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She starts at 2, so she at least needs 98 more to go.

The only number that is divisible by 5 and is the smallest number greater than 98 is 100

So Laverne sends Shirley screaming and running at 2 + 100 = 102

p + r = 9 + p means r = 9.

From q - 2r = -2 + q, we have r = 1.

These two equations are contradictory. Hence, the system has no solutions.

If you drove at 50 mph, you would have arrived  EARLIER, not, LATER

a1 + a3 + a5  =  7

a1 + (a1 + 2d) +( a1 + 4d)  =  7

3a1 + 6d  = 7

a2 + a4  = 6

(a1 + d)   + ( a1 + 3d )  =  6

2a1 + 4d  = 6

3a1 + 6d   =  7    mult through by 4

2a1 + 4d = 6      mult through by -6

12a1 + 24d  = 28

-12a1 - 24d  = -36         adding these we get

0  =  -  8

No solution

Notations: We use the following notation \(\displaystyle \binom{n}k\) to mean \(\displaystyle \binom{n}k = C^n_k\).

Note that \(\displaystyle \binom{20}{10} = 184756\) so all possible paths are exhausted by the penguins.

There is an important observation in all 4 problems, which I will state below:

Observation:

Let \((x, y)\) be a lattice point, with \(0 \leq x \leq 10\), \(0 \leq y \leq 10\).

Each path that passes through this lattice point must start with a sequence of x + y moves,

consisting of x moves to the right, and y moves upwards, in some order.

By observation, to count the number of path passing through (x, y), we are counting the number of permutations of x (indistinguishable) right moves and y (indistinguishable) up moves. That number is \(\displaystyle \binom{x + y}{x}\). From there, we move from (x, y) to (10, 10) by 10 - x right moves and 10 - y up moves. By the same logic, there are \(\displaystyle \binom{(10 - x) + (10 - y)}{10 - x} = \binom{20 - x - y}{10 - x}\) ways to move from (x, y) to (10, 10). 

Therefore, by basic principles of combinatorics, there are \(\displaystyle \binom{x + y}x \cdot \binom{20 - x - y}{10 - x}\) different paths that passes through \((x, y)\).

With this in hand, the problems are easily solved.

Problem 1

The total sum is \(\displaystyle \binom{7 + 10}{7} \cdot \binom{20 - 7 - 10}{10 - 7} + \binom{8 + 9}8 \cdot \binom{20 - 8 - 9}{10 - 8} + \binom{9 + 8}{9} \cdot \binom{20 -9-8}{10-9} + \binom{10 + 7}{10} \cdot \binom{20 -10- 7}{10-10}\), which is 184756. 

Seeing this answer, you may think there is an easier way to do this. In fact, there is. Note that for each penguin, after 17 moves, one of the 4 points must be reached. From each of the 4 points, the other 3 points cannot be reached. Therefore, each penguin passes through one and only one penguin counter. Therefore the answer is exactly the number of penguins we have.

Problem 2

From the observation in Problem 1, if you fix a number k such that \(0 \leq k \leq 20\) and place counters on each point (x, y) such that \(x + y = k\), every penguin will be counted exactly once. (In Problem 1, the case k = 17 is shown. )

The problem then becomes "how many ways are there to choose 2 integers from 0 to 20 inclusively", which is \(\displaystyle \binom{21}2 = 210\).

Problem 3

The expected sum is 11 times the average number of paths passing through a point, i.e., in summation notation, \(11 \cdot \dfrac{\displaystyle \sum_{x = 0}^{10} \sum_{y = 0}^{10} \binom{x + y}x \binom{20 -x-y}{10-x}}{11^2}\)

While it may seem intimidating, there is a clever way to calculate the double summation \(\displaystyle \sum_{x = 0}^{10} \sum_{y = 0}^{10} \binom{x + y}x \binom{20 -x-y}{10-x}\). From the observation in Problem 2, the sum of counters over the lattices points with \(x + y = k\) is just the total number of penguins, where \(0 \leq k \leq 20\) is a fixed constant. Therefore, the sum is equal to \(\displaystyle\sum_{k = 0}^{20} 184756 = 21(184756) = 3879876\). Now we substitute into the formula and get the expected sum to be \(\dfrac{3879876}{11} = \boxed{352716}\).

Problem 4

The answer is just \(\displaystyle \sum_{x = 4}^{4} \sum_{y = 0}^{10} \binom{x + y}x \binom{20 -x-y}{10-x} = \sum_{y = 0}^{10} \binom{4 + y}4 \binom{16 - y}{6}\). I will let you figure out the rest.

W               A   15      X

                                 m

                                 C

                                3m                            

Z         B      20         Y

tan  CAX =  m / 15   

tan  CBY  = 3m/ 20

Since angle ACB   = 60   then   angles CAX + CBY = 120

So

tan  ( CAX + CBY)  =   tan (120)

Identity  

tan (A + B)  =  [ tan A + tan B ] / [ 1 - tan A * tan B ] 

[ m/15 + (3m) / 20 ] / [ 1 - (m/15) [ (3m) / 20] ] =       -sqrt 3

Solving this  for positive m produces  m ≈  18.05  .....3m ≈  54.15

AC =  sqrt [ 15^2 + 18.05^2 ] ≈  23.47

BC =  sqrt [ 20^2 + 54.15^2 ] ≈ 57.73

Law of Cosines

AB  =  sqrt [ 23.47^2 + 57.73^2  - 2 ( 23.47 * 57.73)(1/2) ]  ≈  50.29

No

Because AD is parallel to BE.....then  because CE  is a transversal cutting  parallel lines, angle ADC  = angle BEC

