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Licemer1 [7]
3 years ago
15

What is 1+1 please help irgent

Mathematics
2 answers:
Free_Kalibri [48]3 years ago
7 0
1+1 is equal to 2 as far as im concern
Ugo [173]3 years ago
4 0
Either it is 2, or it is window. Take your pick, chum. Also, I do believe you need to learn how to spell "urgent". 
Happy Thanksgiving, dear. 
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2x+29 find the value of x​
NikAS [45]

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The answer would be x=13.5

Step-by-step explanation: 13.5 times 2 equals 27. 27 plus 2 equals 29.

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Please please please help me guys
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The answer is A hope this helps
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The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?
Lina20 [59]
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\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
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\\2y\ y' + \left(3y+3x\ y'\right)  ((xy)^2+2xy+1) = 0
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\\2y\ y' + \left(3y+3x\ y'\right)  ((xy)^2+2xy+1) = 0
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\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
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\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
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\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y


y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
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7 0
3 years ago
Is the answer a, b, c, or d
Sveta_85 [38]

Answer: 30

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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Hatshy [7]

We have the equation:

9x^2-12x+4=0

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a=9\qquad b=-12\qquad c=4

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This equation has one solution.

5 0
3 years ago
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