CPhill

2)   x = A radial line that is perpendicular to a chord bisects that chord. Thus (1/2) of 12 = 6. So we have a right triangle  with two "legs" 4 and 6 and the radius "x" forms  the hypoteneuse....so...... √(^6^2 + 4^2) = √(52) = 2√13

3) z° = 60°  We have a 30-60-90 triangle.......the angle that intercepts arc measuring "z°" is a 30°

 angle, so it intercepts an arc of twice its measure!!

4) x = 10.5   Products of segments of intersecting chords in a circle...... (7*3) = (2 * x)...

21 = 2x........x = 10.5

5) To find x°, we have to use the theorem that says that the angle formed by drawing two secants from a point outside a circle = (1/2) (the difference of the measures of the arcs they intercept)

40° = (1/2)(106° -x)    multiply both sides by 2

80° =106° - x    ....     add x to both sides and subtract 80 from both sides

x = 106 - 80 = 26°

To find y°  ....its measure is just 1/2 of the arc that reamains after we subtract the two known arcs from 360....thus  ...   360 - 106 - 145 = 109° ....and 1/2 of that is just 54.5°

7) I had to refresh my memory on this one.......it's the "secant-secant" rule that says that a whole secant times its external part = the other whole secant times its external part....I'm assuming that the "20" is the whole length of that one secant...so we have

20*10 = 12*x

200 = 12*x

x = 200/12 = 100/6 = 50/3  = 16 + 2/3 or 16.66

9) The measure of arc AB is 65° as well......(I'm assuming that either 1/2 of the length of both bisected chords is 8, or that a 'partial raidus' of length 8 bisects the chords...either way, they're both congruent chords, interceptig arcs with equal measures)

These problems aren't well-labeled, IMHO!!!

The sides of this triangle form what's known as a "Pythagorean Triple."   It's probably one you should memorize, because it occurs frequently. The remaining side is 5.  Proof:

13^2 - 12^2 = (adjacent side)^2

169 - 144 = (adjacent side)^2

25 = (adjacent side)^2       .......  take the square root of both sides

5 = (adjacent side)

So, the Pythagorean Triple is just 5-12-13.  Note a seemingly odd thing (not really)...all multiples of these numbers will be Pythagorean Triples, too..i.e.,  10-24-26 .......15-36-39, etc.

Excellent, rosala...."points" and "thumbs" from me

8)   The locus of these points is just the angle bisector of ∠ABC......here's a pic... (ray BD is the bisector...at least it's supposed to be a ray...I forgot to include the little "arrow" at the end!!!...use your imagination!!)

10) The locus of these points is just a circle with a radius of 1 "inside" a circle of radius 2.....here's a pic.....(Note that the locus of points all lie on the midpoints of the radii of a circle of 2 cm)

14) I think the answer to this one is given by two perpendicular bisectors of AB and CD drawn from the center of the circle (In effect, radial lines drawn to the edge of the circle that bisect chords AB  and CD....if a radial line bisects a chord it is also perpendicular to that chord!!) .....something like this..(OE and OF are the locus of points, i.e., perpendicular bisectors of the two chords AB and CD....or at least as "perpendicular" as I could make them!!!) ...Note that every point on OE is equidistant from AB and every point on OF is equidistant from CD....to be techically correct, if OE and OF were extended infinitely, all the points on both would be equidistant from the two ends of their respective bisected chords...... 

OK.......we're using the quad formula to find the solutions.

First....let's write out the quad formula........we have:

[-b ± √(b² - 4ac)] /  [2a]  where.... a = 3    b = 4  c = -2     ....  so we have

[-4 ± √(4² - 4(3)(-2))] /  [2(3)]  =

[-4 ± √(16 + 24)] /  [6)]  =

[-4 ± √(40)] /  [6)] =

[-4 ± √(10)*4] /  [6)] =

[-4 ± 2√(10)] /  [6)]         Now....divide the "-4"  the "2" and the "6" all by 2......this gives us

[-2 ± √(10)] /  [3)]           And that's the answer as shown by the on-site solver. It also gives us the decimal approximations for both solutions!!!!

$${\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}{{\mathtt{3}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{1.720\: \!759\: \!220\: \!056\: \!126\: \!4}}\\
{\mathtt{x}} = {\mathtt{0.387\: \!425\: \!886\: \!722\: \!793\: \!1}}\\
\end{array} \right\}$$

4/36 or 1/9

To see this, note that we could have:

1 on first die  + 4 on second die

4 on first die + 1 on second die

2 on first die + 3 on second die

3 on first die + 2 on second die

And there are 6 possibilties on the first die and 6 on the second...so the total number of outcomes is just 6 * 6 = 36

So.....(4 posibilities for a five) / (36 total possibilities) = 1/9

cos^-1(x) ,  arccos (x)  and acos(x)   can all be used to denote the cosine inverse........the on-site calculator uses acos.......just hit the "2nd" button to access it......

Let P₂  = Probabilty that Chris and Melody answer subsequent probability question correctly

Let P₁ =  Probabilty that Chris and Melody answer prior probability question incorrectly

So we have

[ P₂ l  P₁ ] = [P₁ * P₂] /  [P₁]    →  Somewhere between 0  and 1  ....... (probably)

Thanks for the flowers.....!!!!

Notice that if I rearrange the first equation, we have ...... 3y = 11 - 4x

And substituting into the second equation gives us   ........(11 - 4x) + 2x = 13

Simplifying, we have  .........11 - 2x   = 13 

Add 2x to both sides ..........   11 = 13 + 2x

Subtract 13 from both sides  ........11 - 13 = 2x

Simplify ...........  -2 = 2x

Divide by 2 on both sides ............  -1 = x    ...... so there's "x"

To find "y," just substitute -1 into either of the two original equations for "x" .....I'll use the first one

3y + 4(-1) = 11

3y - 4 = 11   add 4 to both sides

3y = 15     divide by 3 on both sides

y = 5

So we have (x , y)  = (-1 , 5)