Question

Let us write by calculating the amount on Rs 80000 for 2½ years at the rate of 5% compound interest per annum.​

Answer 1

Answer:

The functions of the given are : 1) PLASMA MEMBRANE or CELL MEMBRANE : ¤ It separates content of cell from its surroundings... ¤ It regulate the entry of certain solids and ions into the cell.27-Mar-2018$\orange{\bold{\underbrace{\overbrace{❥Answer᎓}}}}$

Integrate the function

$\huge\green\tt\frac{ \sqrt{tanx} }{sinxcosx}}$

⇛$\huge\tt\frac{ \sqrt{tanx} }{sinxcosx}$

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⇛$\huge\tt \frac{ \sqrt{tanx} }{sinxcosx \times \frac{cosx}{cosx}}$

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⇛$\huge\tt \frac{ \sqrt{tanx} }{sinx \times \frac{ {cos}^{2} x}{cosx}}$ ㅤ ㅤ ㅤ

⇛ $\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2} x \times \frac{sinx}{cosx} }$

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⇛$\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2}x \times tanx }$

⇛$\huge\tt {tan}^{ \frac{1}{2} - 1 } \times \frac{1}{ {cos}^{2} x}$ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x⇛(tan)$

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⇛$\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = ∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx⇛(tan)$

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$\bold\blue{☛\: Let tanx=t}$

$\bold\blue{☛ \:Differentiating \: both \: sides \: w.r.t.x}$

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⇛$\huge\tt {sec}^{2} x = \frac{dt}{dx}$

⇛$\huge\tt{dx \frac{dt}{ {sec}^{2}x }}$

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⇛$\huge\tt∴∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx$

⇛$\huge\tt ∫ {(t)}^{ - \frac{1}{2} } \times {sec}^{2} x \times \frac{dt}{ {sec}^{2}x }$

⇛$\huge\tt ∫ {t}^{ - \frac{1}{2} }$ㅤ ㅤ

⇛ $\huge\tt\frac{ {t}^{ - \frac{1}{2} + 1} }{ - \frac{1}{2} + 1 }$

⇛ $\huge\tt \frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + c = 2 {t}^{ \frac{1}{2} } + c = 2 \sqrt{t}$

⇛$\huge2 \sqrt{t} + c = 2 \sqrt{tanx}$

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Answer 2

Rate of Interest(R) = 5% p.a. First, we will calculate the compound interest for the first two years. =4,000. Amount at the end of the first year = ₹ (80,000+4,000) = ₹84,000