All Answers 358098 Answers

3.333333333 has no end.

For example,

$$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...$$

If you add 1/2 that is half to previous number and continue it,

you will see that the sum of those are going closer and closer to 1.

By common sense, they will never reach 1.

But,

$$S_n = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}$$

If I multiply this by 2, then we can see it equals to $$S_n$$.

$$2S_n=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+...+\frac{2}{2^n}
=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{2}{2^n^-^1}
=1+S_n-\frac{1}{2^n}$$

If we subtract $$S_n$$ from both sides,

$$s_n = 1 - \frac{1}{2^n}$$

As n is infinity, $$S_n = 1.$$

I think this proof apply to your question too.

As you said 3.33333+3.33333+3.33333 = 9.99999

But 3.333...+3.333...+3.333...=1

Hope it helped! 

$$\\5000=PV(1.04)^{10}\\\\
PV= 5000\div(1.04)^{10}\\\\$$

In triangle ABC, AB = 8.3cm and BC = 5.3cm.

a) Describe the possible triangles

Point C lies on the circle and Point B is centre of the circle with radius 5.3cm

b) To the nearest degree, determine possible measures of acute angle A

In the right triangle with max. A: Max. A = arcsin(5.3 / 8.3) = 39.6840953824 degrees.

A = 0...39.6840953824 degrees

4^2+12x=27

16+12X=27

Now subtract 16 from both sides

then

divide both sides by 12   

How many equal parts?

you could divide it into half inch squares.

14 by 11

that would give you 154 equal squares :)

$$\\3^x=3^{5*4}\\\\
x=20$$

OABC is a square. M is the mid-point of OA, and Q divides BC in the ratio 1:3. AC and MQ meet at P.  If O to A = a, O to C = c .  Express O to P in terms of a and c.

1.  $$\small{\text{
If $ \vec{Q} = \vec{c} + \frac{3}{4}\vec{a}
$ and $ \vec{M} = \frac{1}{2}\vec{a} $
so line $ \overline{MQ} $ is $\frac{1}{2}\vec{a} + \lambda ( \vec{Q}-\vec{M}) = \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) $
}}$$

2.  $$\small{\text{
Line $ \overline{AC} $ is $\vec{a} + \mu ( \vec{c}-\vec{a}) $
}}$$

3.

$$\\\small{\text{
The intersection of the line $\overline{MQ}$ with the line $\overline{AC}$ is
}}
$\\$\small{\text{
the point $\vec{P} = \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) $.}}
$\\$\small{\text{
We equate:
$
\frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) = \vec{a} + \mu ( \vec{c}-\vec{a})
$
}}
$\\$\small{\text{
so
$
\lambda (\vec{c}+\frac{1}{4}\vec{a}) - \mu ( \vec{c}-\vec{a}) = \frac{1}{2}\vec{a}
$
}}
$\\$\small{\text{
The perpendicular vector of $(\vec{c}-\vec{a})$ is $(\vec{c}+\vec{a})$, because we have a square and
}}
$\\$\small{\text{
$(\vec{c}-\vec{a}) * ( \vec{c}+\vec{a} ) = 0$, because the dot-product of vectors
}}
$\\$\small{\text{
perpendicular to each other is a Zero.
}}
$\\$\small{\text{
We multiply our equation with $( \vec{c}+\vec{a} )$ and see what passed:
}}$$

$$$\\$\small{\text{
$
\lambda (\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a}) - \mu \underbrace{( \vec{c}-\vec{a}) (\vec{c}+\vec{a})}_{dot-product = 0} = \frac{1}{2}\vec{a} (\vec{c}+\vec{a})
$
}}
$\\$\small{\text{
Now we can dissolve after $\lambda$ and receive for $\lambda$:
}}
$\\$\small{\text{
$
\lambda = \dfrac{
\frac{1}{2}\vec{a} (\vec{c}+\vec{a}) }
{(\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a})
}
$
}}
$\\$
\boxed{
\small{\text{
$\vec{P} = \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) $}} = \lambda \vec{c} + (\frac{1}{2}+\frac{\lambda}{4})\vec{a}
$
}}}$$

I don't know the answer yet, I'm gonna ask it to my teacher today.

