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Good job, Melody and guest   !!!!

Here's my solution :

Let  a > b > c > c > d > e       where these are the weights of the girls

And  each is weighed 4 times.....and their total weight  = 1212 lbs

So we have that    4 ( a + b + c + d + e)  = 1212  ⇒  a + b + c + d + e   = 303

And here are the equations that we are sure of

The two heaviest  = a + b  = 129  (1)

Since a > b, then a + c > b + c....so a + c is the second heaviest weight

a + c  =  125  (2)

The two lightest =  d + e  =  114   (3)

Since c > d, then c + e >  d + e  ......so c + e is the second lightest weight

c + e  =  116  (4)

Subtract  (2) from (1)  =  b - c  = 4  ⇒  b  = c + 4

Subtract  (4) from (3)  =  c - d  = 2  ⇒  c  =  d + 2

So this implies that   b =  d + 6

Now.....add (1) and (2)

2a + b + c  =  254   ⇒   b + c    = 154 - 2a  ⇒  (d + 6) + (d + 2)  = 154 - 2a  ⇒

2d + 8  = 154 - 2a    →  2a + 2d  = 246  ⇒  a + d  = 123  ⇒  a = 123 - d

Similarly.....add (3) + (4)  and we have that

c + d  + 2e  =  230  ⇒  (d + 2) + d =  230 - 2e ⇒  2d = 228 - 2e ⇒ d = 114 - e ⇒ e = 114 - d

So

a + b + c + d + e  = 303    and substituitng, we have

(123 - d) + (d + 6) + (d + 2)  + d + (114 - d)  = 303

d + 245  =  303

d = 58

So

a = 123 - d  =  65 lbs

b = d + 6 =  64 lbs

c = d + 2  =  60 lbs

d = 58 lbs

e = 114 - d = 56 lbs

its 0

Solve for L:
2.45 = 2.00607 sqrt(L)

2.45 = 2.00607 sqrt(L) is equivalent to 2.00607 sqrt(L) = 2.45:
2.00607 sqrt(L) = 2.45

Divide both sides by 2.00607:
sqrt(L) = 1.2213

Raise both sides to the power of two:
L = 1.49156

"If a plane flies from south to north with 324 km/h and the wind blows from northeast to southwest with 28.3 m/s what is the angle between the fly direction and the deviating direction if the pilot ignores the wind? "

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Google translate gives:

"There are a few cats and ducks. 48 feet, 32 eyes. How many cats and how many ducks are there?"

c = cats,  d = ducks

4c + 2d = 48

2c + 2d = 32

2c = 48 - 32   so  2c = 16   so c = 8

16 + 2d = 32   so 2d = 32 - 16  so  2d = 16  so d = 8

There are 8 cats and 8 ducks.

有8隻貓和8隻鴨子

Suppose x = 4 and f(x) = x^2 + 1/x

Enter  x=4;x^2+1/x      (i.e. separate the x and f(x) by a semicolon ";")

Then press the "=" sign to get:

.

I am just displaying the question better. 

\(\frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \cdots\)

guest there really is not a lot of point posting if you cannot give any working.

28n^4+16n^3-80n^2

\(28n^4+16n^3-80n^2\\ =4n^2(7n^2+4n-20)\)

I could use the quadrative formula to solve it but I will see if I can figure it out another way.

I need 2 numbers that multiply to 7*-20  and add to give 4

One will need to be positive and the other negative and since 4 is positive, the bigger number will be positive.

7*-20 = -7*2*2*5 = 14* -10  that works, the numbers are  14 and -10

change 4n into  14n-10n

\(=4n^2(7n^2+4n-20)\\ =4n^2(7n^2+14n-10n-20)\\ \text{Now factor in pairs}\\ =4n^2[7n(n+2)-10(n+2)]\\ =4n^2(7n-10)(n+2)\\ \)

I think guest has done something similar to me only more simply but I have not looked at it closely.

My answer was mathematically precise.  There are no other possible answers. 

This is not very rigorous mathematically, but it seems to work !

No kidding?

This is not very rigorous etymologically, either.

