Consider the graph of the quadratic function y = 3x2 – 3x

All of these ODEs model a system with a spring, mass and dashpot.

Wewaii [24]

Answer:

− 3 y ' ' − 3 y ' + 3 y = 0 : over-damped

− 2 y ' ' − 4 y ' + 1 y = 0 : over-damped

1 y ' ' + 7 y ' + 5 y = 0: over-damped

Step-by-step explanation:

Using the characteristic equation you can express a differential equation of order n as an algebraic equation of degree n:

$a_ny^n+a_n_-_1y^{n-1}+...+a_1y'+a_oy=0$

This differential equation will have a characteristic equation of the form:

$a_nr^n+a_n_-_1r^{n-1}+...+a_1r+a_o=0$

Now, you can classify the solution for a differential equation using a simple method. In order to do it, you just need to use the discriminant.

If the discriminant is greater than zero, the solution is over-damped

If the discriminant is less than zero, the solution is under-damped

If the discriminant is equal to zero, the solution is critically damped

So, given the differential equation:

$-3y''-3y+3y=0$

Which has characteristic equation of the form:

$-3r^2-3r+3=0$

The quadratic polynomial of the form:

$ar^2+br+c=0$

Has discriminant:

$Disc=b^2-4ac$

In this case:

$a=-3\\b=-3\\c=3$

So:

$Disc=(-3)^2-4(-3)(3)=9-(-36)=45$

In this case:

$Disc=45>0$

Therefore the solution is over-damped.

Now, given the differential equation:

$-2y''-4y'+1y=0$

Which has characteristic equation of the form:

$-2r^2-4r+1=0$

The quadratic polynomial of the form:

$ar^2+br+c=0$

Has discriminant:

$Disc=b^2-4ac$

In this case:

$a=-2\\b=-4\\c=1$

So:

$Disc=(-4)^2-4(-2)(1)=16+8=24$

In this case:

$Disc=24>0$

Therefore the solution is over-damped.

Finally, given the differential equation:

$1y''+7y'+5y=0$

Which has characteristic equation of the form:

$1r^2+7r+5=0$

The quadratic polynomial of the form:

$ar^2+br+c=0$

Has discriminant:

$Disc=b^2-4ac$

In this case:

$a=1\\b=7\\c=5$

So:

$Disc=(7)^2-4(1)(5)=49-20=29$

In this case:

$Disc=29>0$

Therefore the solution is over-damped.

6 0

3 years ago

On Martin's first stroke, his golf ball traveled 4/5 of the distance to the hole. On his second stroke, the ball traveled 79 met

Sav [38]

Answer:

0.395 kilometre

Step-by-step explanation:

Given:

On Martin's first stroke, his golf ball traveled 4/5 of the distance to the hole.

On his second stroke, the ball traveled 79 meters and went into the hole.

<u>Question asked:</u>

How many kilometres from the hole was Martin when he started?

<u>Solution:</u>

Let distance from Martin starting point to the hole in meters = $x$

On Martin's first stroke, ball traveled = $\frac{4}{5} \ of \ total \ distance\ to\ the\ hole$

$=\frac{4}{5} \times x=\frac{4x}{5}$

On his second stroke, the ball traveled and went to the hole = 79 meters

Total distance from starting point to the hole = Ball traveled from first stroke + Ball traveled from second stroke

$x=\frac{4x}{5} +79\\ \\ Subtracting\ both\ sides\ by \ \frac{4x}{5}\\ \\ x- \frac{4x}{5}= \frac{4x}{5}- \frac{4x}{5}+79\\ \\ \frac{5x-4x}{5} =79\\ \\ By \ cross\ multiplication\\ \\ x=79\times5\\ \\ x=395\ meters$

Now, convert it into kilometre:

1000 meter = 1 km

1 meter = $\frac{1}{1000}$

395 meters = $\frac{1}{1000}\times395=0.395\ kilometre$

Thus, there are 0.395 kilometre distance from Martin starting point to the hole.

8 0

2 years ago

Solve the following question:

qwelly [4]

Answer:

Once

Step-by-step explanation:

Whenever v(t) crosses the x axis. Not just bounce off it.

5 0

2 years ago

Free Points!!....what is the best restaurant

Rainbow [258]

Answer:

ummmmm panda express ig

Step-by-step explanation:

im just in the mood for chinese food rn lol

5 0

3 years ago

Read 2 more answers

Select the correct answer.

Arlecino [84]

I’m pretty sure it’s linear!

7 0

2 years ago

Consider the graph of the quadratic function y = 3x2 – 3x – 6.