All Answers 358077 Answers

FD = 24/5.

why does it say "blocked by moderator"?

The area of the circle is 12*pi.

Here is a picture

this case is the range of numbers choose sides of a triangle. 

so 6 choose 3 = 20

and we want to choose a set of numbers that when you apply the pythagorean theorem (a^2+b^2+c^2) the result is a whole number

only 3, 4, and 5 works in this case between the range 1 to 6 (proof: 3^2+4^2=9+16=25, 25=5^2)

so 1, 2, and 6 doesn't work

so the probablility is 1/2 if no number can be used more than once. 

but if a number can be used more than once in the range 1 to 6 (still only the set 3, 4, and 5 works), then the answer is 1/6 choose 3 = 1/20

3 4 5 triangle is the only one I am aware

6 c 3 = 20   choices for 3 sides

1 choice   3 4 5   works       1/20  or  5%

Since  DB  = CD   then  DB  = 5

If we let   A  = (0,0)  , then using similar triangles   we  have  that   the coordinates of  D   =  (3 , 4)

 Then AD  =   sqrt(3^2 + 4^2)  = sqrt (25)  = 5

And the area of  triangle ADE   = (1/2)  AD  ( ED)  = (1/2) (5) ED = (5/2)ED     ( 1)

Also  the altitude   of  ADE  = 3

So its area also  =   (1/2)  3(EA) =  (3/2) EA   (2)

Equating (1)  and (2)   we  have  that 

5ED = 3EA

ED  =  (3/5) EA

Using the  Pythagorean Theorem

EA^2   - ED^2   = AD^2

EA^2  - [( 3/5) EA ] ^2   = 5^2

EA^2   - (9/25)EA^2  =   25

(16/25) EA^2  =25

EA^2 =  (25* 25)  /16   =  625/16

EA =  sqrt (625/16)  =   25 / 4  

So....the area of  ADE = (1/2)(EA)(altitude)   =   (1/2) (25/4)(3)   = 75/8  units ^2 = 9.375  units^2

Midline of the cos funtion is USUALLY   y = 0 

   this cos fxn is shited  UP 0.5

y = 0.5

the minus 1.5  flips it around the x axis   and   enlarges it by 1.5

   the - pi/3 is a phase shift

2x +  y  = 14      ⇒ y = 14 - 2x

2L  + z =  20   ⇒ z = 20 - 2L

x + y + z  = 12

x + z + L  =  18

Using the last two equations and substituting

x + 14 - 2x + 20 - 2L =  12

-x - 2L   =  -22        (1)

x + (20- 2L) + L  =18

x - L  =  -2          (2)

Add (1)  and (2)

-3L  = -24

L = 8

x - 8  =  -2

x =  6

z = 20  - 2L  =   4

y = 14 - 2x   =  2

x +  z =   10

Consider a square ABCD with side length 2. Let E be the midpoint of AB, F the midpoint of BC, and P and Q the points at which line segment AF intersects DE and DB, respectively.  What is the area of EBQP?

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I'll borrow Phill's diagram.

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Area of Δ ABD = 2

Since DE is a median of Δ ABD, the area of Δ BED = 1

∠ADE = tan^-1(1/2)

DP = cos^-1(∠ADE) * 2

∠EDB = 45 - ∠ADE

PQ = tan(∠EDB) * DP

Area of Δ DPQ = 1/2 (DP * PQ)

[EBQP] = [EDB] - [DPQ] = 7/15     or     0.466666667 square units

Happy 2 help, Chris!

Got to  be careful here.....we might not necessarily  have a right triangle

In general.....The sum of any two sides of a triangle is greater than the remaining side  (this is known as the trinagle inequality)

So

11+ 17   =  28

So   the remaining side   must  be  <  28 =    27

use the good ol' pythagorean theorem

a^2+b^2=c^2

11^2+17^2=410

x^2=410

sqrt410 (can't be simplified any further) 

Thanks, guys   !!!   

Let  the   first angle  = 100

And let  the greatest angle be   ( 100 + 4(n -1) )   where n  is the  number of sides

Sum  of  the  interior  angles  = 180 ( n - 2)

So....using the  formula for the sum of an arithmetic progression  we have that

(100  +  100 + 4 ( n - 1) ) ( n /2)  =  180 ( n - 2)

( 200 + 4n - 4)   ( n/2)   =180 (n  - 2)

(196 + 4n) (n /2)   =180 ( n- 2)

(98 + 2n)n   = 180 n  -  360

98n + 2n^2   = 180n - 360

2n^2  - 82n + 360  =  0

n^2  - 41n + 180   =  0

(n - 45) ( n + 4)  =  0

Setting the first  factor to 0  and solving for  n  gives

n = 45  = the number of sides

after bashing randomly, I got x=6, y=2, z=4, and L=8. 

since x=6 and z=4, 6+4=10

@cryptoaops: I'm pretty sure they are. To the first question: 

add 21 to both sides and divide both sides by -4 to get x=-5

To the second question: subtract 12 on both sides and multiply both sides by 3 to get m=18

To the third question: distribute and subtract 10 on both sides and divide both sides by -12 to get z=-4

To the last question: combine like terms and add the numbers and add 5 to both sides and divide both sides by 7 to get a=5

sniped -_-

1. -4x-21=-1

-4x=20

x=-5

2. m/3+12=18

m/3=6

m=18

3. -2(6z-5)=58

-12z+10=58

-12z=48

z=-4

4. 4a-7+3a-2=30

7a=21

a=3

-hihihi

😎😎😎

GJ @dd0.

Wow, ur dd0... nice... ill use the message center here 2 chat w/ u...

in this case the formula would be what Joe owes subtracted by how many dollars Joe paid Mike, so 49-23=26

also @cryptoaops, I'm pretty sure you know me. I'm dd0

btw, I upvoted your post. 

See  the  following :

Segment   DB   has a slope of    (2-0) / (0-2)  =2/-2  = -1

And the equation of  line containing DB  is    y = -x  + 2      (1)

Segment AF  has a slope of  ( 1-0) /(2 - 0)   =1/2

And the equation of the  line  containing  AF  is

y = (1/2)x     (2)

Find the x  coordinate of  Q  ⇒    (1)  = (2)

-x + 2 =  1/2x

2 =  3/2x

x = 4/3

And the y coordinate  is   (1/2)(4/3)  =  4/6 = 2/3

Thie is the height  of    triangle  AQB....and the  base  is  AB  = 2

So  the area of tis triangle is   (1/2) (2/3) (2)  =  2/3        (3)

And the  slope of the segment containing DE  is  (2 - 0) ( 0-1)  =  -2

And the equation of the line comtaining this segment is   y = -2x + 2     (4)

Find the  x coordinate  of P    ( 2)  = (4)

1/2x =   -2x + 2

5/2x  = 2

x  =   4/5

And the  y coordinate of   P  =  (1/2)( 4/5)  =  4/10  =  2/5

And this is the  height  of triangle APE    and the base = AE =  1

And its area is  (1/2) ( 1) (2/5)  = 1/5     (   5)

The area of   EBQP   =  ( 3)  - ( 5)  =   2/3   -  1/5  =      7/ 15 

EDIT TO CORRECT A  PRIOR ERROR  !!!!

Are these 4 different questions?