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Its money so you can stop after the first nine :))

Thanks Alan    

There are six permutations that work

1 3 2 5 4

1 5 2 3 4

2 3 1 5 4

2 5 1 3 4

3 5 1 4 2

3 5 2 4 1

This simplifies to 1/(x + 2).

using vieta's formulas, \(r+s=12\) and \(rs=22\). Squaring \(r+s\) and subracting \(2rs\), we get that \(r^2+s^2=100\)

1 digit numbers: 0 {}

2 digit numbers: 1 {22}

3 digit numbers: 3 {112,121,211}

4 digit numbers: 1 {1111}

(1, 1, 2, 2, 4, 4, 4)==7! / (3!2!2!)==210 permutations

(1, 2, 2, 2, 4, 4, 4)==7! / (3!3!)   ==140 permutations

(1, 2, 2, 3, 4, 4, 4)==7! / (3!2!)   ==420 permutations

(1, 2, 2, 4, 4, 4, 4)==7! / (4!2!)   ==105 permutations

(1, 2, 2, 4, 4, 4, 5)==7! / (3!2!)   ==420 permutations

(1, 2, 2, 4, 4, 4, 6)==7! / (3!2!)   ==420 permutations

Total =210 + 140 + 420 + 105 + 420 + 420==1,715 possible sequences of rolls.

n / [n - 1] * [n + 1] / n * [n + 2] /[n + 1] * [n + 3] / [n + 2] *..........* [n + 19] / [n + 18]==3, solve for n

They all cancel out except:

[n + 19] / [n - 1] ==3

[n  + 19] ==[3n  -  3]

19 + 3 ==3n - n

22 == 2n

n==22 / 2==11 

Therefore, the first fraction was: 11 / 10

Check: 11 / 10  *  12 / 11  *  13 / 12  *..........* 30 / 29 ==3

Lets call the first integer x 

Integer 1=x 

Integer 2= x+2 becuase it's a consecutive even integer (adding two takes it to the next even integer)

Integer 3=x+4

x+x+2+x+4=-36 

3x+6=-36

3x=-42

x=-14

Substitute x into the integer equation 

x=-14, x=-12, x=-10

A) The first 9 numbers only have 1 digit. So it's 9*1=9. Then the 90 other numbers have 2 digits, 90*2=180. Add those together, so there are 189 digits.

B) There are 9 numbers that end with 0, so 9 is the numerator. We know there are 189 total digits, so that is the denominator. So our fraction is 9/189, which is equal to 1/21.

C) First, the tens digits. Each tens digit repeats 10 times, so we do 10(1+2+3+4+5+6+7+8+9)=450. After that, we calculate the ones digits, which is 10 sets of 1+2+3+...+8+9=45, and 10*45=450. So, we add those two 450s together to get 900 as the sum of the digits.

We can begin by using the Pythagorean theorem to relate the side lengths of the right triangle ABC:

a^2 + b^2 = c^2

We can also express the area of the triangle ABC in two different ways:

Area = (1/2)ab = (1/2)ch

where h is the length of the altitude from C to AB. Solving for h, we get:

h = ab/c

We can now substitute this expression for h into the equation we want to prove:

(x + h)^2(y + h)^2 = (a + b)^4

(x + ab/c)^2(y + ab/c)^2 = (a + b)^4

We can now multiply both sides of this equation by c^4 to simplify the expression:

c^4(x + ab/c)^2(y + ab/c)^2 = c^4(a + b)^4

Using the Pythagorean theorem to substitute c^2 = a^2 + b^2, we get:

((a^2+b^2) + ab)^2(x + ab/c)^2(y + ab/c)^2 = (a + b)^4(a^2+b^2)^2

Expanding both sides of the equation and simplifying, we get:

a^2 b^2 (x^2+y^2) + 2abxy(x+y) + (x^2+y^2)(a^4+b^4+2a^2b^2) + 2ab(a^3+b^3) = 2a^2b^2(a+b)^2

