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3 years ago

A. 23.3 cm B. 16 cm C. 8 cm

Mathematics

1 answer:

Hitman42 [59]3 years ago

6 0

B. 16 cm

Square both 20 and 12

20x20=400

12x12=144

400-144=256

find the square root of 256
and the square root is 16

so b is 16 cm

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2 years ago

Integrate <img src="https://tex.z-dn.net/?f=e%5E%7B4x%7D%5Csqrt%7B1%2Be%5E%7B2x%7D%20%7D%20dx" id="TexFormula1" title="e^{4x}\sq

AnnyKZ [126]

Answer:

$(\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C$

Step-by-step explanation:

Step(i):-

Given that the function

f(x) = $e^{4x} \sqrt{1+e^{2x} }$

Now integrating on both sides, we get

$\int\limits{f(x)} \, dx = \int\limits{e^{4x} \sqrt{1+e^{2x} } dx$

= $\int\limits{e^{2x} e^{2x} \sqrt{1+e^{2x} } dx$

Step(ii):-

Let $1 + e^{2x} = t$

$e^{2x} = t -1$

$2e^{2x}dx = d t$

$e^{2x}dx = \frac{1}{2} d t$

= $\int\limits{( \sqrt{1+e^{2x} }) e^{2x} e^{2x} dx$

= $\int\limits {\sqrt{t}(t-1)\frac{1}{2} dt }$

= $\frac{1}{2} \int\limits {\sqrt{t} (t) -\sqrt{t} ) dt }$

= $\frac{1}{2} \int\limits {(t^{\frac{1}{2} } t^{1} +t^{\frac{1}{2} } ) } \, dx$

$= \frac{1}{2} \int\limits {(t^{\frac{3}{2} } +t^{\frac{1}{2} } ) } \, dx$

= $\frac{1}{2} (\frac{t^{\frac{3}{2} +1} }{\frac{3}{2}+1 } + \frac{t^{\frac{1}{2} +1} }{\frac{1}{2}+1 } )+C$

= $\frac{1}{2} (\frac{t^{\frac{3}{2} +1} }{\frac{5}{2} } + \frac{t^{\frac{1}{2} +1} }{\frac{3}{2} } )+C$

= $\frac{1}{2} (\frac{t^{\frac{5}{2} } }{\frac{5}{2} } + \frac{t^{\frac{3}{2} } }{\frac{3}{2} } )+C$

= $(\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C$

Final answer:-

= $(\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C$

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2 years ago