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3 years ago

A triangle with sides lengths of 17, 21, and 10. is this a right triangle?

Mathematics

1 answer:

USPshnik [31]3 years ago

3 0

Use Pythagorean thrm
17^2+10^2=21^2
289+100=441
389≠441
This is not a right triangle

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Answer:

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Step-by-step explanation:

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Answer:

$\int\limits {e^{2s} cos\frac{s}{4} ds$ = $\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s} ))$

Step-by-step explanation:

Step(i):-

Given that $f(s) = e^{2s} cos\frac{s}{4}$

Now integrating

$\int\limits {f(s)} \, ds = \int\limits {e^{2s} cos\frac{s}{4} ds$

By using integration formula

$\int\limits { e^{ax} cos b x dx = \frac{e^{ax} }{a^{2}+b^{2} } ( a cos b x + b sin b x )$

Step(ii):-

$\int\limits {e^{2s} cos\frac{s}{4} ds$ = $\frac{e^{2s} }{(2)^{2}+(\frac{1}{4}) ^{2} } ( 2 cos (\frac{1}{4} ) s + \frac{1}{4} sin \frac{1}{4} s ))$

= $\frac{e^{2s} }{(4+\frac{1}{16})} ( 2 cos (\frac{1}{4} ) s + \frac{1}{4} sin \frac{1}{4} s ))$

= $\frac{e^{2s} }{(\frac{65}{16} } ( \frac{8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s}{4} ))$

= $16 X\frac{e^{2s} }{65 } ( \frac{8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s}{4} ))$

= $\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s} ))$

Final answer:-

$\int\limits {e^{2s} cos\frac{s}{4} ds$ = $\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s} ))$

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