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"Two real numbers are chosen at random between 0 and 2. What is the probability that their sum is no more than 1?"

What is the length of BD?

Hello Guest!

Euclid's theorem of height says

\(h^2 = q \times p\)

so

\(\overline{BD}^2=\overline{AD}\times \overline{CD}\\ \overline{BD}^2=4\cdot 8\\ \overline{BD}=\sqrt{32}=\sqrt{2\cdot 16}\)

\(\overline{BD}=4\sqrt{2}=5.657\)

The length of BD is 5.657

  !

I think this just means the distance from 0

\(\sqrt{\frac{1}{4}+\frac{9}{64}}\)

etc

"You have 10 boxes with 1000 coins each, either all real or all fake. Real coins are 16 ounces, fake ones are 17. You have a scale, so you put some number of coins on the scale, and then it tells you the weight of those coins. Your job is to figure out which boxes have fake coins and which ones have real coins in one press of the scale. How do you do this?"

Those are very heavy coins

Ok I have this ( and I didn't cheat)    It will help you enormously if you can think in base 2 (which I can)

Let the boxes be called box0, box1........ box9

Take 2^(the box number) coins from each box.  

2^0=1 from box0

2^1=2 from box1

up to

2^9=512 from box 9

altogether you will have 1023    I know this because  2^10-1=1023  but you can just add them all up or use a GP formula if you want.

Let R be the number of real coins and F be the number of fake coins.   And let the weight be W

We know that   F+R=1023    so    R=1023-F  (1)

and we know   17F+16R= W  (2)

solve these simultaneously and we get   F=W-16368

So now you know exactly how many fakes that you have weighed.  But which boxes do they belong to?

Again, understanding base two helps a lot.

Take   F and divide it by 2, the if there is a remainder it has to be 1 and that would mean that the 0 box is Fake

Take the answer and divide it by 2 again and if there is a remainder then box 1 is Fake

keep going until the whole part is zero then you will have worked out which boxes are fake.  The other boxes are real.

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ex

Say the weight is 16660oz

The number of fake coins is 16660-16368= 292

292 divided by 2 = 146 Remainder 0    so box 0 is not fake

146 divided by 2 =  73 Remainder 0     so box 1 is not fake

73 divided by 2 =  36 Remainder 1     so box 2   IS  fake

36 divided by 2 =  18 Remainder 0     so box 3   is not fake

18 divided by 2 =  9  Remainder 0     so box 4 is not fake

9 divided by 2 =  4 Remainder 1     so box 5   IS  fake

4 divided by 2 =  2 Remainder 0     so box 6 is not fake

2 divided by 2 =  1 Remainder 0     so box 7 is not  fake

1 divided by 2 =  0 Remainder 1     so box 8   IS  fake

So boxes 2, 5 and 8 are full of fake coins and

boxes 0, 1, 3, 4, 6, 7 and 9   are full of real coins

I will leave the checking to you 

(y - -7) = 2 (x- - 1)

y+7 = 2x +2

update: I solved part a but cant figure out part b yet .

Nvm, I see now sorry

I'm a bit confused as to how you got that $f(f(f(x)))=x\dots$ I put it into a calculator and didn't get that

We are not here to do all your homework for you.

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I will say that if there is not a squared term under a square root sign then the expression cannot be simplified.

so  

sqrt(14)   There is no squared number (other than one and that does not count) under the square root so you cannot simplify it.

sqrt(72)=sqrt(36*2)  36 is a square number so you can simplify it to   sqrt36 * sqrt2 = 6sqrt2

Please leave Turtle to answer the rest for him/herself

Thanks Guest, you have utilized C++ (I assume) and answered for a specific case but this was reasonable.

I have given a more general answer.  But I liked your answer    

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Let n be a positive integer with exactly 3 positive prime divisors. If n^2 has 27 divisors, how many does n have?

