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= sqrt 2*2 * 3*5    * 2*2 * 3   * 7*3*3  * 7*3*2   x⁴

= x² sqrt  2^5 3^5 *  5 * 7^2

 = 252x²  sqrt( 30)

You need to 'complete the square for x '

2x^2 + 6x + 11

2 (x^2 + 3x)       + 11            take 1/2 of the 'x' coefficient :    1.5

2 ( x+1.5)²       + 11          if you expand this out , you will find that you have added   2 *  1.5²   to the expression....

                                                so you need to subtract that amount to keep the original equation

2(x+1.5)^2    -   2 * 1.5²      +11

2 ( x+1.5)²   + 6.5         is the final form.....  take it from here .....

When x = 1    y = 3

           x = 2   y = 5

            x = 3   y = 7                 y = 2x + 1

Professor Grok's sequence is 8, 20, 32, so their first terms are 8.

The 987th term is 1^2 + 2^2 + ... + 987^2 = (987)(988)(987 + 988)/6 = 320988850.

Twenty eight

5 1/2 = 11/2

 11/2 mod 1/3 = 0.1666666666666667    there will be 1/6 pound left over

11/2  / 1/3 =   11/2  *   3 = 33/2 =   16  1/2           sixteen bags    there willbe 1/2 of a 1/3 leftover = 1/6 lb

If everybody meets at 

A, the total distance walked by A is 0, by B is 5, by C is 9, by D is 11, by E is 14, total 39  

B, the total distance walked by A is 5, by B is 0, by C is 4, by D is 6, by E is 9, total 24  

C, the total distance walked by A is 9, by B is 4, by C is 0, by D is 2, by E is 5, total 20  

D, the total distance walked by A is 11, by B is 6, by C is 2, by D is 0, by E is 3, total 22  

E, the total distance walked by A is 14, by B is 9, by C is 5, by D is 3, by E is 0, total 31 

Let's try to improve on the 20 total  

If everybody meets halfway between  

B and C, the total distance walked by A is 7, by B is 2, by C is 2, by D is 4, by E is 7, total 22  

C and D, the total distance walked by A is 10, by B is 5, by C is 1, by D is 1, by E is 4, total 21  

It appears that if you move left of C, the total distance increases, 

and likewise, if you move right of C, the total distance increases.  

I don't know if this is the right way to find the answer, but I'd put my money on P is at the home of C  

Therefore A to P is 9      I'm not going to highlight this in red because I'm not relatively confident.  

_.

A card is drawn from a standard deck, it is replaced and another card is drawn. What is the probability that the first card is a jack and the second card is a heart?      

To get a jack the first card:      4 / 52  

To get a heart the next card:  13 / 52  

For both to occur:   (4 / 52) times (13 / 52)  =  52 / (52 • 52)  =  1 / 52  

_.

four jacks      and  13 hearts in a deck of 52

4/52 *  1/4

2x-5 ≤ 2x+6

-5≤6                              this is true for all 'x'      (-∞ , +∞)

Here is a picture....can you calculate the distance between the two labeled points?

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

(a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5

5a^4b = 15

10a^3b^2 = 60

Can you take it from here?

=^._.^=

Sorry.....here is my correction:

b - 1/4 b =  j + 1/4 b

b = j + 1/2 b

1/2 b = j

b = 2j

b + 1/4 j = 3 ( j - 1/4j)

b + 1/4 j = 3j - 3/4 j

b = 3j - j 

b = 2j

Bob has twice as many marbles as Joe.    But we cannot determine EXACTLY how many marbles Bob has.....could be something like 

                                     60:30       or         80:40       or     92:46     etc .....    

1000 cm³ = 1 litre

1 cm³ = 10^-3  litres

8.74×10^-24 cm³ = 10^-3*8.74×10^-24 = 8.74×10^-27 litres

 8.74×10^-27​ litres =  8.74×10^-27​*10⁶ =  8.74×10^-21​ microlitres 

Find the set of all rea numbers x so that x^2 <= 3. Express your answer in interval notation     

[–sqrt(3),+sqrt(3)]      although I prefer –sqrt(3) < x < +sqrt(3)    

This is the bit that went wrong.

sqrt((3432*385)/(14875*60333)) = 0.038370552862457

sqrt((2^3*3*5*7*11^2*13)/(3*5^3*7^2*13^2*17^2)) = 0.038370552862457

\(\sqrt{\frac{2^3*3*5*7*11^2*13}{3*5^3*7^2*13^2*17^2}}\\~\\ =\sqrt{\frac{2^3*11^2}{5^2*7*13*17^2}}\\~\\ =\frac{2*11}{5*17}\sqrt{\frac{2}{7*13}}\\~\\ =\frac{22}{85}\sqrt{\frac{2}{91}}\\~\\\)

22/85*sqrt(2/91) = 0.038370552862457

LaTex

This is true only if  AD : DC = 1 : 2

Let's call this Civonamzuk's theorem. 

We will need to use these ideas:  If  x >= 0  then  |x| = x.

                                                      If  x < 0    then  |x| = -x.

                                                      If  y >= 0  then  |y| = y.

                                                      If  y < 0    then  |y| = -y.

If  x >= 0  and  y >= 0,  then:  |x| + x + y  =  14   --->   x + x + y  =  14   --->   2x + y  =  14

                                               x + |y| - y  =  18   --->     x + y - y  =  18   --->           x  =  18

Solving this system, we find that  x = 18  and  y = -22  (but this answer is impossible because y was assumed to be >= 0.

If  x >= 0  and  y < 0, then  |x| + x + y  =  14   --->   x + x + y  =  14   --->   2x + y  =  14

                                           x + |y| - y  =  18   --->      x - y - y  =  18   --->    x - 2y  =  18

Solving this system, we find that  x = 9.2  and  y = -4.4.

Now, do a similar analysis for the other two possibities (when x < 0) to get your final answer.

9. AB-1=9

oh really? I thought I got it wrong.... Thanks!!

Not sure what you are trying to accomplish here..... final result is   - .00194344359   according to my calcs

If I calculate the top line     I get this answer      if I calculate your bottom line   I get the same anwer  

Equate the two equations then solv for 'x' using the quadratic formula to find the two

x coordinates  of the two points of intersection

x = 3 +- sqrt 11       use these values in either of the equation to calculate the corresponding y coordinate of the points of intersection.

Just using the log rules    k = 2