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Really??? I would NEVER copy anyone, though. I know how serious plagiarism is. My equations are formatted differently with different steps...Chris has way more equations than me!

x = ky^2         y = k/z^3      combined      x = k /z^6      

-16 = k/ 2^3

k = -128    

x = -128 / z^6

x = -128 / (1/2)^3   = -1024

Thank you CTG  :-) 

I know, I'm just saying "if". However, the equations are formatted the exact same way(the amount of spacing is the exact same which caught my attention), with the logic and steps being the same? I'm not too sure.....

x = lesser number      other number  = 5 + 3x    sum to 37

x   +   5+3x = 37

4x = 32

x = 8 = lesser number

Nicely done, Guest!

The sum of two numbers is 37. The greater number is 5 more than three times the lesser number. What is the lesser of the two numbers?   

Call the lesser number X 

Call the greater number Y 

given                                 X + Y = 37     which means that Y = 37 – X   substitute this for the Y In the next given equation

given                                 Y = 3X + 5

                                          37 – X = 3X + 5   

add X to both sides            37       = 4x + 5 

subtract 5 from both sides       32 = 4X  

divide both sides by 4                8 = X      the lesser number is 8  

by the way, X + Y = 37 which means that the greater number is 29 but the problem didn't ask for that.

_.

Nice, EP!

Honestly... this is one of the only ways to solve the problem....... it might be similar...

? I didn't even know that this was posted before...........

Sorry, I forgot to Include the circle, this is the circle

Not sure if this is a coincidence, but CTG's explanation goes exactly in tandem with Cphill's previous explanation of this problem. If you want to see for yourself, the original link is here:

https://web2.0calc.com/questions/pls-help-due-tmrow_10

Now that I take a closer look, it seems that CTG's explanation is the exact same as Cphills. @CalTheGreat if you're going to use Cphill as a reference, I'd recommend linking his explanation first and foremost! Again, this is just a big "if"

No prob :)

If you want more reference info, search up "stars and bars combinatorics" on the internet. That's the formula that CPhill used in order to reach his answer for the second problem. Stars and bars only works when the "balls" or things you want to distribute are indistinguishable, and the "boxes" or groups you want to divide those things into, are distinguishable(if it was flipped around, you need to use stirling numbers like Cphill used for the first question). 

Here's a wikipedia link to it, if you really want to learn more:

https://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29

So, what exactly is the length CD? There is no point "C" ever mentioned within the problem itself.

BBBBUUUUUTTTT   there is ANOTHER possibility

if  9  and 41 are the LEGS of the triangle

9^2 + 41^2 = hyp^2      Hypotenuse = 41.976    is the third side  

                      (the 40 third side answer is likey the correct one,,,,but you should realize that this solution is a correct answer TOO )

Logan's marbles at the end are only 6 marbles....becuase Alex has 14 of the original 20 marbles.

10 marbles      +   4 wins  X 1 marble/win    -   losses (1 marble/loss)  =  6 marbles

10                   + 4                                         -   losses                    =6

14    - losses = 6  

          losses = 8

he won four games and lost 8 games for a total of   12 games played   

Thank you so much! 

Here is a desmos graph:   (edited....had a typo in the last solution)

Thank you for the formula, though. I appreciate it and it will help a lot. Will the formula work on all types of problems like this? e.g the 3rd one I posted.

edit: I don't think it will.

AC^2  =  2^2  + 3^2 - 2(3)(2)cos(ABC)  

AC^2  = 13 -12cos(ABC)  

 cos(ADC) =-cos(ABC)

We get...

AC^2 = 6^2  + 10^2 - 2(6)(10)cos(ADC)

AC^2 = 6^2 + 10^2+120cos(ABC)   

AC^2 = 136 + 120cos(ABC)   (2)

Use the subtraction method to get...

0=123+(120+12) cos(ABC)

0=123+132cos(ABC)

DB^2  = 6^2+2^2-2(6)(2)cos(DAB)

DB^2  = 40-24cos(DAB)

Therefore, DAB is obtuse

Second step!

DB^2 =10^2 + 3^2  - 2(10)(3)cos(DCB)

DB^2 =109 + 60cos(DAB)

Use the subtraction method again...

0 = 69+(60 + 24)cos(DAB)

-23/28 = cos(DAB)

We get that sin (DAB)  = √ [ 1 - (23/28)^2 ]  = √(255)/28 = sin PCB

(PB + AB) * PB = PC(PC+CD)

(PB + 2) * PB= 7/11)PB((7/11)PB + 10)

PB^2 + 2PB=(49/121)PB^2 + (70/11)PB

Now, let

PB  = x

x^2+2x = (49/121)x^2 +(70/11)x

(121 - 49) x^2 / 121 + (2 - 70/11)x  = 0

(72/121)x^2  - (48/11)x  = 0

(72/11)x = 48

So the answer is 616/9

THAT IS ONE TOUGH PROBLEM!!!!

Look at your negative z-score table for .0500 (or closest to it) for 5 %

z = -1.65     and   - 1.64    are equally apart from .0500

I'll take the average of these two (you may have been instructed to just choose one)

-1.645   standard deviations from the mean of 41

-1.645 ( 9) + 41 = 26.195  is 5th percentile

Also, sorry about the misrenderance. It was also supposed to be 4 and 3, and not just random blanks :(

Thanks so much for your time and effort. Also, is C in this case the choose function?