Question

hey if anyone willing to help me with some precal questions please send me a message . i need helpp please be my saviorrr

Answer 1

Sure! Just close this question and post a new one

Answer 2

do you think we can message privatly

Answer 3

Hmm well I would rather you post the questions so other users can also comment and add to the discussion (also, medals ;P )

Answer 4

lol i have attachments

Answer 5

but ok

Answer 6

It is easier to attach files on questions anyways :D

Answer 7

here is the first 2 questions

Answer 8

@lolamona
Okay so if
\[f(x)=x^2-4x-5\]&
\[g(x)=x^2-25\]
& they're asking about:
\[\left( \frac{ f }{ g } \right)(x)=\frac{ f(x) }{ g(x) }\]
Than:
\[\left( \frac{ f(x) }{ g(x) } \right)=\frac{ (x-5)(x+1) }{ (x+5)(x-5)}\]

Answer 9

*Notice how i factored the top & bottom.. because a term will cancel @lolamona what cancels?

Answer 10

o yes i see

Answer 11

and then you make x+5 and x+1 equal to zero right

Answer 12

well x-5 on both the top and bottom

Answer 13

cancel out

Answer 14

yep which is interesting...

anywho the domain are all real numbers *** except when the denominator is zero...
so what do you think it is ....
if
\[\left( \frac{ f(x) }{ g(x) } \right)=\frac{ (x-5)(x+1) }{ (x-5)(x+5) }\]

Answer 15

Second one --->

Answer 16

cant equal cant equal -1 and 5 or am i wrong

Answer 17

no it can! heres why

Answer 18

sooo.. do you recall that you can't divide by zero?

Answer 19

x-5 cause it cancels out

Answer 20

basically it's against the law in math lol & yeah i didn't mean to confuse you! if they were asking for the most simplified version of f/g(x) than yeah... the x-5 would cancel out.

BUT it is still in the domain

Answer 21

so the domain is all real numbers EXCEPT 5 & -5

Answer 22

aka ZEROS of the denominator

Answer 23

ohhhhhhhh lol

Answer 24

Yep XD now the next one

Answer 25

Recall that: \[(f og )(x)=f(g(x))\]
Sooo basically this means f composed with g...
If:
\[g(x)=\frac{ 1 }{ x-1 }\]
\[f(x)=x^2+6x+5\]
Than:
\[f(g(x))=(g(x))^2+6(g(x))+5\]

Answer 26

so plugging in f and g

Answer 27

Yep would look something like this...
\[f(g(x))=\left( \frac{ 1 }{ x-1 } \right)^2+6\left( \frac{ 1 }{ x-1 } \right)+5\]

Answer 28

ok let me write this down lol

Answer 29

simplify-->
oh shoot messed up a sign ah

Answer 30

x+1**

Answer 31

\[f(g(x))=\left( \frac{ 1 }{ x+1 } \right)^2+6\left( \frac{ 1 }{ x+1 } \right)+5\]
\[f(g(x))=\left( \frac{ 1 }{ x+1 } \right)\left( \frac{ 1 }{ x+1 } \right)+\left( \frac{ 6 }{ x+1 } \right)+5\]
\[f(g(x))=\left( \frac{ 1 }{ x^2+2x+1 } \right)+\left( \frac{ 6 }{ x+1 } \right)+5\]

Answer 32

& alternate forms of that ^

\[f(g(x))=\frac{ 5x^2+16x+12 }{ x^2+2x+1 }\]

Answer 33

yay i got that right

Answer 34

Woo :D good job!

Answer 35

lol think you can help me with a couple more you explain very well

Answer 36

yeah sure i would love too :D

Answer 37

Okay so the first one-->
we are working with an arithmetic sequence:
If the sequence is 18,54,90,126
*Notice that each therm is increasing by 36.
If:
\[a_{n}=a_{n-1}+d\]
&
an= nth term in sequence
an-1=the term before the nth term in the sequence
&
d=the common difference
than the formula is :
\[a_{n}=a_{n-1}+36\]

Answer 38

So: they're giving us up to the 4th term in the sequence & they're asking about the eighth term:
Basically there's the logic ^
But you can just multiply 36*4 and add it the the 4th term and you arrive at the same results

Answer 39

@lolamona
sooo... what is the eighth term in the sequence? aka the distance the projectile traveled in 8 seconds

Answer 40

hold on

Answer 41

im lost -.-

Answer 42

Hmm.. okay so lets work with the arithmetic sequence actually

Answer 43

With @sunnnystrong, you are in good hands, @lolamona ....

Answer 44

\[a_{n}=a_{n-1}+36\]

\[a_{1}=18\]
\[a_{2}=18+(2-1)36\]
\[a_{2}=18+(1)36=54\]
\[a_{3}=18+(3-1)36=90\]
\[a_{4}=18+(4-1)36=126\]
\[a_{5}=18+(5-1)36=162\]
\[a_{6}=18+(6-1)36=194\]
\[a_{7}=18+(7-1)36=226\]
\[a_{8}=18+(8-1)36=258\]

Answer 45

@lolamona there's the sequence.. up to the eighth term.\[a_{n}=a_{1}+(n-1)d\]
d= common difference (32)
a1=18 (aka first term in sequence)

Answer 46

so 18 is the first number so a1 and what ever number you are trying to find you put subtracted by 1 and the common difference is 32

Answer 47

Yep :D

Answer 48

okay i messed up somewhere ^^ one sec haha

Answer 49

for 8 its 270

Answer 50

from the 6th term --> guess i messed up simple math ok lol
a6=198
a7=234
a8=270

okay lets do the next :D

Answer 51

kk

Answer 52

\[s_{n}=\frac{ 1 }{ 2 }(a_{1}+a_{n})\]
sn= the sume of the first n #'s inn the sequence (in this case up to 24 & prior)
a1= first # in secuence
an= nth number in sequence

Answer 53

\[s_{24}=\frac{ 1 }{ 2 }(a_{1}+a_{24})\]

Answer 54

we will have to find a1 no

Answer 55

pretty much yea :P

Answer 56

so if we know the common difference (aka d) is 3.7 &
\[a_{13}=1.9\]
than:
\[a_{13} =a_{1}+12d=1.9\]
Solve for a1
a1=-42.5

find a24:
\[a_{24} =-42.5+(24−1)3.7=42.6]

Answer 57

omg thank you so much for explaining it so well i really appreciate your help and patience

Answer 58

can i message you anytime i need help with precalc you have really helped me alot

Answer 59

no problem! :D always happy to help!
so
a24=42.6
a1= -42.5

Than-->
\[S_{24}=.5(24)((-42.5)+(42.6))=1.2\]

Answer 60

phewwww finally done with that sequence stuff :D good job & yeah sure if i can i will try to help :)

Answer 61

lol thank you , you have been an amazing help

Answer 62

thanks you :) !