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3 years ago

Helppppppppppppppppppppp

Mathematics

2 answers:

katrin2010 [14]3 years ago

8 0

Answer:

Step-by-step explanation:

You use subtraction property of equality to get rid of that 4.

Subtract 4 from both sides basically.

You get 2x > 6

Then you divide both sides by 2 to get x > 3

Then you graph it.

So it starts on point 3, and since x is greater than 3, it goes to points greater than 3, so 4, 5, 6, ect. but it never goes below or on 3.

The graph starts on three, as 3.00000000000000001 is a viable solution, but since it says > instead of the underlined >, it is an open circle like o---->

So it is E.

babymother [125]3 years ago

5 0

Answer: its B

Step-by-step explanation:

because the circle is open and its the greater than sign

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

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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