3 years ago

Please help me answer these. Its all about upper and lower bounds.

1 answer:

5 0

Answer:

red question= 8.5 yellow question=4.25

Step-by-step explanation:

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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

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

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3 years ago

rjkz [21]

Yea I’m gonna go watch my internet and then go to go home to go to

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2 years ago

nikdorinn [45]

I think I may be wrong check

5(x^2n-1)×(2x^3n-1) ^2

=20n3 x8 -20n2 x5 +20n x3 +20n x3 +5n x2-5

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3 years ago

liberstina [14]

Answer:

Step-by-step explanation:

8,6,2, and 1

(2^6 / 8) * 1 = (64/8) * 1 = 8 * 1 = 8

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3 years ago

GuDViN [60]

A x B x C = 0 Where A and B are integers > 10 Any value multiplied by 0 = 0 Therefore C = 0

5 0

3 years ago