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morpeh [17]
3 years ago
11

A car is moving along Highway 138 according to the given equation, where x meters is the directed distance of the car from a giv

en point P at t hours. Find the values of t for which the acceleration is zero, and then find the position of the car at this time.
x = 1/4t^3 + 1/6t^3 - t^2+ 1
Mathematics
1 answer:
loris [4]3 years ago
8 0

Answer:

1111¹5&-78+&%%567;!9098888

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Help!!!!!!!!!!!!!!!!!!
Mrrafil [7]

Answer:

d. a f

Step-by-step explanation:

i hope this helps :)

5 0
3 years ago
Read 2 more answers
20. Solve 3(5x − 4) &lt; 15x.<br> All Real Numbers<br> x&gt;-4 <br>x&lt;4<br>no solution ​
Natasha2012 [34]

Answer:

No solution

Step-by-step explanation:

3(5x - 4) < 15x \\ 15x - 12 < 15x \\ 15x - 15x < 12 \\ 0 < 12

8 0
2 years ago
Anyone help with my homework plz?
iren2701 [21]
Hi!

This is definitely a trapezoid, as it has two equal sides and two inequal sides. It's not a parallelogram, because two sides eventually intersect. It is a quadrilateral, because it has four sides.

None of the others apply here.

If you found this especially helpful, I'd appreciate if you'd vote me Brainliest for your answer. I want to be able to assist more users one-on-one, as well as to move up in rank! :)
3 0
3 years ago
Solve:<br> 22,440=6,200e^0.06T<br><br> 22,440=6,200e^0.1T
Tomtit [17]

Answer:

22440=6200e^{0.06T}

\implies \dfrac{561}{155}=e^{0.06T}

\implies \ln\dfrac{561}{155}=\ln e^{0.06T}

\implies \ln\dfrac{561}{155}=0.06T

\implies T=\dfrac{50}{3}\ln\dfrac{561}{155}

\implies T=21.43826314...

22440=6200e^{0.1T}

\implies \dfrac{561}{155}=e^{0.1T}

\implies \ln\dfrac{561}{155}=\ln e^{0.1T}

\implies \ln\dfrac{561}{155}=0.1T

\implies T=10\ln\dfrac{561}{155}

\implies T=12.86295789...

3 0
2 years ago
Solve for v<br><br> 3v+9-8v= -31<br> Simplify your answer as much as possible.
balandron [24]
Let's solve your equation step-by-step.
<span><span><span><span>3v</span>+9</span>−<span>8v</span></span>=<span>−31
</span></span>Step 1: Simplify both sides of the equation.
<span><span><span>−<span>5v</span></span>+9</span>=<span>−31
</span></span>Step 2: Subtract 9 from both sides.
<span><span><span><span>−<span>5v</span></span>+9</span>−9</span>=<span><span>−31</span>−9</span></span><span><span>−<span>5v</span></span>=<span>−40
</span></span>Step 3: Divide both sides by -5.
<span><span><span>−<span>5v</span></span><span>−5</span></span>=<span><span>−40</span><span>−5
</span></span></span><span>v=<span>8</span></span>
6 0
3 years ago
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