geno3141

Use sin:  sin(39)  =  x/10

Because you did not include a picture, I don't know which of the 16 triangles are shaded.

Use the formula  Area  =  ½·pi·radius²  to find the areas of the following semi-circles.

Step 1)  Find the area of the largest semi-circle; it would be grey if it were competely

                     shaded and its radius is 8.

Step 2)  Find the area of the next-largest semi-circle; it would be white if it were completely

                     shaded and its radius is 6.

Step 3)  Find the area of the second-smallest semi-circle; it would be grey if it were completely

                     shaded and its radius is 4.

Step 4)  Find the area of the smallest semi-circle; it would be white if it were completly

                     shaded and its radius is 2.

Start with the answer of Step 1; subtract the area of Step 2; add the area of Step 3; and, finally,

                     subtract the area of Step 4.

Since  x - 1/x  =  5     --->                          (x - 1/x)³  =  5³

Multiplying out:  

--->                        x³ - 3(x)²(1/x) + 3(x)(1/x)² - 1/x³  =  125

--->               [x³ - 1/x³] + [ - 3(x)²(1/x) + 3(x)(1/x)² ]  =  125

But:  -3(x)²(1/x) + 3(x)(1/x)²  =  -3x + 3/x  =  -3(x - 1/x)  =  -3(5)  =  -15

So:    [x³ - 1/x³] + [ - 3(x)²(1/x) + 3(x)(1/x)² ]  =  125

--->   [x³ - 1/x³] + [ - 15 ]  =  125

--->   x³ - 1/x³  =  140

Step 1)  Write as:  x =  1 + sqrt(2) / x

Step 2)  Multiply each term by x.

Step 3)  Gather all terms on one side.

Step 4)  Use the quadratic formula.

Let me give you some further hints:

Step 1)  Use Heron's formula to find the area of triangle(DBC).

              Heron's formula:  Area  =  sqrt( s · (s - a) · (s - b) · (s - c) )

              where a, b, and c are the sides of the triangle:  e.g., a = 8, b = 7, and c = 9

              and s is the semi-perimeter:  s = (a + b + c) / 2

Step 2)  Draw BP perpendicular to DC with P on DC.

              This makes BP the height of triangle(DBC).

              Find the value of BP by using the formula for the area of a triangle:  A = ½·base·height

               You know the area from Step 1, let the base be 8, and find the value of the height.

Step 3)  Look at triangle(BPC).

              You know the value of BC and, from step 2, you know the value of BP.

              Use the Pythagorean Theorem to find the value of PC.

Step 4)  Draw AQ perpendicular to DC with Q on DC.

              DQ = PC; so you know the value of DQ since you found the value of PC in Step 3.

              AB = DC - DQ - PC.

Step 5)  Now use the formula to find the area of a trapezoid; you now have all the parts.

If f(x) is of degree 5 and the sum of f(x) + g(x) is of degree 3, then the degree of g(x) must also be 5.

g(x) = -2x⁵ + x⁴ + a·x³+ (don't really care about the x²-term, the x-term, and the consonant; they can be anything) where a is not zero.

If g(x) is created in this way, the fifth-degree term of the sum will disappear as will, also, the four-degree term of the sum, but a third-degree term will remain.

f(x)  =  1 - 12x + 3x² - 4x³ + 5x⁴         g(x)  = 3 - 2x - 6x³ + 9x⁴

Find  c  such that the polynomial  f(x) + c·g(x)  has degree 3.

To have degree 3 means that the result must have a term or degree 3 and no term with a deree larger than 3.

So, we need a value for  c  such that when multiplied by the fourth-degree term of  g(x)  and added to the

fourth-degree term of  f(x)  those fourth-degree terms will cancel.

Try  - 5/9   --->

( 1 - 12x + 3x² - 4x³ + 5x⁴ )  +  (-5/9) · ( 3 - 2x - 6x³ + 9x⁴ )

=  1 - 12x + 3x² - 4x³ + 5x⁴ - (5/9)(3) - (5/9)(-2x) - (5/9)(-6x³) - (5/9)(9x⁴)

=   ........

(I'll let you finish.)

Since it starts in the upper left-hand corner of the graph and ends in the upper right-hand corner

of the graph it must be both an even degree polynomial and the coefficeint of the even degree must be

positive.

Therefore, the answer is ........

Since the ratio of team assistants to player is 1/6, there are 1/6^th as many assistants as players.

(1/6) · (36) = 6     --->     there are 6 team assistants.

The ratio of coaches to assistants is  3/6  =  1/2.