2 years ago

Derivatives of sin(logx) from first principle

Mathematics

1 answer:

Nina [5.8K]2 years ago

5 0

Answer:

cos(logx)*(1/x)

Step-by-step explanation:

(sin(logx))'=cos(logx)*logx'=cos(logx)*(1/x)

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15 inches = 1.25 feet

Ratio = 12 : 1.25

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Checking...

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In triangle $ABC,$ point $D$ is on $\overline{AC}$ such that $AD = 3CD = 12$. If $\angle ABC = \angle BDA = 90^\circ$, then what

Neporo4naja [7]

Answer:

$BD=4\sqrt{3}\ units$

Step-by-step explanation:

we know that

$AD=12\ units$

$3CD=12$ ----> $CD=4\ units$

see the attached figure to better understand the problem

Triangles ABD and BCD are similar by AA Similarity Theorem

Remember that

If two figures are similar, then the ratio of its corresponding sides is proportional

$\frac{BD}{AD}=\frac{CD}{BD}$

substitute the given values

$\frac{BD}{12}=\frac{4}{BD}$

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$BD=\sqrt{48}\ units$

simplify

$BD=4\sqrt{3}\ units$

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2 years ago

Derivatives of sin(logx) from first principle​

Derivatives of sin(logx) from first principle