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Thank you! I always wondered how to go about these types of questions. 

6, 6, 6, 6, 6, 6, 6, 6

196^213 last digit is 6

3, 9, 7, 1, 3, 9, 7, 1

213^196 last digit is 1

6*1 = 6

=^._.^=

Using similar triangles. 

8/15x+x = 8

(23/15)x = 8

15/8x+x = 15

(23/8)x = 15

Solve either of those equations. 

=^._.^=

Subtract 14 from both sides to get x=11.

Thank you so much!

I would like to see an answer to this too   

The set S contains 72 numbers. Since the restriction is that the first term must be a prime, only 2, 3, and 5 work.  3 * 4 * 3 * 2 * 1 = 72

Now, if we need the third term to equal 2, we can get 2 * 3 * 1 * 2 * 1, which equals 12. 

12/72 = 1/6

:>

At  x  = 0,  y =  4

We  can find the first (and only) positive root  as

(x^2  - 2)^2   =  0

x^2  - 2   =  0

x^2   =  2          take  the positve root  

x  =   sqrt (2)

This root is of  a  multiplicity of  2   which  means  that  the  graph  touches  the  x  axis  at x = sqrt 2  but   does not pass through the   x axis.....and  the  graph to  the right of this  point is ever-increasing

So  the  range  =   the  domain  =  [0, inf )

See the  graph  here  :  https://www.desmos.com/calculator/6ary7m5ryp  

Compute \(999^{-1} \pmod{1000}\).

Express your answer as an integer from 0 to 999.

Modular multiplicative inverse using Euler's theorem

\(\small{ \begin{array}{|rcll|} \hline && 999^{-1} \pmod{1000} \\ &\equiv& \dfrac{1}{999} \pmod{1000} \quad & | \quad \gcd(1000,999)=1! \\ &\equiv& 999^{\phi(1000)-1} \pmod{1000} \quad & | \quad 999 \equiv -1 \pmod{1000} \\ &\equiv& (-1)^{\phi(1000)-1} \pmod{1000} \quad & | \quad \phi(1000) = 400~\text{ (Euler torent function)} \\ &\equiv& (-1)^{400-1} \pmod{1000} \\ &\equiv& (-1)^{399} \pmod{1000} \\ &\equiv& -1 \pmod{1000} \\ &\equiv& -1+1000 \pmod{1000} \\ \mathbf{ 999^{-1} \pmod{1000} }&\equiv& \mathbf{ 999 \pmod{1000}} \\ \hline \end{array} }\)

This  triangle  is  isosceles....the  shortest  angle  bisector  is  that of  angle BAC  =

sqrt [ AC^2  -  (BC/2)^2  ]  =

sqrt [  14^2  - 10^2 ]  =

sqrt  96  =

4 sqrt 6

Bryan and Harold had 108 marbles altogether. Bryan gave away 1/2 as many marbles as Harold gave away. The number of marbles Harold had left was the same as the number of marbles he had given away. If Harold had 1/2 as many marbles left as Bryan, how many marbles did Harold have at first?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

B + H = 108

B gave = 1/2x marbles

H gave =  x  marbles

H had left x  marbles

B had left 2x marbles

1/2x + x + x + 2x = 108

x = 24

Harold had 48 marbles 

If z is a complex number satisfying z+1/z=1
calculate z^10+z^(-10)

Answer see:

Very  nice,  EP    !!!!!

See   #2   here  :   https://web2.0calc.com/questions/2-problems

C  =  .70t   +  9

The  slope =  .70   =  7/10

This  means  that  for every minute of use,  it will  cost  Carter  7/10  dollars   =  70 cents

z=zoe's money

l=luke's money

z-50=l+50

z=l+100

\(\frac{2l+200}5=\frac{2l}3+12\)

\(6l+600=10l+180\)

\(4l=420\)

\(l=105\)

so Luke had 105 dollars at first

Let  Z  =  what Zoe started with   and  L  =  what Luke started  with

(2/5)Z  =  (2/3)L  +  12          Multiply through by  15

6Z  = 10L  + 180

Z   =  (10L  + 180)  / 6          (1)

So

Z - 50   =  L  +  50    (2)         sub (1)  into (2)

(10L + 180)  / 6   -  50   =  L +  50

(5/3)L  + 30  - 50  = L + 50

(5/3)L - 20  =  L +  50

(5/3)L  - L  =  50 + 20

(2/3)L  =  70

L  =  70  (3/2)   =  $105   =  what Luke started with

Thank you so much!!

- 2   + x

_______   =   1             multiply  through  by  2

      2

-2 + x   =  2

x + 2 + 2

x  =  4

-1  +  y

_____    =  -  6            multiply  through  by  2

     2

-1  +  y   =  -12

y  =  -12 + 1

y  =  -11

B  =   (  4 , - 11)        sum of  the coordinates =  -7

The area is 6*pi - 2.

sec   θ  =     r / x

So

r    =  r  / x                  divide  out  r

1 =  1/ x

x  =  1

Janet spent 30% of her money on 7 cupcakes and 4 cookies on Monday. The cost of each cupcake was twice the cost of each cookies. She bought some more cupcakes and another 10 cookies with 6/7 of her remaining money on Tuesday. How many cupcakes did she buy on Tuesday?

Let  the  amount  she  started  with  =  A

Let the  cost of a  cookie  =  C    and  the  cost of  a cupcake   =  2C

So

.30A  =  4C  +  7(2C)  ⇒   .30A  =  18C   ⇒   A  =  (18C)  / .30   =   60C   

So   she  had    .70A  left  after  spending  .30A   and   6/7  of this  =  (6/7) .70A =   .60A  = .60(60) C  =  36C

So  with  this   the   amount  she spent  on  cupcakes  on Tuesday  =

36C  -  10C    =     26C  =   13 (2C)

So...she  bought   13  cupcakes

sin    =   sqrt   (25^2  - 7^2)   / 25    =  sqrt  ( 625   -  49)  / 25   =     sqrt  (576)  /25   =  24 / 25

Let   the  number  that  John started  with  = J

And  let  the  number  that  Kaden  started  with  =  K

Let  the  number  that Kaden  gave  away   =  G

Let  the  number  that John gave  away  =   (3/5)G

So

J  -  (3/5)G  =   (3/5)G  - 4        (1)

And

K  -  G    =   (2/5) [ (3/5)G  - 4  ]       (2)

And

[(3/5)G  - 4]  -  (2/5) [ (3/5) G - 4 ]   =  21      (3)

Simplifying  the  3rd  equation, we  have  that

(3/5)G  - 4  - (6/25)G  + 8/5  =  21                  multiply  through  by 25

15G   -  100  - 6G  +  40  =  525

9G   -  60  =  525

9G   =  585

G =  585 /  9   =  65

Sub  this  into  the  second  equation  for   G

K  -  65    = (2/5)  [  (3/5) 65  - 4 ]

K  -  65  =  (2/5)  [ 39 - 4 ]

K - 65  =  (2/5) ( 35)

K  -  65   =  14

K   =  14  + 65   =    79   =  what Kayden started  with

Proof   

Kayden    is left with  (2/5)(35)   =  14

John    is  left  with (3/5)G - 4  = (3/5)(65) - 4  =  39 - 4 =    35 =  21  more  than Kayden