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Effectus [21]
3 years ago
9

What is the solution of the equation 5x – 1 = 45? Round your answer to the nearest ten-thousandth

Mathematics
1 answer:
trasher [3.6K]3 years ago
4 0
The correct question is
<span>What is the solution of the equation (5^x)-1=45? Round your answer to the nearest ten- thousandth

we have that
</span>(5^x)-1=45--------> (5^x)=45+1----------> (5^x)=46
<span>we apply logarithm both members
</span>ln(5^x)=ln(46)---------> x*ln(5)=ln(46)-----------> x=ln(46)/ln(5)-----------> 2.37886

the answer is x=2.3789
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Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by s(t) = -t^3+3t^2+7t +4, where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

s(t) = -t^3+3t^2+7t +4

Derivate w.r.t. time,

v(t)=s'(t) = -3t^2+6t+7

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

x=-\frac{b}{2a}

Where, a=-3, b=6 and c=7

t=-\frac{6}{2(-3)}

t=-\frac{6}{-6}

t=1

As 1 lie between interval [0,4]

Substitute t=1 in the function,

v(t)= -3(1)^2+6(1)+7

v(t)= -3+6+7

v(1)=10

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

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