A. A farmer irrigates her fields with 120,000 cubic yards of water. She has a 1,000-acre farm. How

Mathematics

1 answer:

horsena [70]2 years ago

8 0

Answer:

120 cc yards of water

Step-by-step explanation:

A farmer has 1,000-acre farm. And she irrigates her fields with 120,000 cubic yards of water. We are asked to find how many cubic feet of water does she apply per acre?

1000 acre is irrigated by = 120000 cc yard of water

Hence 1 Acre is irrigated by

$= \frac{120000}{1000}$ cc yards of water

Hence 1 Acre is irrigated by 120 cc yards of water .

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Using your own words, explain direct and inverse proportionality. (Be sure to use complete sentences.)

JulijaS [17]

In an inverse proportion two veriables are inversely proportional to each other.You can write the equation as P=n/x, and N is the constant of proportionality.In the equation the X increases and the P decreases.

8 0

3 years ago

Read 2 more answers

Compute the sum:

Nady [450]

You could use perturbation method to calculate this sum. Let's start from:

$S_n=\sum\limits_{k=0}^nk!\\\\\\\(1)\qquad\boxed{S_{n+1}=S_n+(n+1)!}$

On the other hand, we have:

$S_{n+1}=\sum\limits_{k=0}^{n+1}k!=0!+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=0}^{n}(k+1)!=\\\\\\=1+\sum\limits_{k=0}^{n}k!(k+1)=1+\sum\limits_{k=0}^{n}(k\cdot k!+k!)=1+\sum\limits_{k=0}^{n}k\cdot k!+\sum\limits_{k=0}^{n}k!\\\\\\(2)\qquad \boxed{S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n}$

So from (1) and (2) we have:

$\begin{cases}S_{n+1}=S_n+(n+1)!\\\\S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\end{cases}\\\\\\
S_n+(n+1)!=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\\\\\\
(\star)\qquad\boxed{\sum\limits_{k=0}^{n}k\cdot k!=(n+1)!-1}$

Now, let's try to calculate sum $\sum\limits_{k=0}^{n}k\cdot k!$ , but this time we use perturbation method.

$S_n=\sum\limits_{k=0}^nk\cdot k!\\\\\\
\boxed{S_{n+1}=S_n+(n+1)(n+1)!}\\\\\\
$

but:

$S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!=\\\\\\=
\sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!=\\\\\\=
\sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!=\\\\\\=\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\
\boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}$

When we join both equation there will be:

$\begin{cases}S_{n+1}=S_n+(n+1)(n+1)!\\\\S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\end{cases}\\\\\\
S_n+(n+1)(n+1)!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\\\
\sum\limits_{k=0}^{n}(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n=\\\\\\=
(n+1)(n+1)!-\sum\limits_{k=0}^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]=\\\\\\=(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1=\\\\\\=
n(n+1)!+1$

So the answer is:

$\boxed{\sum\limits_{k=0}^{n}(1+k^2)k!=n(n+1)!+1}$

Sorry for my bad english, but i hope it won't be a big problem :)

8 0

3 years ago

A city bus collected $780 in fares on one day.

Brilliant_brown [7]

Answer:

450 people paid the discounted fare and 750 people paid the regular fare.

Step-by-step explanation:

let r be regular fares paid and d be discounted fares paid

Total fares = 0.8r + 0.4d = 780

Since 1200 people paid the fares,

r + d = 1200 = Total people

Rearrange this formula: