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Algorithm

(1)  drop the decimal

(2)  In the numerator :  Subtract  the  non-repeating part from the  repeating part

(3) In the denominator : Write the same  number of 9's  as  the digits in the  repeatnig part  and the  same  number of 0's  as  the non-repeating part

.12  (repeating )     =  (12  - 0)  / 99  =      12  / 99  =   4/33

.132 ( repeating)   = (321 - 0 )  / 99  =     321 / 99     107/33

4 / 33

______  =                 4   / 107  

107 33

As follows:

UTS....   there are no mistakes in my answer that I can find.....

2.

            .1875

16   l   3000

           16

         ______

            140

            128

            ______

              120

              112

             ____ 

                  80

                  80

                ____

Here's  1

2(8)^2   + 4(8)    +   5   =

128  +  32   +  5

165₁₀

a.      N   = 10000 ( 2)^(t/40)        where N is the number of  bacteria after t  minutes

b.  N  =  10000 (2)^(60/40)    ≈  28284

c.    5000 +  8 =  10000 (2)^(t/40)

50008  =  10000  (2) ^(t/40)

50008  /10000 =  2^(t/40)          take the log of each  side

log   (50008 / 10000)  =  log 2^(t/40)       and we  can write

log (50008 / 10000)   =  (t/40)  log (2)

t =    40 log ( 50008  / 10000) / log (2)  ≈    92.9  minutes

d.  Here's  the graph   ( subbing y for N  and x  for t)

https://www.desmos.com/calculator/t5egnmgcks

a) let t be the number of bacteria as its instructing you, A be the starting point, r be the rate of increment and T be time.

 lets use the formula t=A\times r^T

T itself can be broken down into 2 components -- the 40 minutes were stated, and another unknown value which would be the time you want it to be; so: $ t=\frac{T_m}{40} $

respectively, substitute the values in:

$ t=1000\times 2^{\frac{T_m}{40}} $

b) we are working in minutes so one hour is equal to 60 minutes ; just plug that as T_m

$ t=1000\times 2^{\frac{60}{40}} $

$t\approx 1000\times 2.828 $

$t\approx 2828 $

c)  its stated that $ t=(50000 +m)$ where $m=8$

$(50000 +8)=1000\times 2^{\frac{T_m}{40}} $ thus,  

$50008=1000\times 2^{\frac{T_m}{40}}$

get rid of the base

$  2^{\frac{T_m}{40}}=\frac{6251}{125} $

$ \frac{T_m}{40}\ln \left(2\right)=\ln \left(\frac{6251}{125}\right) $

$ T_m=225.7  $

d) use desmos, or try manually graphing it.
 

We can solve  with a proportion

1  / 4  =     x   /  13200           cross-multiply

1320    =  4x

x = 13200  /  4    =     3300  sq ft   

(10x - 3y)  / ( 3x - y)   =   3/5     cross-multiply

5 (10x  - 3y)  =   3(3x  - y)

50x  -  15y   =   9x   -  3y

50x  -  9x   =  15y  -  3y

41x  =  12y

x / y   =   12  /  41

                A

       P

B                              C

Since  APC   =115

Then angle  BPC  = 180  -115   =   65

And in triangle   BPC,  CP   = BC

So  angles  BPC  and PBC   =  65

So  angle  PBC  =  angle ABC 

And AB = AC

So angle ACB = angle ABC =  65

So angle  BAC = 180   - 2(65)  =  50 = angle  PAC

So   angle   APC + angle PAC  + angle ACP    =180

So

115  +  50  + angle ACP   =180

165    +  angle ACP    =180

angle ACP =   180  -  165   = 15°

The roots of x^2 + px + q = 0 are real, and the roots of (x + a)(2x + p) = 0 are real, so the roots of x^2 + px + q + (x + a)(2x + p) = 0 are real.

Rearrange  as

2x^2  -2x  + 2y^2  +  4y   =  20           divide through by  2

x^2  -  x  +  y^2   +  2y  =   10              complete the  square on x and  y

x^2  -  x  +  1/4  + y^2  + 2y  + 1 = 10  +  1/4 + 1

(x   -  1/2)^2   +  ( y + 1)^2   = 11.25

The  radius  =  r^2  =  11.25

The  area of the square  =  (2r)  ( 2r)  =  4r^2  =   4 (11.25)   =  45  units ^2

