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3 years ago

jennifer bikes 7 miles south and then turns to bike 13 miles east how far away is she from where she started?

Mathematics

1 answer:

Marina CMI [18]3 years ago

3 0

Answer:

She is about 14.765 miles ( $\sqrt{218}$ miles) from where she started

Step-by-step explanation:

There is a relation between the three sides of the right triangle

The side opposite to the right angle is called hypotenuse and it is the longest side

The other two sides called legs of the right angle

The relation between them is: (hypotenuse)² = (leg1)² + (leg2)²

∵ Jennifer bikes 7 miles south

∵ She turns to bike 13 miles east

∵ South and East are perpendicular

→ That means the distance from her start point to end point represents

a hypotenuse of a right triangle, whose legs are 7 and 13

∴ (hypotenuse)² = (leg1)² + (leg2)², where

hypotenuse is the distance between her start and end points

leg1 is her distance in south direction

leg2 is her distance in east direction

∵ Leg1 = 7 miles

∵ leg 2 = 13 miles

∴ (hypotenuse)² = (7)² + (13)²

∴ (hypotenuse)² = 49 + 169

∴ (hypotenuse)² = 218

→ Take √ for both sides

∴ hypotenuse = $\sqrt{218}$

∴ hypotenuse ≅ 14.76482306

∴ She is about 14.765 miles ( $\sqrt{218}$ miles) from where she started.

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Answer: The required answer is

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Step-by-step explanation: We are given to find the inverse Laplace transform of the following function as a function of x :

$F(s)=\dfrac{2s^2}{(s-1)(s^2+1)}.$

We will be using the following formulas of inverse Laplace transform :

$(i)~L^{-1}\{\dfrac{1}{s-a}\}=e^{ax},\\\\\\(ii)~L^{-1}\{\dfrac{s}{s^2+a^2}\}=\cos ax,\\\\\\(iii)~L^{-1}\{\dfrac{1}{s^2+a^2}\}=\dfrac{1}{a}\sin ax.$

By partial fractions, we have

$\dfrac{s^2}{(s-1)(s^2+1)}=\dfrac{A}{s-1}+\dfrac{Bs+C}{s^2+1},$

where A, B and C are constants.

Multiplying both sides of the above equation by the denominator of the left hand side, we get

$2s^2=A(s^2+1)+(Bs+C)(s-1).$

If s = 1, we get

$2\times 1=A(1+1)\\\\\Rightarrow A=1.$

Also,

$2s^2=A(s^2+1)+(Bs^2-Bs+Cs-C)\\\\\Rightarrow 2s^2=(A+B)s^2+(-B+C)s+(A-C).$

Comparing the coefficients of x² and 1, we get

$A+B=2\\\\\Rightarrow B=2-1=1,\\\\\\A-C=0\\\\\Rightarrow C=A=1.$

So, we can write

$\dfrac{2s^2}{(s-1)(s^2+1)}=\dfrac{1}{s-1}+\dfrac{s+1}{s^2+1}\\\\\\\Rightarrow \dfrac{2s^2}{(s-1)(s^2+1)}=\dfrac{1}{s-1}+\dfrac{s}{s^2+1}+\dfrac{1}{s^2+1}.$

Taking inverse Laplace transform on both sides of the above, we get

$L^{-1}\{\dfrac{2s^2}{(s-1)(s^2+1)}\}=L^{-1}\{\dfrac{1}{s-1}\}+L^{-1}\{\dfrac{s}{s^2+1}+\dfrac{1}{s^2+1}\}\\\\\\\Rightarrow f(x)=e^{1\times x}+\cos (1\times x)+\dfrac{1}{1}\sin(1\times x)\\\\\\\Rightarrow f(x)=e^x+\cos x+\sin x.$