Can someone please help me

Mathematics

1 answer:

laiz [17]3 years ago

8 0

Answer:

(a)0,2 1,-1 2,-4 3,-7 (b) yes

You might be interested in

I need help with this

Contact [7]

7/10= 28/40

5/8=25/40

0.8=4/5=32/40

32>28>25

So 0.8>7/10>5/8

8 0

3 years ago

Find the position vector R(t) and velocity vector V(t), given the acceleration A(t) and initial position and velocity vectors R(

valentina_108 [34]

The fundamental theorem of calculus tells us that

$\vec v(t)=\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du$

$\vec r(t)=\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du$

So we have

$\vec v(t)=(\vec\imath-\vec\jmath-2\,\vec k)+\displaystyle\int_0^t(u^2\,\vec\imath-2u^{1/2}\,\vec\jmath+e^{3u}\,\vec k)\,\mathrm du$

$\vec v(t)=(\vec\imath-\vec\jmath-2\,\vec k)+\left(\dfrac{u^3}3\,\vec\imath-\dfrac{4u^{3/2}}3\,\vec\jmath+\dfrac{e^{3u}}3\,\vec k\right)\bigg|_0^t$

$\vec v(t)=\dfrac{t^3+3}3\,\vec\imath-\dfrac{4t^{3/2}+3}3\,\vec\jmath+\dfrac{e^{3t}-7}3\,\vec k$

and

$\vec r(t)=(2\,\vec\imath+\vec\jmath-\vec k)+\displaystyle\int_0^t\left(\frac{u^3+3}3\,\vec\imath-\frac{4u^{3/2}+3}3\,\vec\jmath+\frac{e^{3u}-7}3\right)\,\mathrm du$

$\vec r(t)=(2\,\vec\imath+\vec\jmath-\vec k)+\left(\dfrac{u^3+12u}{12}\,\vec\imath-\dfrac{8u^{5/2}+15u}{15}\,\vec\jmath+\dfrac{e^{3u}-21u}9\,\vec k\right)\bigg|_0^t$