Answer: Infinite solutions
Step-by-step explanation:
Answers:
- a) 15000 represents the starting amount
- b) The decay rate is 16%, which means the car loses 16% of its value each year.
- c) x is the number of years
- d) f(x) is the value of the car after x years have gone by
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Explanation:
We have the function f(x) = 15000(0.84)^x. If we plug in x = 0, then we get,
f(x) = 15000(0.84)^x
f(0) = 15000(0.84)^0
f(0) = 15000(1)
f(0) = 15000
In the third step, I used the idea that any nonzero value to the power of 0 is always 1. The rule is x^0 = 1 for any nonzero x.
So that's how we get the initial value of the car. The car started off at $15,000.
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The growth or decay rate depends entirely on the base of the exponential, which is 0.84; compare it to 1+r and we see that 1+r = 0.84 solves to r = -0.16 which converts to -16%. The negative indicates the value is going down each year. So we have 16% decay or the value is going down 16% per year.
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The value of x is the number of years. In the first section, x = 0 represented year 0 or the starting year. If x = 1, then one full year has passed by. For x = 2, we have two full years pass by, and so on.
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The value of f(x) is the value of the car after x years have gone by. We found that f(x) = 15000 when x = 0. In other words, at the start the car is worth $15,000. Plugging in other x values leads to other f(x) values. For example, if x = 2, then you should find that f(x) = 10584. This means the car is worth $10,584 after two years.
Answer:
So, solution of the differential equation is
Step-by-step explanation:
We have the given differential equation: y′′+4y=5xcos(2x)
We use the Method of Undetermined Coefficients.
We first solve the homogeneous differential equation y′′+4y=0.
It is a homogeneous solution:
Now, we finding a particular solution.
we get
So, solution of the differential equation is
How to solve your problem
9−4+3
9x−4+3x9x-4+3x9x−4+3x
Simplify
1
Combine like terms
9−4+3
9x−4+3x{\color{#c92786}{9x}}-4+{\color{#c92786}{3x}}9x−4+3x
12−4
12x−4{\color{#c92786}{12x}}-412x−4
Solution
12−4
SO D
