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3 years ago

The area of square pond is 5625meter square and 2 meter width path is made around the pond

Mathematics

2 answers:

algol133 years ago

7 0

Answer:

Step-by-step explanation:

good luck hoped it helped

Katyanochek1 [597]3 years ago

4 0

Answer:

I think it would be b I dont know for sure

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Andrew has already played 32 minutes on his video game this morning and wants to play an additional 15 minutes on each of x game

belka [17]

Given :

Andrew has already played 32 minutes on his video game this morning and wants to play an additional 15 minutes on each of x games after school today.

To Find :

Which equation can be used to find y, the total minutes Andrew will play video games today.

Solution :

Number of minutes he play after school time, n = 15x .

Number of minutes already played, c = 32 .

So, total number of minutes he play games for today is :

y = n + c

y = 15x + 32

Therefore, equation which is useful to find number of minutes video game played per day is y = 15x + 32 .

8 0

3 years ago

Hi! I would really appreciate it if you can help me with this problem.

miss Akunina [59]

GCF: 2v5
2v5(4v+1+5v4)

7 0

3 years ago

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

3 0

2 years ago

Describe the main parts of a proof.

Vsevolod [243]

Answer:

Is there a picture?

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Working together to palms can drain a certain pool in three hours if it takes the older pump nine hours to drain the pool by its

andreyandreev [35.5K]

Answer:

It would take the newer pump 4.5 hours to drain the pool

Step-by-step explanation:

Let's investigate first what is the fraction of the job done in the unit of time (hour in this case) by each pump if the work individually:

older pump: if it takes it 9 hours to complete the job, it does $\frac{1}{9}$ of the job in one hour.

newer pump: we don't know how long it takes (this is our unknown) so we call it "x hours". Therefore, in the unit of time (in one hour) it would have completed $\frac{1}{x}$ of the total job.

both pumps together: since it takes both 3 hours to complete the job, in one hour they do $\frac{1}{3}$ of the job.

Now, we can write the following equation about fractions of the job done:

<em>The fraction of the job done by the older pump plus the fraction of the job done by the newer pump in one hour should total the fraction of the job done when they work together.</em> That is in mathematical terms:

$\frac{1}{9} +\frac{1}{x} =\frac{1}{3}$

and solving for x:

$\frac{1}{9} +\frac{1}{x} =\frac{1}{3} \\x+9=3x\\2x=9\\x=\frac{9}{2} \,\,hours\\x = 4.5\,\,hours$

4 0

3 years ago