Question

EASY!!!! CAN SOMEONE HELP ME PLEASE. WILL FAN N MEDAL. PLEASE

Answer 1

http://prntscr.com/4l8mly

Answer 2

just take the cube root of 1/8 to solve for x

Answer 3

ok. what would it be then? please ill fan n medal you. @e.chado

Answer 4

do you have a calculator you can type it into? or do you have to do this problem without a calculator?

Answer 5

i can do it with a calculator. what should i type into it?

Answer 6

which calculator do you have? TI-84 or TI-83?

Answer 7

umm im just using google's calculator lol

Answer 8

do you have either of the ones i mentioned? it'll be WAY easier with one of those

Answer 9

@e.chado pls dont leave......im so lost.................

Answer 10

@babbygirl12345 @sammixboo

Answer 11

@surjithayer

Answer 12

@Broskishelleh @brianaBieber @braden_foster_1

Answer 13

@babbygirl12345 can u help me please

Answer 14

yea sure i can help

Answer 15

ok can u tell me the answer n explain it a little?

Answer 16

ill medal n fan u

Answer 17

kk hold on

Answer 18

okay

Answer 19

\[x^3=\frac{ 1 }{ 8 }\]
\[x^3-(\frac{ 1 }{ 2 })^3=0\]
\[a^3-b^3=\left( a-b \right)\left( a^2+ab+b^2 \right)\]
\[\left( x-\frac{ 1 }{ 2 } \right)\left( x^2+\frac{ 1 }{ 2 }x+\frac{ 1 }{ 4 } \right)=0\]
either \[x=\frac{ 1 }{ 2 }\]
\[or~x^2+\frac{ 1 }{ 2 }x+\frac{ 1 }{ 4 }=0\]
\[or~4 x^2+2 x+1=0\]
\[x=\frac{ -2\pm \sqrt{4-4*4*1} }{ 2*4 }\]
x=?

Answer 20

:o

Answer 21

what do u think it is @babbygirl12345

Answer 22

@mathstudent55

Answer 23

@paki @fateal @sammixboo

Answer 24

PLEASE HELP

Answer 25

x=0.5

Answer 26

http://prntscr.com/4l8zdh thank u. what answer choice would it be?

Answer 27

@fateal

Answer 28

id say the second tho

Answer 29

thank you so much!!!!!! can u help me with another?

Answer 30

anytime ^^ ya sure :)

Answer 31

ok ill open another thread. ^^

Answer 32

kk ^^