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That is a good question Dragonlance.

I have laid it out a little better now so hopefully it is a little clearer    

$$f(\pm1)=\pm1-3\mp2+6\;\;=\;\;3\pm1\mp2\;\;\rightarrow\;\;2\ne0\;\;or\;\;4\ne 0\\$$

The signs on the top belong with f(+1)

and the signs on the bottom belong with f(-1)

FYI, as usual, im the one who has access to make this story perfect sense

HAVE BEEN meant that Rosala and 7UP were released outside the bag BUT the Joker hld on to them both

anyways, theyre fine........for now.................

y/8 - 3/4 = 7/12     add 3/4 to each side

y/8  = 7/12 + 3/4     get a common denominator on the right

y/8 = 7/12 + 9/12

y/8 =  16/12

y/8 = 4/3      multiply both sides by 8

y = 32/3

I thought I put more than two up there.

Miss Melody is there a difference between these plus minus things?

$$\pm \;\mp$$

I not ever see the one on the right before.

LOL!!!....Melody always says that she likes to "complicate" things.....actually, both approaches are good .... the "Rational Zeros" approach is always helpful if it's not immediately obvious how to factor a polynomial [if it's actually capable of being factored, that is ]

$$\\f(x)=\frac{Log[1 + 3*Sin[x]^2]}{2}\\\\
f(\pi/2)=\frac{Log[1 + 3*Sin[\pi/2]^2]}{2}\\\\
$now is this meant to be $[sin(pi/2)]^2\;\;or\;\;\;sin[(pi/2)^2] \;\;??\\\\
$I will assume $\;\;[sin(pi/2)]^2\\\\
f(\pi/2)=\frac{Log[1 + 3*[sin(pi/2)]^2]}{2}\\\\
f(\pi/2)=\frac{Log[1 + 3*1]}{2}\\\\
f(\pi/2)=\frac{Log[2]}{2}\\\\
$ now you can do it on a calc$\\\\\\
f(0)=\frac{Log[1 + 3*[sin(0)]^2]]}{2}\\\\
f(0)=\frac{Log[1 + 0]}{2}\\\\
f(0)=0\\\\$$

Yes, CPhill's way is better :)

Here's another approach:

(x^3-3x^2)-(2x-6)=0     factor  by grouping

x^2(x - 3) - 2(x -3)  = 0

(x^2 - 2) (x - 3) = 0        set each factor to 0  ....     and  we find that x = ±√2, 3

Here's a graph.......

Dragonlance, you usually give good answers.......we mods aren't always correct, either....I seem to remember a few days ago that you spotted one of my errors on a Physics question....!!!!

$$\\(x^3-3x^2)-(2x-6)=0\\
x^3-3x^2-2x+6=0$$

Now use remainder theorem to find a root

$$\\let\\
f(x)=x^3-3x^2-2x+6\\
f(\pm1)=\pm1-3\mp2+6\;\;=\;\;3\pm1\mp2\;\;\rightarrow\;\;2\ne0\;\;or\;\;4\ne 0\\
f(3)=27-27-6+6=0\qquad therefore\;\;x=3 \;\;$is a root$$$

 (this has been edited a little)

I used polynomial division to determine that

$$\\(x^3-3x^2-2x+6)\div(x-3)=(x^2-2)\\\\
so\\\\
x^3-3x^2-2x+6=0\\\\
(x-3)(x-\sqrt2)(x+\sqrt2)=0\\\\
x=3\quad or\quad x=\sqrt2\quad or \quad x=-\sqrt2$$

minus temperatures are very cold.  Positive ones are hot.

If you take away cold air from a room then the room gets hotter.   

So        - - = +

3 - - 4 = 3+4 = 7

Sorry. I thought maybe it do have three because there are three angles in a triangle. But I look at about 20 questions on maths sites that have altitude or height as part of the problem and it only ever have one altitude part, so I didn’t know. But I do now. :)

A good thing to know is if a mod have a answer different from someone else then the mod is usually right. They always say so if they make a mistake and they not make very many anyway.

Hallo anonymous!

solve 3 x 2^a-1 = 96

3 x 2^a-1 = 96    

lg 3 + (a - 1)*lg 2 = lg 96

a -1 = (lg 96 - lg 3) / lg 2

a = (lg 96 - lg 3) / lg 2 + 1 

a = 6

test:

3 * 2^(6-1) = 96

3 * 2^5 =96

3 * 32 = 96   okey

:- )

Here you go, I love thissite.

OH I thought if he said 7UP  had been in that would be past tense. I though have been in is the same as is in

I still learing this verb tense thing. I think I know how to read it ok but I not know how to write it correctly yet. 

Any way I like Rosala's singing but not as much as 7UP like it :)

LOL  Dragonlance   

I don't thinks TR had you all in the bag at the same time.  It was the same bag but  the others had each been dumped out of it already.   

"joker puts Dragonlance in the same kidnapping bad Rosala and 7up have been in"

See, it is past tense, you get the bag all to yourself!  Wow, what luxury.  :D

How can you write all that TR?

Anyway you stick me in the kidnap bag with Rosala and 7UP and it was really crowded. Sometime three is a crowd and it always is a crowd inside a kidnap bag.

When you put me in there Rosala was singing to 7UP. She was singing “I only have fangs for you.”

I know 7UP like it because he listen closely and not scratch any werewolf fleas while she was singing, and after she was finished he howl and say “bite me.” Hahahaha

That depends,

If it is exact than it is rational.

If it is an approximation because your calc ran out of digits, then it most likely represents an irrational number but i would have to see the origonal calc input to give you a more definite answer.  :)

@@ End of Day Wrap   Tues 25/8/15     Sydney, Australia Time   7:37pm       ♪ ♫   

Hi all.

Today our fabulous answers were provided by Heureka, asinus, Dragonlance, Geno3141, CPhill, TitaniumRome, Jcersy, Matsunnymat and Exacto.    Thanks everyone  

Huh,

There are only 2 questions and I did both of them.

The axes need labels but i could not work out how to do that :/

Thanks Chris :)))

 poor u 

$$\\\frac{8}{3}*x-1=\frac{3}{x}-1\\\\
\frac{8x}{3}-1=\frac{3}{x}-1\\\\
\frac{8x}{3}=\frac{3}{x}\\\\
3x*\frac{8x}{3}=3x*\frac{3}{x}\\\\
8x^2=9\\\\
x^2=\frac{9}{8}\\\\
x=\pm\frac{3}{\sqrt{8}}\\\\
x=\pm\frac{3}{2\sqrt{2}}\\\\
x=\pm\frac{3\sqrt{2}}{4}\\\\$$