At this point, we can use the fact that a^2+b^2=c^2 and a+b=c, and simplify the equation further:

a^2b^2(x^2+y^2) + 2abxy(x+y) + c^4(c^2-2ab) + 2ab(c^3-3abc) = 2a^2b^2c^2

Finally, we can substitute the expression we derived earlier for h, namely h = ab/c, to simplify the equation further:

a^2b^2(x^2+y^2) + 2abxy(x+y) + a^2b^2(a^2+2ab+b^2) + 2ab(ab(a+b)-3ab(a+b)) = 2a^2b^2(a+b)^2

a^2b^2(x^2+y^2) + 2abxy(x+y) + a^4b^2 + 2a^3b^3 + b^4a^2 = 2a^2b^2(a+b)^2

This last equation is equivalent to the equation we started with, so we have shown that (x+h)^2(y+h)^2 = (a+b)^4 as desired.

B==Black,  W==White,  R==Red

W / B ==2 / 7..........................................(1)

W / R ==3 / 11........................................(2)

B + W + R ==343....................................(3), solve for B, W, R

Use substitution to get:

B ==147 - Number of black cars

W==42 - Number of white cars

R ==154 - Number of red cars

Where is "the diagram" ?

Here are just a few:

10 - 1.85 =8.15
10 - 2.75 =7.25
10 - 3.65 =6.35
10 - 4.55 =5.45
10 - 5.45 =4.55
10 - 6.35 =3.65
10 - 7.25 = 2.75
10 - 8.15 =1.85
10 - 9.05 =0.95
10 - 4.5455=5.4545
10 - 1.4585 =8.5415
10 - 2.4575=7.5425
10 - 3.4565 =6.5435
10 - 4.4555 =5.5445
10 - 6.4535 =3.5465
10 - 7.4525=2.5475
10 - 8.4515 =1.5485

I think it means you just keep putting the number you calculate back into the function until the result is 1.  i.e. C(9) = 28, C(28) = 14, C(14) = 7, ...etc.

Here is a short computer program that does the calculations:

Thanks guest answerer,

a=13

You are certainly right that |r|<1

\(S_{\infty}=\frac{a}{1-r}\\~\\ S_{\infty}=\frac{13}{1-r}\\~\\\)

Now for this question  \(S_{\infty}\)    must be an integer so I will let it be N

r must be between -1 and 1   which means that  1-r  is between 0 and 2

so N is between (not including)  13/2   and  infinity

So the smallest is infinitesimally bigger than 6.5   

Which means ths smallest integer value is 7

\(S_{\infty}=\frac{13}{1-r}\\~\\ N=\frac{13}{1-r}\\~\\ 7=\frac{13}{1-r}\\~\\ 7(1-r)=13\\~\\ -7r=6\\~\\ r=\frac{-6}{7} \)

Has anyone seen that notation before?    I haven't.

Maybe it just means to the power of ???

C(9) = 27

27^m=1

m=0

The only counting you need to do is

1) all sides are the same,        2         

2) two sides are the same        2

=4   

Consider the terms \(a^{2},ab,b^{2}.\).

Their arithmetic mean will be greater than or equal to their geometric mean, (equality when \(a=b),\)

so

\(\displaystyle \frac{a^{2}+ab+b^{2}}{3} \geq\sqrt[3]{a^{2}.ab.b^{2}}\\ a^{2}+ab+b^{2}\geq3ab \\\sqrt{a^{2}+ab+b^{2}}\geq\sqrt{3}\sqrt{ab}.\)

Repeat for the other two and add together all three.

Equality when a = b = c.

I'll head you in the right direction.

Its easier if you think in cents

275+725=1000

Let a,b and c represent the 3 digits

abc+bac=1000

Now just do it as an addition puzzle and you will bet all the possible numbers this can work with.

Immunodeficiency, unspecified · D84.9 is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes. 

This  was asked and answered a few days ago. Probably by you or someone in our class.

Try looking for the earlier post.