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3 years ago

Help and explain pls and ty

Mathematics

1 answer:

agasfer [191]3 years ago

8 0

$\displaystyle\bf (3-5\sqrt{5} )-(5\sqrt{3} -3)=3+3-5\sqrt{5} -5\sqrt{3} =\boxed{6-5\sqrt{5} -5\sqrt{3} } \\\\\\\\\sqrt{5} (\sqrt{5} -2x)=\boxed{-2x\sqrt{5} +5}$

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finlep [7]

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The velocity function is $v(t)=10t+5$ .

The acceleration function is $a(t)=10$ .

When t = 44, the velocity is $v(44)=445 \:\frac{ft}{s}$ .

When t = 44, the acceleration is $a(44)=10\: \frac{ft}{s^2}$ .

Step-by-step explanation:

We know that the position function is given by

$s(t)=5t^2+5t$

Velocity is defined as the rate of change of position or the rate of displacement. If you take the derivative of the position function you get the instantaneous velocity function.

$v(t)=\frac{ds}{dt}$

Acceleration is defined as the rate of change of velocity. If you take the derivative of the instantaneous velocity function you get the instantaneous acceleration function.

$a(t)=\frac{dv}{dt}$

The instantaneous velocity function is given by

$v(t)=\frac{d}{dt} s(t)=\frac{d}{dt}(5t^2+5t)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(5t^2\right)+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}\\\\v(t)=10t+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dt}\left(t\right)=1\\\\v(t)=10t+5$

The instantaneous acceleration function is given by

$a(t)=\frac{dv}{dt} =\frac{d}{dt}(10t+5)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\a(t)=\frac{d}{dt}\left(10t\right)+\frac{d}{dt}\left(5\right)\\\\a(t)=10$

To find the velocity and acceleration when t = 44, we substitute this value into the velocity and acceleration functions

$v(44)=10(44)+5\\v(44)=445 \:\frac{ft}{s}$

$a(44)=10\: \frac{ft}{s^2}$

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4 years ago