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https://web2.0calc.com/questions/power-of-a-point-circle-question

The answer is 6.

mn = 22.

The answer is 1.

Arc PS = 52 degrees.

Arc CE = 95 degrees.

(a,b,c) = (1,-3,5).

Ooops....forgot it was sideways      vertex is at   (13.5,1.5)

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 ...

Vertex form of the parabola is    y = a ( x-1)^2 -6       sub in the point on the graph  3, -2 to calculate   a

                                                      a = 1

y =  (x-1)^2 -6       expand to

                  y =    x^2 - 2x - 5         <===== you can finish from here I hope !

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

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

Find the center of the circle.

Hello proyaop!

\(f(x)=x^2\\ \frac{dy}{dx}=2x=2\cdot 2=4\\ m_g=- \frac{1}{4}\ |\ P_g(2,4)\\ g(x)=-\frac{1}{4}(x-2)+4\\ g(x)= -\frac{1}{4}x+\frac{9}{2}\)

\(h(x)=\frac{4-1}{2-0}x+1\\ h(x)=\frac{3}{2}x+1\)

\(m_i=-\frac{2}{3}\ |\ P_i(1,2.5)\\ i(x)=-\frac{2}{3}(x-1)+\frac{5}{2}\\ i(x)= -\frac{2}{3}x+\frac{19}{6}\)

\(\color{blue}i(x)=g(x)\\ -\frac{2}{3}x+\frac{19}{6}=-\frac{1}{4}x+\frac{9}{2}\\ \frac{5}{12}x=-\frac{8}{6} \)

\(x_C=-\frac{16}{5}\)

\(y= -\frac{1}{4}x+\frac{9}{2}\\ y= -\frac{1}{4}\cdot (-\frac{16}{5})+\frac{9}{2}\\ \color{blue}y_C=\frac{53}{10}\)

The center of the circle is \(P_C(-\frac{16}{5},\frac{53}{10})\)

  !

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).

Re-arrange    x = 2y^2 -6y + 18     this is a sideways parabola  

    arrange into vertex form

              x = 2 (y^2-3y) + 18

              x = 2 (y-1.5)^2   + 13.5         Vertex is at   (1.5,13,5 )

(a,b,c) = (2,-5,7).

The x-intercepts of the graph are 5 and 8.

Using the symmetry of the parabola, the other x-intercept is at (10,0), so the answer is 10.

Since it is equilateral, then AD bisects CB, so CD = 4. Using the pythagorean theorem, AD is 8^2 - 4^2 = AD^2

So AD = \(4\sqrt{3}\)

sqrt 3 / 2  * AB  = sqrt 3 / 2 * 8 = 4 sqrt 3

32761 = 181^2

Add them   = -45 n⁵

y = 110

110 = -16t^2 + 26t + 105

0 = -16t^2 + 26t -5                     Now use Quadratic Formula to calcualte 't' values,,,

                                                        the smaller t will be on the way up...the larger will be on the way down

a = -16  b = 26  c = -5

\(t = {-26 \pm \sqrt{26^2-4(-16)(-5)} \over 2(-16)}\)