** All the pronumeral I have used are positive integers.  a,b and c are also prime numbers

\(n=a^pb^qc^r\qquad \text{This will have }(p+1)(q+1)(r+1)\;\;factors\\ n^2=a^{2p}b^{2q}c^{2r} \qquad\text{This will have }(2p+1)(2q+1)(2r+1)\;\;factors \)

We are told that

\((2p+1)(2q+1)(2r+1)=27\)

Since p,q and r are all positive integer the smallest value that any of those brackets can take is 3

3^3=27

so that means that  p=q=r=1

so n must have 2*2*2=8 factors.

LaTex:

n=a^pb^qc^r\qquad \text{This will have }(p+1)(q+1)(r+1)\;\;factors\\
n^2=a^{2p}b^{2q}c^{2r} \qquad\text{This will have }(2p+1)(2q+1)(2r+1)\;\;factors

\((x-h)^2+(y-k)^2=r^2 \)

Sub in your three points to get three equations and then solve them simultaneously.

Have you tried that? 

If you are stuck then show us what you have done so far.

Note: I have not worked through this question but this is how I would start.

X = original amount

   spending 1/6 th leaves   5/6 ths of the origianl

(5/6 x   -5 ) 3/5  -12  = 24                         ( spending 2/5ths leaves 3/5th s)

1/2x - 15 = 24

1/2 x = 39

x = 78  dollars originally    

a rail road crew can replace 450m of rail in three days. How many kilometers of real can they repair in 24 days?    

8 times as many days, therefore 8 times as much rail.   8 x 450 = 3600 m = 3.6 km  

First the problem said how much they can replace, then it asked how much can they repair. 

We're a rail road crew can replace 450m of rail in three days. How many kilometers of real can they repair in 24 days?   

We're having to assume that replacing the rail is the way they repair it.  

_.

Assuming  ' repair '  and ' replace '  are the same  

450 m /3  day   *   24 days   *  1 km/1000m =  3.6 km

3.6 kilometers unless this is a trick question

Let the the number of liters of Drink B==B
The numbers of liters of drink A ==B + 30

0.15B  +  0.20 * [B + 30] ==100.5, solve for B
0.15B  +  0.20B +  6 ==100.5
0.35B ==100.5 - 6 ==94.5

B ==94.5 / 0.35 ==270 Liters of Drink B was made
270 + 30 == 300 Liters of Drink A was made.

oh yeah forgot this was here i solved it already its 11 and 20 btw

19th term  -  6th term ==13 terms

7.8  -  39 ==- 31.2 sum of difference of the 13 terms

- 31.2 / 13 == - 2.4 - this is the common difference.

39  -  ( - 2.4 * 5) ==39  +  12 ==51 - this is the first term.

Your sequence looks like this:

51 , 48.6 , 46.2 , 43.8 , 41.4 , 39 , 36.6 , 34.2 , 31.8 , 29.4 , 27 , 24.6 , 22.2 , 19.8 , 17.4 , 15 , 12.6 , 10.2 , 7.8 , 5.4 , 3 , 0.6 , -1.8 , -4.2 , -6.6 , >>Number of terms = 25>>Total Sum == 555

Write a system of equations that could be used to determine the number of liters of Drink A made and the number of liters of Drink B made. Define the variables that you use to write the system. 

Hello Guest!

The volume of the fruit drink from company A is a.

The volume of the fruit drink from company B is b.

\(20\%a+15\%b=105l\\ \underline{a=b+30l}\)

\(20\%(b+30l)+15\%b=105l\\ 0.2b+0.2\cdot 30l+0.15b=105l\\ 0.35b=105l-0.2\cdot 30l\\ 0.35b=105l-6l\\0.35b=99l \)

\(b=\dfrac{99}{0.35}\cdot \dfrac{20}{20}=\dfrac{1980}{7}\)

\(b=282\frac{6}{7}\ l=282.\overline {857142}\ l\\ a=312\frac{6}{7}\ l=312.\overline {857142}\ l\\\)

  !

Find side length of the cube

s * s* s = 6859

s = 19 ft

Sq footage of floor =   19 ft * 19 ft =  361 ft²

5 dollars/ft²   *  361 ft² = _________    dollars

Find least common multiple of   45 and 75       =   225 minutes

t = 0 at initial height      =   480 feet  !