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8 sides (octagon) 

Total no. of balls \(=3+4+5\) 

                           \(=12\)

3 balls are drawn without replacement. 

⇒ No. of ways of choosing 3 balls \(=\binom{12}{3}\)

                                                       \(=220\)

Case 1: All are same color. 

Favourable outcomes \(=\binom{3}{3}+\binom{4}{3}+\binom{5}{3}\)

                                   \(=1+4+10\)

                                   \(=15\)

∴ P(same color balls) \(= {15\over 220}\)

                                   \(={3 \over 44}\)

Thus, the probability that they're same color is \(3 \over 44\).

Case 2: All are different colors.

Favourable outcomes \(=\binom{3}{1}*\binom{4}{1}*\binom{5}{1}\)

                                   \(=3*4*5\)

                                    \(=60\)

∴ P(different color balls) \(={60 \over 220}\)

                                       \(={3 \over 11}\)

Thus, the probability that they're of different colors is \(3 \over 11\).

a) In order to plot g(x) we translate the endpoints of f(x)  as follows :- 

       \((-4,4)\) in f(x) ⇔ \((-13,2)\) in g(x) 

       \( (0,2)\) in f(x) ⇔ \((-1,1)\) in g(x) 

       \((-1,0)\) in f(x) ⇔ \((-4,0)\) in g(x)

       \((4,-4)\) in f(x) ⇔ \((11,-2)\) in g(x)

   ∴ The graph of g(x) is plotted below -

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b) Each point \((a,b)\) in  f(x) becomes \((3a-1,{b \over 2})\) in g(x).

    Now, that implies a horizontal stretch by a factor of 3 and a horizontal shift by 1 unit left for a. 

    So, that gives us \(f({1 \over 3}.(x+1))\)

    As for b, the graph shrinks vertically by a factor of \(1\over 2\). 

    So that gives \({1\ \over 2}f({1 \over 3}.(x+1))\)

    So that completes, 

     \(g(x) = {1 \over 2}f({x+1 \over 3})\)

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c) In order to get the graph \(y=g(x)\), the original graph can be stretched horizontally  by a factor of 3, then shrunk vertically by a factor of 1/2 and shifted left by 1 unit. 

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~ It's a pleasure! :) 

Let's assume Dave had \(x\) apples at first. 

⇒ Alex had = \(2x\) apples

When Dave gave 10 apples to Alex, he had 8 times as many apples as Dave in the end.

∴ According to question, 

\(2x+10=8(x-10)\)

\(2x+10=8x-80\)

\(6x=90\)

⇒\(x=15\)

Thus, Alex had 30 apples at first.

First of all, I think there's a small error in the expression given, 

However minute it may seem it can confuse easily... So the correction is 

                      \(3(-10)+2[3/2(-10)]+4[1/4(20)]+5(20) \)

Now, after referring the steps we see that Sal made his first error in step 2. 

∴ The correction is here as follows - 

\(3(-10)+2[{3\over 2}.-10]+4[{1\over 4}.20]+5(20) = 3(-10)+2(-15)+4(5)+5(20) \)

                                                                       \(=-30-30+20+100\)

                                                                       \(=60\)

∴ The final answer is 60.

The ratio of Alan's apples to Don's apples was 3:5. When Alan gave 30 of his apples to Don, the ratio of Alan's apples to Don's apples became 1:5. Find the number of apples Alan had at first?

A : D = 3 : 5

(3x - 30)/(5x + 30) = 1/5

x = 18

Alan:    3x = 54

Don:    5x = 90

The ratio of the length of a rectangle to its width is 4:3. Given that the perimeter of the rectangle is 70 cm, find the area of the rectangle.

2(4x + 3x) = 70

8x + 6x = 70

x = 5

L = 4x = 20           W = 3x = 15

Area = L * W  

The ratio of apples Andrew, Ben, and Declan had was 3 : 8 : 4. When Ben gave 21 apples to Andrew and Declan, all of them have the same number of apples in the end. How many apples did Ben have at first?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

A : B : D = 3 : 8 : 4

3x + 4x +21 = 2(8x - 21)

x = 7

8x = 56

Ben had 56 apples  

The answer of \(17^7 / 35\) would be 11723962.1. The remainder would be 3 

Sorry for the first answer, I didn't read the problem correctly 

Excellent, CentsLord   !!!!!

Look more closely.  The problem says –7.   :D  

nice problem. congrats on finding your solution! here is another i may present.

theres 243 arrangements to put 5 distinguishable balls in 3 distinguishable boxes. nice thinking with that, though :)

in these arrangements, theres one thats being counted three times - (5, 0, 0) i.e. put all the balls in one box. leaves us with 240.

for the other 240 cases, we divide by 3! instead of 3. thus, theres 240 / 6 = 40 cases. and 40 + 1 = 41. :)

Write down the quadratic equation whose roots are x = -7 and x = 1 and the coefficient of x^2 is 1.

Enter your answer in the form "x^2 + bx + c = 0".  

x = –7  means one factor is (x + 7)  

x =   1  means one factor is (x – 1)  

Multiply them together                        (x + 7)(x – 1)  =  x² + 6x – 7 = 0  

_.

Aww thank you. :))

=^._.^=

oh no, those formulas do not look fun to use/memorize. 

I shall try to study them. 

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OMG 

Thank you both for answering my question with such detailed solutions... i upvoted yourposts! :)

I'll have to admit  that I am not  familiar  with  these....but....these two websites  helped me  answer  this question

https://www.careerbless.com/aptitude/qa/permutations_combinations_imp7.php

https://keisan.casio.com/exec/system/1292214964#!

Good job, catmg   !!!!

What is a Stirling number?

=^._.^=

As long as  we  have no restrictions ( some boxes  may  be empty)  the  number of  ways  =

S ( 5 , 1)    +  S (5, 2)  + S (5,3)   =  1  + 15  + 25   =    41

Where  "S"  represents  a Stirling number of the   2nd kind 

In this case, case work might just be easier. 

Each number shows the number of balls in that box. 

0 0 5     1 way

0 1 4     5 ways

0 2 3    10 ways

1 1 3    10 ways

1 2  2   15 ways

1+5+10+10+15 = 41

Hopefully it's 41. :))

=^._.^=

Edit: The more I think, the more unsure I hate this answer. There has to be a nicer way to solve this. Casework wasn't too bad since the numbers were small, but I don't like using it too often since calculation mistakes occur. I hope someone else also answers. 

You're welcome. :))

=^._.^=

yayy

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tysm!

Can confirm. also, the number is 685528825939742403772959575428195371930634×13 + 11, so the remainder is 11 :D :P