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Let \(\displaystyle x=m+n, \text{ where }m\text{ is an integer, and }0

Then, \(\displaystyle \lfloor x \rfloor =m,\text{ and } x- \lfloor x \rfloor=n.\)

If the common ratio of the GP is \(\displaystyle r,(0

\(\displaystyle (m+n)r=m,\text{ and } mr=n.\)

Substitute the second equation into the first and we have, \(\displaystyle n(1+r)=m.\)

 \(\displaystyle r<1, \text{ so }(1+r)<2,\) 

and since n is also less than 1,

it follows that the only possible (positive) integer value for m is m = 1.

That implies that r = n, and substitution into the equation \(\displaystyle (m+n)r=1,\text{ gets us }n^{2}+n-1=0,\)

from which

\(\displaystyle n=\left(-1+\sqrt{5}\right)/2\approx0.618,\)

(the golden ratio).

\(\displaystyle x\approx1.618.\)

Tiggsy

The bases of trapezoid ABCD are line AB and line CD. We are given that AB < CD, CD = 8, AD = BC = 7, and BD = 9. Find the area of the trapezoid.

Here is a correction to you answer guest :)

Simplify the following:

(-125)^(1/3)

(-125)^(1/3) = (-1×125)^(1/3) = (-1×5^3)^(1/3):

(-5^3)^(1/3)

(-5^3)^(1/3) = (-1)^(1/3) (5^3)^(1/3) = (-1)^(1/3)×5^(3/3) = (-1)^(1/3)×5:

= -1*5

= -5

(-1)^(1/3) = -1 becasue  (-1)^3=-1

(-125)^1/3

since 

 (-5)^3=-125

(-125)^(1/3) = -5    

A fair, twenty-faced die has 19 of its faces numbered from 1 through 19 and has one blank face. Another fair, twenty-faced die has  of its faces numbered from 1  through 8 and10 through 20 and has one blank face. When the two dice are rolled, what is the probability that the sum of the two numbers facing up will be 24.  Express your answer as a common fraction.

P(sum is 24)= 15/(20*20) = 15/400 = 3 /80

There are: 7C2 =21  2-number combinations.

There are: 12, 23, 34, 45, 56, 67, or 6 combinations where the difference between them is < 2

21 - 6 = 15 combinations where the difference between them is 2 or > 2

15 / 21 = 5/7 probability that the difference between 2 randomly drawn numbers will be 2 or greater.

The letter 'e' ?

-41.6666666667

Simplify the following:

(-125)^(1/3)

(-125)^(1/3) = (-1×125)^(1/3) = (-1×5^3)^(1/3):

(-5^3)^(1/3)

(-5^3)^(1/3) = (-1)^(1/3) (5^3)^(1/3) = (-1)^(1/3)×5^(3/3) = (-1)^(1/3)×5:

5(-1)^(1/3)

sorry I misread the question

Thank you so much! This is correct!

Tell me if this is right !

5+4 = 9 total number of plates.

5/9 x 4/8 = 20 / 72 = 5 /18 - probability that the 2 plates are RED.

4/9 x 3/8 =12 /72 =1 / 6 - probability that the 2 plates are BLUE.

5 / 18  +  1 / 6 = [5 + 3] / 18 = 4 / 9 - probability that the 2 plates are the SAME COLOR.

Some numbers are missing from your question.

Correct! Thx!

Let the number of 9th grade students = N

Let the number of 10th grade students = T

Then the number of 11th grade students = 2[N+T]

The number of 12th grade students = 2N

2[N+T] = 10[2N]................................(1)

N + T + 2[N+T] + 2N = 32................(2), solve for N, T

N = 1 - number of students in 9th grade.

T = 9 - number of students in 10th grade.

2[1+9] = 20 - number of students in 11th grade.

2 x 1 = 2 - number of students in 12th grade.

If the number of dimes = D, then:

The number of nickels = D - 1, and:

The number of quarters= 3[D - 1] - 1

D + D - 1 + 3[D - 1] - 1 = 45, solve for D

D = 10 - number of dimes

10 - 1 = 9 - number of nickles

3 x 9 - 1 = 26 - number of quarters.

Can I make this a shorter answer?    YES I CAN      

Pat writes all the 7-digit numbers in which all the digits are different and each digit is greater than the one to its right (so the tens digit is greater than the units, the hundreds greater than the tens, and so on). For example, 9,865,320 is one of the numbers that Pat writes down.

(a) How many numbers does Pat write down?

9,8,7,6,5,4,3,2,1,0     3 of the 10 must be removed    10C3 = 120

(b) One of Pat's numbers is chosen at random. What is the probability that the tens digit is a 1?

(9,8,7,6,5,4,3,2) 1,0       3 of  the 8 must be left out     8C3= 56

P(1 in the 10s column) = 

(c) One of Pat's numbers is chosen at random. What is the probability that the middle (thousands) digit is a 5?

(9,8,7,6 one of these four must be left out) 5 ( 4,3,2,1,0 two iof thse 5 must be left out) 4C1*5C2=4*10=40

P(5 is in the middle) =  

Wow, great solution, hectictar! 

4/25 x 3/24 =12 / 600 = 1 / 50 = 2%

3+6 = 9 total number of balls

3/9 x 2/8 =6 / 72 = 1/12 =~8.3%

Surface area of one face =1/2*4^2*sin(45) =5.657 in^2

Height =5.657 / 2 =2.828 inches.

Volume =1/3 * 4^2 * 2.828 = 15.08 in^3.

@CPhill You are correct. Thank you!