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Sine is \(\frac{opposite}{hypotenuse}\)

Cosine is \(\frac{adjacent}{hypotenuse}\)

Tangent is \(\frac{opposite}{hypotenuse}\)

Take logarithms on both sides

log 0.254 = X log 0.5

Therefore X = log 0.254 / log 0.5 ≈ 1.977099598.........

600 + 14.36%

= 600 + 0.1436

= 600.1436

3^4=81

x/(x-7) = y

x = y(x-7)

x = xy-7y

x-xy = -7y

xy-x = 7y

x(y-1) = 7y

x = 7y/(y-1)

72 * 3 = 216

Calculators are always your best friend

any number except 0 to the power of 0 is 1

because \(0^0 = 0{}^{-0}=1/0^0 = infinity\)

2^0 = 1

\((a+1)(a+2)=(a)(a+2)+1(a+2)=(a)(a)+2a+a+2=a^2+3a+2\)

\(\frac{-1\frac{3}{5}}{-\frac{2}{3}}=\frac{\frac{8}{5}}{\frac{2}{3}}=\frac{8}{5}\times\frac{3}{2}=\frac{12}{5}=2\frac{2}{5}\)

=45(14)(n)

=630n

Usage 1: Arithmetric operations

such as

5*[23+(5/9)]

Usage 2: Matrices

\(\begin{bmatrix} 1 && 2 \\ 3 && 4 \end{bmatrix}\)

Using polynomial division

\(\frac{x^3-7}{x+5}= x^2-5x+25-\frac{32}{x+5}\)

-4x + 8 = -8 

FACTORIZATION

-4(x-2) = -8

x-2 = -8/-4

      = 2

x=4

\(\frac{7(23-18)}{(12-5)} = \frac{7(5)}{7} = 5\)

As the guest said,

F(n) = Fibonacci[n] + Prime[n]

The next term of 40 would be the 9th term

The next 5 terms are:

F(9) = 34 + 23 = 57

F(10) = 55 + 29 = 84

F(11) = 89 + 31 = 120

F(12) = 144 + 37 = 181

F(13) = 233 + 41 = 274

No,

sqrt(1457786467 = 38180.9699588682529321

1/5 times (5+2)

=7/5

=1.4

Thx 

Equate the two functions to get x^2 = m*x - 2   or  x^2 - m*x + 2 = 0

This quadratic has a single point solution if (-m)^2 - 4*1*2 = 0  or m^2 = 8  

Therefore there are two possibilities:  m = sqrt(8) and m = -sqrt(8)

These mean that the linear function represents tangent lines to the x^2 function.

It's the area of a semicircle of radius 3.  This is (1/2)*pi*3^2   or just 9pi/2.

Essentially correct User101, though the result should be expressed in terms of x rather than t.

Not all possible functions are implemented in the calculator.  You tend to get TO DO when you try a function that hasn't yet been implemented (and possibly, that Admin intends to implement at some time in the future?).

Not quite Chris!  The temperatures must be absolute;  T1 = 273+25 and T2 = 273+125,  so the final volume will be smaller than the initial volume because the pressure increses far more than tha temperature.

Using

P1 V1       =     P2 V2

  T1                   T2

We have

155 * 1     =   605 * V2

  25                  125

Solving for V2, we have

(155 * 125)     =   V2    =  about 1.281 L

 (25 * 605)