Solve the homogeneous linear odes with constant coefficients: y" - 6y' + 8y = 0. m^2 - 6m + 8 = 0 (m - 2)(m - 4) m_1 = 2 m_2 = 4 y = c_1 e^2x + c_2 x e^4x y" - 6y' + 9y = 0. m^2 - 6m + 9 = 0 (m - 3)^2 = 0 m_1 = m_2 = 3 y = c_1 e^3x + c_2 x e^3x y" - 6y' 10y = 0. y"' + 3y" + 3y' + y = 0 y_1 = x^4 is a solution to the ode x^2 y" - 7xy' + 16y = 0, use reduction of order to find another independent solution.

a:

step-by-step explanation:

a

here is me explaining it:

the pizza would be delivered around 5, give or take 15 minutes. this means it could be delivered anywhere from 4: 45 to 5: 15

let's say 5: 00 is equivalent to 0 on a number line and x represents time. 4: 45 would be equal to -15x because we are 15 minutes behind our starting point, 5: 00. 5: 15 would be equal to 15x because we are 15 minutes ahead of your starting point (which means you're late on the pizza delivery! )

absolute value is very confusing. the basic definition is how far you are from your starting point. if we deliver the pizza at 4: 45 (15 minutes before the starting point) we would be -15. this means we are 15 away from our starting point, 5: 00. that's the absolute value. |-15| (absolute value of negative fifteen) is equal to 15. if you did not understand this i suggest rereading this paragraph.

the normal way you would write this scenario would be 4.75 (4 and 3 quarters, or 4: 45. you can also write as a fraction.) is less than x and x is less than 5.25 (5 and 1 quarter or 5: 15). 4.75< x< 5.25 the reason why i am using numbers is because you cannot represent typical time as values on a graph, number line, or in an equation.

another way to write this scenario is with absolute value. we can write this as |x| being less than or equal to .25 (remember, this just means 15 minutes).           |x| ≤ .25.

sorry if the end was confusing. because we are talking about time this whole question is 10x harder. like i said earlier, you cannot represent time as, well, time, in a graph, equation, or number line. i set the values as follows: 1/4 or .25 for 15 minutes, and 1 for 1 hour. this is just a common way to set up graphs involving time.

|x| ≤ .25

i hope this ! : )

tell me if you need more assistance

the answer i think is 1 exit.

a is the answer to ur question