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Zielflug [23.3K]
3 years ago
11

Please help question eleven

Mathematics
1 answer:
saw5 [17]3 years ago
5 0
The answers D, the explanation is that it’s basically common sense just look for the same numbers and make it make sense :)
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Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention
Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

a=\frac{d^2s}{dt^2}

Differentiate the position vector with respect to t.

\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}

Differentiate both sides of the obtained equation with respect to t.

\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}

The initial position is obtained at t=0. Substitute t=0 in the given position function.

\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}

8 0
1 year ago
100 POINTS!!!!!! HELP
irga5000 [103]
A - 1766! I hope this helps
7 0
3 years ago
Can I get cubes and cube roots examples?
Serjik [45]

Answer:

2^3=2×2×2=8

27=3×3×3 => cuberoot(27)=3

6 0
3 years ago
Need help asap giving 20 points for right answer!!
andrey2020 [161]

Answer:

I think it's 600 I think it might be wrong

4 0
2 years ago
For what values of x is the inequality (x-4)2 > true?
Komok [63]

Answer:

do it yoself

Step-by-step explanation:

xdddddd

4 0
3 years ago
Read 2 more answers
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