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Here :

https://web2.0calc.com/questions/probability_23388

Simplify as

9a^4 + 9a^3  =  9a^3 (a  -1)

Equation of line

y = 2 -x

Sub this into equation of circle

x^2 + ( 2 - x)^2  = 6

x^2 + x^2 - 4x + 4   = 6

2x^2 - 4x   = 2

x^2 -2x  =  1

x^2  -2x  + 1 =  1 + 1

(x -1)^2   = 2       

x - 1  = sqrt (2)         and  x  -1 = -sqrt (2)

x =  1 + sqrt (2)      and  x = 1  -sqrt (2)

y = 1 -sqrt (2)        and y =  1 +sqrt (2)

A = (1 + sqrt (2) , 1 -sqrt (2) )          B =  (1-sqrt (2) , 1 + sqrt (2) )

Midpoint = (1,1)

For convenience, let the sides of the square =  2

Let

B = (0,0)

C= (2,0)

D =(2,2)

A =(0,2)

M = (2,1)

Line through AC has the slope  -1

The equation of this  line  is  y= -1 (x -2)  =  -x + 2

Line through BM  has a sloe of (1/2)

Equation of  this  line is   y = (1/2)x

x intersection of these two lines 

-x + 2 = (1/2)x

2 = (3/2)x

x = 4/3

y = (1/2)(4/3) = (4/6)  = (2/3)

So the coordinates  of P   =( 4/3 , 2/3)

Area of BPC = (1/2)(BC)(2/30 =  (1/2)(2)(2/3)  =  2/3

Area of APB =  (1/2) (AB)(4/3) =  (1/2)(2)(4/3) =  4/3

Area of MPC =  (1/2)(MC) (2 -4/3)  =( 1/2) (1) (2 - 4/3)  =  (1/2)(2/3)  =  2/6  = 1/3

Area of APMD =  area of square less the other areas =   4 - 2/3 - 4/3  - 1/3 =  4 - 7/3  = 12/3 - 7/3 = 5/3

The problem wants to  find the ratios  of the areas to each other ???....if so

MPC : BPC : APB : APMD =   1 : 2 : 4 : 5

Using the circle centered at (0,0) with a radius  of 4  we can find the intersection of this  circle with the lines y = 8x

And  y= 11x

For a line  with a slope of  8, the tangent = 8 / 1   so   r = sqrt ( 1^2 + 8^2)  =  sqrt 65)

cos =  1/sqrt (65)     sin  8/ sqrt (65)

P  will  have the coordinates  (  4 * 1/sqrt (65) , 4* 8 /sqrt (65)) =  ( 4/sqrt (65) , 32/ sqrt (65)

For a line with  a slope of  11, the tangent  =11/1 so r  = sqrt (1^1 + 11^2)  = sqrt (122)

cos =  1/ sqrt (122)    sin = 11/sqrt (122)

Q will have coordinates  ( 4 * 1 /sqrt (122) , 4 * 11/sqrt (122) ) =   (4/sqrt(122) , 44 /sqrt (122) )

Slope  between P , Q   =    [ 44/sqrt (122) - 32/sqrt (65) ] /  [ 4/sqrt (122) - 4/ sqrt (65) ]  ≈ -.108

                        2x^2

x^2 - 2x   -5    [  2x^4  - 4x^3  +   kx^2  ]

                    -  [  2x^4 -  4x^3  -  10x^2 ]

k + 10   =  0

k  = -10

At first, Adam had A dollars and  Ben had 600 - A dollars

A 

B = 600 - A

After  Adam gave (1/4) of his  money to  Ben

Adam had  (3/4)A

Ben had   600 - A + (1/4) A  =  600 - (3/4)A

After Ben gave (1/3)  of his  money to Adam

Adam had  (3/4)A +  [ 600 - (3/4)A ] / 3  =    200 + (1/2)A

Ben had   (2/3) [ 600 - (3/4)A ]  =  400 - (1/2)A

And in the end  Ben had (3/2)as much as Adam

400 - (1/2)A  =  (3/2) (200 + (1/2)A )

400 - (1/2)A  =  300 + (3/4)A

100 =  (5/4)A

100 (4/5) =  A  =  $80   (Adam  started with $80 )

Proof :

At first   :   A = 80      B =  520

After the first transaction :     A = 60      B =  540

After the second transacyion :   A = 240  B = 360   ⇒   B/ A = 360 / 240  = 3 / 2

In a rhombus, all the side lengths are $12,$ and one of the angles is equal to $150^\circ.$  Find the area of the rhombus.     

                             Area  =  base • height    

                             Area  =  12 • 6  

                             Area  =  72  

Where did that value for the height come from?  