So, the presumed age of the 5 girls could be:

square root of 11(3 - 4 square root of 7)

\(\sqrt{11}(3-4\sqrt{7})\\ =3\sqrt{11}-4\sqrt{77}\)

The equation was  a/(sqrt 1-x^2) = b/c.   Make x the subject of this equation

This is what you have written... Is it what you want or do you need to use more brackets.

It is important that you learn to use the brackets properly.

\(\frac{a}{\sqrt 1-x^2} = \frac{b}{c}\\ \frac{a}{1-x^2} = \frac{b}{c}\\\)

This is what I think you actually want.         a/(sqrt (1-x^2) )= b/c

This is :

\(\frac{a}{\sqrt {1-x^2}}=\frac{b}{c}\\ \qquad \qquad \text{If this formula is for the real number system then you need to see that }\\ \qquad \qquad 1-x^2>0\quad \quad and \quad c\ne0\\ \qquad \qquad -1

4^(7)×4^(3)÷4^(6) → 4^(7+3-6) → 4^4 → 256

(1 + x²)i/(1 + xi) + x- 6i → (1 + x²)i(1 - xi)/((1 + xi)(1 - xi)) + x - 6i

→ (1 + x²)i(1 - xi)/(1 + x²) + x - 6i → 1 - xi + x - 6i 

→ 1 + x - (x + 6)i 

Like so

\(\frac{1}{27}=3^{-y}\\ \frac{1}{27}=\frac{1}{3^y}\\ 3^y=27\\y=3\)

Hi Rudi,

You might find this old post interesting too.

It is a visual representation of what I was trying to say.

it is   

|3n|>15

This is saying that the distance of 3n from the origin must be more than 15.  

Have a look at what I have done on the number line. :)

| x - 3 | < 7   We are looking for the solution.

Thank you! Your answers are very useful to me.

LG   !

Thanks CPhill and thank you guest!  

If any of you are interested in his riddles, there's a site full of his riddles!

Site: https://www.mathsisfun.com/puzzles/sam-loyd-puzzles-index.html

Go online and use this Stat. Calculator to get the answers you want:

https://www.hackmath.net/en/calculator/quartile-q1-q3

Challenge accepted     

Let the girls be a,b,c,d, and e   where a is the lightest and e is the heaviest.

If I add all those paired weithgts together I get  1212 pounds

Each girl is included 4 times so

4(a+b+c+d+e) = 1212 pounds

a+b+c+d+e     =   303

The average weigt of the children is  60.6 pounds

The average weight of each pair of girls is 121.2 pounds

So now it is important to know how far each pair  of girls deviate from the average

A+B = -7.2  So   A= -7.2 -B

D+E= 7.8   so     B+6+E = 7.8   so      E = 1.8 - B

These are all deviations so they have to add up to 0.

-7.2 - B +B +B+2 +B+6+1.8-B=0

2.6+B=0

B=-2.6

So the girls weigh  56, 58, 60, 64 and 65 pounds respectively.       

Is that how you did it Chris?

∑[2/(n(n+1)(n+2)), n=1 to 1000] =0.49999.......converges to 1/2.

Let us designate the girls as: a, b, c, d, e
Each girl is weighed 4 times in such manner:
a+b, a+c, a+d, a+e. Similarly for b, c, d, e 
So if  we add their total weights given in the puzzle, we get:
129+ 125+124+123..........+114 =1,212 Pounds.
Since this is the total of 4 weighing for 5 girls, we, therefore, have: 1212 / 4 =303 pounds. This is the combined weight of the 5 girls. Since 2 of them weigh 129 pounds, then 129/2 =64.5 pounds. Will round these 2 as 65 and 64 pounds.
Then the weight of the 3 remaining girls must be: 303 - 65 - 64 =174 pounds. And 174/3 =58 pounds. To get an average of 58 pounds for the 3 girls, we could distribute them as:
56, 58, 60 =174 pounds. So, the presumed age of the 5 girls could be: 56, 58, 60, 64, 65 pounds. This combination seems to balance ALL the paired weighings given in the question. That is:65+64=129. 65+60=125, 64+60=124.........etc. all the way down to 56+58=114.
This is not very rigorous mathematically, but it seems to work !!.