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Consider ΔABC. Through each vertex draw a line perpendicular to the corresponding angle bisector. These three lines will form a triangle - say ΔA'B'C'. Note that, since A'B' is perpendicular to CLc, ∠BCA' = ∠ACB'. The same is true of the pairs of angles at vertices A and B. Let's call this a mirror property. As we know, the orthic triangle of ΔA'B'C' has the mirror property. We'll make use of this observation shortly.

1. 

Since X is the intersection of the angle bisector of angle P with side QR, then QX:XR = PQ:QR = 9:10. Also, since Y is the foot of the perpendicular from X to side PR, then XY:YP = QR:PR = 10:17. Combining these two ratios, we get QX:XY:YP = 9:10:17. Let QX = 9x, XY = 10x, and YP = 17x. Then by the Pythagorean Theorem on triangle PQX, we have PQ^2 = QX^2 + XQ^2 = 26x^2. Since PQ = 9, then 26x^2 = 9^2 = 81. Then x^2 = 9, so x = 3. Therefore, XY = 10x = 30.

2. 

Since △DEF is equilateral, we have DE=DF=5.

Let x=AE. Then AF=5−x.

By the Law of Cosines on △DCB,

\begin{align*} BC^2 &= CD^2 + CB^2 - 2CD \cdot CB \cos 60^\circ \ &= 5^2 + 2^2 - 2 \cdot 5 \cdot 2 \cdot \frac{1}{2} \ &= 7. \end{align*}By the Law of Cosines on △CEA,

\begin{align*} AC^2 &= AE^2 + CE^2 - 2AE \cdot CE \cos 60^\circ \ &= x^2 + 2^2 - 2x \cdot 2 \cdot \frac{1}{2} \ &= x^2 - 2x + 4. \end{align*}By the Law of Cosines on △ABC,

\begin{align*} AB^2 &= AC^2 + BC^2 - 2AC \cdot BC \cos ACB \ &= (x^2 - 2x + 4) + 7 - 2(x^2 - 2x + 4) \cdot \cos 90^\circ \ &= x^2 - 2x + 11. \end{align*}Since AB=AF+BC=5−x+7​, we have

[5 - x + \sqrt{7} = \sqrt{x^2 - 2x + 11}.]Squaring both sides, we get

[25 - 10x + 7 + x^2 - 2x + 4 = x^2 - 4x + 12.]Solving for x, we find x=5/3.

3.

Since the circumcircles of triangles ABX and ACX have the same radius, then triangle AXB is similar to triangle ACX. Let r be the radius of the circumcircles of triangles ABX and ACX. Then the circumradius of triangle ABC is also r. By Power of a Point,

[AX^2 + BX^2 = r^2] and [AC^2 + CX^2 = r^2]. Adding these equations, we get

[279 = 3r^2]. Then r=93

The area of triangle ABC is

[\frac{1}{2} \cdot BC \cdot r = \frac{1}{2} \cdot 10 \cdot \sqrt{93} = \boxed{5 \sqrt{93}}].

4.

The areas of regions R1, R2, R3, and R4 can be calculated as follows:

Region R1: This region is a sector of the circle with angle measure 90 degrees. The area of a sector with angle measure θ and radius r is 360∘θ​πr2, so the area of region R1 is 360∘90∘​π(10)2=25π.

Region R2: This region is a triangle with vertices (0,0), (6,0), and (6,4). The area of this triangle is 21​⋅6⋅4=12.

Region R3: This region is a triangle with vertices (0,0), (0,4), and (6,4). The area of this triangle is 21​⋅0⋅4=0.

Region R4: This region is a rectangle with dimensions 6 by 8. The area of this rectangle is 6⋅8=48.

Therefore, the area of region R1 is 25π, the area of region R2 is 12, the area of region R3 is 0, and the area of region R4 is 48. Hence,

[R1] - [R2] - [R3] + [R4] = 25π - 12 - 0 + 48 = 25π + 36.

In conclusion, if [R1]>[R2]>[R3]>[R4] then [R1]-[R2]-[R3]+[R4] = 25π + 36.

3pi / 16

Wait hold on I'm a bit confused ... In what way is this geometry???

It's 43.

Note that each member of the team paints 2000/20=100 square feet per minute at this rate. It will take them 5000/100​ = 50 minutes to paint a 5000 square foot wall.

20-2000-35

20-1-35/2000

20-5000- 87.5min

When the distance covered back and forth is the same, then:

Average speed ==Harmonic mean

Harmonic mean==3 / [100^-1  +  120^-1  +  150^-1] ==120 m/hour

Average speed ==120 meters/hour.

What would be the fractional answer though?

