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Igoryamba
2 years ago
13

14 deliveries in 7 hours = deliveries per hour?

Mathematics
2 answers:
kykrilka [37]2 years ago
5 0
Answer: 2 deliveries per hour.

Explanation: 14 ÷ 7 is 2. You can also check it by doing 2 x 7 to equal the 14 deliveries in 7 hours.
jolli1 [7]2 years ago
4 0

Answer:

2

Step-by-step explanation:

14/7=2

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3 years ago
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Fudgin [204]

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6 0
3 years ago
Read 2 more answers
Given the recursive formula:<br> a1=3<br> An=2(an-1+1)<br> Find a2 a3 a4
fenix001 [56]

Answer:

a_2 = 8

a_3 = 18

a_4 = 38

Step-by-step explanation:

We are given the following recursive formula:

a_n = 2(a_{n-1}+1)

And consider a_1 = 3

a2

a_2 = 2(a_1 + 1) = 2(3 + 1) = 8

a3

a_3 = 2(a_2 + 1) = 2(8 + 1) = 18

a4

a_4 = 2(a_3 + 1) = 2(18 + 1) = 38

6 0
3 years ago
Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions
monitta
I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take z=y', so that z'=y'' and we're left with the ODE linear in z:

y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2

Now suppose y has a power series expansion

y=\displaystyle\sum_{n\ge0}a_nx^n
\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}
\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}

Then the ODE can be written as

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0

\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0

All the coefficients of the series vanish, and setting x=0 in the power series forms for y and y' tell us that y(0)=a_0 and y'(0)=a_1, so we get the recurrence

\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}

We can solve explicitly for a_n quite easily:

a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}

and so on. Continuing in this way we end up with

a_n=\dfrac{a_1}{n!}

so that the solution to the ODE is

y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x

We also require the solution to satisfy y(0)=a_0, which we can do easily by adding and subtracting a constant as needed:

y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}
4 0
3 years ago
What is the answer to this? Will someone help me?<br> 5+(9)(6)
sineoko [7]
Answer: 59
explanation: PEMDAS, first multiply 9x6 as they are multiplication then add

good luck!
4 0
3 years ago
Read 2 more answers
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