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3 years ago

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Q28. In a sale, the normal price of a book is reduced by 30%. The sale price of the book is £2.80 Work out the normal price of t

he book.

1 answer:

lisov135 [29]3 years ago

8 0

Answer:

0.90

Step-by-step explanation:

090. = 0.90

30/2.80

28

00

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Forces of 9 lbs and 13 lbs act at a 38º angle to each other. Find the magnitude of the resultant force and the angle that the re

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Answer: $R=20.84\ lb\quad 22.57^{\circ},15.43^{\circ}$

Step-by-step explanation:

Given

Two forces of 9 and 13 lbs acts $38^{\circ}$ angle to each other

The resultant of the two forces is given by

$\Rightarrow R=\sqrt{a^2+b^2+2ab\cos \theta}$

Insert the values

$\Rightarrow R=\sqrt{9^2+13^2+2(9)(13)\cos 38^{\circ}}\\\Rightarrow R=\sqrt{81+169+184.394}\\\Rightarrow R=\sqrt{434.394}\\\Rightarrow R=20.84\ lb$

Resultant makes an angle of

$\Rightarrow \alpha=\tan^{-1}\left( \dfrac{b\sin \theta}{a+b\cos \theta}\right)\\\\\text{Considering 9 lb force along the x-axis}\\\\\Rightarrow \alpha =\tan^{-1}\left( \dfrac{13\sin 38^{\circ}}{9+13\cos 38^{\circ}}\right)\\\\\Rightarrow \alpha =\tan^{-1}(\dfrac{8}{19.244})\\\\\Rightarrow \alpha=22.57^{\circ}$

So, the resultant makes an angle of $22.57^{\circ}$ with 9 lb force

Angle made with 13 lb force is $38^{\circ}-22.57^{\circ}=15.43^{\circ}$

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A game lasts 5 hours. Ali played 4 of these games,

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Compute the second partial derivatives ∂2f ∂x2 , ∂2f ∂x ∂y , ∂2f ∂y ∂x , ∂2f ∂y2 for the following function. f(x, y) = 2xy (x2 +

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Answer with step-by-step explanation:

We are given that a function

$f(x,y)=2xy(x^2+y^2)^2$

Differentiate partially w.r.t x

Then, we get

$\frac{\delta f}{\delta x}=2y(x^2+y^2)^2+8x^2y(x^2+y^2)=(x^2+y^2)(2x^2y+2y^3+8x^2y)=2(5x^2y+y^3)(x^2+y^2)$

Differentiate again w.r.t x

$\frac{\delta^2f}{\delta x^2}=2(10xy)(x^2+y^2)+4x(5x^2y+y^3)=20x^3y+20xy^3+20x^3y+4xy^3=40x^3y+24xy^3$

Differentiate function w.r.t y

$\frac{\delta f}{\delta y}=2x(x^2+y^2)^2+2xy\times 2(x^2+y^2)\times 2y$

$\frac{\delta f}{\delta y}=(x^2+y^2)(2x^3+2xy^2+8xy^2)=2(x^2+y^2)(x^3+5xy^2)$

Again differentiate w.r.t y

$\frac{\delta^2f}{\delta x^2}=2(2y)(x^3+5xy^2)+20xy(x^2+y^2)=4x^3y+20xy^3+20x^3y+20xy^3=24x^3y+40xy^3$

Differentiate partially w.r.t y

$\frac{\delta^2f}{\delta y\delta x}=2(2y(5x^2y+y^3)+(x^2+y^2)(5x^2+3y^2))=10x^4+36x^2y^2+10y^4$

$\frac{\delta^2f}{\delta y\delta x}=10x^4+36x^2y^2+10y^4$ $\frac{\delta^2f}{\delta x\delat y}=2(2x(x^3+5xy^2)+(3x^2+5y^2)(x^2+y^2))=10x^4+36x^2y^2+10y^4$

$\frac{\delta^2f}{\delta x\delat y}=10x^4+36x^2y^2+10y^4$

Hence, if f(x,y) is of class $C^2$ (is twice continuously differentiable), then the mixed partial derivatives are equal.

i.e $\frac{\delta^2f}{\delta y\delta x}=\frac{\delta^2f}{\delta x\delta y}$

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The domain is {3, -1, 0, 3}.

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