Use place value to solve Pablo has a piece of wire that is 0.08 meters long Beth has a piece of wire that is 10 times as long as problems wired leaves wire is one-tenth as long as problems how long is Beth wire how long is li's

wire

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review the conditions for the series to exist. In this case, it will exist for x=0, so we want

y = ∑anx^n
y' = ∑nanx^(n-1)
y" = ∑n(n-1)anx^(n-2)

The DE then says
∑n(n-1)anx^(n-2) + x*∑nanx^(n-1) + ∑anx^n = 0

∑n(n-1)anx^(n-2) + ∑nanx^n + ∑anx^n = 0

∑n(n-1)anx^(n-2) + ∑(n+1)anx^n = 0

Now shift y" so the powers of x add up

∑(n+2)(n+1)an+2x^n + ∑(n+1)anx^n = 0

∑[(n+2)(n+1)an+2 + (n+1)an]x^n = 0

For the series to be zero for all x, we have the recurrence relation

(n+2)(n+1)an+2 + (n+1)an = 0

an+2 = -an/(n+2)

At this point, just start listing the coefficients and see where you go. There is a similar DE used as an example at

http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx

To solve this problem using place value, we need to understand the relationship between the given lengths of wire.

Pablo has a piece of wire that is 0.08 meters long. We can represent this as 0.08 m.

Beth's wire is 10 times as long as Pablo's wire. This means that Beth's wire is equal to 10 multiplied by Pablo's wire length.

To calculate the length of Beth's wire, we can multiply 0.08 m by 10:

0.08 m × 10 = 0.8 m

Therefore, Beth's wire is 0.8 meters long.

Next, we are given that Li's wire is one-tenth as long as Pablo's wire. This means that Li's wire length is equal to one-tenth of Pablo's wire length.

To calculate the length of Li's wire, we can multiply 0.08 m by one-tenth (0.1):

0.08 m × 0.1 = 0.008 m

Therefore, Li's wire is 0.008 meters long.