Which steps can be used to solve StartFraction 6 Over 7 EndFraction x + one-half = StartFraction 7 Over 8 EndFraction for x? Che

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)

The answer is in the image provided.

Please refer to the accompanying image.

Answer:

Step-by-step explanation:

The probability of an event quantifies the likelihood of that event occurring.

Definition: The probability of an event is given as

1. An event that is sure to occur (Number of favorable outcomes = Total number of outcomes) has a probability of 1.

2. An event that is impossible (Number of favorable outcomes = 0) has a probability of 0.

3. If there's a possibility that an event can occur, its probability will fall between 0 and 1.

Answer:

Parallel

corresponding angles theorem

NYM

AA

Step-by-step explanation:  I DID IT ON EDGE!: )

We know that side NM is  _________________

✔ parallel

to side XZ. If we consider side NY as the transversal, angle pairs are formed. According to the  _____________

✔ corresponding angles theorem

, it follows that ∠YXZ is congruent to ∠YNM. Also, angle XYZ is congruent to angle  _____________

✔ NYM

due to the reflexive property. Hence, triangle XYZ is similar to triangle NYM by the  _____________

✔ AA

 similarity theorem.