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denpristay [2]
3 years ago
11

Round 251031 to nearest thousand

Mathematics
2 answers:
antiseptic1488 [7]3 years ago
6 0
251031=251,031
251,000 is the rounded form.

vladimir1956 [14]3 years ago
4 0

Answer:

251000

Step-by-step explanation:

Round 251031 to nearest thousand

Lets write the place value of numbers from left to right

1 is in ones place

3 is in tens place

0 is in hundreds place

1 is in thousands place

5 is in ten thousands place

2 is in hundred thousands place

To round to nearest thousand , we look at the number in hundreds place

0 is in hundreds place

251031 to nearest thousand is 251000

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300/200 as a terminating decimal.
Tomtit [17]

Answer:

1 100/200

Step-by-step explanation:

3 0
3 years ago
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10 points <br><br> slope of (-2,-4) and (6,7)<br> (explain)
Finger [1]

Answer:

11/-8

Step-by-step explanation:

-2-6=-8

-4-7=11

then flip

4 0
2 years ago
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
HELP ASAP PLSSSSSSSSSSSS
ludmilkaskok [199]

Answer:

f = .39

Step-by-step explanation:

d (v) = 2.15 v^2 / (64.4 f)

v = 40

d = 138

substitute these in

138 = 2.15 (40)^2 / (64.4 *f)

simplify

138 = 2.15 (*1600) / (64.4 *f)

138 = 3440 / (64.4 *f)

multiply each side by 64.4f

138 * 64.4 f = 3440

8887.2 f = 3440

divide by 8887.2

8887.2 f /8887.2= 3440/8887.2

f=.38707

to the nearest hundredth

f = .39

8 0
3 years ago
Read 2 more answers
Determine if the given value makes the equation true or false. For x3 – 8 = 15, if x = 69
MrRa [10]

Answer:

(69)(3)−8 = 15

199≠15

False

Step-by-step explanation:

hope it helps, 69(3) is already bigger than 15, then -8 makes it 199, which is definitely not equal to 15.

6 0
2 years ago
Read 2 more answers
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