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bearhunter [10]
2 years ago
10

QUICK will give brainliest to correct answer please and thank you.​

Mathematics
1 answer:
hammer [34]2 years ago
7 0

Answer:

b

Step-by-step explanation:

because  if  you square root 30, the value is: 5.477225575, it is neither 5.4 and 5.5

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Bennett has 107 photographs to place in the school yearbook. He will put the same number of photos on each of the 13 pages. if h
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3

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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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\\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
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\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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\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
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7 0
3 years ago
At a meeting, there were 18 men and 24 women. What is the ratio of men to women at the meeting?
sergey [27]

Answer:

men = 12

women = 8

Step-by-step explanation:

COuld i have branliest and heart with 5 star rating thanks!

8 0
3 years ago
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N.p + p for n = 3 and p = 6
zheka24 [161]

Answer:

Hey there!

3.6+6

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Let me know if this helps :)

4 0
4 years ago
Can someone help me with this?
stepladder [879]
Answer: -4F and 3

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5 0
3 years ago
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