geno3141

For this problem the "three-dimensional" Pythagorean Theorem states:

             x² + x² + x²  =  20²  

If the equation is in vertex form:

1)  If the vertex is on the x-axis, there is one root (a double root)

2)  If the vertex is above the x-axis and the parabola rises, there are no real roots.

3)  If the vertex is above the x-axis and the parabola falls, there are two real roots.

4)  If the vertex is below the x-axis and the parabola rises, there are two real roots.

5)  If the vertex is below the x-axis and the parabola falls, there are no real roots.

xy  =  1/9

x(y + 1)  =  7/9   --->   xy + x  =  7/9   --->   1/9 + x  =  7/9   --->   x  =  6/9   --->   x  =  2/3

y(x + 1)  =  5/18   --->   y( 2/3 + 1 )  =  5/18   --->   y(5/3)  =  5/18    --->   y  =  1/6

Now you know the values of both x and y, so (x + 1)(y + 1) = ...

There are 6 choices for each question; one of which is correct.

To get 3 incorrect:  5/6

To get 7 correct:     1/6

To get e incorrect  and  7 corect:  (5/6)  x  (1/6)  =  5/36.

If  (4a + 3b) / (a - 2b)  =  5

then              4a + 3b  =  5(a - 2b)

                     4a + 3b  =  5a - 10b

                            13b  =  a

Therefore, substituting 13b for a:

       (a + 11b) / (a - b)  =  (13b + 11b) / (13b - b)  =  24b / 12b  =  2

sin(x)  =  cos(2x) - 9/8

Using the identity:  cos(2x)  =  1 - 2sin²(x) ...'

                                                   sin(x)  =  [ 1 - 2sin²(x) ] - 9/8

                          2sin²(x) + sin(x) + 1/8  =  0

                         16sin²(x) + 8sin(x) + 1  =  0

                                      [ 4sin(x) + 1 ]²  =  0

                                            4sin(x) + 1  =  0

                                                   4sin(x)  =  -1

                                                     sin(x)  =  -¼

Try again ...

What do you mean by "x2100"?

14A6₁₂ - 5B9₁₂  =  ?                                           A₁₂ = 10_ten     B₁₂ = 11_ten

You can first change each of these into base ten:

     14A6₁₂  =  (1·12³) + (4·12²) + (10·12) + 6  =  2430

       5B9₁₂  =  (5·12²) + (11·12) + 9                 =     861

    2430 - 861  =  1569

Now change 1569₁₀  into base twelve:

     1569 / 12  =  130, remainder 9

       130 / 12  =  10,   remainder A

          10 /12  =  0,     remainder A

Answer:  AA9

If you have 7 gumballs, they might consist of 2 red, 2 blue, 1 white, and 2 green.

The eighth gumball must be the third of a color.