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Let that constant be k.

\((120 + k)^2 = (60 + k)(140 + k)\\ k^2 + 240k + 14400 = k^2 + 200k + 8400\\ 40k + 6000 = 0\\ k = -150\)

Common ratio = \(\dfrac{120 + k}{60 + k} = \dfrac13\)

\(\quad \text{Required sum}\\ = (3 + 13 + 23 + \cdots + 993) + (4 + 14 + 24 + \cdots + 994) + (6 + 16 + 26 + \cdots + 996) + (9 + 19 + 29 + \cdots + 999)\\ = 40(1 + 2 + \cdots + 99) + 100(3 + 4 + 6 + 9)\\ = 200200\)

Rearrange as

4x + 4y - xy + 115  =  0

4y - xy  =  -4x - 115     multiply through by -1

xy - 4y  = 4x + 115

y [ x - 4]  =  [ 115 + 4x ] 

y  =  [ 115 + 4x  ] / [ x - 4]

Note that when  x = 5,   y =  135

x + y =   5 + 135  =   140

Let \(\langle \cdot, \cdot \rangle\) denote the inner product. Then 

\(\dfrac{\langle \mathbf u, \mathbf v \rangle}{\|\mathbf u\|\|\mathbf v\|} = \cos 60^\circ = \dfrac12\\ \langle \mathbf u ,\mathbf v \rangle = \dfrac{3\cdot 2}2 = 3\)

\(A = \begin{pmatrix}\mathbf u^T\\\mathbf v^T\end{pmatrix}\\ A \mathbf u = \begin{pmatrix}\mathbf u^T\\\mathbf v^T\end{pmatrix} \mathbf u = \begin{pmatrix}\langle \mathbf u,\mathbf u \rangle\\\langle \mathbf v, \mathbf u\rangle\end{pmatrix} = \begin{pmatrix}\|\mathbf u\|^2\\\langle \mathbf u, \mathbf v\rangle\end{pmatrix} = \begin{pmatrix}9\\3\end{pmatrix}\\\)

Similarly we have

\(A \mathbf v = \begin{pmatrix}\langle \mathbf u, \mathbf v\rangle \\\|\mathbf v\|^2\end{pmatrix} = \begin{pmatrix}3\\4\end{pmatrix}\)

If this is not the notation you are used to, \(\langle \mathbf u,\mathbf v \rangle \) is just \(\mathbf u \cdot \mathbf v\).

r = (1/4) / 16  =  1 / 64

30th term  = 

(1/4)r^6

(1/4) (1/ 64)^6   =

(1/2)^2 * ( 1/ 2^6)^6  =

(1/2^2)  * ( 1/2^36)  =

1/ 2^38

1 / 274877906944

Here's a very similar problem

https://web2.0calc.com/questions/help-pls_105

14 + 19 > x             14 + x >  19

33 >  x                       x >  19 - 14

x = 32                        x >  5   →  x = 6

32 - 6 =   26

x = 31^2  =  961

I found the answer: (a,b,c,d,e) = (4,2,-3,-1,8).

Expand 10.2^3 and 10.3^3 and computing the difference, we find that there are 60 integers between 10.2^3 and 10.3^3.

Let m=37. We can show that n is divisible by 37 if and only if the result of this process is divisible by 37 as follows:

If n is divisible by 37, then the alternating sum of the two-digit chunks is divisible by 37.

This is because the alternating sum of the two-digit chunks is just the sum of the digits of n, modulo 37. Since n is divisible by 37, the sum of its digits is also divisible by 37, so the alternating sum of the two-digit chunks is also divisible by 37.

If the alternating sum of the two-digit chunks is divisible by 37, then n is divisible by 37.

This is because the sum of the digits of n is congruent to the alternating sum of the two-digit chunks, modulo 37.

Therefore, if the alternating sum of the two-digit chunks is divisible by 37, then the sum of the digits of n is also divisible by 37, so n is divisible by 37.

Therefore, the alternating sum of the two-digit chunks is a divisibility test for 37.

To see this in action, let's consider the number 354764. The alternating sum of the two-digit chunks of 354764 is 64−47+35=52. Since 52 is divisible by 37, we know that 354764 is also divisible by 37.

We can solve this problem by considering the two ways to win and summing their probabilities:

Matching at least two white balls:

Favorable outcomes: We need to calculate the number of winning tickets where we match at least two of the chosen white balls.

Overcounting: We initially consider all possibilities where we pick three white balls (220 ways) and any red ball (8 ways). However, this overcounts tickets where we match all three white balls (counted three times for each red ball).

Correcting the overcount: There are 220 ways to pick three white balls and only 1 way to pick the red ball (doesn't matter which) for a total of 220 tickets where we match all three white balls. Subtracting this from the overcount gives us the actual number of winning tickets with at least two white ball matches.

Matching the red SuperBall:

Favorable outcomes: Here, we can pick any three white balls (220 ways) as long as they don't match the chosen red SuperBall (8 ways).

