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3 years ago

find the surface area of a rectangular prism that is 16 inches long ,12 inches wide, and 5 inches high.

Mathematics

1 answer:

mrs_skeptik [129]3 years ago

5 0

Length * Width * Height. 16*12*5=960.

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alexdok [17]

Answer:

$f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}....$

Approximate error = 0.4426

Step-by-step explanation:

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Maclaurin series formula used is given below

$f(x)=\sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)x^{n}}{n!}=f(0)+f'(0)x+\dfrac{f''(0)}{2!}x^{2}+\dfrac{f'''(0)}{3!}x^{3}+....$

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$f'(x)=sec^{2}x\\f'(0)=sec^{2}0=1\\f''(x)=2sec^{2}xtanx\\f''(0)=2sec^{2}0tan0=0\\f'''(x)=-4sec^{2}x+6sec^{4}x\\f'''(0)=-4sec^{2}0+6sec^{4}0=-4+6=2\\$

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$f(x)=0+x+0+\dfrac{2x^{3}}{3!}+\dfrac{16x^{4}}{4!}\\$

$f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}\\$

Hence, the Taylor series for f(x)=tanx is given by

$f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}....$

Maclaurin series upper bound error formula used is given as

R_n(x)=|f(x)-T_n(x)|

R_3(x)=|tanx-T_3(x)|

$R_3(x)=|tanx-x-\dfrac{x^{3}}{3}-\dfrac{2x^{4}}{3}|$

Plugging this value x=1

$R_3(x)=|tan(1)-1-\dfrac{1}{3}-\dfrac{2}{3}|\\$

R_3(x)=|1.5574-1-0.333-0.666|

R_3(x)=|-0.4426|=0.4426

Hence, upper bound on the error approximation

tan(1)=0.4426

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