7) The area of rectangle A is twice the area of square B. Find x. 12

Mathematics

1 answer:

Veronika [31]4 years ago

6 0

Answer:

Step-by-step explanation:

A=2B

AreaA= Area of a rectangle=Base* height = base (12-x)*height(4) = 48-4x

AreaB = Area of rectangle=base(x) *Height(x) =x^2

Since A=2B

48-4x =2x^2

Divide both sides of the equation by 2

24-2x=x^2

Put everything all on one side

X^2-2x-24=0

(x+4)(x-6)=0

X=-4 and x=6

Only x makes sense since it’s not negative or else your base or one of your sides would be negative

So x=6

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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

fiasKO [112]

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

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