                                C = 120

A = 30          D         M            E                 B = 30

            x                   x                     x

Let M  be  the  midpoint of   AB

Then angle ACM  =  60        angle AMC   = 90     angle  CAM  = 30      

Then   ACM  is A  30 - 60-90  right triangle

AB  = 3x

AM  =  (1/2) AB    =    (3/2)x

AM - AD =  DM  

(3/2)x  -  x  = DM

x/2   = DM

CM =  AM  / sqrt 3   =  (3/2)x / sqrt (3)   =  [sqrt (3)  / 2] x

tan (angle CDM) =   CM /DM

tan  ( angle CCM)  =  [(sqrt (3) / 2) x ] /  ( x / 2) 

tan (angle CDM)  =   sqrt 3

arctan (sqrt (3) )  =  angle CDM   = angle CDE   =  60°

-3 + 3i

Let  the 2  right triangles  contained  within ABCD  have legs  of   2 and  x

Their  combined  areas =   (2) (1/2) ( 2x)  =  2x

The shaded  area is a parallelogram with  a height of 2  and bases of  sqrt (2^2 + x^2)  = sqrt (4 + x^2)

So....these combined areas = 2 * 8    =  16

So

2x   +  sqrt  (4 + x^2)  * 2    =  16

x   +  sqrt ( 4  + x^2)    = 8

sqrt ( 4 + x^2)  = 8 - x               square both sides

4 + x^2   =  x^2  - 16x  +  64

16x   - 64  + 4  =   0

16x  - 60  =   0

4x  -  15 = 0

4x  = 15

x = 15/4

So....the  area of  the  parallelogram  (shaded area )   =   

sqrt  [  4 +  (15/4)^2 ] * 2  =

sqrt  [ 64  + 225 ]  / 4   *   2   =

17  /  2   =

8.5  units ^2   

1. The smallest possible area of triangle OAB is 60.

2. The minimum value of f(1) is 144.

a  +  bi  = z

l z  l  ≤  5                square  both sides

l z l ^2  ≤  25

a^2  + b^2   ≤  25

This is a  circle  centered at  (0,0)  with a radius of 5

The  number of  integer coordinates  =

1  +  4 (5)   +   4 [ floor  sqrt (5^2 -1^2)  + floor sqrt (5^2 - 2^2)  + floor sqrt (5^2 - 3^2) + floor sqrt (5^2 - 4^2)   ]  =

1 +  20  +   4  [ floor  sqrt 24   +  floor sqrt (21)  + floor  sqrt (16)  +  floor sqrt ( 9)  ]   =

1  +  20   +  4 [  4    +  4    +   4   + 3 ]   =

1 + 20   +  4 *15  =

1 +  20  +  60  =

81

The total length of the lines is 140.

Ohhh thank you so much, so the three terms that were used in the Am Gm thing are the 3 things that sum to b, Thanks so much :)

Solution without the use of the quadratic formula:

Combine all terms to the right side of the equation:

99x^2 - 24x - 3 = 0

Divide by the common factor of 3:

33x^2 - 8X - 1 = 0

Factor the equation into two linear terms:

(11x + 1)(3x - 1) = 0

Since the product is zero, one or the other of the terms must be zero.  Setting each to zero and solving for x gives the two solutions:

11x + 1 = 0

11x = -1

x = -1/11

3x - 1 = 0

3x = 1

x = 1/3

The two solution (both real) are: -1/11, 1/3

$x^2+4x+4=100x^2-20x+1$

$x^2+4x+3=100x^2-20x$

$   x^2+24x+3=100x^2 $

$ -99x^2+24x+3=0  $
you can chose many ways to solve this, but i will do quadratic formula $ _2 x_1=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $ where $ a=-99 \ ; \ b=24 \ ; \ c=3  $
$  _2 x_1=\frac{-24\pm\sqrt{576+1188} }{2\left(-99\right)}  $

$  _2 x_1=\frac{-24\pm \:42}{2\left(-99\right)}  $

$_2 x_1=\frac{-24\pm \sqrt{24^2-4\left(-99\right)\cdot \:3}}{2\left(-99\right)}  $

$  x_1= \frac{18}{-198} $  and  $  x_2 \frac{66}{198} $

which  you can simpify:

$ x_1=\frac{-1}{11} $ and  $x_2= \frac{1}{3}   $

the reply above had a little mistakes :D

re-arrange to :

99x^2 -24x-3 = 0      Use quadratic formula   

Simplify a bit by dividing by 3

33x^2 - 8x -1=0

a = 33    b = -8    c = -1

\(x = {-(-8) \pm \sqrt{(-8)^2-4(33)(-1)} \over 2(33)}\)

Distance  between  B  and C =    8

Distance  between   A  and B  =  sqt  [ (-6-9)^2  + (-6-6)^2  ]=  sqrt  [ 225  + 144 ]   =  sqrt [ 369 ] = 3sqrt (41)

Distance  between   A  and C  = sqrt  [ (-6-9)^2  +  (6 - 2)^2  ]=   sqrt  [ 225 + 16 ] = sqrt [ 241]

Perimeter =   8  +  3sqrt (41)   +  sqrt [241 ] ≈   42.7