Question

f (x) = 2x + 1

Answer 1

ok now we have something
\[f(x)=2x+1\] what we don't have this time is a question!!

Answer 2

any chance you can scan this or link to it somehow?

Answer 3

how?

Answer 4

are you supposed to find \(f(3)\) for example? or graph the function? there must be some question here
\[f(x)=2x+1\] is just a statement, a definition of this particular function

Answer 5

it is like saying "i am satellite73"
it is not a question, just a statement

Answer 6

i think im supposed to find f(3)

Answer 7

oh, i see

Answer 8

oooooh!!! ok now we have a question!!

Answer 9

seems rather off to me as well

Answer 10

\[f(x)=2x+1\]
\[f(3)=2\times 3+1=6+1=7\]

Answer 11

i.e. replace \(x\) by \(3\) and compute

Answer 12

i don't know if it's "3" specifically though!!! ughh

Answer 13

write the entire question, i am sure we can answer it

Answer 14

this is such bull****. the teacher can't even clarify.

Answer 15

I kid you not, there's no so-called "question"

Answer 16

there must be more words somewhere
for example, now we know \(f(x)=2x+1\) right?
do we have another function, say \(g(x)\) ?

Answer 17

do you have gmail

Answer 18

i am dying to see what is in front of you

Answer 19

it's a google doc

Answer 20

ok send it via gmail

Answer 21

i mean
satellite73.openstudy@gmail.com

Answer 22

https://docs.google.com/file/d/1XqM16ad5Du-6x8wJSe7jGbPhyVwSsck_2YC4dlaTRK0V1iKNjm2T8JG6x1ZZ/edit?usp=sharing

Answer 23

can you see that?

Answer 24

i sent it via email too

Answer 25

it says i don't have permission to access it, but send it to the email

Answer 26

okay got it

Answer 27

man this is a real piece of crapola isn't it?

Answer 28

but we can do it!!

Answer 29

yes, it really is

Answer 30

lets start with 4

Answer 31

you have \(f(x)=2x+1\) and \(g(x)=2x-5\) and you want
\[f(g(2))\]
first we find \[g(2)=2\times 2-5=4-5=-1\]

Answer 32

then we find
\[f(-1)=2\times (-1)+1=-2+1=-1\]

Answer 33

so the answer to number 4 is
\[f(g(2))=f(-1)=-1\] done !

Answer 34

perfect. youre perfect.

Answer 35

what about the rest of 4?

Answer 36

sorry if im annoying. i just need this within the next 20 min

Answer 37

the rest of 4 is some sort of typo, or else has to do with the problem beneath it

Answer 38

this whole thing is ridiculously written junk, sorry

Answer 39

no worries. not your error.

Answer 40

thank you

Answer 41

we can go one to 5 i you like

Answer 42

yes

Answer 43

you are asked to solve
\[-x+y=3\]\[-3x+2y=1\] the solution is
\[x=5,y=8\] i can explain if you like

Answer 44

i got the solution down, is that all i have to do? i can't graph virtually..

Answer 45

this sheet has not actual questions on it, so it is kind of hard to figure out what the instructions are
i am just guessing that the question was to solve for \(x\) and \(y\) and you don't need to graph to do it

Answer 46

module 6 has two expressions
the first one is an addition, and since they are not like terms you cannot do anything with it at all

Answer 47

what is this madness

Answer 48

the second is \(3x^3y^2\times 4x^5y^8\)

Answer 49

the second on module 5?

Answer 50

this we can do easily by adding the exponents
\[12x^8y^{10}\]

Answer 51

no, the second on module 6
we did module 5, got \(x=5,y=8\)

Answer 52

there are two parts to module 6, the first you can do nothing with, the second you can multiply and get \(12x^8y^{10}\)

Answer 53

see its written so oddly i thought it was two separate ones

Answer 54

i can see why you are confused
frankly i am pretty much guessing at what whoever wrote this wants
he/she should be embarassed

Answer 55

im so confused as to which part youre doing

Answer 56

module 6 has two expressions right?

Answer 57

actually scratch that, it has 6 expressions

Answer 58

yes

Answer 59

the first one \(9x^2 + 3x^3y^2z\) you can do nothing with

Answer 60

the second one \[(3x^3y^2)( 4x^5y^8)=12x^8y^{10}\]

Answer 61

third one by subtracting exponents you get
\[\frac{5a^3c^3}{b^2}\]

Answer 62

with me so far?

Answer 63

yes sir

Answer 64

fourth one when you square you get
\[16w^{14}v^8\]

Answer 65

couldve had a v8 haha get it

Answer 66

the third one is the number 3,000,000 and i am going to make a big guess,which is that you are supposed to write this in scientific notation as
\[3,000,000=3\times 10^6\]

Answer 67

yeah , with vodka and hot sauce

Answer 68

hahahaha i'm russian so you're not too far off with the vodka

Answer 69

that is really a guess because there are literally no instructions
just the number 3,000,000

Answer 70

what "third one"

Answer 71

this teacher is either the laziest on the face of the earth or a moron

Answer 72

sorry i meant fifth one

Answer 73

hahahaha she doesn't even teach math; she's a bio teaher

Answer 74

the one were you are looking at the number 3,000,000 and have to decide what the question is

Answer 75

yes

Answer 76

then it is the schools fault for letter her teach math

Answer 77

*letting

Answer 78

i have to call her on the phone fairly soon and itll be awkward

Answer 79

and finally the last one is \(2.6\times 10^{-2}=0.026\)

Answer 80

move the decimal point two places left in other words

Answer 81

how about module 3, you good with that?

Answer 82

thank you. you're really great for all of this.

Answer 83

yw
you should put her on the phone with me so i could straighten her out

Answer 84

hahahahaha omg bad idea

Answer 85

like the skull
good luck with the phone call (phone call??)

Answer 86

It's to catch up on math-related things is all

Answer 87

Much appreciated

Answer 88

i have an idea for the rest of number 4, the second part

Answer 89

I'll try not to bash my head against a wall

Answer 90

oh?

Answer 91

no don't do that
you have \(y=7x-2\) and underneath is written \(y=-2\) so maybe you are supposed to solve
\[-2=7x-2\] for \(x\) and get \(x=0\)

Answer 92

then you also have \(x=3\) so maybe you are supposed to write
\[y=7\times 3-2=21-2=19\]

Answer 93

really i am just guessing
i have never seen such a sheet with a bunch of stuff and not one word of instruction
must be her first year teaching outside of the esp institute

Answer 94

we can do module 3 if you like, up to you

Answer 95

It should not be this difficult, and no worries, module 3 is complete. it was the easiest

Answer 96

okay good luck
with the phone call