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1.

re:  there is no '6' in base 5

omg, what a bone-headed mistake I made.  Thank you. 

.

Awesome, thank you for helping!!!!!

Thank you so much! The explanations refreshed my memory for the other ones as well!!

By the Law of Cosines:

\(a^2 = c^2 + b^2 - 2cb \cos A\\~\\ (\overline{\text{BC}})^2 \,=\,(7\sqrt2)^2+(6)^2-2(7\sqrt2)(6)\cos(45°)\\~\\ (\overline{\text{BC}})^2 \,=\,98+36-(84\sqrt2)(\frac{\sqrt2}{2})\\~\\ (\overline{\text{BC}})^2 \,=\,98+36-84\\~\\ (\overline{\text{BC}})^2 \,=\,50\\~\\ \overline{\text{BC}} \,=\,\sqrt{50}\\~\\ \overline{\text{BC}} \,=\,5\sqrt{2}\)

THX, EP  !!!!!

1.

Select Not independent or Independent for each situation.

A spinner is divided into three equally sized sections. The spinner is spun three times. = Independent

A number cube is rolled twice. = Independent

A bag contains 8 white balls and 3 yellow balls. Two balls are selected from the bag without replacement. = Not Independent

Thank you soo much for your help and explanations!!!!

(8)^5/2=   ( 2*2*2)^5/2 =   (4*2)^5/2 =   2^5  * 2^5/2 = 32 (2*2*2*2*2)^1/2 = 32 * 4 sqrt2 = 128 sqrt 2

Don't understand the second one

Last one

(8x^7y^4)^(2/3)  =

8^(2/3) *(x^7)^(2/3) * (y^4)^(2/3) 

[8^(1/3)[^2  * [x^14/3]  * [y^8/3]  

2^2  *  x^4 * x^(2/3)  *  y^2 * y^(2/3)

4x^4y^2 ∛ [ x^2 y^2 ] 

1.  all whole numbers

2.

Below are all possible outcomes:

2, 2       4, 2       7, 2       9, 2

2, 4       4, 4       7, 4       9, 4

2, 7       4, 7       7, 7       9, 7

2, 9       4, 9       7, 9       9, 9

The ones colored blue are the ones which have a product less than 30 and contain at least one odd number.

There  are  6  of them.

3.

y  =  x + 6

y  =  x² + 8x + 16                Substitute  x + 6  in for  y  in this equation.

x + 6  =  x² + 8x + 16

                                            Subtract  x  and subtract  6  from both sides.

0  =  x² + 7x + 10

                                            Factor the right side.

0  =  (x + 2)(x + 5)

                                            Set each factor equal to zero and solve for  x .

x + 2 = 0     or     x + 5 = 0

 x  =  -2       or       x  =  -5

When  x  =  -2 ,     y  =  x + 6  =  -2 + 6  =  4       So  (-2, 4)  is a solution.

When  x  =  -5 ,     y  =  x + 6  =  -5 + 6  =  1       So  (-5, 1)  is a solution.

Here's the first one, PF :  https://www.desmos.com/calculator/ysfv3ogzwy

The x values are - 2  and -1

(Guest:   Note....you cannot have a   64₅     .....  there is no '6'   in base 5 )

See other answers here:

https://web2.0calc.com/questions/help-please_24894

1234      9876     5432

1357      9753     4321

2345      9630     3210 

2468      8765    

3456      8642     

3579      7654      

4567      7531      

5678      6543    

6789      6420     

I count 21......anyone have a different result  ????

thanks but what about the second problem?

Very nice, EP  !!!!

1 - 7^2010 mod 100 = 49

Using hectictar's diagram..and switching the base and height per CU...we have that

triangle   CBA  is similar to triangle DBC

Therefore

AC / AB  =  DC / BC  subbing we have that

15/ sqrt ( 15^2 + 8^2 ) =  r  /  8

8 * 15  / sqrt (289)  = r

8*15 / 17  =  r  =   120 / 17

You didn't post all of the question, but this may be the same question answered a few days ago:

https://web2.0calc.com/questions/can-i-have-some-help-on-this-question_1#r1

(Guest:   Note....you cannot have a   64₅     .....  there is no '6'   in base 5 )

You're welcome !   I hope you actually understand the answers and how they were derived.....   

Thanks!

Compute $29^{13} - 5^{13}$ modulo 7

2)

    12 sections  * x pages   + 11 pages   must be less than 100 and a multiple of 5    + 1

       12(x) + 11 < 100       11 is already a multiple of 5     + 1

                                       so we need 12x to be a multiple of 5       60 will work nicely    so x = 5    (the next multiple of 5  would be 120....too big)

12(5) + 11 = 71 pages     (assuming 2nd to last page = next to last page)   

1)    554 in base b      5*(b^2) + 5(b) + 4

            must equal   24_b² =    [2(b)+4) ]²

^{Eqaute the two of them:}

5b^2 + 5b + 4 = 4b^2 +16b+16

b^2 -11b-12 = 0

(b-12)(b+1)  = 0        b = 12      (throw out the b = -1)