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iVinArrow [24]
3 years ago
6

Approximate: log 5 ^4/7

Mathematics
2 answers:
wolverine [178]3 years ago
6 0

Do you mean the following:

log(5^(4/7))=around 0.4

log(5)^(4/7)=around 0.815

log(5^4)/7= same thing as top one

log((5^4)/7)=around 1.95

vagabundo [1.1K]3 years ago
5 0

Answer:

–0.3477

Step-by-step explanation:

just took it

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8 0
3 years ago
Please help it’s urgent
katrin2010 [14]

\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)}}

5 0
3 years ago
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