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lorasvet [3.4K]
3 years ago
14

Pls help me!!!! I'm stuck on how to solve this!!??

Mathematics
1 answer:
Luda [366]3 years ago
6 0
Let A represent the value of the car after each year.
A= initial value (P)×(1+percent increase(r)) ^time
A=P×(1+r)^t
A=18710×(1+(-12%))^8
A=18710×(1-12%)^8
A=18710×(1-0.12)^8
A=18710×(0.88)^8
A= 6728.7619591115
The best approximation is 6729
Therefore the value of the car will be about $6729 after 8 years
Your answer is B.
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Step-by-step explanation:

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A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par
lyudmila [28]

Answer:

a(\frac{1}{5e})=5e

Step-by-step explanation:

we are given equation for position function as

s(t)=tln(5t)

Since, we have to find acceleration

For finding acceleration , we will find second derivative

s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)

=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t

=1\cdot \ln \left(5t\right)+\frac{1}{t}t

s'(t)=\ln \left(5t\right)+1

now, we can find derivative again

s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)

=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)

=\frac{1}{t}+0

a(t)=\frac{1}{t}

Firstly, we will set velocity =0

and then we can solve for t

v(t)=s'(t)=\ln \left(5t\right)+1=0

we get

t=\frac{1}{5e}

now, we can plug that into acceleration

and we get

a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}

a(\frac{1}{5e})=5e


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