2 years ago

HELP PLSSSSSSS ILL GIVE YOU BRAINLIST

1 answer:

5 0

Answer:

ewdhyredhkiydg

Step-by-step explanation:

ggdsdhudgjjthfd

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Ilia_Sergeevich [38]

a. The $\vec k$ component tells you the particle's height:

$3t=16\implies t=\dfrac{16}3$

b. The particle's velocity is obtained by differentiating its position function:

$\vec v(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=-\sin t\,\vec\imath+\cos t\,\vec\jmath+3\,\vec k$

so that its velocity at time $t=\frac{16}3$ is

$\vec v\left(\dfrac{16}3\right)=-\sin\dfrac{16}3\,\vec\imath+\cos\dfrac{16}3\,\vec\jmath+3\,\vec k$

c. The tangent to $\vec r(t)$ at $t=\frac{16}3$ is

$\vec T(t)=\vec r\left(\dfrac{16}3\right)+\vec v\left(\dfrac{16}3\right)t$

4 0

2 years ago

Andrej [43]

40º

7) In this problem, we can see that both tangent lines to that circle come from the same point O.

So, we can write out the following considering that there is one secant line DO and one tangent line to the circle AO

$\begin{gathered} m\angle1=\frac{1}{2}(160-80) \\ m\angle1=\frac{1}{2}(80) \\ m\angle1=40^{\circ} \end{gathered}$

3 0

1 year ago

Alla [95]

1.6m − 4.8 = −1.6m
- 1.6m -1.6m
 -4.8 = -3.2m

divide both sides by -3.2

m = 1.5

7 0

2 years ago

olganol [36]

Answer:

C) 35

Step-by-step explanation:

3 0

2 years ago

Afina-wow [57]

We need to see the picture of the problem

8 0

3 years ago