A homogeneous second-order linear differential equation, two functions y 1 and y 2, and a pair of initial conditions are given. First verify that y 1 and y 2 are solutions of the differential equation. Then find a particular solution of the form y equals c 1 y 1 plus c 2 y 2 that satisfies the given initial conditions. Primes denote derivatives with respect to x.

i think is 3 and 6. this took me 10 min to think.hope to and hope that my answer is brainliest.(follow me )

im here

step-by-step explanation:

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places

3/6 chance

step-by-step explanation:

half of the nubers are odd numbers (1,3,5) which in that case if you would be told to simplify it would be 1/2