One of the angles is 150^o so the one next to it is 30^o.  

Drop a perpendicular from the 150^o angle at the top vertex.   

Use trig.  You know the hypotenuse is 12 and the angle is 30^o.  

sin(30) = h / 12   (opposite over hypotenuse)  

0.5 = h / 12   ====>  h = 6  

^.

Thank for helping, you got it correct!

Thank You for trying but that was incorrect. The correct answer is 12. Thanks again for telling me that 210 was incorrect and why!

The y intercept  = 0

See here :

From the center of the  small circle  draw a perpendicular to  the bottom  edge of the  square

  .....the distance = 4-r   = A

From  the  center of the semi-circle draw a segment to the point where the first segment intersects the base

.....this distance is 2-r  = B

Connect the center of the  small circle to the center of the semi-circle....the distance is 2 + r = C

We have a right triangle  such that

C^2 = A^2 + B^2

(2 + r)^2  = (4- r)^2  + (2-r)^2    simplify

r^2 + 4r + 4 =  r^2 -8r + 16  + r^2 - 4r + 4

r^2 + 4r + 4  =  2r^2 - 12r + 20

r^2  - 16r + 16  =  0

r^2 -16r   =  -16

r^2 -16r  + 64   = -16 + 64

(r - 8)^2  =  48

r - 8  =   sqrt (48)           or      r - 8 = -sqrt (48)

The first value for r is too large

The second value produces   r =  8 -sqrt (48)  ≈  1.07

\(f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\right\rfloor \)

Write this  in another way

floor  [  (-3x + 2) / (3x + 1) ] 

Evaluating this  function (without the floor) at f(1)  = -1/4    and floor (-1/4)  = -1

And any positive integer  value of  x produces a function value (without the floor)   which is greater than -1  but less than 0  

So the floor of all such values =  -1

So

f(1)  + f(2)  +......+ f(999) + f(1000)   =

1000 (-1)  =   -1000

{ 11, 13 , 17, 19, 23 ,29, 31}

11 *  13  = 143                 13 * 17 = 221                     17 *  19 =  323

11 *  17  = 187                 13 * 19 = 247                     17 *  23  =  391

11 *  19  = 209                13 *  23 = 299

11 *  23  = 253                 13 * 29 = 377

11 *  29 =  319

11 *  31 =  341

12 different products

The prime numbers greater than 10 and less than 20 are 11, 13, 17, and 19. The product of any two of these primes is less than 400.

Let’s list the possible products:

11 * 13 = 143

11 * 17 = 187

11 * 19 = 209

13 * 17 = 221

13 * 19 = 247

17 * 19 = 323

So, there are 6 different products that Syd could have ended up with. Please note that 210 is not a product of two prime numbers greater than 10, which is why it was incorrect. I hope this helps! 😊

Let’s solve this step by step:

First, let’s consider the case where

t

is an integer. In this case,

⌊t⌋=t

and

⌊2t⌋=2t

. Substituting these into the equation gives

t=3t+4−2t

, which simplifies to

t=−4

. However,

−4

is not a solution because when we substitute

−4

into the original equation, we get

−4=−8+4−(−8)

.

Now, let’s consider the case where

t

is not an integer. In this case,

t

can be written as

t=n+r

where

n

is an integer and

0≤r<1

. Substituting this into the equation gives

n=3n+3r+4−2n−2r

, which simplifies to

n=−4

and

r=0

. However,

r=0

implies that

t

is an integer, which contradicts our assumption that

t

is not an integer.

Therefore, there are no values of

t

that satisfy the given equation. The equation

⌊t⌋=3t+4−⌊2t⌋

has no solutions.

The function you’ve given is

f(x)=⌊3x+12−3x​⌋

where the notation

⌊y⌋

represents the greatest integer less than or equal to

y

.

Let’s evaluate this function for a few values:

For

x=1

,

f(1)=⌊3(1)+12−3(1)​⌋=⌊4−1​⌋=−1

because the greatest integer less than

−1/4

is

−1

.For

x=2

,

f(2)=⌊3(2)+12−3(2)​⌋=⌊7−4​⌋=−1

because the greatest integer less than

−4/7

is

−1

.

You can see a pattern here. The function

f(x)

is always equal to

−1

for all positive integers

x

. This is because the numerator

2−3x

is always negative and the denominator

3x+1

is always positive for all positive integers

x

. The fraction is thus a negative number between

0

and

−1

, and the floor of this number is always

−1

.

Therefore, the sum

f(1)+f(2)+f(3)+⋯+f(999)+f(1000)

is simply

−1

added to itself

1000

times, which equals

−1000

. So, the value of the given expression is

−1000

.