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3 years ago

5

Hudson is already 40 miles away from home on his drive back to college. He is driving 65 mi/hr. Write an equation that models th

e total distance d travelled after h hours.

1 answer:

sergejj [24]3 years ago

6 0

He is 40 mins away form home because he is going 65 mi/hr = 1 min per mile

hopefully that helps you make an equation

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What is the product of -13/8 and 24/5

IgorC [24]

Answer:

-7 4/5 or -7.8

Step-by-step explanation:

Product means multiplication

( - 13/8 )( 24/5)

= (-13 · 24)/(8 · 5) negative · positive = negative

= - 312/40

= -7 32/ 40 = -7 4/5 or 7.8

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3 years ago

The value of polynomial 5x-4x^2+3, when x=-1<br>

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Step-by-step explanation:

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2 years ago

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2 years ago

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3 years ago

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.

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

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1 year ago