geno3141

Another way:   (doing a lot of factoring)

Numerator:  100! + 99!  =  99!(100 + 1)  =  99! · 101

                        98! + 97!  =  97!(98 + 1)  =  97! · 99

                        96! + 95!  =  95!(96 + 1)  =  95! · 97

                                  ...

                                2! + 1!  =  1!(2 + 1)  =  1! · 3

Total numerator:  99! · 101 · 97! · 99 · 95! · 97 · ... · 1! ·3

Denominator:  100! - 99!  =  99!(100 - 1)  =  99! · 99

                            98! - 97!  =  97!(98 - 1)  =  97! · 97

                            96! - 95!  =  95!(96 - 1)  =  95! · 95

                                     ...

                                  2! - 1!  =  1!(2 - 1)  =  1! · 1

Total denominator:  99! · 99 · 97! · 97 · 95! · 95 · 1! · 1

Notice that all the factorials of the numerator and the denominator cancel each other.

Notice that all the single numbers cancel each other except for the 101 in the numerator

and th 1 in the denominator.

Answer  =  101 / 1  =  101

PA and PB are tangents to circle(O).  Draw PO. Draw AB. Let X be the point where PO and aB intersect.

PA = 12   and   OA = 9

Because triangle(OPA) is a right triangle, PO = 15.

Triangle(AXO is similar to triangle(PAO)   --->   AX / AO  =  PA / PO   --->   AX / 9  =  12 / 15

--->   AX  =  7.2

Since X is the midpoint of AB   --->   AB  =  14.4.

There are two externally tangent circles, circle(A) with radius = 10 and circle(B) with radius = 7.

Therefore, AB = 17.

There is an externally tangent line (with both circle(A) and circle(B) on the same side of the line)

XY, where X is the point of intersection with circle(A) and Y is the point of intersection with circle(B).

Therefore, AX = 10 and BY = 7.

Extend both AB and XY so they intersect at point C.

Triangle(ACX) will be similar to triangle(BCY) because they have corresponding congruent angles.

Call BC = x.  This makes AC = x + 17.

This means that:  AX / BY  =  AC / BC.   --->   10 / 7  =  (x + 17) / x

--->   10x  =  7(x + 17)   --->   10x  =  7x + 119   --->   3x  =  119   --->   x  =  119/3

BC  =  119/3

1)  The line  x + 2y + 4  =  0   --->   2y  =  -x - 4   --->   y  =  -½x - 2   --->   m  =  -½.

      Since the line  6x - ny + 5  =  0  is perpendicular to the other line, its slope must be  +2.

      6x - ny + 5  =  0   --->   -ny  =  -6x - 5   --->   ny  =  6x + 5   --->   n  =  (6/n)x + 5/n

      The slope of this line is  6/n.

      This slope must equal +2   --->   6/n  =  2   --->   6  =  2n   --->   n  =  3

2)  x + 2y  =  10  has these intercepts  (0,5)  and  (10,0).

     The line with an infinite number of x-intercepts is the x-axis.

     The line with an infinite number of y-intercepts is the y-axis.

      The vertices of the triangle are  (0,0),  (0,5),  and  (10,0).

      The area  =  ½·5·10  =  25.

Draw the regular hexagon and the inscribed circle.

Draw a triangle from the center of the circle, along a radius to a point of tangency of the circle and the 

hexagon, and along the tangent to a corner.

This triangle is a 30^o-60^o-90^o triangle with the length of the side from the point of tangency to the

corner = 10.

This will make the radius = 10·sqrt(3).

For a circumscribed quadrilateral, the sum of a pair of opposite sides must equal the sum of the other pair.

Therefore:  6 + ?  =  5 + 13  or  ? = 7.

This follows from the theorem that the two tangents draw from a point outside the circle are congruent.

The first problem:  log₅(8x) + 2log₅(x)  =  3

--->   log₅(8x) + log₅(x²)  =  3        (property of logs: a multiplier goes in as an exponent)

--->              log₅(8x ·x²)  =  3        (property of logs: adding logs = multiplying numbers)

--->                   log₅(8x³⁾ = 3         (simplify)

--->                         8x³  =  5³        (rewriting in exponential form)

--->                          x³  =  5³/8       (divide)

--->                          x³  =  5³/2³      (rewrite 8 as 2³)

--->                            x  =  5/2        (find the cube root)

To solve     2cos²(x) - sin(x) - 1  =  0     use the identity:  cos²(x)  =  1 - sin²(x):

--->   2[ 1 - sin²(x) ] - sin(x) - 1  =  0

             2 - 2sin²(x) - sin(x) - 1  =  0

                -2sin²(x) - sin(x) + 1  =  0

                  2sin²(x) + sin(x) - 1  =  0

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

So:  either  2sin(x) - 1  =  0   --->   2sin(x)  =  1   --->   sin(x)  =  ½   --->   x  =  sin^-1(½)

              or  sin(x) + 1  =  0   --->     sin(x)  =  -1   --->                                x  =  sin^-1(-1)

From here, you can get the answers ...

The problem arose when you set  n² - 2  =  14   --->   n²  =  16

Normally, you would get two answers,  -4  and  4.

However, in this case there is a restriction upon using  n² - 2 

---   you are only supposed to use  n² - 2  when  n < 0

---   so, for this case, the answer is only -4  (you need to exclude the answer 4)

Everything else that you did was "textbook perfect" ...

Thanks --  I read it as 2, not -2:

First:      9x + 16  =  -2   --->   9x  =  -18   --->   x  =  -2

Second:  2x - 14  =  -2   --->   2x  =  12    --->   x  =  6

To "find the sum of all possible values of x"

-- there are only two ways; either x  =  -2  or  x  =  6

So, to find the sum, the answer is:  -2 + 6  =  4.