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Hit the "2nd" button and find the "asin" key.....this is the inverse sine function......

answer me

thx :D

Wavelength = c/v = 3 x10^8/6x10^8 =  5 x 10^-1 meters      1/2 meter

2x + 3 = 15       subtract 3 from both sides

2x  = 12            divide both sides by 2

x  = 6

Oops.

sin(theta) bigger than 1 is impossible xD

Summarizing the 2 answers above:

Theta is a Greek letter which can represent an unknown angle 

Example :\(sin \theta = \frac{3}{2}\)

See if this helps, Max :   http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx

1 x 5 x 12 = 60

3 x 4 x 5 = 60

1x 2 x 30 = 60

all we do is to make one number significantly bigger than the other two

first i will consider 1 as the first number

as we have to make the 3 numbers different, we make the second number the smallest possible factor of 60, which is 2

then we divide 60 by 2 and get the 30

The largest possible sum = 1 + 2 + 30 = 33

Just replace fractions with decimals or division

okey so found the maximum present value = 1.96666.... 7 or 23 months.

Would this also be the optimal harvesting time?

I see the profitt at 2 is actually bigger than 1.9666 but, pi ' is negative here.

"Of"   means "Multiply"   ..... so we have

40% of 66  =

.40 x 66   =

26.4

x = 4 + y    →   x - y  = 4   (1)

y = 3 + x   →   -x - y  = 3  (2)

Add (1)  and (2)   and  we get

0 = 7

Therefore, as EP said........no solutions exist

If a 40° sector is taken out......1/9 of the volume is removed......therefore.....the remaining volume =

(8/9)(4/3)pi (18)^3  = 21714.7 units^3

X

---  = 10

-4

\(\frac{x}{-4}=10\\ \frac{x}{-4}*-4=10*-4\\ x=-40\)

bump!

3 e3 y−1 = 5 x3+7 can u solve it in terms of y?

do you mean this- if you do not mean this then you need to put some brackets in.

3 e^3 y−1 = 5 x^3+7 

\(3 e^3 y−1 = 5 x^3+7 \\ 3 e^3 y = 5 x^3+8 \\ y =\frac{ 5 x^3+8}{ 3 e^3 } \\\)

Divide by 10^6

.

Try the Wikipedia article: https://en.wikipedia.org/wiki/Theory_of_relativity

inverse sine is the same as arc sine so use  [2nd] [asin]

Hello Guest!

A triangular prism has a base that is an equilateral triangle .The smaller prism has a base that is also an equilateral triangle.How many smaller prisms will fit inside the larger prism.

The prisms are visually. {nl} You have an infinitesimal skin. {nl} That's why: {nl} It fit an infinite number of smaller prisms in this prism. It does not matter which position they occupy.

Greeting asinus :- )

!

0/0 is undefined.

You cannot divide any number by 0

Solve for n:

1000= 50 x(1.00125)^n + 10 x(1.00125^n - 1/1.00125 - 1)

Please show your steps.

Let a = 1.00125

\(\small{ \begin{array}{rcll} 1000 &=& 50 \cdot (1.00125)^n + 10 \cdot (1.00125^n - \frac{1}{1.00125} - 1) \qquad & | \qquad a = 1.00125\\ 1000 &=& 50 \cdot a^n + 10 \cdot (a^n - \frac{1}{a} - 1) \qquad & | \qquad :10\\ 100 &=& 5 \cdot a^n + (a^n - \frac{1}{a} - 1)\\ 100 &=& 5 \cdot a^n + a^n - \frac{1}{a} - 1\\ 100 &=& 6 \cdot a^n - \frac{1}{a} - 1 \qquad & | \qquad +1\\ 101 &=& 6 \cdot a^n - \frac{1}{a} \\\\ 6 \cdot a^n - \frac{1}{a} &=& 101 \qquad & | \qquad + \frac{1}{a}\\\\ 6 \cdot a^n &=& 101 + \frac{1}{a} \qquad & | \qquad :6\\\\ a^n &=& \dfrac { 101 + \frac{1}{a} }{ 6 } \qquad & | \qquad \log_{10}()\\\\ \log_{10}(a^n) &=& \log_{10} \left( \dfrac { 101 + \frac{1}{a} }{ 6 } \right)\\\\ n\cdot \log_{10}(a) &=& \log_{10} \left( \dfrac { 101 + \frac{1}{a} }{ 6 } \right) \qquad & | \qquad \log_{10}(a)\\\\ n &=& \dfrac{\log_{10} \left( \dfrac { 101 + \frac{1}{a} }{ 6 } \right) }{ \log_{10}(a)} \qquad & | \qquad a = 1.00125\\\\ n &=& \dfrac{ \log_{10} \left( \dfrac { 101 + \frac{1}{1.00125} }{ 6 } \right) }{ \log_{10}(1.00125)} \\\\ n &=& \dfrac{ \log_{10} \left( \dfrac { 101 + 0.99875156055 }{ 6 } \right) }{ \log_{10}(1.00125)} \\\\ n &=& \dfrac{ \log_{10} \left( \dfrac { 101.998751561 }{ 6 } \right) }{ \log_{10}(1.00125)} \\\\ n &=& \dfrac{ \log_{10} ( 16.9997919268 ) }{ \log_{10}(1.00125)} \\\\ n &=& \dfrac{ 1.23044360575 }{ 0.00054252909} \\\\ \mathbf{n} & \mathbf{=} & \mathbf{2267.97718911} \end{array} }\)

One question, heureka.......did you you find the number of partitions by "trial and error"  or did you use some algorithm for this???

Hallo CPhill,

i did it by computing program. I wrote a algorithm for this in c++.

Here is the counting for all sums from 1 to 60:

   1 = 0
   2 = 0
   3 = 0
   4 = 0
   5 = 0
   6 = 0
   7 = 0
   8 = 0
   9 = 0
  10 = 1
  11 = 10
  12 = 55
  13 = 220
  14 = 715
  15 = 2002
  16 = 4995
  17 = 11340
  18 = 23760
  19 = 46420
  20 = 85228
  21 = 147940
  22 = 243925
  23 = 383470
  24 = 576565
  25 = 831204
  26 = 1151370
  27 = 1535040
  28 = 1972630
  29 = 2446300
  30 = 2930455
  31 = 3393610
  32 = 3801535
  33 = 4121260
  34 = 4325310
  35 = 4395456
  36 = 4325310
  37 = 4121260
  38 = 3801535
  39 = 3393610
  40 = 2930455
  41 = 2446300
  42 = 1972630
  43 = 1535040
  44 = 1151370
  45 = 831204
  46 = 576565
  47 = 383470
  48 = 243925
  49 = 147940
  50 = 85228
  51 = 46420
  52 = 23760
  53 = 11340
  54 = 4995
  55 = 2002
  56 = 715
  57 = 220
  58 = 55
  59 = 10
  60 = 1

Genral form  y = x^2 + 2x -3

Vertex form     y = x^2 + 2x + 1 -1 -3

                          = (x+1)^2  - 4                 vertex = -1, -4