geno3141

|x - 2|  =  (x - 2)  if  x >= 2

|x - 2|  =  (2 - x)  if  x < 2

Since every point in the triangle has an x value >= 0 and <= 5/3, this means that we should take the second definition.

We can write it as:  "2 - x".

I'm assuming that "·" means "multiplication", not "composition".

f(x)  =  x - 1          g(x)  =  4x² + 3x - 5

(f · g)(x)  =  f(x) · g(x)  =  (x - 1) · (4x² + 3x - 5)

                                   =  (x)(4x²) + (x)(3x) - (x)(5) + (-1)(4x²) + (-1)(3x) - (-1)(5)

                                   =  4x³ + 3x² - 5x - 4x² - 3x + 5

                                          =   4x³ - x² - 8x + 5

(f · g)(-2)  =  4(-2)³ - (-2)² - 8(-2) + 5  =  -15

The next problem is like this problem

A formula to find the sum of an arithmetic sequence is:  Sum  =  N(F + L)/2

   N = number of numbers = 2197

   F = first number

   L = last number = F + 2196

   Sum = 65⁴ = 17,850,625

--->     17,850,625  =  2197( F + (F + 2196) ) / 2

           35,701,250  =  2197( 2F + 2196)

                   16250  =  2F + 2196

                   14054  =  F

                     7027  =  F

You can find the last term ...

To have a vertical asymptote of x = 7, the function needs a factor of (x - 7) in the denominator.

The function then can look like this:  f(x)  =  a / (x - 7)  for any non-zero value of a.

This function will have a horizontal asymptote at y = 0.

For this function to have a horizontal asymptote of y = -3, the function can be modified to be:  f(x)  =  a / (x - 7) - 3.

Rewriting the function, using the common denominator:  f(x)  =  [ -3x + (21 + a) ] / (x - 7)

Since the y-intercept is (0,6):                                              6    =   (21 + a) / -7

Solving for a:                                                                         a  =  -63

The function is now:  f(x)  =  (-3x - 42) /(x - 7)

                                                  3sqrt(x)  =  x + k

Squaring both sides:                          9x  =  x² + 2xk + k²

Rewriting:                   x² - 9x + 2kx + k²  =  0

                                   x² + (2k - 9)x + k²  =  0

This is a quadratic; it will have a unique solution when the discriminant of the quadratic formula is zero.

For the quadratic formula:  a = 1     b = 2k - 9     c = k²

Discriminant:                  b² - 4·a·c  =  0

                       (2k - 9)² - 4·(1)·(k²)  =  0       

                       4k² - 36k + 81 - 4k²  =  0

                                       -36k + 81  =  0

                                               -36k  = -81

                                                    k  =  9/4

Guest is correct -- but there is an easier way ...

Using the Law of Cosines:

[You can't use the Pythagorean Theorem because it isn't a right triangle.]

Use the Law of Cosines to find the third side of the triangle:  c²  =  a² + b² - 2·a·b·cos(C)

   --->   c²  =  (2.6)² + (5.6)² - 2·(2.6)·(5.6)·cos(70)

   --->   c²  =  28.160

   --->    c  =  5.3

Now, use Heron's formula:

s  =  (2.6 + 5.6 + 5.3) / 2  =  6.75

Area  =  sqrt( 6.75 · (6.75 - 2.6) · (6.75 - 5.6) · (6.75 - 5.3) )  =  6.83

But, there is an easier way:  use this formula for area:  Area  =  ½ · a · b · sin(C)

                                                                                        Area  =  ½ · (2.6) · (5.6) · sin(70)

                                                                                        Area  =  6.84

The difference between the two answers is due to a rounding error in my using Heron's formula.

Draw a picture!

It's in Quadrant IV, so the x-value (cos) will be positive and the y-value (sin) will be negative.

sec(theta)  =  5/2   --->   1/cos(theta)  =  5/2   --->   cos(theta)  =  2/5

In your reference triangle, the x-value  =  2  and the hypotenuse  =  5

Using the Pytagorean Theorem, the y-value  =  - sqrt(21)

sin  =  y / r   --->   sin(theta)  =  - sqrt(21) / 5

I think that the best way to approach these problems is to draw the picture ...

Factor both the numerator and the denominator:

                  f(x)  =  [ 2(x + 5)(x - 5) ] / [ (x + 5)(x - 3) ]

Since the factor (x + 5 ) is found in both the numerator and the denominator, it produces a hole in the function at x = -5.

Cancelling out that term, we have:  f(x)  =  [ 2(x - 5) ] / [ (x - 3) ]

There will be a vertical asymptote at x = 3 (because it makes the denominator zero).

y  =  (x - 3) / (x + 7)

Multiply by (x + 7):     y(x + 7)  =  x - 3

Simplify:                      xy - 7y  =  x - 3

                               xy - 7y - x  =  -3

                               y(x - 7) - x  =  -3

                         y(x - 7) - x + 7  =  4

                        y(x - 7) - (x - 7)  =  4

                             (y - 1)(x - 7)  =  4