Study the solutions of the three equations on the right. Then, complete the statements below. There are two real solutions if th

Answer:

15.6

Step-by-step explanation:

1. 
2. Insert 39/4: 39/4 ÷ 5/8
3. 39/4 ÷ 5/8 = 39/4 × 8/5
4. 39/4 × 8/5 = 312/20
5. 

I hope this information is helpful!

Answer:

A. True

Step-by-step explanation:

The Triangle Inequality Theorem establishes that the sum of two sides must exceed the length of the third side. Let’s verify this.

a + b

9 + 1 = 10>9

a + c

9 + 9 = 18>1

c + b

9 + 1 = 10>9

The flea experiences an acceleration of 1,000 m/s².

Detailed explanation

This scenario involves motion with constant acceleration.

The relevant variables include the following.

We know the flea attains a takeoff velocity of 1.0 m/s over a distance of 0.50 mm.

The flea starts from rest, so the initial velocity is zero. The question asks for the flea's acceleration during leg extension.

The equation we apply is:

• a = acceleration (m/s²)
• u = initial speed = 0 m/s
• v = takeoff velocity = 1.0 m/s
• d = displacement = 0.50 mm

First, convert 0.50 mm to meters:

Solution steps:

Rearrange the formula to isolate acceleration (a).

Insert the given values into the rearranged equation.

This yields the flea's acceleration as 1,000 m/s².

Additional resources

1. Scientific notation: brainly.com/question/7263463
2. Determining substance mass: brainly.com/question/4053884
3. Conversion into cubic units: brainly.com/question/1446243

Keywords: flea, jumping, leg extension, takeoff velocity, initial speed, displacement, acceleration, constant acceleration, unit conversion

Answer:

u(t)  = -(3 + w^2 ) cos t /(1- w^2)cos t + 7 sin t + 8 cos wt /(1- w^2)

Step-by-step explanation:

The characteristic equation is k² + 1 = 0, which leads to k² = -1, resulting in k = ±i.

The roots are k = i or -i.

The general solution takes the form  u(x)=C₁cosx+C₂sinx.

Applying the method of undetermined coefficients, we have

Uc(t) = Pcos wt  + Qsin wt

Calculating the derivatives gives us Uc’(t) = -Pwsin wt  + Qwcos wt

And differentiating again yields Uc’’(t) = -Pw^2cos wt  - Qw^2sin wt

With the equation U’’ + u = 8cos wt, we substitute:

-Pw^2cos wt  - Qw^2sin wt + Pcos wt  + Qsin wt = 8cos wt.

This simplifies to (-Pw^2 + P) cos wt   + (-Qw^2 + Q) sin wt = 8cos wt.

From -Pw^2 + P = 8, we find P= 8  /(1- w^2).

From -Qw^2 + Q = 8, we can conclude Q = 0.

Thus, Uc(t) = Pcos wt  + Qsin wt = 8 cos wt /(1- w^2).

Combining gives us U(t) = uh(t ) + Uc(t)

     = C1cos t + c2 sin t + 8 cos wt /(1- w^2).

Initial conditions yield:

U(0) = C1cos(0) + c2 sin (0) + 8 cos (0) /(1- w^2)

Which leads us to C1 + 8 /(1- w^2) = 5

So C1 = 5 - 8 /(1- w^2) = -(3 + w^2 ) /(1- w^2).

Next, taking the derivative:

U’(t) = -C1 sin t + c2 cos t - 8 w sin wt /(1- w^2).

Evaluating at t = 0 gives us:

U’(0) = -C1 sin (0) + c2 cos (0) - 8 w sin (0) /(1- w^2) = 7.

Thus, c2 = 7.

  u(t)  = -(3 + w^2 ) cos t /(1- w^2)cos t + 7 sin t + 8 cos wt /(1- w^2)