geno3141

Because angle(SAP) and angle(SOP) each has its vertex on the circle and they cut off the

same arc, arc(SP), they must be equal in size.

Let  x  =  angle(SAP)  =  angle(SOP)

angle(SAP)  =  4 · angle(SOP) - 178^o

                 x  =  4 · x - 178^o

              -3x  =  - 178^o

Now, finish ...

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

For this problem:  Area  =  ½ · AB · BC · sin(ABC).

Put the values into this equation and finish ...

The equation of this polynomial is:  x³ + 12x² + 44x + 48.

There are three zeros; they are (-6,0), (-4,0), and (-2,0).

The axis of symmetry is the line:  x = 6.

Since one of the x-intercepts is (8,0) which is to the right of the line of symmetry,

the other intercept will be to the left of the line of symmetry:  (4,0).

If the first of 19 consecutive positive integers is  x  then the 19th integer is  x + 18.

Using the formula:  Sum  =  number · (first + last) / 2

                               Sum  =  19( x + (x + 18) ) / 2

                               Sum  =  19( 2x + 18 ) / 2

                               Sum  =  19( 2(x + 9) ) / 2

                               Sum  =  19( x + 9 )

So, I think that the answer is 19.

I believe that you mis-stated the problem because angle(DBA) is angle(ABD).

If  x  is the answer, this problem reduces to:  x  =  6 + 1/x.

Simplify and use the quadratic formula.

On the calculator, hover below the (log) key to get the (ln) key;

press that key and enter 171

Some hints:

(a) Domain = set of all the x-values that have been graphed;

                                list as:  ( least x-value, greatest x-value )

(b) Range = set of all the y-values that have been graphed;

                                list as:  ( least y-value, greatest y-value )

(c) Open interval(s) on which f(x) is strictly increasing = 

                the interval which produces a graph from lower-left to upper-right

                         (there are two of these)

(d) Open interval(s) on which f(x) is strictly decreasing =

                the interval which produces a graph from upper-left to lower-right

                          (there is one of these)

(e) Open interval(s) of which f(x) is constant =

                the interval which produces a flat graph

                           (there are two of these)

Let O be the origin.

Triangle(OAC) is similar to triangle(SRC).

OA = 8 and OC = 8 making OA = OC.

Because of similar triangles, SR = SC; call this length "x".

Area(triangle(SRC)) = ½·x·x = 12.5     --->     x² = 25     --->     x = 5

The distance from S to R is 5.

The distance from C to S is also 5.

Can you take it from here?