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-2r^2  - 6r   =  0

Factor out the -2

-2 ( r^2  + 3r )  =  0

We are going to complete the square on  "r"

Take   1/2 of 3   =  3/2    .....square it  =9/4.....add and subtract it inside the parentheses....so we have...

-2 ( r^2  + 3r  +  9/4   -  9/4)        factor the first three terms

-2 [  (  r + 3/2)^2  - 9/4)  ]      distribute the -2 over each term in the parentheses

-2(r + 3/2)^2  +  18/4

-2(r + 3/2)^2  +  9/2

This is vertex form

The vertex  is  (-3/2,  9/2)

-

*nvm

nvn i got it

Solution:

$\text {The easy parts first: }\\ \text {There are }\dbinom{52}{5} = 2598960 \text{ ways to select 5 from 52 cards. } \\ $

$\text{Single suit probability – five cards of the same suit. }\\ \dbinom{4}{1}\dbinom{5}{13} = 5148\\ \rho(1) = \dfrac{5148}{2598960} = 0.19808 \% \\ -------------------$

..

$\text{Four suit probability – five cards four suits. }\\ \text {A five-card hand will have at least two cards of the same suit. } \\ \text {Select two cards with matching suits. }\\ \text{There are }\underbrace {\dbinom{4}{1}}_ {\small \text { ways to choose suit }} * \underbrace {\dbinom{13}{2}}_ {\small \text { ways to choose 2 of 13 }} \\ \text{Then for the 3 remaining cards }\\ \text{ there are } \dbinom{13}{1}*\dbinom{13}{1}*\dbinom{13}{1} \text { ways to choose unique suit for each of remaining cards. } \\ \text {The product of these counts gives the total number of five-card hands with four unique suites. }\\ \dbinom{4}{1} * \dbinom{13}{2} * \dbinom{13}{1} * \dbinom{13}{1}* \dbinom{13}{1} = 685464 \\ \rho(4) = \dfrac {685464}{2598960} \text { 26.3746}\% \\ $

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$\text {Two suits. Five cards }\\ \text {There are } \dbinom{4}{2} \text { ways to chooses two suits. Twenty-six (26) cards represent these two suits. }\\ \text{There are } \dbinom{26}{5} \text { ways to choose five cards from the 26. }\\ \text{These selections include single suits, }\\ \text{so subtract the single-suit counts of three for the five-card selections. } -(3)\dbinom{4}{1} * \dbinom{13}{5}\\ \dbinom{4}{2} * \dbinom{26}{5} – (3)\dbinom{4}{1} * \dbinom{13}{5}= 379236\\ \rho(2) = \dfrac{379236}{2598960} = 14.5918\%$

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$\text{For three suits, subtract the counts for 1, 2, and 4 suits from } \dbinom{52}{5}\\ \dbinom{52}{5} - 5148 - 379236 - 685464 = 1529112\\ \rho(3) = \dfrac{1529112}{2598960} = 0.14.5918\% \\ $

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$\rho(1) = \dfrac{5148}{2598960}\\ \rho(2) = \dfrac{379236}{2598960}\\ \rho(3) = \dfrac{1529112}{2598960}\\ \rho(4) = \dfrac {685464}{2598960}\\ -----------------\\ \text{The probability of a five-card hand having at least three (3) unique suits is }\\ \dfrac{1529112}{2598960} + \dfrac {685464}{2598960} = \dfrac {2214576}{2598960} = \dfrac {507}{595}\\ \text{ }\\ \text{ }\\ \small \text { Source: Lancelot Link & Co. Solutions for AoPS probability questions. }$

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The result matches Melody’s answer.  This proves we monkey around until we get it right.

GA

2.

4p+5c=6.71

5p+3c=7.12

Multiply the first equation by 3    and the second by - 5

12p  + 15c   =  20.13

-25p  - 5c   =  -35.60            add these

-13p   =  -15.47      divide both sides by -13

p  =  1.19

Using  any equation to find c, we have

4(1.19)  + 5c  = 6.71  

4.76  + 5c  = 6.71     subtract 4.76 from both sides

5c  = 1.95     divide both sides by 5

c  = 0.39

So   .... { p, c }  =  { 1.19, 0.39}

1. 

