geno3141

His cost is:  26 kg x $20.00 / kg  +  30 kg x $36.00 / kg  =  $1600.00

He creates:  26 kg  +  30 kg  =  56 kg.

His income:  56 kg x $30.00 / kg  =  $1680.00

His profit:  $1680.00 - $1600.00  =  $80.00

His percentage profit:  $80.00 / $1600.00  =  0.05  =  5%

There are 360^o around each point.

There are 60^o in each vertex angle of an equilateral triangle.

There are 90^o in each vertex angle of a square.

To find the number of degrees in each vertex angle of a regular pentagon, use this formula:

    degrees  =  (n - 2)·180^o / n  (where n = number of sides)  =  (5 - 2)·180^o​ / 5  =  108^o.

360^o - 60^o - 90^o - 108^o  =  ..... 

( x + 1/x ) / ( x - 1/x )  =  5 / 4

Cross-multiply:       4( x + 1/x )  =  5(x - 1/x )

                                  4x + 4 / x  =  5x - 5/x

Multiply by x:                4x² + 4  =  5x² - 5

Add 5:                           4x² + 9  =  5x² 

Subtract 4x²:                          9  =  x²

Two possible answers:  3  and  -3

If you place these numbers back into the original equation, you will find that 3 works and that -3 doesn't work.

1)  A square, a regular pentagon, and a regular n-gon surround a point.

     The number of degrees in each vertex angle of a regular n-gon is:  (n - 2) · 180^o / n

     We know that each vertex angle of a square has 90^o.

     Each vertex angle of a regular pentagon has  (5 - 2) · 180^o / 5  =  108^o.

     There are 360^o around each point.

     Subtracting the vertex angle of a square and the vertex angle of a regular pentagon,

     we are left with:  360^o - 90^o - 108^o  =  162^o.

     Now we use the formula:   (n - 2) · 180^o / n  =  162^o  to find the value of n.

                  (n - 2) · 180^o  =  162^o · n

               180^o · n - 360^o  =  162^o · n

          180^o · n - 162^o · n  =  360^o 

                            18^o · n  =  360^o 

                                     n  =  20

The areas of similar polygons have the same ratio as the squares of the corresponding sides.

[Perimeters count as lengths.]

Since Area ( polygon A )  +  Area( p0lygon B )  =  65     --->     Area( polygon B )  =  65  -  Area( polygon A )

Area( polygon A) : Area( polygon B )  =  Perimeter( polygon A )² : Perimeter( polygon B )² 

Area( polygon A) : [ 65 - Area( polygon A ) ]  =  12 : 18

To make the algebra easier, call Area( polygon A ) = x

--->   x : ( 65 - x )  =  12 : 18     

Cross-multiply:  12( 65 - x )  =  18x

                            780 - 12x  =  18x

                                     780  =  30x

                                          x  =  26

--->   65 - x  =  65 - 26  =  39

Team A will win the tournament if the games go like this:

A and A:                              (0.60) · (0.60)  =  .....

A and B and A:       (0.60) · (0.40) · (0.60)  =  .....

B and A and A:       (0.40) · (0.60) · (0.60)  =  ..... 

Find each of the individual probabilities and add those values together  .....

The two triangles, triangle(APB) and triangle(CPD) are similar by AA.

The corresponding lengths of sides of similar figures are in the ratio of the square roots of the areas.

So:  AP / PC  =  sqrt(4) / sqrt(9)   --->   AP / PC = 2 / 3.

Let's call the length of AP = 2x, which makes the length PC = 3x; also, BP = 2x and PD = 3x.

A formula for the area of a triangle is:  A  =  ½·a·b·sin(C).

Using this formula on triangle(APB):  a = AP = 2x     b = BP = 2x     A = 4 cm² 

     --->     A  =  ½·a·b·sin(angle( APB) )     --->     4  =  ½·(2x)·(2x)·sin(angle( APB )     

     --->     sin( angle(APB) )  =  2 / x² 

Similarly, sin( angle(CPD)  =  2 / x² 

Area of triangle(APD)  =   ½·(2x)·(3x)·sin(angle( APB )  =  (3x²)·sin(angle( APB )

(since sin( angle(CPD)  =  2 / x² )     --->     Area of triangle(APD)  =  (3x²)·(2 / x²)  =  6 cm²

Therefore, the total area will be  4 cm² + 9 cm² + 6 cm² + 6 cm²  =  25 cm² 

Solve:       ( x² + x + 1 ) ( x² + x + 2 )  =  12

Multiply out:  x⁴ + 2x³ + 4x² + 3x + 2  =  12

Rewrite:       x⁴ + 2x³ + 4x² + 3x - 10  =  0

There are two rational solutions; you can find them by trying all the factors of -10:

     1, -1, 2, -2, 5, -5, 10, -10          (two of these will work)

After you have these two solutions, you can divide their factors out of the original problem.

This will give you a quadratic that can be solved by using the quadratic equation.

sin(20^o)[ tan(10^o) + cot(10^o) ] 

1)            =  sin(20^o) · tan(10^o)  +  sin(20^o) · cot(10^o)

2)           =  sin(20^o) · sin(10^o) / cos(10^o)  +  sin(20^o) · cos(10^o) / sin(10^o )

3)            =  sin(20^o) · sin²(10^o) / [ sin(10^o) · cos(10^o ) ]  +  sin(20^o) · cos²(10^o) / [ sin(10^o) · cos(10^o ) ] 

4)            =  [ sin(20^o) · sin²(10^o) + sin(20^o) · cos²(10^o) ] / [ sin(10^o) · cos(10^o ) ] 

5)            =  [ sin(20^o) · ( sin²(10^o) + cos²(10^o) ) ] / [ sin(10^o) · cos(10^o ) ] 

6)            =  [ sin(20^o) · 1 ] / [ sin(10^o) · cos(10^o ) ] 

7)            =  [ 2 · sin(10^o) · cos(10^o) ] / [ sin(10^o) · cos(10^o ) ] 

8)            =  2

1)  Distribute

2)  Rewrite tan as sin/cos; rewrite cot as cos/sin

3)  Write with the common denominator of sin(10^o) · cos(10^o )

4)  Write with a single denominator

5)  Factor out sin(20^o)

6)  Use the fact that sin² + cos²  =  1

7)  Use the fact that sin(2x)  =  2sin(x)cos(x)

8)  Cancel

The slope of  y  =  x + k  is 1.

We need to find where the slope of  y  =  x²  is 1.

Find the first derivative of y:  y'  =  2x     This formula will give us the slope of  y  =  x²  at every point.

When does  2x  =  1?     --->     x  =  1/2.

So, when  x  =  1/2,  the slope of   y  =  x²  is 1.

When  x  =  1/2,  y  =  x²   --->   y  =  (1/2)²  =  1/4.

So the point of intersection of the line and the parabola is the point  (1/2, 1/4).

The equation of the line will be:  y - 1/4  =  1(x - 1/2)   --->   y  =  x - 1/4