CPhill

Subbing x for t......look at the graph here, Julius

https://www.desmos.com/calculator/6wqntzjcax

The graph has a positive slope  from  [0, 3]....so the max is reached at 3 months  ≈  512.733 (thousands)

\(f(x)=\frac{x+6}{\sqrt{x^2-3x-4}}\)

Note that     x^2 - 3x - 4    must be >  0

Set  this to 0

x^2  -3x - 4  = 0

(x - 4) ( x + 1)  = 0

Setting each factor to  0   and solving for x gives us

x = -1    and x  = 4

We have three  possible intervals

(-inf , -1)  U (-1, 4)   U (4, inf)

Testing x  = 0   in the middle interval makes the polynomial negative....so....this interval does not work

The domain is the other two intervals ⇒    (-inf, -1)  U  ( 4, inf )

Here's a graph  :   https://www.desmos.com/calculator/axfthxpp5t

Great answers, Omi....!!!!

\(w(x) = \sqrt{x-2} + \sqrt[3]{x-3}~\)

The domain is that which is the most restrictive for both parts of the function

The second term  has a domain of all real numbers because we can take the cube root of all real numbers

However.....the frist term is only defined from  [2, inf)

So.....this is the domain

The diagonals of a rhombus bisect each other.......this bisection forms a right triangle with each leg (1/2) the length of its respective diagonal.....the hypotenuse forms one of the equal sides

So.....the perimeter is given by  

4√ [ 9^2 + 6^2  ] =

4 √117  =

4 √ (9 * 13 )  =

12√13 ft

Sounds OK to me.....

2.

Position the lower left corner, A , at (0,0)

The equation of AD  is   y  = x

And  let  B  be  positioned at (4,0)   

And the equation of  BC    is   y = (-5/4) ( x  - 4)  =  (-5/4)x + 5

Setting  these two equations equal and solving for x we have

x  = (-5/4)x + 5

(9/4) x  = 5

x  = 20/9  =   y  =  height of triangle ABE

So....the area of ABE  = (1/2) (base) (height)  = (1/2) (4) (20/9)  =

40/9  units^2  

4:  The number of inches in the perimeter of a square is equal to the number of square inches in its area. Find the length, in inches, of a side of the square.

Let S  = the side of the square

P  = 4S

A  = S^2

This impies that

4S  = S^2

S^2  - 4S  = 0     factor

S ( S - 4)  = 0

Setting both factors to 0  and solving for S  produces  S  = 0  and S  =4

Reject the first....accept the second

S  = 4 in

1.Let the side  of the smaller triangle   = S

Since the area  of the larger triangle is 16 times as much....each dimension must be 4 times that of the smaller triangle...so the length of its side must be  4S

And the sum of ther perimeters = 45....so

3S  + 3(4S)  = 45

15S  = 45

S  = 3 in

So....the side of the larger trinagle  = 4S  = 4(3)  =12 in

And its area  =  (1/2) 12^2 sin(60)  = 72√3 / 2   =    36√3  in^2

g(x)   = y

y  =       3                    2 (x - 4)

        _______    +      _________

        x^2 - 16             (x + 4)(x - 4)

y  =    3 + 2x - 8

        __________

          x^2  - 16

y  =  2x -  5

       _______

       x^2  - 16

y (x^2 - 16)  = 2x - 5

yx^2 - 16y  = 2x - 5

16y  =  yx^2 - 2x + 5

16y  = y (x^2 - (2/y)x + 5/y)

16 =  x^2  - (2/y)x + 5/y      complete the square on x

16  = x^2 - (2/y)x + 5/y + 1/y^2 - 1/y^2 

Add  1/y^2  to both sides....subtract 5/y from both sides

16 + 1/y^2 -5/y   = x^2 -(2/y)x + 1/y^2      factor the right side

16 + 1/y^2 - 5/y  = (x - 1/y)^2

1/y^2 - 5/y + 16  =  (x - 1/y)^2      get a common denominator on the left

[16y^2 -5y + 1] / y^2  =  (x - 1/y)^2      take both roots

±√[ (16y^2 -5y + 1) / y^2 ]  =  x - 1/y      add 1/y to both sides

1/y ±√  (16y^2 -5y + 1) / y

[ 1 ±√  (16y^2 -5y + 1)  ] / y  = x       "swap" x and y

[ 1 ±√ [ 16x^2 - 5x + 1) ] / x  =  y  =  g^-1(x)

Thus....this is the inverse.....but it is not one-to-one.....we have two values for g^-1(5)

g^-1 (5)  =   ( 1 + √ [ 16(5)^2 - 5(5) + 1 ] )  / 5   = ( 1 + √ 376 ) /  5  ≈ 4.078

g^-1(5) =  ( 1 - √ [ 16(5)^2 - 5(5) + 1 ] )  / 5   = ( 1 - √ 376 ) /  5  ≈ -3.678

So   g-1 (5)   =   ≈4.078    and   ≈ -3.678