Calculations:

Matching at least two white balls:

Total outcomes (overcounted): 220 (white balls) * 8 (red balls) = 1760

Overcounted matches (all three white balls): 220 (white balls) * 1 (red ball) = 220

Corrected favorable outcomes: 1760 (total) - 220 (overcounted) = 1540

Matching the red SuperBall:

Favorable outcomes: 220 (white balls) * 8 (red balls that don't match) = 1760

Total Probability:

We win if either of these scenarios occurs. So, the total probability is the sum of the probabilities of each scenario:

Probability (matching at least two white balls): 1540 favorable outcomes / (12C3 * 8) total outcomes = 1540 / (220 * 8) = 44 / 248

Probability (matching the red SuperBall): 1760 favorable outcomes / (12C3 * 8) total outcomes = 1760 / (220 * 8) = 11 / 31

Total Probability (winning the SuperLottery):

(44 / 248) + (11 / 31) = 33/62.

Therefore, the probability that you win a super prize is 33/62.

a) \(\dfrac{8 \cdot 3 + 2 \cdot 8}{10} = \dfrac{24 + 16}{10} = \dfrac{40}{10} = \boxed{4}\)

b) \(\dfrac{8 \cdot 3 + 3 \cdot 8}{11} = \dfrac{24 + 24}{11} = \boxed{\dfrac{48}{11}}\)

c) \(\dfrac{8 \cdot 3 + 4 \cdot 8}{12} = \dfrac{24 + 32}{12} = \dfrac{56}{44} = \boxed{\dfrac{14}{11}}\)

d) \(\begin{align*} \dfrac{8 \cdot 3 + (2 + x) \cdot 8}{10 + x} &= 6 \\ \dfrac{24 + 16 + 8x}{10 + x} &= 6 \\ \dfrac{40 + 8x}{10 + x} &= 6 \\ 40 + 8x &= 60 + 6x \\ 40 + 2x &= 60 \\ 2x &= 20 \\ x &= \boxed{10} \end{align*}\)

e) \(\begin{align*} \dfrac{8 \cdot 3 + (2 + x) \cdot 8}{10 + x} &= 7 \\ \dfrac{24 + 16 + 8x}{10 + x} &= 7 \\ \dfrac{40 + 8x}{10 + x} &= 7 \\ 40 + 8x &= 70 + 7x \\ 40 + x &= 70 \\ x &= \boxed{30} \end{align*}\)

Thank you, using what you told me I arrived at p and t values of 5, and found that the problems the mathematician solved the day he drank coffee was \(\boxed{50}.\)

Simplifying gives x^2 - mx + 14 = 0. By Vieta's formula, x_1 x_2 = 14 and m = x_1 + x_2. Which integers multiply to 14? (-7, -2), (-14, -1), (7, 2), and (14, 1), assuming |x_1| > |x_2|. So the possible values of m are -9, 9, -15, and 15.

All numbers divisible by 8 has the property that the last 3 digits are also divisible by 8. But note that since the 4-digit number is a palindrome, we are actually counting the number of 3-digit multiples of 8 that has the same first two digits, and does not end with 0. We can then make put the last digit in front to get back the 4-digit palindrome we wanted. The other case is that the number ends with 00(digit), so we need to take that into account as well, but this case is simple because the only possibility is 8008. 

For the first case, we first count the number of 3-digit multiples of 8 that has the same first two digits. They are 112, 224, 336, 440, 448, 552, 664, 776, 888, and 992. So in total 10, but 440 ends with 0 so we exclude that case. Hence, we have only 9 cases here, which gives the palindromes 2112, 4224, 6336, 8448, 2552, 4664, 6776, 8888, 2992 respectively.

Including 8008, the answer is 10.

Let \(P = (x, -x+6)\). We have

Note that \(m_{AO} = \dfrac{8 - (-12)}{2 - 10} = -\dfrac 52\). So the slope of perpendicular bisector of AO is \(\dfrac25\).

Midpoint of AO is \(\left(\dfrac{10+2}2, \dfrac{-12+8}2\right) = (6,-2)\). Since PA = PO, P lies on the perpendicular bisector of AO. Then,

\(\dfrac{(-x+6) - (-2)}{x - 6} = \dfrac{2}5\\ 5(8 - x) = 2(x - 6)\\ 40 - 5x = 2x - 12\\ 52 = 7x\\ x = \dfrac{52}7\)

The point is \(\left(\dfrac{52}7, -\dfrac{52}7 + 6\right) = \left(\dfrac{52}7, -\dfrac{10}7\right)\).

Let \(d = a_2 -a_1\). Note that \(S_5 = 5a_1 + (d + 2d + 3d + 4d) = 5a_1 + 10d\), and \(S_{10} = 10a_1 + (d + 2d + \cdots + 9d) = 10a_1 + 45d\), and \(S_{15} = 15a_1 + (d + 2d + \cdots + 14d) = 15a_1 + 105d\).

So, the given conditions are \(\begin{cases} 5a_1 + 10d = \dfrac15\\ 10a_1 + 45d = \dfrac1{10} \end{cases}\). Solving gives \(d = -\dfrac3{250}\) and \(a_1 = \dfrac8{125}\).

So \(S_{15} = 15a_1 + 105d = -\dfrac3{10}\).

Suppose \(d = a_2 - a_1\). Note that the given conditions are equivalent to:

\(\begin{cases} a_1 + d = -1\\ a_1 + 3d = \dfrac13 \end{cases}\)

Solving gives \(d = \dfrac23\) and \(a_1 = -\dfrac53\).