All Answers 358116 Answers

I just did it and deleted it accidently.  start by multiplying both sides by 5.  That will get rid of all the fractions.

You try it and see how you go.  

$$\begin{array}{rll}
-3(x-1)&=&9\\
(x-1)&=&-3\qquad \mbox{divide both sides by minus 3}\\
x-1+1&=&-3+1\\
x&=&-2
\end{array}$$

now you should sub x=-2 into the original equation to check that it works.

Interesting!

$$^7C_0(-1)^0(i)^7+
^7C_1(-1)^1(i)^6+
^7C_2(-1)^2(i)^5+
^7C_3(-1)^3(i)^4+
^7C_4(-1)^4(i)^3+
^7C_5(-1)^5(i)^2+
^7C_6(-1)^6(i)^1+
^7C_7(-1)^7(i)^0\\\\$$

$$=(i)^7+
7(-1)^1(i^2)^3+
^7C_2*1*(i^4)i+
^7C_3*-1*(i^2)^2+
^7C_4*1*(i^2)i+
^7C_5*-1*(i^2)+
7*1(i)^1+-1\\\\$$

$$=(-1)i+
7*-1*(-1)^3+
^7C_2*1*(-1)i+
^7C_3*-1*(-1)^2+
^7C_4*1*(-1)i+
^7C_5*-1*(-1)+
7*1*(i)^1+
-1\\\\$$

$$=-i+7+^7C_2(i)+
^7C_3(-1)+
^7C_4(-i)+
^7C_5(-1)+
7i
-1\\\\$$

$$=-i+7+21i
-35
-i35
+21
+7i
-1\\\\
=i(-1+21-34+7-1)+(7-35+21-1)\\\\
=i(-8)-8\\\\
=-8-8i$$

NOTE: I would actually be surprised if this is not riddled with careless errors. (I have fixed one little error)

BUT that is the technique that i would use.

Thanks admin.  

It is the distance from zero on the number line.

examples:

|6|=6

|-8|=8

-3*|-6+5|=-3*|-1|=-3*1=-3

I believe i already answered this, this morning (Australia time) just look through the other questions and answers.

Sorry but there is no question here.

IDK sorry. :( 

On the web2.0calc calculator put it in as: log(16, 3) 

The result should give you: $${{log}}_{{\mathtt{3}}}{\left({\mathtt{16}}\right)}$$

---------------

The  answer above is correct as well.

try putting it in as log16/log3

I believe the answer to your question is $1.30.

How I did it: 

65%=65/100=0.65 this is the percentage in decimals

0.79x0.65=0.5135$ this the amount (of money, in this case) it is increasing by, so add it to the original price:

0.79$+0.5135$=1.3035 round that to 1.30$ that's the answer

ta-DA! :P

Find 65% of 79 c/= (Yay rudimentary cent sign!) to begin with.

$${\mathtt{79}}{\mathtt{\,\times\,}}{\mathtt{0.65}} = {\frac{{\mathtt{1\,027}}}{{\mathtt{20}}}} = {\mathtt{51.35}}$$

Add that number to the original.

$${\mathtt{79}}{\mathtt{\,\small\textbf+\,}}{\mathtt{51.35}} = {\frac{{\mathtt{2\,607}}}{{\mathtt{20}}}} = {\mathtt{130.35}}$$

The candy bar would then cost 130 c/= (Or, if you prefer, $1.30)

1,1x-6x = -4,9x

-4,9x = -34,3

34,3/4,9 = 7

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{60}}^\circ\right)} = {\frac{{\mathtt{x}}}{{\mathtt{12}}}}$$

$${\mathtt{12}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{60}}^\circ\right)} = {\mathtt{x}}$$

$${\mathtt{6}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}} = {\mathtt{x}}$$

If you want to check the solution, refer back to the unit circle. $$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{60}}^\circ\right)}$$ on the unit circle is $${\frac{{\sqrt{{\mathtt{3}}}}}{{\mathtt{2}}}}$$.  When you put $${\mathtt{6}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}$$ in place of x for the $${\frac{{\mathtt{x}}}{{\mathtt{12}}}}$$, you get $${\frac{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\right)}{{\mathtt{12}}}}$$. Simplify that number further and you get $${\frac{{\sqrt{{\mathtt{3}}}}}{{\mathtt{2}}}}$$. 

cotx/sinx.cosx

 $$cotx\div sinx\times cosx\\\\
=\frac{cosx}{sinx}\div sinx\times cosx\\\\
=\frac{cosx}{sinx}\times \frac{1}{sinx}\times cosx\\\\
=\frac{cos^2x}{sin^2x}\\\\
=Cot^2x$$

it is barely 4

I'll just finish it.

In 500 rolls he can expect to get this oucome about  27.77...%* 500 =  139 times.

$${\mathtt{0.277\: \!777\: \!777\: \!77}}{\mathtt{\,\times\,}}{\mathtt{500}} = {\frac{{\mathtt{1\,250}}}{{\mathtt{9}}}} = {\mathtt{138.888\: \!888\: \!885}}$$

It goes from left to right.

In primary school you learned about number lines.  They go from left to right across the page.

X comes before Y in the alphabet.

so the one you learned first is the X axis.

$$\sqrt{\frac{9}{49}}=\frac{3}{7}$$

Your username does not inspire a lot of confidence failed.

Your method is reasonable but it would only be correct if the order of selection mattered and i don't think that it does in this case.

The answer is 12C5 = $${\left({\frac{{\mathtt{12}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)} = {\mathtt{792}}$$

this is

$$\frac{12*11*10*9*8}{5*4*3*2*1}$$           

So first you select your 5 players just as you did.  

but then you think about how many ways those 5 players can be ordered.  And you divide by this number.

does this make sense?

26*25*24=15600

or

$$^{26}P_3 =$$     $${\left({\frac{{\mathtt{26}}{!}}{({\mathtt{26}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{15\,600}}$$

It is also     $$^8P_8$$

That is, there are 8 different objects.  In how many ways can you select 8 and order them.

$${\left({\frac{{\mathtt{8}}{!}}{({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{8}}){!}}}\right)} = {\mathtt{40\,320}}$$

Alan's method is probably best but if you want to start playing with LaTex you can open it up and type

log_3(x+1)=4

and this will be your output

$$log_3(x+1)=4$$

-6+12=

Use the number line.

Start at -6 then go 12 places in the positive direction.  What is the answer?

If you haven't already seen this clip, it is definitely worth a look!

(It has been cycling throught the forum for a long time)