All Answers 358150 Answers

$$cos(2x) \\
= cos^2x-sin^2x\\
=1-sin^2x-sin^2x\\
=1-2sin^2x$$

Congratulations KimLee,

I am very glad we could help 

I have added this to the logic puzzles thread  :)

That is good MathsGod1

BUT

can you show me how you can cancel earlier please ?

I used a slightly different approach from heureka's.......we still get the same answers....!!!

1, 5, 13, 25.....

It appears that the series is defined by ......

1 = 1 +  0   =  1 +  4(0)

5 = 1 +  4   =  1 + 4(1)     

13 = 1 + 12  = 1 +  4(3)

25 = 1 + 24  =  1 + 4(6)

41= 1 + 40   = 1 + 4(10)

So, the nth total is defined as........

1 + 4[sum of the first n whole numbers] =

1 + 4[(n)(n - 1)/2]  =

1 + 2[(n)(n-1] =

So....the 20th figure will have  ...

1 + 2[(20)(19)]   =  1 + 2[380] = 1 + 760 = 761 turquoise squares.....

congrats, Kim!

How many turquoise squares will be required to build the twentieth figure in this pattern?

n = turquoise squares

n-1 = white squares

n + n - 1  = all squares = $$\small{\text{
$
(2i-1)^2
$}}$$    i is the number of the figure

$$\begin{array}{rcl}
n+n-1 &=& (2i-1)^2\\
2n-1 &=& (2i-1)^2\\
2n &=& 1+(2i-1)^2\\\\
n &=& \dfrac{1+(2i-1)^2}{2}\\
\end{array}\\\\\\
\boxed{~n_i = \dfrac{1+(2i-1)^2}{2} \qquad \rm{or} \qquad$ n_i = 1 + 2i(i-1)$~}\\\\
n_1 = \dfrac{1+(2*1-1)^2}{2} = 1\\\\
n_2 = \dfrac{1+(2*2-1)^2}{2} = 5\\\\
n_3 = \dfrac{1+(2*3-1)^2}{2} = 13\\\\
n_4 = \dfrac{1+(2*4-1)^2}{2} = 25\\\\
\cdots\\\\
n_{20} = \dfrac{1+(2*20-1)^2}{2} = 761 \small{\text{$\qquad \rm{or} \qquad n_{20}= 1+2\cdot 20(20-1) = 1+40\cdot 19=761 $}}$$

I think so, Z Man !!!!!

^ Is this really the answer........

Fill the 5 cup container full....pour 3 cups into the 3 cup container....throw these 3 cups out.........there will be 2 cups now in the 5 cup container....pour these 2 cups into the 3 cup container.....

Fill the 5 cup container full, again......pour 1 cup into the 3 cup container which is now full.......the 5 cup container will now contain exactly 4 cups.....!!!!

Alan has given a proof for the 7* one here      

For seven:

Take the last digit, double it, and subtract it from the rest of the number. If the answer is divisible by 7 (including 0), then the number is also divisible by 7:

861:86-1x2=84

84/7=12

1.2*10^-3+2.7*10^-4 = 0.00147

I don't know what you mean by "correctly rounded".......it depends upon what you want to round to.......nearest integer???.....nearest thousandth???.......nearest ten thousandth.....???

your welcome.

That's great, KimLee......glad we could help.....!!!!

That is a good question MG  :)

There is no real difference.  You must always cancel ONE thing on the top with ONE thing on the bottom.

Eg. you might start by divided the 8 by two on the bottom and the 34 by two on the top.  

That would be one step  

Cos(2x) is just a "double angle" trig function......there's nothing to "do" on a calculator, unless we were trying to evaluate the cosine for some value of 2x where x is an angle and 2x is twice the measure of that angle.......

The equilibrium price will be found where the "x's" are equal.....so.....

25p - 2625   = -3p + 875     rearrange

28p  = 3500    divide both sides by 28

p = $125   ....this is the equilibrium price......

do you mean the quadratic forumla? if so its and usually its done by for example. 12+14=84

12=A

14=B

84=c

Then insert into formula if you need more help let me know i guess.

2 hr 35 min 00 sec can be written as   1hr 95min = 1hr 94min 60sec

So

   1 hr 94 min  60 sec

-  1 hr 42 min  25 sec

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

           52 min  35 sec

You were correct with your first sentence, MG!!!!

It IS undefined  {not, 0 } ........

0,0,0,1,2,2,2,2,3,4,5,5,5,5,5   The median is 2 days

The average is

(3(0) + 1(1) + 4(2) + 1(3) + 1(4) + 5(5)) / 15  = 41/15   days

So ....... 41/15 - 2 =  [41 - 30]/15 = 11/15 days

125/1000 

Divide both numerator and denominator by 125:

125/1000  =   1/8

The denominator is 8.

Replace "is" with "=", "of" with "times", and "what" with "x":

5000  =  .333 · x

Divide both sides by .333:

x  =  5000/.333

x  =  15015 (approximately)

If you are using .333 as an approximation of 1/3, the answer is 15000.