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1%3 = 2 is wrong expression dude

If you were a member then your answer would have been automatically saved in your personal watchlist and you would have no problem finding it.   

$$\\1500000=25000(1+r)^{13}\\
60=(1+r)^{13}\\
60^{1/13}=1+r\\
r=60^{1/13}-1\\$$

$${{\mathtt{60}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{13}}}}\right)}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0.370\: \!190\: \!226\: \!566\: \!422}}$$

$$\\37\%\\$$

I will  assume that the coupon rate is annual 

I am not sure but i think you are just asking what $900 will grow to in 10 years if interest is compounded.

FV=900(1+0.05)^10

You need to explain how you did it selena     

That was a great piece of trivia - thanks anon  

There you go.

I am practicing my philosophical approach.

Why do it the easy way if there is a long way that works just as well.    LOL

Thanks Alan.  

24x 3rd power + 375

$$\\24x^3+375\\
=3*8x^3+3*125\\
=3(8x^3+125)$$

To answer your question Melody, use the following notation;

$$A=ae^{i\alpha}$$

$$B=be^{i\beta}$$

then

$$\frac{A}{B}=\frac{a}{b}e^{i(\alpha - \beta)}\rightarrow \frac{a}{b}(\cos(\alpha -\beta)+i\sin(\alpha - \beta))$$

.

$${\mathtt{0.7}} = {\frac{{\mathtt{7}}}{{\mathtt{10}}}}$$

i just typed in 0.7 and this was the calculators ouput :)

Very impressive Chris :)

Use v² = u² + 2as, where v = final speed (0), u = initial speed (10000/3600 m/s), a = acceleration (-0.8m/s²) and s = distance (metres)

0 = (10000/3600)² -2*0.8*s

s = (100/36)²/1.6 m

$${\mathtt{s}} = {\frac{{\left({\frac{{\mathtt{100}}}{{\mathtt{36}}}}\right)}^{{\mathtt{2}}}}{{\mathtt{1.6}}}} \Rightarrow {\mathtt{s}} = {\mathtt{4.822\: \!530\: \!864\: \!197\: \!530\: \!9}}$$

s ≈ 4.8m

.

(5-8+13)mn = 10mn

For the first part use s = ut + (1/2)at², where s = distance, u = initial speed, a = acceleration, t = time:

$${\mathtt{s}} = {\mathtt{0}}{\mathtt{\,\times\,}}{\mathtt{3.5}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{2.8}}{\mathtt{\,\times\,}}{{\mathtt{3.5}}}^{{\mathtt{2}}} \Rightarrow {\mathtt{s}} = {\mathtt{17.15}}$$

So distance = 17.15m

For the second part find the speed from v = u + at:

$${\mathtt{v}} = {\mathtt{0}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.8}}{\mathtt{\,\times\,}}{\mathtt{3.5}} \Rightarrow {\mathtt{v}} = {\mathtt{9.8}}$$

v = 9.8m/s

At this speed the sprinter will cover the remaining distance in time t2 = (100 - 17.15)/9.8 seconds

$${\mathtt{t2}} = {\frac{\left({\mathtt{100}}{\mathtt{\,-\,}}{\mathtt{17.15}}\right)}{{\mathtt{9.8}}}} \Rightarrow {\mathtt{t2}} = {\mathtt{8.454\: \!081\: \!632\: \!653\: \!061\: \!2}}$$

t2 ≈ 8.45 s

So, total time for the 100m is 3.5 + 8.45 seconds = 11.95 seconds

.

Yes happy - I wonder that too   

 Binomial expansion with fractional power.  I've seen this before (by Bertie) but i have never actually used it.  Thanks Alan     (25/11/14)

find the exact value of the inverse of cos (-1/2)

for starters it is negative so it is the second quadrant.

Now I'll look for the first quad equivalent

This is just an equilateral triangle of side lenght 2 that has been split down the middle.

You can see that inverse cos (1/2) is 60degrees

The second quad equivalent 180-60=120°

$$cos^{-1}(-1/2)=120^0$$

Take a look at this post then if you have questions feel free to ask.    

Let me see.

Is 46^78+89^67 dividable with 5?

If a number is divisable  by 5 it must end in a 0 or a 5

6^1=6

6^2=36 which ends in a 6

663=36*6 which ends in a 6

6^(any positive integer) must end in 6

6+4=10 (ends in 0) 

6+9=15 (ends in 5)

so the question becomes  does 89^67 end in a 4 or a 9

what could it end in?  9,81,..9,..1..   Ok, it could end in a 1 or a 9  it cannot end in a 4

89^1 ends in a 9

89^2 ends in a 1

89^3 ends in a 9

89^(an odd number) ends in a 9

SO     89^(67) ends in a 9

9+6=15 

So yes that number will be divisable by 5     

im sure its: c=-(7/2)

your answer: 3.877272 * 10^4

Is 46^78+89^67 dividable with 5  ?

yes !

$$\boxed{46^{78} + 89^{67} \stackrel{?}{=} 0 \mod 5} \\\\
\small{
\underbrace{(46^{78}\mod 5)}_{ 1^{78} \mod 5} + \underbrace{(89^{67}\mod 5)}_{-1^{67}\mod 5} \stackrel{?}{=} 0 \quad | \quad (46 \mod 5 = 1) \ and \ (89 \mod 5 = 4 = -1)
} \\\\\\
\small{
1^{78} = 1 \ and \ (-1)^{67} = -1}\\\\
\small{
( 1 \mod 5) + (-1 \mod 5) = [ (1-1) \mod 5 ]=( 0 \mod 5) = 0 \quad q.e.d.
}$$

4/5      

If you want more info there is a sticky note - How do I type in fractions (bottom left of page)

The Quadratic Formula is used when you have a quadratic equation in the form: $${{\mathtt{ax}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{bx}}{\mathtt{\,\small\textbf+\,}}{\mathtt{c}} = {\mathtt{0}}$$

This page explains the concept more: http://www.purplemath.com/modules/quadform.htm

1+1 In order to solve this we need to put it on brackets. So we have: (1+1)=x

Then you have to expand the brackets:

(1+1)=x

(  1  +  1   )=x

(           1         +       1             )=x

After you have expanded the  a lot the plus is somewhere lost in the hole of space and time

(                                                     1                                               1                                  )=x

so you get it all together and you have:

(    1      1 )=x

(  1   1  )=x

( 1  1 )=x

( 1 1)=x

( 11)=x

So 

x=11

The answer is 1+1=11