We can  consider  just a 1 x 1 square and  two  circles  with  equations  x^2 + y^2 =1  and x^2 + y^2 = 1/4

The area of the  smal square= 1

The area between  the  circles is   (1/4) pi ( 1^2  - (1/2)^2)  =  (3/4)pi / 4  = (3/16) pi

The probability  is  (3/16)pI / 1  ≈ .589 ≈  58.9%

The discriminant is absolutely useful here. The discriminant gives insight about the number of solutions for a particular quadratic equation in standard form. When the discriminant equals zero, then there is one and only one solution.

\(4x^2 + nx + 25 = 2x^2 + 5 \\ 2x^2 + nx + 20 = 0\)

Now, the equation is in standard form, so we can use the information from the discriminant to determine values of n that yield one and only solution in the x-variable.

\({\color{red}2}x^2 + {\color{blue}n}x + {\color{green}20} = 0 \\ \Delta = {\color{blue}b}^2 - 4{\color{red}a}{\color{green}c} \\ \Delta = {\color{blue}n}^2 - 4*{\color{red}2}*{\color{green}20} \\ \Delta = n^2 - 160\)

Set the discriminant to zero to find the n-values that lead to one and only one solution.

\(n^2 - 160 = 0 \\ n^2 = 160 \\ n = \pm \sqrt{160} \\ n = 4\sqrt{10} \text{ or } n = -4\sqrt{10}\)

The question asks for the positive value of n only, so we reject the negative answer. Therefore, \(n = 4 \sqrt{10}\)

Triangles EAC , ABF and BCD  are all 45 - 45 - 90 right triangles

Their legs are    6/sqrt 2 ,  8/sqrt 2  and 10/sqrt 2   respectively   

R =  (1/2) ( 6/sqrt 2) ^2  =   9

S =  (1/2) (8/sqrt 2)^2  =    16

T =  (1/2) (10/sqrt 2)^2  =  25

(R + S)  / T  =   (9 + 16)  / 25  =    1

It looks like  RS should be 28  and PQ = 7

Area of  trapezoid =  (1/2)(height of triangle QSR) ( sum of  base lengths)

12 =  (1/2)(height of triangle QSR) ( 28 + 7)

12 = (35/2)  ( height of triangle QSR)

12 ( 2 / 35)  = height of  triangle QSR

24/35 =  height of  triangle QSR

Area of triangle QSR = (1/2)RS (height of QSR) =  (1/2)(28)(24/35) =  (14/ 35)*24 = (2/5) (24) =  48/5  = 9.6

[PQS ] =  area of trapezoid - area of triangle QSR =  12 - 9.6 =  2.4

Let the side of the  square = S

QT = S/2

Triangle QUT is a  45 -45 -90  right triangle

Each leg is   (S/2) (1/sqrt 2) =  S / [ 2sqrt 2 ]

Its area =  (1/2)(product of the  legs) =  (1/2) (S/[2sqrt 2 ] )^2  =  (1/16)S^2

And  triangle QSR =  1/2 area of the square=  S^2/2

So  [ RTUS ]  =  S^2/2 - S^2/16 =  (7/16)S^2

So   [ QUT ] / [ RTUS ]   =   [ (1/16)S^2 ] / [ (7/16)S^2 ]  =  1 / 7

Let G be the total  number of  games   you  play

Let .75 G be the number of  games  you  have played

Let . 45 (.75)G be the number of  games  you  have won

Let .25G be the games  left

Let  .50G = the number of  wins  you  must have

And number of wins you  need - Number of  wins  you  have =  No. of  wins you need in  your remaining  schedule

So

[ No. of wins you  need - No, of  wins you have  ] / [ No. of Games Left ]

[ .50G - .45 (.75)G ] /  .25G  =

(.1625 G )  / ( .25G) =  .65    (you need to  win  65%  of your remaining  games )

Since 75% of the season has been completed, 25% of the season remains. Thus, we have 25 games left to play.

We must win 55% of the remaining games in order to have a 50-50 win-loss record. To find this, we convert the percentage to a decimal and divide by 4: 55%=55/100​=11/20​. So, we must win 44%​ of the remainder of our games in order to finish with a 50-50 win-loss record.

\( \frac{-3 \pm 3 \sqrt{29}}{14} \)

The discriminant =  3^2 * 29  =   9*29 =  261

So

3^2  - 4 (7) (k)  =  261

9 - 28k = 261

-28k =  252

k = -252 / 28   =    -9 

Start with your variable, x. Now begin writing an equation. We start with 72 - since Jessica baked that amount. After she sold x cookies, she had some amount left (so, we have 72 - x =), which is supposedly equal to the amount Max had left, thus the same things go for Max, so we have 72 - x = 36 - x/2.

Adding x to both sides, we get 72 = 36 + x/2, and subtracting 36 from both sides we get 36 = x/2. Multiplying by two we get 72 = x. Thus, the answer is D. 72. And Bosco is correct.