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3 years ago

17 cm 12 cm 5 cm pls help

Mathematics

1 answer:

miskamm [114]3 years ago

7 0

Answer:

30 cm²

Step-by-step explanation:

A = bh/2

A = (12 cm)(5 cm)/2

A = 30 cm²

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Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

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2 years ago

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2 years ago

A lighthouse is built on an exposed reef, 4.5 miles off-shore. The shoreline is perfectly straight, and a town is located 6 mile

Alisiya [41]

Answer: 3.94 hours

Step-by-step explanation:

From the question, the following parameters are given:

Distance offshore = 4.5 miles

Distance downshore = 6 miles

Speed when running = 3.4 mph

Speed with boat = 1.8 mph

distance of boat rowing = sqrt(4.5^2 + x^2)

Where

Speed = distance/time

Time = distance/speed

Time = distance of boat rowing/1.8

distance of running = 6 - x

Time = (6-x)/3.4

total travel time

t = sqrt(4.5^2 + x^2)/1.8 + (6 - x)/3.4

dt/dx = x/(1.8×sqrt(4.5^2 + x^2) - 1/3.4

d^2t/dx^2 = +ve at any x

x is at its minimum when dt/dx=0

x/(1.8×sqrt(4.5^2 + x^2) = 1/3.4

x = 6/sqrt(253) × 4.5 = 1.697 miles

Substitute x in

t = sqrt(4.5^2 + x^2)/1.8 + (6 - x)/3.4

We obtaine

t = sqrt(4.5^2 + 1.697^2)/1.8 + (6 - 1.69)/3.4

t = 2.672+ 1.266 = 3.94 hours

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