Question

Am i correct?

Answer 1

yeah you know why?

Answer 2

why

Answer 3

it's hard to explain, but 4xy³ can go into

8xy⁵
-16x²y³
20x⁴y⁴

Answer 4

it can go into them all evenly

Answer 5

huh so its a?

Answer 6

i mean d

Answer 7

your answer was right. B

a greatest common factor is a factor that you can pull out of a set of numbers\[8xy^5 - 16x^2y^3+20x^4y^4 =\]\[4xy^3(2y^2-4x+5x^3y)\]

Answer 8

i divided the whole thing by 4xy³ and had nothing else i could divide it by without having all whole numbers.

so 4xy³ is the gcf (greatest common factor)

Answer 9

ok you were right this time

Answer 10

omgggg

Answer 11

im always right

Answer 12

ur hereee

Answer 13

ok idk this one

Answer 14

you want satellite to help? i could save his time by showing you

Answer 15

Well if satelite is still online, probably looking up rolex's

Answer 16

@satellite73

Answer 17

just an fyi: this has to do with gcf

Answer 18

a?

Answer 19

i am finding myself a nice rolex

Answer 20

the gcf of 8 and 12 is 4

Answer 21

how much time do u spend on ebay

Answer 22

the gcf of \(x^2\) and \(x\) is \(x\)

Answer 23

hours

Answer 24

k is it a

Answer 25

or d

Answer 26

yes

Answer 27

A

Answer 28

wearing a nice LeCoultre Memovox today

Answer 29

says "slow down" lol

Answer 30

b?

Answer 31

link doesnt work

Answer 32

its another factoring problem

Answer 33

yes B

Answer 34

aint tht expensive

Answer 35

now it works

Answer 36

i am not a poor kid like you who can't even buy sephoro makeup
but actually no, i got it at a watch show for not too much $

Answer 37

ha ha ha

Answer 38

what did you guess for the last one?

Answer 39

nothing

Answer 40

to many numbers

Answer 41

then do it, this one is easier than the others

Answer 42

only common factor of those "too many numbers" is 4 (there are only three numbers there)

Answer 43

c?

Answer 44

no, there is no \(x^2\) in each term, only a 4

Answer 45

b

Answer 46

no, \(4\times 4x^2=16x^2\) but you only have \(4x^2\)

Answer 47

a

Answer 48

finally

Answer 49

haha

Answer 50

kk hold on

Answer 51

and your guess?

Answer 52

b..

Answer 53

on no \[(x+4)(x+4)=x^2+8x+16\] not \(x^2-4x+4\)

Answer 54

d

Answer 55

closer, but \[(x-4)(x-4)=x^2-8x+16\]

Answer 56

a

Answer 57

\[(x+2)(x+2)=x^2+4x+4\] but you want \[x^2-4x+4\]

Answer 58

so the remaining guess is ...

Answer 59

c

Answer 60

finally \[(x-2)(x-2)=x^2-4x+4\] yes

Answer 61

u cant even see numbers

Answer 62

lol because there aren't any

Answer 63

if i were u id take a sharpiie and put the numbers

Answer 64

like a boss

Answer 65

eeeew

Answer 66

this is not mine, but mine is almost identical

Answer 67

wow k

Answer 68

this time give me sec to guess

Answer 69

it is a "wrist alarm" that is what the second crown is for
before they had apps

Answer 70

ok take your time

Answer 71

b?

Answer 72

or c not sure

Answer 73

lets go slow
which one has the constant (last number at the end) as a "perfect square"
hint 60 is not a perfect square

Answer 74

100?

Answer 75

yes, but also \(144=12^2\) and \(36=6^2\) so there are three of them

Answer 76

a?

Answer 77

the next question is, which one has a middle term that is twice the number whose square is at the end?

Answer 78

for a, \(6^2=36\) but \(2\times 6=12\) so not that one \[(x+6)^2=x^2+12x+36\]

Answer 79

ohh

Answer 80

and \[(x+10)^2=x^2+20x+100\]

Answer 81

so not that one either

Answer 82

ddddd

Answer 83

yes because \(12^2=144,2\times 12=24\) and so \[(x+12)^2=x^2+24x+144\]

Answer 84

huh i was thinking of d first but then the others looked more like perfect square

Answer 85

\[(x+a)^2=x^2+2ax+a^2\]

Answer 86

a?

Answer 87

@satellite73