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natali 33 [55]
3 years ago
10

Can some one help me please fast

Mathematics
2 answers:
frez [133]3 years ago
4 0
10.
1/2 and 3/5 have the denominators 2 and 5. The smallest number that both 2 and 5 can go into is 10.
Mariana [72]3 years ago
4 0
I belive 20 is the answer hope this helps.!!

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Write the equation of a line perpendicular to y = -5x + 1 that goes through (10. 4).
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y = 1/5x + 2

Step-by-step explanation:

y = 1/5x + b

4 = 1/5(10) + b

4 = 2 + b

2 = b

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Given s left parenthesis t right parenthesis equals 5 t squared plus 5 ts(t)=5t2+5t​, find
finlep [7]

Answer:

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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