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(2x + 17)/(x^2 - x - 2) = a/(x - 2) + b/(x + 1), (x² - x - 2) = (x - 2)(x + 1)

(2x + 17) = a(x + 1) + b((x - 2) => (2x + 17) = (a + b)x + (a - 2b)

Equating terms: a + b = 2, a - 2b = 17 => 3b = -15 => b = -5, a = 7

Consider.     (BB)     G     (BB)         G      (BB)        G     (BB)

The 2 boys can occupy 4 spaces as shown above.

There are 4 ways to place that group of boys in the line, 4!= 24. 3!=6 ways of arranging the girls after that and 2!=2 ways of arranging the boys within their cluster, so the answer must be 2×4!×3!=2x24x6 = 288.

A common stratergy you can use to tackle these problems is to treat the two boys, as a single person, and then just count the possible ways to arrange the people.

Lets think of these 2 boys, as a single person. Then, there are 4! ways to arrange the single person, and the 3 other girls in a line.

There is also usually a mistake made, because the two people mixed in as a single person, can also be re-arranged.

4! * 2! = 48

This problem is similar to the AMC 8 2020 #10 problem.

You can view the problem here:

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10

x²+5x+c=0 => a=1, b=5, c=?  => b² = 25, 4ac = 4c => b² - 4ac = 17 => 25 - 4c = 17 => c = 2.

 \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Let father's age be 9v, son's age be v.

9v + 3 = 6(v + 3)

9v + 3 = 6v + 18

3v = 15

v = 5

Father; 45, Son: 5.

\(2+3(4x+2)=3x-4\)

\(12x+8=3x-4\)

\(12x+8-8=3x-4-8\)

\(12x=3x-12\)

\(12x-3x=3x-12-3x\)

\(9x=-12\)

\(\frac{9x}{9}=\frac{-12}{9}\)

\(x=-\frac{4}{3}\)

so when you copy paste on a thing called alcumus is gives a $ sign

scroll up and you will see a question like this. Hopefully that helps you!

wow.. a much easier way!

\(5x^2-11x+4=0\)

\(x_{1,\:2}=\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\cdot \:5\cdot \:4}}{2\cdot \:5}\)

\(x_{1,\:2}=\frac{-\left(-11\right)\pm \sqrt{41}}{2\cdot \:5}\)

\(x_1=\frac{-\left(-11\right)+\sqrt{41}}{2\cdot \:5},\:x_2=\frac{-\left(-11\right)-\sqrt{41}}{2\cdot \:5}\)

\(x=\frac{11+\sqrt{41}}{10},\:x=\frac{11-\sqrt{41}}{10}\)

1/(a+1)=\(\frac{21-\sqrt{41}}{40}\)

1/(b+1)=\(\frac{21+\sqrt41}{40}\)

total=\(\frac{21}{20}\)

 Find 1/(a + 1) + 1/(b + 1).

Hello Guest!

\(5x^2 - 11x + 4 = 0\\ x^2-2.2x+0.8=0\\ x=1.1\pm \sqrt{1.21-0.8}\\ x=1.1\pm \sqrt{0.41}\\ a=1.1+ \sqrt{0.41}\\ b=1.1- \sqrt{0.41}\\\)

\(\frac{1}{a+1}+\frac{1}{b+1}=0.364921+0.685078=1.05\)

  !

mhm Guest is correct, so next time you come across these type of problems, you multiple the top and the botttom by the same amount to get to your answer.

Thank you!

Kelvin and Patrick had some money at a 3:5 ratio. The ratio of their expenses was 3:4. If Kelvin spent 1/3 of his money and Patrick has $55 left, how much money did Kelvin start with?

Hello Guest!

\(k:p=3:5\\ \frac{1}{3}k:(p-55\ \$)=3:4\\ 5k=3p\\ \underline{\frac{4}{3}k=3p-165\ \$}\\ \frac{11}{3}k=165\ \$\)

\(k=45\ \$\)

\(With\ 45\ \$\ did\ Kelvin\ start .\)

  !

for the first box, 4 choices

for the second box, 3 choices

for the third box, 2 choices

4*3*2=24

x + 0.085(x) = 35

   1.085x = 35

   1.085x/1.085 = 35/1.085

            x = 32.26

    --> 32.26 (0.085)

    --> $2.74

Initial tokens = 60

left tokens = 24

used tokens = 60 - 25

                    = 35

% of used tokens = used token/inital tokens * 100

                             = 35/60 * 100

                             = 58.3%

Ans: 58%

I think it was easier than the 2020 AMC 8

Two odds always add up to an even, and an even number raised to any power will aways be even. Thus, your answer is 0.

\( \sqrt{60x} = \sqrt{4} * \sqrt{15x} = 2\sqrt{15x}\) 

\(\sqrt{12x} = \sqrt{4} * \sqrt{3x} = 2\sqrt{3x}\)

\(\sqrt{63x} = \sqrt{9} * \sqrt{7x} = 3\sqrt{7x}\)

\(\sqrt{42x} = \sqrt{42x}\)

Now, multiply all the radicals and you get \(252x^2 \sqrt 30\).

I think...

There are:
9 x [10^(int(3 - 1)) / 2 ]
9 x [10^(int(2/2)) ]
9 x [10^1] ==9 x 10 ==90 3-digit palindromes.

All palindromes that begin with an EVEN number, will end with an even number, such as: 202, 414, 626, 838. There are 10 palindromes for each even number for a total of: 10 x 4 ==40 - which are divisible by, or are multiples of 2.

Therefore, the probability is: 40 / 90==4 / 9 

1/4 = 3/12 times 3 on top and bottom 

MELODY 

Please don't give up on me yet.  The light came on over my head.  I know what you mean now.   

You're looking at REDA REDB as "There are 2 lots of 4 letters here" 

referring to the four individual letters.  My answer wasn't clear enough.  

for example, REDA means the red A card ... the A card   

                     YELA means the yellow A card ... the A card (yellow didn't show up, I had to use gold)

So REDA REDB means two cards:  A and B   

Would it have been more clear if I had called them A_red and B_red?  Or Red_A and Red_B?  

_.

If you want to get something right you must display it in a form suitable for the media.

Are you incapable of comprehending that?

It says, in Latex, (you haven't bothered to write it in English) "In the digatram below"

There is no diagram, it is just a bunch or unrendered code.

I would not bother with this if I were you.  Stick to 2+2.  You should not try to move on too quickly.