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dexar [7]
4 years ago
8

I need help doing subtracting fractions please help me ​

Mathematics
1 answer:
forsale [732]4 years ago
3 0
I hope this helps you




1/2-3/7



7.1/2.7-2.3/2.7



7/14-6/14


1/14


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D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx
sineoko [7]

Answer:

\frac{1}{2\sqrt{5} }

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Let, \text{sin}^{-1}(\frac{x}{6}) = y

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\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})

\frac{d}{dx}\text{sin(y)}=\frac{1}{6}

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\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}

\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}

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          = \sqrt{1-(\frac{x}{6})^2}

          = \sqrt{1-(\frac{x^2}{36})}

Therefore, from equation (1),

\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

Or \frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

At x = 4,

\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}

\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}

                   =\frac{1}{6\sqrt{\frac{36-16}{36}}}

                   =\frac{1}{6\sqrt{\frac{20}{36} }}

                   =\frac{1}{\sqrt{20}}

                   =\frac{1}{2\sqrt{5}}

4 0
3 years ago
What is the solution to the equation fraction 1 over 4x = 5?
AVprozaik [17]
\displaystyle  \frac{1}{4x}=5\quad \Big|\cdot 4x\\\\1=20x\quad\Big|:20\\\\x= \frac{1}{20}
7 0
3 years ago
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