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1 year ago

Solve the equation using square roots. 6x2 = 54

Mathematics

2 answers:

Aneli [31]1 year ago

7 0

Answer:

The solution of the equation is $x=\pm 3$

Step-by-step explanation:

Given : Equation $6x^2=54$

To find : Solve the equation using square roots?

Solution :

The given equation is $6x^2=54$

Step 1 - First we divide the equation by 6

$\frac{6x^2}{6}=\frac{54}{6}$

$x^2=9$

Step 2 - Taking square root both side,

$\sqrt{x^2}=\sqrt{9}$

$x=\pm 3$

Therefore, The solution of the equation is $x=\pm 3$

Mars2501 [29]1 year ago

5 0

6x^2 = 54

6(x^2) = 54

6(x^2)/6 = 54/6

x^2 = 9

rt(x^2) = rt(9)

x = 3

hope this helps

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Option D:

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2 years ago

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Answer:

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2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

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$j'(t)=95-18t$

and we set the first derivative equal to zero so we get:

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and solve for t

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$j(t)=95t-9t^{2}$

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