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3 years ago

an Olympic sized pool which holds 660000 gallons of water is only 63% full. how many gallons of water are in the pool

Mathematics

2 answers:

user100 [1]3 years ago

5 0

415800 gallons of water will be in the pool when it is 63% full

irinina [24]3 years ago

4 0

660,000*63%=660,000*0.63=415,800
There are currently 415,800 gallons of water in the pool.

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Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

tekilochka [14]

Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$

Applying the ratio test:

$\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}$

$\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

Applying n → ∞

$\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

The numerator as well denominator of $\frac{a_{n}}{a_{n+1}}$ are polynomials of fifth degree with leading coefficients:

$(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}$

4 0

2 years ago

For the month of October Lisa plan to go running every fifth day and walk every sixth day. If she begins on October 1, on what d

Tomtit [17]

Answer:

October 30 is the day of the month she will run and walk during the same day.

Step-by-step explanation:

To determine what day of the month will she run and walk during the same day, we will list the days she plans to go running and the days she plans to walk.

From the question, Lisa plan to go running every fifth day and walk every sixth day, then

If she begins on October 1, the days she will run will be

October 5, October 10, October 15, October 20, October 25, and October 30;

while the days she will walk will be

October 6, October 12, October 18, October 24, and October 30.

Since October 30 occurred among the days she will run and walk, then October 30 is the day of the month she will run and walk during the same day.

4 0

3 years ago

Simplest form equivalent to 3y + y + 2y - 4y

bixtya [17]

Answer:

4y+6y

Step-by-step explanation

PUT ME THE BRAINLIEST PLEASE

7 0

3 years ago

Read 2 more answers

Write a slope-intercept equation for a line perpendicular to f(x) = −2x + 4 and

Ierofanga [76]

Answer:

y-(-1)=-2(x-(-4))

Step-by-step explanation:

6 0

2 years ago

Hi, the answer is K but can anyone show why

dlinn [17]

Let's do 51 and 52.

51. The contrapositive has the same truth value as the original statement. That's opposed to the converse, which may or may not be true independent of the original statement.

The contrapositive of IF P THEN Q is IF not Q THEN not P. They're equivalent. Here that's If the cat is not female then it is not tricolor.

Answer: C

52.

$(x^3)^{(4-b^2)}=1$

$x^{3(4-b^2)} = 1$

For the statement to be true, the exponent must be zero:

$3(4-b^2) = 0$

$b^2 = 4$

$b = \pm 2$

Both positive 2 and negative 2 have a square of 4.

Answer: K

By the way, usually we assume $0^0=1$ so the restriction that $x \ne 0$ isn't really necessary. Think of the definition of a polynomial or the binomial expansion:

$\displaystyle f(x)=\sum_{k=0}^n a_k x^k$

$\displaystyle(x+y)^n=\sum_{k=0}^n {n \choose k} x^{k}y^{n-k}$

For these common equalities to work when $x=0$ we need to define $0^0=1$

3 0

3 years ago