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irinina [24]
3 years ago
14

How long is pie? And what are all of the numbers of pie.

Mathematics
2 answers:
Flauer [41]3 years ago
6 0

Answer:

pi is infinite. i only know up to 3.14159265358979323846264338327950288419716939937510.

qwelly [4]3 years ago
6 0
Pie is infinite meaning it never ends. Like it goes on FOREVERRR.


Some of the numbers are: 141592653589793238462643383279502884197169399375105820974944592307816406286 208998628034825342117067982148086513282306647093844609550582231725359408128481 117450284102701938521105559644622948954930381964428810975665933446128475648233 786783165271201909145648566923460348610454326648213393607260249141273724587006 606315588174881520920962829254091715364367892590360011330530548820466521384146 951941511609433057270365759591953092186117381932611793105118548074462379962749 567351885752724891227938183011949129833673362440656643086021394946395224737190 702179860943702770539217176293176752384674818467669405132000568127145263560827 785771342757789609173637178721468440901224953430146549585371050792279689258923 542019956112129021960864034418159813629774771309960518707211349999998372978049 951059731732816096318595024459455346908302642522308253344685035261931188171010 003137838752886587533208381420617177669147303598253490428755468731159562863882 353787593751957781857780532171226806613001927876611195909216420198938095257201 065485863278865936153381827968230301952035301852968995773622599413891249721775 283479131515574857242454150695950829533116861727855889075098381754637464939319 255060400927701671139009848824012858361603563707660104710181942955596198946767 837449448255379774726847104047534646208046684259069491293313677028989152104752 162056966024058038150193511253382430035587640247496473263914199272604269922796 782354781636009341721641219924586315030286182974555706749838505494588586926995 690927210797509302955321165344987202755960236480665499119881834797753566369807 426542527862551818417574672890977772793800081647060016145249192173217214772350 141441973568548161361157352552133475741849468438523323907394143334547762416862 518983569485562099219222184272550254256887671790494601653466804988627232791786 085784383827967976681454100953883786360950680064225125205117392984896084128488 626945604241965285022210661186306744278622039194945047123713786960956364371917 287467764657573962413890865832645995813390478027590099465764078951269468398352 595709825822620522489407726719478268482601476990902640136394437455305068203496 252451749399651431429809190659250937221696461515709858387410597885959772975498 930161753928468138268683868942774155991855925245953959431049972524680845987273 644695848653836736222626099124608051243884390451244136549762780797715691435997 700129616089441694868555848406353422072225828488648158456028506016842739452267 467678895252138522549954666727823986456596116354886230577456498035593634568174 324112515076069479451096596094025228879710893145669136867228748940560101503308 6179286809208747609178249385890097149096759852613655497818931

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Write the following<br> as an exponential expression.
gogolik [260]

Answer:

x^3/7

Step-by-step explanation:

In exponential form the number on the outside of the √ will be the denominator of the fraction.

3 0
3 years ago
Cedar Point has 17 roller coasters. If you want to ride at least 15 of them, how many different combinations of roller coasters
stellarik [79]
<h3>Answer:   154</h3>

================================================

Explanation:

I'll be using the formula

_n C _r = \frac{n!}{r!*(n-r)!}

which is the nCr combination formula. We use nCr instead of nPr because the order of coasters doesn't matter. The whole time, the value of n stays fixed at n = 17 which is the total number of coasters to pick from.

While n is constant, the value of r will vary. It ranges from r = 15 to r = 17 inclusive of both endpoints. In other words, r will take on values from the set {15,16,17}. So we have three cases to consider. The r value is how many coasters we select. This is due to the "at least 15" which means "15 or more".

-----------------------

If you ride r = 15 coasters, then we have the following number of combinations

_n C _r = \frac{n!}{r!*(n-r)!}\\\\_{17} C _{15} = \frac{17!}{15!*(17-15)!}\\\\_{17} C _{15} = \frac{17!}{15!*2!}\\\\_{17} C _{15} = \frac{17*16*15!}{15!*2!}\\\\_{17} C _{15} = \frac{17*16}{2!}\\\\_{17} C _{15} = \frac{17*16}{2*1}\\\\_{17} C _{15} = \frac{272}{2}\\\\_{17} C _{15} = 136\\\\

There are 136 ways to ride exactly 15 coasters (when selecting from a pool of 17 total). The order doesn't matter.

We'll use this result later, so let x = 136.

-----------------------

If r = 16, then we follow the same steps as above. You should get 17C16 = 17

There are 17 ways to ride 16 coasters where order doesn't matter. Effectively this is the same as saying "there are 17 ways of picking a coaster that you won't ride"

Let y = 17 so we can use it later

-----------------------

Lastly, if r = 17, then nCr = 17C17 = 1 represents one way to ride all 17 coasters where order doesn't matter.

Let z = 1

-----------------------

Add up the values of x, y, and z to get the final answer

x+y+z = 136+17+1 = 154

4 0
3 years ago
Write an equation in slope-intercept form of the line that passes through (2,0) and is perpendicular to y=1/5 x-2.
Advocard [28]

Answer:

y = -5x+10

Step-by-step explanation:

Given

(x_1,y_1) = (2,0)

y = \frac{1}{5}x - 2

Required

Determine an equation that perpendicular to the equation

An equation has the form:

y = mx + b

Where

m = slope

By comparison:

m = \frac{1}{5}

Next, we determine the slope of the new line.

When two lines are perpendicular, the following relation exist:

m_2= -\frac{1}{m_1}

Substitute 1/5 for m1

m_2= -\frac{1}{1/5}

m_2= -5

The equation of the line is then calculated using:

y - y_1 = m(x - x_1)

Where:

m_2= -5 and (x_1,y_1) = (2,0)

This gives:

y - 0 = -5(x - 2)

y = -5(x - 2)

y = -5x+10

8 0
3 years ago
Which table shows a proportional relationship?
Nady [450]
I think the first one
4 0
4 years ago
15
xxTIMURxx [149]
The error is 65 and it need to be added by 60 so the answer is -5
3 0
3 years ago
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