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Taya2010 [7]
2 years ago
12

Peter invests R4000 for 6years at an interest rate of 16% per annum compounded monthly.calculate the value of the investment at

the end of the six year period
Mathematics
1 answer:
AlexFokin [52]2 years ago
7 0

Answer:

principle \:  \times rate \times time \div 100\%

4000×16%×6÷100%=3920

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Evaluate the iterated integral 2 0 2 x sin(y2) dy dx. SOLUTION If we try to evaluate the integral as it stands, we are faced wit
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Answer:

Step-by-step explanation:

Given that:

\int^2_0 \int^2_x \ sin (y^2) \ dy dx \\ \\ \text{Using backward equation; we have:} \\ \\  \int^2_0\int^2_0 sin(y^2) \ dy \ dx = \int \int_o \ sin(y^2) \ dA \\ \\  where; \\ \\  D= \Big\{ (x,y) | }0 \le x \le 2, x \le y \le 2 \Big\}

\text{Sketching this region; the alternative description of D is:} \\ D= \Big\{ (x,y) | }0 \le y \le 2, 0 \le x \le y \Big\}

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\int^2_0 \int^2_0 \ sin (y^2) dy dx = \int \int _o \ sin (y^2) \ dA  \\ \\ = \int^2_o \int^y_o \ sin (y^2) \ dx \ dy \\ \\ = \int^2_o \Big [x sin (y^2) \Big] ^{x=y}_{x=o} \ dy  \\ \\=  \int^2_0 ( y -0) \ sin (y^2) \ dy  \\ \\ = \int^2_0 y \ sin (y^2) \ dy  \\ \\  y^2 = U \\ \\  2y \ dy = du  \\ \\ = \dfrac{1}{2} \int ^2 _ 0 \ sin (U) \ du  \\ \\ = - \dfrac{1}{2} \Big [cos  \ U \Big]^2_o \\ \\ =  - \dfrac{1}{2} \Big [cos  \ (y^2)  \Big]^2_o  \\ \\ =  - \dfrac{1}{2} cos  (4) + \dfrac{1}{2} cos (0) \\ \\

=  - \dfrac{1}{2} cos  (4) + \dfrac{1}{2} (1) \\ \\  = \dfrac{1}{2}\Big [1- cos (4) \Big] \\ \\  = \mathbf{0.82682}

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Dividing the </span><span>∆x, we get 
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chur answer would be X=1

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