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Scorpion4ik [409]
3 years ago
14

Please help it’s urgent

Mathematics
1 answer:
katrin2010 [14]3 years ago
5 0

\bold{\text{Answer:}\quad \dfrac{-48x^4-42x^3-15x^2-5x}{(8x+7)(3x+1)}}

<u>Step-by-step explanation:</u>

.\quad \dfrac{-5x}{8x+7}-\dfrac{6x^3}{3x+1}\\\\\\=\dfrac{-5x}{8x+7}+\dfrac{-6x^3}{3x+1}\\\\\\=\dfrac{-5x}{8x+7}\bigg(\dfrac{3x+1}{3x+1}\bigg)+\dfrac{-6x^3}{3x+1}\bigg(\dfrac{8x+7}{8x+7}\bigg)\\\\\\=\dfrac{-15x^2-5x}{(8x+7)(3x+1)}+\dfrac{-48x^4-42x^3}{(8x+7)(3x+1)}\\\\\\=\large\boxed{\dfrac{-48x^4-42x^3-15x^2-5x}{(8x+7)(3x+1)}}

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See attached image. For part b, my answer is <img src="https://tex.z-dn.net/?f=h%28t%29%3D%5Cfrac%7Bt%5E2%7D%7B100%7D-%5Cfrac%7B
ArbitrLikvidat [17]

Answer:

a) shown

b) h = [sqrt(17) - (5/2)t]²

c) t = 2sqrt(17)/5 seconds

Step-by-step explanation:

V = pi × r² × h

V = pi × 5² × h

V = 25pi × h

a) dV/dt = dV/dh × dh/dt

-5pi × sqrt(h) = 25pi × dh/dt

dh/dt = -sqrt(h)/5

b) 1/sqrt(h) .dh = -5. dt

2sqrt(h) = -5t + c

t = 0, h = 17

2sqrt(17) = 0 + c

c = 2sqrt(17)

2sqrt(h) = -5t + 2sqrt(17)

sqrt(h) = [2sqrt(17) - 5t] ÷ 2

sqrt(h) = sqrt(17) - (5/2)t

Square both sides

h = [sqrt(17) - (5/2)t]²

c) empty: h = 0

0 = [sqrt(17) - (5/2)t]²

sqrt(17) - (5/2)t = 0

(5/2)t = sqrt(17)

t = 2sqrt(17)/5

t = 1.64924225 seconds

sqrt: square root

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