Question

can some one please help me with these problems? i'll fan and medal

Answer 1

1png answer A

Answer 2

@escamer do you think you could help?

Answer 3

hmmm, @imqwerty

Answer 4

eh

Answer 5

i went back in and finished attachments one and 3

Answer 6

okay so you need help with 2,4,5,6,7?

Answer 7

yes please, and im willing to do the work if you explain it

Answer 8

okay :)

Answer 9

qwerty is here :)

Answer 10

wt

Answer 11

okay so in the 2nd image

6)
if you wanna check if \((x+l)\) is a factor of quadratic equation say \(ax^2+bx+c\)
then just equate \((x+l)\) to 0 and get value of \(x\) like this-
\(x+l=0\)
\(x+l-l=0-l\)
\(x=-l\)
now put this value of \(x\) in your expression which is -> \(ax^2+bx+c\)
if on substituting \(x=-l\) in \(ax^2+bx+c\) you get 0
then this means that \((x+l)\) is a factor of \(ax^2+bx+c\)

now try this one :)

Answer 12

im a bit confused, could you break it down a bit and explain each step?

Answer 13

you gotta try the options if you go by this method
i can tell another method which may be easier

Answer 14

oh so im putting
the options in as you have listed?

Answer 15

okay
i this method which i told is a bit length
lets do it the other way?

Answer 16

so if i do that i get -14,-2,-7,3 and the factor that would be corect is A

Answer 17

(x+14)

Answer 18

no thats not correct
ima show you 1 example

if \((x+14)\) [the 1st option] is a factor of \(x^2+15x-4\) then-
\(\large\color{violet}{Step~1-}\)
1st we gotta equate \((x+14)\) to 0 and get value of x
\((x+14=0\)
\(x+14-14=0-14\)
\(x=-14\)

\(\large\color{violet}{Step~2-}\)
now we have to put this value of x in our equation which is -> \(x^2+5x-14\)

we substitute \(x=-14\) in it we get this-
\((-14)^2+5(-14)-14\)
simplifying this we get-> \(112\)

if this value would have been 0 then \((x+14)\) would be considered as a factor
but its not 0
so (x+14) is not a factor

so like this you check the options
the one which gives the value as 0 is correct

Answer 19

ooo okayyy

Answer 20

yea now get the answer lol

Answer 21

so the answer will be (x+7) ?

Answer 22

yeah!! :D

Answer 23

yay!

Answer 24

okay now next one

Answer 25

i got 7 qalready

Answer 26

oh okay so 8th

Answer 27

i think its c but i wanted to check

Answer 28

yeah thats correct!! :D

Answer 29

yay!

Answer 30

okay now next one
11th

Answer 31

yeah, i cant figure out how to do this one

Answer 32

okay :)
the roots are the x coordinate of points where the graph cuts the x axis

1st locate the points where the graph cuts x axis

2nd note down the x coordinate of those points

these x coordinates are you roots :)

Answer 33

so where x=-2 and x=3?

Answer 34

yeah correct!! :D

Answer 35

now next one
question 15

Answer 36

and that answers 16 tooo!

Answer 37

yes it does!! :D

Answer 38

okay so for question 15

1st notice that x=3 is a root
okay?

Answer 39

yep!

Answer 40

also x=-1 is a root then

Answer 41

yes it is :)

Answer 42

i could theoretically graph each equasion and see what matches the graph as a last chance on a test right?

Answer 43

yea you can do that :)

you can also do this-
try putting x=3 and x= -1 in the options
the one which satisfies is the correct one

Answer 44

and they need to equal 0

Answer 45

yeah :D

Answer 46

and i got c as correct

Answer 47

you are close
but c is not correct
it does becomes 0 when you put x=3 in it

Answer 48

thats weird i actually half did it and only put in -1 and it was the only one to equal 0

Answer 49

putting x=-1 in option c gives this-

\(-3x^2+6x-9\)
\(-3(-1)^2+6(-1)-9\)
\(-3-6-9\)
\(-18\)
so its wrong

Answer 50

oops i must have done something wrong when i typed it in my calculator then

Answer 51

okay try again :)

Answer 52

d i got d this time

Answer 53

yeah its d :)
btw question asks us to find the answer using the intercepts and maximum wanna do it that way?

Answer 54

yeah! so i can do it in the future

Answer 55

yeah :)

okay 1st we know that any quadratic equation lets say->\(ax^2+bx+c\) can be written in factored form like this-> \(a(x-\alpha)(a-\beta)\)
where \(\alpha\) and \(\beta\) are roots of equation

now we know the roots by looking at the graph
but we donno "\(a\)"

so we gotta find a we do this->

\(a(x-\alpha)(x-\beta)=h\)

here we know \(\alpha\) and \(\beta\) and \(h\) is the y coordnate of the maxima(maximum) and \(x\) is the x coordinate of the maximum

just substitute these things and get \(a\)
when you get a just put it in this original form-> \(a(x-\alpha)(x-\beta)\) to get the equation

Answer 56

okay! i remember learning that!

Answer 57

okay so now next question 17

Answer 58

try to figure it out :)
it similar to question 16 and question 14

Answer 59

we know how to find roots by looking at the graph
the 4 options give 4 different approaches out of which only 1 is correct and we have been using it

Answer 60

d right?

Answer 61

yeah correct \(\Huge\color{green}{✓}\)

Answer 62

and 18 is b i think when i graphed and looked

Answer 63

yeah its correct!! :)

Answer 64

okay can you help me with one more little problem?

Answer 65

yeah sure :)

Answer 66

okay :)
just try to options out
simplify the options and the one which matches is correct

Answer 67

oh that sounds easy!!

Answer 68

(B
if A,B,C doesn't match then its D

Answer 69

i got c

Answer 70

thas not correct

opening option C-
\((3x+2)(x+1)\)
\(3x(x+1)+2(x+1)\)
\(3x(x)+3x(1)+2(x)+2(1)\)
\(3x^2+3x+2x+2\)
\(3x^2+5x+2\)

but this is not our original equation which is-> \(3x^2+3x+2\) so it is wrong

Answer 71

okay so try again?

Answer 72

yes :)

Answer 73

b

Answer 74

noo its not b

Answer 75

then it has to be a, i just dont see where im messing up

Answer 76

lol what did you do? :) maybe you are doing a silly mistake somewhere

Answer 77

i was freaking adding!! thats what i was messing up

Answer 78

haah so whats the final answer you got

Answer 79

a, i hope, i think i did it right

Answer 80

no its wrong :D

Answer 81

wait if A B and c are wrong then is the answer D, it cant be factored?

Answer 82

yaeh :) d is the correct one

Answer 83

okay, thank you very much for explaining these to me

Answer 84

:) np

Answer 85

wanna know what i got?

Answer 86

what lol

Answer 87

79%!!!!!!!! I passed!

Answer 88

:) congrats!!

Answer 89

it was a 100 question extra credit so i got 79 right!!!!!

Answer 90

\(\huge\cal\color{blue}{ヾ(⌐■‿■)ノ♪}\color{pink}{*:･ﾟ}\color{gold}{✧*:}\color{red}{･ﾟ✧*}\color{green}{:･ﾟ:･ﾟ}\color{violet}{✧*:･ﾟ✧*:･ﾟ✧*:･ﾟ✧*:･ﾟ✧*:･ﾟ✧}\)

Answer 91

@imqwerty are you good at chem? i just started a practice test and im having some troubles

Answer 92

yeah i can help but i gtg in few mins