But I think you are on right line.

Anyway, I will post the full answer here today.

Thanks for answering though 

$$\\4=a*32^r\qquad(1)\\\\
1=a*2^r\qquad(2)\\\\
\frac{4}{1}=\frac{a*32^r}{a*2^r}\\\\
4=16^r\\\\
4^1=4^{2r}\\\\
2r=1\\\\
r=0.5$$

.

@@ End of Day Wrap      Thurs 11/12/14       Sydney,   Australia         Time 6:10 pm          ♬

Hello everyone, 

It has been another busy day.  Great answers from Heureka, Alan, CPhill, Happy7, flflvm97, Nickdavis, shaniab29544, JMT3137681, hugekase, sasini, Alexa4life, geno3141, NinaTinasTwin, saseflower, SevenUp, Sorasyn and Zegroes.  Thanks everyone.

Here are some interest posts:

1) SevenUp tells us we can expect a play  

Fri 12/11/14

1) This is hilarious.  Thanks happy.

ok

a nickel is 5c

a dime is 10c and a quarter is 25c    

I had to look that up - we are not all from the USA you know!

(I got it wrong last time so I thought I better check LOL)

x nickels

y=4.00+0.05x

12=4+0.05x

8=0.05x

8/0.05=x

x=160

160 nickels

What is the question?

Use the calc on the home page.

$$\\7876+1.2*7876+1.2^2*7876+.......<50000\\
7876+1.2*7876+1.2^2*7876+.......<50000\\
7876(1+1.2+1.2^2+.....................) <50000\\
$In the brackets is the sum of a GP$\\\\
7876(\frac{1.2^n-1}{0.2})>50000\\\\
\frac{1.2^n-1}{0.2}>50000/7876\\\\
1.2^n-1>(50000/7876)*0.2\\\\
1.2^n>\{[(50000/7876)*0.2]+1\}\\\\
log(1.2^n)>log\{[(50000/7876)*0.2]+1\}\\\\
nlog(1.2)>log\{[(50000/7876)*0.2]+1\}\\\\
n>\frac{log\{[(50000/7876)*0.2]+1\}}{log(1.2)}\\\\$$

$${\frac{{log}_{10}\left(\left(\left({\frac{{\mathtt{50\,000}}}{{\mathtt{7\,876}}}}\right){\mathtt{\,\times\,}}{\mathtt{0.2}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{log}_{10}\left({\mathtt{1.2}}\right)}} = {\mathtt{4.495\: \!567\: \!527\: \!544\: \!848\: \!3}}$$

So it will take 5 months

It would have been easier to do it without the formulas LOL.

I always do everything the long way.

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

Here is an easier method

Divide.  (6i)/(5-4i)=  ?

$$\small{\text{
$ \dfrac{6i}{5-4i} \times \dfrac{5+4i}{5+4i} $
}}$\\\\$\small{\text{
$ = \dfrac{6i*(5+4i)}{(5-4i)(5+4i)} $
}}$\\\\$\small{\text{
$ = \dfrac{30i+24i^2}{25-16i^2} \quad \boxed{ i^2 = -1 } $
}}$\\\\$\small{\text{
$ = \dfrac{30i-24}{25+16} $
}}$\\\\$\small{\text{
$ = \dfrac{30i-24}{41} $
}}$\\\\$\small{\text{
$ = -\dfrac{24}{41} + \dfrac{30}{41}i $
}}$$

4.8/1.2=4

So 4 units in the diagram = 1 unit real life

scale is  4:1

inverse cos is the same as arc cos

On the site calc it is acos

1cm=1/4cm or 0.25cm

or

$$7(cos330+ i sin330)\; = \;7(cos(-\pi/6)+isin(-\pi/6)) = 7e^{-\pi i/6}$$

You can do it one the calc on the home page

The symbol for "to the power of" is ^