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Ivahew [28]
3 years ago
10

What are these fractions in simplest form?

Mathematics
1 answer:
Anton [14]3 years ago
8 0
\frac{16y^{3}}{20y^{4}} =  \frac{4}{5y}
\frac{6xy}{16y} =  \frac{3x}{8}
\frac{abc}{10abc} =  \frac{1}{10}
\frac{mn^{2}}{pm^{5}n} =  \frac{n}{pm^{4}}
\frac{12h^{3}k}{16h^{2}k^{2}} =  \frac{3h}{4k}
\frac{8x}{10y} =  \frac{4x}{5y}
\frac{24n^{2}}{28n} =  \frac{6n}{7}
\frac{30hxy}{54kxy}  =  \frac{5h}{9k}
\frac{5jh}{15jh^{3}} =  \frac{1}{3h^{2}}
\frac{20s^{2}t^{3}}{16st^{5}} =  \frac{5s}{4t^{2}}
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you are correct it is, 4a+100b

Step-by-step explanation:

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If sin ⁡x=725, and 0 ∠ x ∠ pi/2, what is the tan (x - pi/4)
krok68 [10]

Tan(x-\frac{\pi }{4}) = \frac{-17}{31}

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

Here we have ,  sin ⁡x=7/25( given sin x = 725 which is not possible ) , 0 . Let's find tan (x - pi/4):

⇒ Tanx = \frac{sinx}{cosx}

⇒ Tanx = \frac{sinx}{\sqrt{1-(sinx)^{2}}}

⇒ Tanx = \frac{\frac{7}{25}}{\sqrt{1-(\frac{7}{25})^{2}}}

⇒ Tanx = {\frac{7}{25}}{\sqrt{(\frac{625}{625-49})^{}}}

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Now , Tan(x-\frac{\pi }{4}) = \frac{Tanx - Tan(\frac{\pi }{4} )}{1+ Tanx(Tan(\frac{\pi }{4} )}

⇒ Tan(x-\frac{\pi }{4}) = \frac{Tanx -1}{1+ Tanx(1)}

⇒ Tan(x-\frac{\pi }{4}) = \frac{\frac{7}{24}  -1}{1+\frac{7}{24} }

⇒ Tan(x-\frac{\pi }{4}) = \frac{\frac{7-24}{24} }{\frac{7+24}{24} }

⇒ Tan(x-\frac{\pi }{4}) = \frac{-17}{24} (\frac{24}{31} )

⇒ Tan(x-\frac{\pi }{4}) = \frac{-17}{31}

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4 years ago
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