfrac{\tan2x+\tan x}{1-\tan2x\tan x}=\dfrac{\tan(x+x)+\tan x}{1-\tan(x+x)\tan x}\\\\=\dfrac{\frac{\tan x+\tan x}{1-\tan x\tan x}+\tan x}{1-\frac{\tan x+\tan x}{1-\tan x\tan x}\tan x}=\dfrac{\frac{2\tan x}{1-\tan^2x}+\tan x}{1-\frac{2\tan x}{1-\tan^2x}\tan x}$

$=\left(\dfrac{2\tan x}{1-\tan^2x}+\dfrac{\tan x(1-\tan^2x)}{1-\tan^2x}\right):\left(\dfrac{1-\tan^2x}{1-\tan^2x}-\dfrac{2\tan^2x}{1-\tan^2x}\right)\\\\=\dfrac{2\tan x+\tan x-\tan^3x}{1-\tan^2x}:\dfrac{1-\tan^2x-2\tan^2x}{1-\tan^2x}\\\\=\dfrac{3\tan x-\tan^3x}{1-\tan^2x}:\dfrac{1-3\tan^2x}{1-\tan^2x}=\dfrac{3\tan x-\tan^3x}{1-\tan^2x}\cdot\dfrac{1-\tan^2x}{1-3\tan^2x}\\\\=\dfrac{3\tan x-\tan^3x}{1}\cdot\dfrac{1}{1-3\tan^2x}=\dfrac{3\tan x-\tan^3x}{1-3\tan^2x}$