$\bf \cfrac{2}{7}(x-2)=4x\implies \cfrac{2(x-2)}{7}=4x\implies 2(x-2)=28x\implies 2x-4=28x \\\\\\ -4=26x\implies \cfrac{-4}{26}=x\implies -\cfrac{2}{13}=x$

6 0

3 years ago

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Please Help! I'm stuck on this question. Show all work!

Valentin [98]

Add all the displacement vectors together, then find the magnitude of this vector sum.

$\vec r_1=(72\,\mathrm{km})(\cos60^\circ\,\vec\imath+\sin60^\circ\,\vec\jmath)$

$\vec r_2=(48\,\mathrm{km})(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)$

$\vec r_3=(100\,\mathrm{km})(\cos150^\circ\,\vec\imath+\sin150^\circ\,vec\jmath)$

Summing these vectors gives the resultant displacement

$\vec r=\vec r_1+\vec r_2+\vec r_3\approx(-50.6\,\vec\imath+64.4\,\vec\jmath)\,\mathrm{km}$

and we have

$\|\vec r\|=\sqrt{(-50.6)^2+64.4^2}\,\mathrm{km}\approx\boxed{81.9\,\mathrm{km}}$

so the plane ends up about 81.9 km away from its starting position.

6 0

4 years ago