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labwork [276]
3 years ago
8

The standard unit of mass is the : A.gram B.kilogram C.pound D.cubic centimeter

Mathematics
1 answer:
arlik [135]3 years ago
5 0

Answer:

kilogram

Step-by-step explanation:

   

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Mr. Jones bought a cloth of length 3890 cm. How much is the length in m and cm?
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Answer:

38m and 90cm

1m = 100cm

3890cm = 38m and 90cm

Step-by-step explanation:

1m = 100cm

3890cm = 38m and 90cm

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The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?
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\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
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\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
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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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\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right)  (xy+1)^2 = 0
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\\2y\ y' + \left(3y+3x\ y'\right)  (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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\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
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