Which is another way to name angle ust

What is the square root of 7? Decimal form please. Person with the most decimal places gets brainliest.

max2010maxim [7]

Answer:

2.64575131106 2.6457513110645905905016157536392604257102591830824501803683344592010688232302

83627760392886474543610615064578338497463095743529888627214784427390555880107722

71715072972832389229968959486508726070097805420372382802371594110034193911600157

85255963059457410351523968027164073737990740415815199044034743194536713997305970

05051399692237545616097119027378154991633288287704000657570674651963497752083793

81811461309087647378659562433057994798128163230705483650107715617946361191553454

53647749482059309049484983403398900210478616673327950369392462257170537164925787

54832290732492671346980298949908037748251109227895568897919808814834090831685251

33535829539172211770714414974576907081989444441458972284741400303502353203719487

07382629318519364090832280596462783761021959794197089096354695861341181793067816

21360849101677835321255633463490021898146042255929503669562418692737327715022087

52309966469813203212818945478568020950635962446628550076190504139350447437123488

52233277362510045059621080672334698120004300514490251206257311759115423194459672

60497833404344683725636178255048160879406205511486397833055016369470076387773745

19511658095220474202064982102444056001996590904100285534069110102261166807200774

83070332613842826296809502470794656496344336830519708820348050486447262317930876

95722489377521119788376007845603903658809421599292501057753682356001720074065236

92473209967419778002535708540617794442886681589731184938101638162073015642647016

29841099690071341323798188477647769954391111945967670336972452475367740100245826

51593077528163005479339717508017899520565284447796194304512502890432970610508746

04492966897542257133380633247034508888234275409806214511439553803393190705392738

14520604335238180389333294731479058055670704161068722878104529957696633035133295

04704245502510500789645502552271857923370921821991843932537781386785650143790894

13002099040860186605228131329821735251716419633617853080179022873400337109527257

36428650922035916343820273976336550966714177816292894279884853951076819102168116

99677176780863793410667897509217050421558644370687652934899322442240273649792906

66842070305782257110537944215790013280304078494579431518120466309284152592404683

65282802722458530778415894238973955590959563491357495109757674781975590533726939

36521431268222909570986506137179452721677198255442288046052307455488310813325041

49932783468182066583447383366862454456169497251774270254688591746231942309355060

38899235355634465872922507362228284298215287903623847595555861643755472027361518

00826144981691174794610631845652307625691344410868551999982851633960745717970829

72776495645142655658205126240305961633868338746134237463526431296594527010213721

72174899143552932049575097114259427207988929665230498875939525378121455723193310

30690840194942118708515205734116214105775416917245804962369512932926647583857193

34312550616097509003957512063487575979549295499224726634074018815634940531052960

4465875564620943Step-by-step explanation:

3 0

3 years ago

We're working with single variable equations and I'm okay at them, but I'm rusty when it comes to using it with fractions. If an

jasenka [17]

Answer: v=-18/5

Step-by-step explanation:

Assuming you want to solve for v, we need to use our algebraic properties.

$\frac{v}{3}+2=\frac{4}{5}$ [subtract both sides by 2]

$\frac{v}{3}=-\frac{6}{5}$ [multiply both sides by 3]

$v=-\frac{18}{5}$

Now, we know that v=-18/5.

3 0

3 years ago

(ab^2 x -b^4a^3)^2<br><br> What is the answers

Umnica [9.8K]

Answer:

$\left(ab^2x-b^4a^3\right)^2:\quad a^2b^4x^2-2a^4b^6x+a^6b^8$

Step-by-step explanation:

Given the expression

$\left(ab^2\:x\:-b^4a^3\right)^2$

solving the expression

$\left(ab^2\:x\:-b^4a^3\right)^2$

$\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a-b\right)^2=a^2-2ab+b^2$

$a=ab^2x,\:\:b=b^4a^3$

so the expression becomes

$=\left(ab^2x\right)^2-2ab^2xb^4a^3+\left(b^4a^3\right)^2$

$=a^2b^4x^2-2a^4b^6x+a^6b^8$

Thus,

$\left(ab^2x-b^4a^3\right)^2:\quad a^2b^4x^2-2a^4b^6x+a^6b^8$

3 0

2 years ago

HELP WILL MARK BRAINLYEST WoW

Varvara68 [4.7K]

Answer:

the x-cordinate for p is -6

Step-by-step explanation:

5 0

2 years ago

What is three tenths times two and five sixths?

Virty [35]

To multiply fraction do top x top and bottom x bottom
so convert $2 \frac{5}{6}$ into $\frac{17}{6}$
then do $\frac{3}{10} * \frac{17}{6} = \frac{3*17}{10*6} = \frac{51}{60} = \frac{17}{20}$

3 0

3 years ago

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