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To paint the 100 ft^2 wall takes 10 members 10 minutes =   100 member_minutes for the wall 

      Now divide by 15 members 

               100 member_minutes / 15 members = 100/15  minutes = 6  2/3 minutes     

This is AoPS homework.  Do not post AoPS homework.

This is AoPS homework.  Do not post AoPS homework.

Simplify as

x^2 + 15x + k   

Double Root means discriminant = 0

15^2  - 4 (1)(k) = 0

225 - 4k  = 0

225 = 4k

k = 225 / 4

Impossible

Tangent - Secant Theorem

XY^2 =  XB ( XB + BA)

8^2  = 10 (10 + BA)

64  = 100 + 10BA

-36 = 10BA

BA   -36/10 =  =  -3.6       (BA is a  distance and  must be positive)

https://web2.0calc.com/questions/sequence-question_5

Write as

-(1/4)x^2 + (5/2)x  =  0

Find the roots

Factor as

-x [ (1/4)x - (5/2)  ]  = 0

Set the factors  to  0 and  solve  for  x

-x  =  0            (1/4)x - (5/2)   =0

x  = 0              (1/4)x = (5/2)

                              x = (5/2)(4/1)  = 10

The roots are

x = 0   and  x  = 10

The horizontal dstance =  10 -  0    =    10 (meters)

Simplify as

x^2 +10x +  p

Take 1/2 of 10 = 5

Square it =  25  =   p

x^2 + 10x + 25 =   (x + 5) (x + 5)  = (x + 5)^2

                        20 A

                                      E                    

                     144 H        

        B                  D               C

ABH forms a triangle

Angle AHB = 144

Angle BAH = 20

So angle ABH =  180 - 144 - 20  =  16° 

1,2,3,4,5,6,6

Number of  identifiable sequences = 

(Number of rolls)   ! / ( Number of Repeated 6's ) ! =   7! / 2! =    2520

Semi-perimieter  = (15 + 9 + 10)  / 2 =    17  = s

Area = sqrt [ s ( s -15) (s- 10) (17-9) ] =  sqrt [ 17 * 2 * 7 * 8 ]  = sqrt (1904)

Shortest altitude, sa,  is  always drawn  to  the  longest base

Area = (1/2) (AB) (sa)

sqrt (1904) = (1/2(15) ( sa)

sqrt (1904) / 7.5  = sa ≈ 5.82

This is AoPS homework.  Do not post AoPS homework.

The point  (0,3)  is  on the line = the y intercept

The equation of the  line is

y= 2x + 3

The x intercept is

0 = 2x + 3

-3 =2x

x = -3/2

Area of triangle   =   (1/2) (3/2) (3)  = 9/4

A

                X

                         5

B            15                 C

If BXA = 90  then BXC =  90

So triangle  BXC is  right 

So

BC^2  = BX^2 + XC^2

15^2  = BX^2 + 5^2

15^2 - 5^2 = BX^2

200 = BX^2

BX  = sqrt (200)   =  10sqrt(2)

\(a^3 - \dfrac{1}{a^3}$ if $a - \dfrac{1}{a} = 0 \)

a^3 - 1/a^3  = ( a  - 1/a) ( a^2 + 1 +1/a^2)  =  (0)(a^2 + 1 + 1/a^2)   =  0

                 45 B 45

A 45              D                    45 C

The triangle is  isosceles  AB = BC =  12/2 = 6

Angle ABC  = 90

So....Angle ABD = 45

So BD = AD =  6

Area ABD =  (1/2) (AD) ( BD)  = (1/2) (6)(6)  =  18

We can set this  up in a  convenient fashion  to make the math a little easier

Let W = (0,0)

X = (0,10)

Let WXY  be a right triangle such that

Let XY  be the hypotenuse

XY = 15  WX = 10

WY =  sqrt (15^2 - 10^2)  =  sqrt (125)  =  5sqrt (5)

Y = (5sqrt (5) , 0)

A is  the circumcenter  = (5sqrt(5)  /2 , 5) =  (sqrt 125) / 2 , 5)

The equation of the circumcircle is (x - sqrt (125)/2)^2 + (y -5)^2  = 7.5^2

The slope   of XY  =  -10/ [5sqrt (5)] =  -2 / [sqrt (5) ]

The slope of the tangent line to the circumcircle =  sqrt (5) / 2

The equation of the  tangent line = y =[sqrt (5) / 2] x + 10

The x intercept of this  line can be  found as

0 = (sqrt (5) / 2) x  + 10

-10 = [sqrt (5) / 2] x

x = -20/sqrt (5) =  -4sqrt (5)

B = (-4sqrt (5) , 0 )

BY = 4sqrt (5) + 5sqrt (5)  = 9sqrt (5)

Because WZ is parallele to BX ,  triangles  XBY  and ZWY  are similar

WY / BY  = ZY / XY

(5sqrt (5)) / (9sqrt (5) ] =  ZY/ 15

(5/9) = YZ /15

YZ  = 15 (5/9)  = 75/9  =  25/3

Use the law of cosines to find  AC

AC^2 = 6^2 + 8^2 - 2 (6)(8)cos135  

AC = 12.957   units

Then use Heron's formula  to find the area

Area = sqrt ( s(s-a)(s-b)(s-c) ) 

      s = 1/2 ( 6 + 8 + 12.957) 

              a = 6     b = 8    c = 12.957    units 

   Plug in the values to find area = 16.97 units^2 

This is AoPS homework.  Do not post AoPS homework.

This is AoPS homework.  Do not post AoPS homework.