hectictar

How do you know the value of sin( 60° )   ? 

It is given that m || n , m∠1 = 65° , m∠2 = 60° , and BD bisects ∠ABC .

Because of the triangle sum theorem,  m∠3 = 55°.

By the definition of bisector, m∠ABC = 110°. (unsure of this one, but no other choice seems right)

m∠5 = 110° because vertical angles are congruent.

Because of the same-side interior angles theorem, m∠5 + m∠6 = 180°.

Substituting gives 110° + m∠6 = 180° .

So, by the subtraction property of equality, m∠6 = 70°

midpoint   =   ( [-3 + 2]/2 ,  [-5 + 7]/2 )   =   ( -1/2 ,  2/2 )   =   ( -0.5 ,  1 )

the  x  coordinate of the minimum   =   -(-15) / [ 2(3) ]   =   15/6   =   5/2

the  y  coordinate of the minimum   =   3(5/2)^2 - 15(5/2) + k

1   =   3(5/2)^2 - 15(5/2) + k

1   =   3(25/4) - 75/2 + k

1   =   75/4 - 75/2 + k

1  =   -75/4 + k

1 + 75/4   =   k

79/4   =   k

5y(2x + 7)  +  4(2x + 7)

Notice the common factor in both terms.

5y(2x + 7)  +  4(2x + 7)

Factor  (2x + 7)  out of both terms.

(2x + 7)(5y + 4)

the set of all points between point F and point G, including point F and point G

slope   =   [difference in  y  values] / [difference in  x  values]

2   =   [k - 3] / [5 - 1]

2   =   [k - 3] / 4

                               Multiply both sides of the equation by  4 .

8   =   k - 3

                               Add  3  to both sides of the equation.

11   =   k

To find  sin( pi/3 ) ,  we can start with an equilateral triangle with side length  1 .

Now draw a line that bisects an angle and the opposite side. The length of this line is sin(pi/3) .

By the Pythagorean theorem...

(1/2)²  +  ( sin(pi/3) )²   =   1²

                                                       Subtract  (1/2)²  from both sides of the equation.

( sin(pi/3) )²   =   1²  -  (1/2)²

                                                       Take the positive sqrt of both sides.

sin(pi/3)   =   √[1²  -  (1/2)²]

sin(pi/3)   =   √[1  -  1/4]

sin(pi/3)   =   √[3/4]

sin(pi/3)   =   √3 / 2      

Here are some examples:

To find  log₁₀(100)  ,   type in:     log(100)

To find  log_e(100)   ,   type in:     ln(100)

To find  log₃(100)   ,   type in:    log(100,3)

Whatever  x  value we choose to plug in to  (g + f)(x)  has to be in the domain of both  g(x)  and  f(x) .

For instance, we can't find  g(1) + f(1)  because we can't find  g(1) .

The domain of  g + f  is all real  x  values such that   x ≥ -14  and  x ≤ -11

The domain of  g + f  is all real  x  values such that   -14 ≤ x ≤ -11

The minimum value in that domain is  -14  .