All Answers 356464 Answers

add these binomials

\(^{99}C_3 +~ ^{99}C_5 +~ ^{99}C_7 + \ldots + ~ ^{99}C_{97}\)

\(\small{ \begin{array}{|rcll|} \hline ^{99}C_0+~^{99}C_1+~^{99}C_2+~^{99}C_3 + \ldots + ~^{99}C_{97}+~^{99}C_{98}+~^{99}C_{99} &=& 2^{99} \\\\ ~^{99}C_1+~^{99}C_3+~^{99}C_5 + \ldots + ~^{99}C_{97}+~^{99}C_{99} &=& \dfrac{2^{99}}{2} \\\\ ^{99}C_3 +~ ^{99}C_5 +~ ^{99}C_7 + \ldots + ~ ^{99}C_{97} &=& \dfrac{2^{99}}{2} -~^{99}C_1-~^{99}C_{99} \\\\ ^{99}C_3 +~ ^{99}C_5 +~ ^{99}C_7 + \ldots + ~ ^{99}C_{97} &=& 2^{98} -99-1 \\\\ \mathbf{ ^{99}C_3 +~ ^{99}C_5 +~ ^{99}C_7 + \ldots + ~ ^{99}C_{97} } &=& \mathbf{2^{98} -100 } \\ \hline \end{array} }\)

If each dimension of a rectangle decreases by 1,
its area will decrease from 2017 to 1917.
What will be the area of the rectangle
if each of its dimensions increases by 1?

\(\text{Let $ab=2017$ }\)

\(\begin{array}{|lrcll|} \hline \text{decreases by $1$ :} & 1917 &=& (a-1)(b-1) \\ & 1917 &=& ab-(a+b) +1 \qquad (1) \\\\ \text{increases by $1$ :} & x &=& (a+1)(b+1) \\ & x &=& ab+(a+b) +1 \qquad (2) \\\\ \hline (1)+(2):& 1917+x &=& ab-(a+b) +1 + ab+(a+b) +1 \\ & 1917+x &=& 2ab+ 2 \\ & x &=& 2ab+ 2 -1917 \\ & x &=& 2ab -1915 \quad | \quad \mathbf{ab=2017} \\ & x &=& 2*2017 -1915 \\ & x &=& 2119 \\ \hline \end{array}\)

The area of the rectangle
if each of its dimensions increases by 1 is 2119

First of all, I do understand the advantages of the metric system. It has a standardized convention for naming units. It works in base 10 for simple manipulation and use with decimals. It was designed to fit with certain properties of the physical world (a gram of water is a milliliter of water, at normal temperature and pressure, for example).

99C0+99C99=1+1=2

99C1+99C98=99+99 = 198

99C2+99C97=4851+4851=9792

If I add all those together I get   9900

So the sum is  2^99 - 9900

I think it is anyway.

The discriminant in the quadratic formula must equal 0

\(\triangle=0\\ b^2-4ac=0\\ 400-28a=0\\ 400=28a\\ a=14\frac{2}{7}\)

Thanks for your response.   

I have opened the original thread.

If you do not like my answer then state why in a new post on the original thread.

Oh since it is back a bit, I might not see it.  Post here a little message saying you have done so.

Using generating functions, the sum is 2^99.

Completing the square, we get (20/7*x + 7)^2 = 0, so a = (20/7)^2 = 400/49.

tan(R) = 6 * sin(R)           ∠R ≈ 80.40593º

PQ ≈ 14.79

30 cans 40 percent are Pepsi = 12 cans    

   12 cans are 75% of what ?

  .75 x = 12

 x = 16 cans   are left in the fridge                 30-16 = 14 cans of Coke were consumed by ' Diabetes Noah'  !     

You're welcome !  

According to WolframAlpha: https://www.wolframalpha.com/input/?i=17%5E1993

The tens digit is 9

Positive solution: \(x = \sqrt{2}-1\)

Multiply the left side by x+2/x+2 to get, \(x=\frac{x+2}{2x+5} \). Multiply both sides by 2x+5,

\(2x^2+5x= x + 2 \)

Subtract x+2 on both sides

\(2x^2+4x-2= 0 \)

Use the quadratic formula to find solutions of, \(x = -1 + \sqrt{2}, -1 - \sqrt{2}\)

We want the positive solution, which is \(-1 + \sqrt{2}\)

The tens digit is 3

There is no information on the sequence of A, but I presume it would be the same as B. Going off this, we can calculate the first few terms of both sequences to see, 

A = {0, 1, 2, 4, 6, 9, ...}

B = {1, 2, 2, 3, 5, 9, ...}

We could keep going, but we don't need to, we already can see a pattern. The problem is asking for the remainder when a_50+b_50 is divided by 5, and we can see a pattern by dividing some of the a_n + b_n by 5. We can see these remainders form a pattern,

1, 3, 4, 2, 1, 3, ...

The pattern repeats on the fifth term, meaning that every 4th term will be a 2. (a_48 + b_48)/5 will have a remainder of 2, as it's a multiple of 4, (a_49 + b_49)/5 having 1, and finally, (a_50 + b_50)/5 having a remainder of 3.

thank you awesome guy and EP

If you put 1/37 into a calculator, you get \(0.\overline{027} \), so every third digit will be a 7.

291 mod 3 results in 0, meaning that 291 is divisible by 3, so the 291st digit will be 7.

1/37= .027 027 027 027....     every multiple of 3 is a 7   291 is a  multiple of 3      the 291st digit is 7

If 60% of them were coke and the rest were pepsi, then 3/5 of 30 is 18 and 2/5 is 12, so at the beginning, there are 18 cans of coke and 12 cans of pepsi. 

Since the percentage after Noak drank the coke of pepsi is 75%, that means it is a 3:1 ratio, pepsi:coke. Bringing this into the ratio of the 12 cans of pepsi, that would mean there would be 4 cans of coke left. 18 - 4 = 14, so Noak drank 14 cans of coke.

g(x) = ?

h(x) =x^8   =    g(x)^4            g(x) is degree 2     then raised to the fourth power by f(x) makes h(x) degree eight.

Sorry, it doesn't 

Yeah that works, thanks so much!