12o+7a=5.36

 8o+5a=3.65 

Multiply the first equation through by 5    and the second through by -7

So we have

60o  + 35a   =  26.8

-56o -  35a  =  - 25.55          add these

4o   =  1.25  =  5/4    divide both sides by  4

o =  5/16  =  .3125

We can find a with any of the equations

8(5/16) + 5a  =  3.65

5/2 +  5a  = 3.65

2.5  + 5a  = 3.65      subtract  2.5 from each side

5a  =  1.15       divide both sides by 5

a  =  .23

So... {o, a}    =   { .3125, .23 }

Agree with Cphil...missed the 6 mw part

p(n) = POWER = 6 (.98)^n  milliwatts      where n = number of reflections....

   Sorry!

p(n)  =  6 (.98)ⁿ

Thank you

One reflection mirror results in new beam of   (1) (.98)    (since .02 is lost)

    next reflectoin of THIS beam (1)(.98)  results in    (1) (.98)(.98)

         Next time    it reflects    (1) (.98)(.98)(.98)

              etc

so  b^n    would be beam stength:  (1) (.98)^n = (.98)^n      Where n is the number of mirrors ....

0.33333333 is bigger. The decimals add just a little bit more to the number. 

84  x  1.12  = 94.08

820 x (1- .25) = 615

0.33333...... > 0

that website had the equations i was looking for

Thank you

Thank you Cphill

Here is the formula used by USGS(United States Geological Survey) to determine modern Magnitude of Eartquakes:

10^((1.5*M) + 4.8) / (4.184E15) =Energy Release in Megatons eqivalent of TNT. M=Magnitude.

10^((1.5*8.2 + 4.8) / (4.184E15)=~30 Megatons of TNT

10^((1.5*M + 4.8) / (4.184E15)=30*5, solve for M

M = 8.67, so:

8.67 - 8.2 =0.47 - difference in Magitude which will release 5 times as much as 8.2 Magnitude.

In other words, 8.67 Magnitude will release 5 times as much energy as 8.2 Magnitude.

Note: Wolfram/Alpha will give you the same answer if you input 8.67 Earthquake Manitude. See here:

https://www.wolframalpha.com/input/?i=Earthquake+magnitude+8.67

1. Sales Tax  =  Just multiply   98  by  .06

2.  Pct Change   =    

[  (  New Price  -  Old Price )  / Old Price  x 100 ]  %  =

[  (35 - 28) / 28   x  100 ] %  =

[ 7 / 28  x 100 ] %  =

[ 1/4 x 100 ] %   =

25%

3.    We want to find this

1000  = 500 * .05 * t      simplify

1000   =  25 * t            divide both sides by 25

40  = t  =   40 years  

BTW..... surprising fact....the time to double any amount at 5% would be the same  !!

Question 1:

This question definitely suggests a percent increase since the number of books increased. In other words, 15 plus some percent of 15 equals 21 books. 

15 + some percent of 15 = 21

Let's let x = some percent because we do not know what that is. 

15 + x of 15 = 21

"Of" in mathematics means multiplication. 

15+15x=21

Now, solve for x:
 

Question 2:

The same concept can be used to calculate the percent decrease. I know that this problem involves that since the original amount is greater than the ending amount. 

60 cars - some percent of 60 cars =  24 cars

Let's let x equal some percent.

60 cars - x of 60 cars = 24 cars

As aforementioned, "of" is a direct indicator of multiplication in mathematics. 

60-60x=24

Note: I realize that I could have used the formula $\frac{y_2-y_1}{y_1}*100$ to get the percent changed, but I think that the following methods allow you to understand what is occurring; the formula, on the other hand, does not.

Not an expert on this....but...I think we can use this formula to find the result :

Magnitude  =  log 5  + 8.2  =

.699  + 8.2   ≈

8.9

Here's a website that goes into more detail :

http://www.sosmath.com/algebra/logs/log5/log56/log56.html

You are so incredibly close to the right answer. Let me see if I can steer you in the right direction, if I can.

$6.9^a=4.7^b\\ \log\left(6.9^a\right)=\log\left(4.7^b\right)\\ a\log 6.9=b\log4.7\\ \frac{a}{b}=\frac{\log 4.7}{\log 6.9}\approx0.80$

Your answer is wrong because you solved for something other than what the question asked for. The above is the work that you provided. In the above, you solved for $\frac{a}{b}$, but what is the question asking you to solve for? Adjust your solving so that you are solving for the correct expression.

Sorry, just realized this had nothing to do with XSquaredFactor. Sorry!

Excellent  !!!!!

for Question 1 I got $270 

and for Question 2 I got $2090

is that okay?