Question

PLEASE HELP! I WILL MEDAL! it's a tricky question

Answer 1

I don't know how to solve it, so Please try to explain it to me.

Answer 2

@iGreen

Answer 3

I think it could be C.

Answer 4

you are correct it is C

Answer 5

Well, in general; suppose that you are given,

\(\color{#000000 }{ \displaystyle x =\frac{a}{b}z +c }\)

and suppose you want to solve for z, then,

\(\color{#000000 }{ \displaystyle x\color{red}{-c} =\frac{a}{b}z +c\color{red}{-c} }\)

\(\color{#000000 }{ \displaystyle x-c =\frac{a}{b}z }\)

multiply both sides times b/a.
(On the right side, (a/b)•(b/a)=1)

\(\color{#000000 }{ \displaystyle \frac{b}{a}(x-c) =z }\)

Answer 6

\(\color{#000000 }{ \displaystyle x =\frac{a}{b}z+c\quad \Longrightarrow \quad z=\frac{b}{a}(x-c) }\)

right?

Answer 7

Yeah.

Answer 8

your claim was,

\(\color{#000000 }{ \displaystyle {\rm T_F}=\frac{9}{5}{\rm T_C}+32\quad \Longrightarrow \quad {\rm T_C}=\frac{9}{5}({\rm T_F}+32) }\)

Answer 9

So, C is, correct, right?

Answer 10

I don't think so.

Answer 11

Let's once again try to solve for \({\rm T_C}\) ok?

Answer 12

Okay.

Answer 13

You are given that:

\(\color{#000000 }{ \displaystyle {\rm T_F}=\frac{9 }{5} {\rm T_C}+32 }\)

Answer 14

What do you think you first step should be?
(What would you first try to isolate if you are solving for \(\color{#000000 }{ {\rm T_C}}\)) ?

Answer 15

I'm not sure, I would think it is re-arrange the problem?

Answer 16

Yes, but if you are solving for \(\color{#000000 }{ \displaystyle {\rm T_C} }\) you would first try to put the term(s) that contains \(\color{#000000 }{ \displaystyle {\rm T_C} }\) on one side of the equation and the rest of the terms on the other side of the equation.

Answer 17

Okay,

Answer 18

Which term, ((on the side where there's a term containting "\(\color{#000000 }{ \displaystyle {\rm T_C} }\)")), does not contain "\(\color{#000000 }{ \displaystyle {\rm T_C} }\)"?

(hope this question isn't too hard to read)

Answer 19

They all have Tc on one side though..?

Answer 20

Recalling the equation:
\(\color{#000000 }{ \displaystyle {\rm T_F}=\frac{9 }{5} {\rm T_C}+32 }\)

There are two terms on the right side, and namely,
(1) \(\color{#000000 }{ \displaystyle \frac{9 }{5} {\rm T_C} }\)

(2) \(\color{#000000 }{ \displaystyle 32 }\)

(And the term \(\color{#000000 }{ \displaystyle {\rm T_F} }\) is already separated - on a different side - from \(\color{#000000 }{ \displaystyle {\rm T_C} }\))

Answer 21

So, (ca you answer my question ?)

Which term - on the side where \(\color{#000000 }{ \displaystyle {\rm T_C} }\) containing terms are located - does not contain \(\color{#000000 }{ \displaystyle {\rm T_C} }\)?

Answer 22

I got no clue, I'm sorry! I suck at this

Answer 23

Would you say that \(\color{#000000 }{ \displaystyle \frac{9 }{5} {\rm T_C} }\) or \(\color{#000000 }{ \displaystyle 32 }\) has a "\(\color{#000000 }{ \displaystyle {\rm T_C} }\)" inside of it?

Answer 24

Yes..?

Answer 25

Was that a yes or no question?

Answer 26

Lol. I thought it was.

Answer 27

Ok, ... can you please answer the last question I asked you?

Answer 28

(Well, if you don't understand what I'm asking let me know)

Answer 29

I don't understand what you are asking,
I mean, Yes, 9/5 Tc has has a 'Tc' in it.

Answer 30

Yes, that is all I asked.

Answer 31

And "32" does not have the '\(T_c\)' in it.

Answer 32

So, we would like to isolate the term that does have the "\(T_C\)" on one side.

Answer 33

Okay, if the answer isn't C, it has to be A..

Answer 34

However, there also a 32 on the same side, and that creates a little problem.

(reminding, your equation is, \(\color{#000000 }{ \displaystyle {\rm T_F}=\frac{9 }{5} {\rm T_C}+32 }\))

So, what would you add/subtract from both sides of this equation, to isolate/separate the term that does have the \(\rm {T_C}\) in it?

Answer 35

(((If you ask these kind of questions it might seem as though you are just trying to get the answer, rather than learn)))

Answer 36

I'm trying to get the answer :P
And learn a little too. xD

Okay, Why do we need to add/subtract from 32? The choices all have 32 in them...?

Answer 37

we are not just removing the 32...

Answer 38

\(\color{#000000 }{ \displaystyle {\rm T_F}=\frac{9 }{5} {\rm T_C}+32 }\)

\(\color{#000000 }{ \displaystyle {\rm T_F}\color{red}{-32}=\frac{9 }{5} {\rm T_C}+32\color{red}{-32} }\)

what did I do to the equation?

Answer 39

Subtracted 32 from both sides

Answer 40

yes, good

Answer 41

What will your right side simplify to, can you tell me?

Answer 42

(you know that 32-32=0, and R+0 is same thing as just R)

Answer 43

There isn't a solution for that

Answer 44

if I gave you the following to simplify,

x+4-4

then, what would you simplify this to?

Answer 45

x

Answer 46

Yes, exactly

Answer 47

\(\color{#000000 }{ \displaystyle {\rm T_F}=\frac{9 }{5} {\rm T_C}+32 }\)

\(\color{#000000 }{ \displaystyle {\rm T_F}\color{red}{-32}=\frac{9 }{5} {\rm T_C}+32\color{red}{-32} }\)

What will your right side simplify to?

Answer 48

9/5 TC?

Answer 49

Yes, good

Answer 50

\(\color{#000000 }{ \displaystyle {\rm T_F}=\frac{9 }{5} {\rm T_C}+32 }\)

\(\color{#000000 }{ \displaystyle {\rm T_F}\color{red}{-32}=\frac{9 }{5} {\rm T_C}+32\color{red}{-32} }\)

\(\color{#000000 }{ \displaystyle {\rm T_F}\color{red}{-32}=\frac{9 }{5} {\rm T_C} }\)

Answer 51

\(\color{#000000 }{ \displaystyle ({\rm T_F}\color{red}{-32})=\frac{9 }{5} {\rm T_C} }\)

by putting parenthesis, I didn't change anything in the equation .. just noting that i didn't change the equation, rather I just take it as one term.

Answer 52

So A is correct, rightttt?

Answer 53

@solomonZelman

Answer 54

no, A is not correct.
(I had a problem with my internet connection)

Answer 55

\(\color{#000000 }{ \displaystyle ({\rm T_F}-32)=\frac{9 }{5} {\rm T_C} }\)

\(\color{#000000 }{ \displaystyle \color{red}{ \frac{5}{9}\times} ({\rm T_F}-32)=\frac{9 }{5} {\rm T_C}\color{red}{\times \frac{5}{9}} }\)

Answer 56

It would be B than?

Answer 57

yes