But angle ACD  >  angle BCE

So  in the two triangles

ADC + ACD  + CAD  =  BEC  + BCE +  CBE

ACD  +  CAD  =  BCE  + CBE

Since  angle ACD  > angle BCE, then angle  CAD  must be  <  angle CBE

The only way that angle CAB = angle CBE  is  if A is on  BC

2T + 1B =  3.15     →  2T   = 3.15   - 1B    →   [3.15 - B ]  / 2  =  T         (1)

1M + 1B  = 3.50  →   M = 3.50 - B          (2)

1T + 2B + 3M  = 8.15     (3)

Sub (1) and (2)  into (3)

[ 3.15 - B ] / 2  +  2B  + 3 [ 3.50 - B ]  = 8.15               mutiply through by 2

[3.15 -  B ] + 4B  + 6 [ 3.50 -B ] = 16.30

-3B + 3.15 + 21  = 16.30

-3B  =  16.30 - 3.15 - 21

-3B = -7.85

B =  7.85 / 3  ≈  $2.62 = 262 cents

https://web2.0calc.com/questions/help-me_40934

6000 / 2000  =  x  /40

3 = x /40

120  = x  = 120 min

a2 + a4 + a6 + a8 + a10 + a12  = 1

a1 + a3  + a5 + a7 + a9 + a11  = -2          subtract these and we  get

Note :   a_n - a_{n -1}  =  d

d + d + d + d + d + d= 3

6d  =3

d = 1/2

Sum of terms  from  a1 to an  =  n * a1 +  (n)(n-1) / 2  *  d

S12  =  

12a1  +  66d =  3

12a1 +  66 (1/2)  =  3

12a1 + 33  = 3

12a1 =  -30

a1 = -30 / 12 =   - 5/2

Sum of an arithmetic series  from  a1 to an  inclusive =   n * a1  +  (n)(n-1) / 2 * d

S5 =      5a1 + 10d =   1/5         (1)

S10 =   10a1 + 45d =  1/10      (2)

Multiply  (1)  through by -2   and add it to (2)

25d = -3 /10

d =  -3 / 250

5a1 + 10 (-3/250)  = 1/5

5a1 + -30/250  = 1/5

5a1  = 1/5 + 3/25

5a1 = 8/25

a1 = 8/125

S15 =    15 ( a1)  +  105 d =    15(8/125) + 105 ( -3/250)   =  -3/10

nice work, sir!

Analyzing the Given Information:

Convex Pentagon: ABTCD is a convex pentagon, meaning all interior angles are less than 180 degrees.

Equal Sides: AB = CD implies sides AB and CD are congruent.

Internally Tangent Circles: The circumcircles of triangles TAB and TCD touch each other inside the pentagon.

Right Angle: Angle ATD is a right angle (90 degrees).

Angles and Sides: Angle BTC is 120 degrees, BT = 4 (length of side BT), and CT = 5 (length of side CT).

Key Points for Area Calculation:

Right Triangle: Triangle TAD is a right triangle due to angle ATD being 90 degrees.

Power of a Point: Since the circumcircles of TAB and TCD are internally tangent, point T is the power of point A with respect to circle BTC.

This means the product of TA and its projection onto BT (let's call it AP) is equal to the square of BT (which is 4).

Solving for AP and Area of Triangle TAD:

Projecting AP: Since triangle BTC is isosceles with angle BTC = 120 degrees, line BT bisects angle ABC. This means projection point P lies on line BT.

Using Power of a Point: Since T is the power of point A with respect to circle BTC, we have:

TA * AP = BT^2 (given)

Substitute known values: TA * AP = 4^2 = 16

Pythagorean Theorem in Triangle TAP: We know AP and need to find TA (the hypotenuse) and the area of triangle TAD (which is 1/2 * base * height).

Use the Pythagorean theorem: TA^2 = AP^2 + TP^2 (where TP is the leg opposite the right angle)

Substitute known values from step 2: TA^2 = 16 + TP^2

Finding TP: Since BT bisects angle ABC, triangle ABT is a 30-60-90 triangle with BT = 4 (half of the hypotenuse) and CT = 5 (the other leg). This implies AB = 2 * BT = 8 (hypotenuse).

Using Pythagoras in triangle ABT: TP^2 = AB^2 - BT^2 = 8^2 - 4^2 = 48

Finding TA and Area of Triangle TAD:

Substitute TP^2 from step 4 into the equation from step 3: TA^2 = 16 + 48 = 64

Take the square root of both sides to find TA: TA = 8

Now calculate the area of triangle TAD: Area = 1/2 * base * height = 1/2 * 4 (base TD) * 8 (height TA) = 16

Answer:

Therefore, the area of triangle TAD is 16.

The key to this problem lies in the fact that Cindy can have 22 different pile sizes (Y).

Here's why this information helps us find the smallest number of coins:

Divisibility: Since Cindy can have 22 different pile sizes, the total number of coins (let's call it T) must be divisible by 22 different numbers.

Smallest Divisors: The smaller the pile sizes (Y), the more divisors T will have. In other words, for T to have the smallest number of coins, we want the first 22 positive integers to divide T.

Smallest Multiple: The smallest number of coins (T) that satisfies all these conditions will be the least common multiple (LCM) of the first 22 positive integers.

Finding the LCM of all 22 numbers can be computationally expensive. However, we can make some observations:

All the first 11 positive integers divide any larger positive integer.

We only need to focus on the remaining prime numbers less than or equal to 22 (which are 2, 3, 5, 7, 11, 13, 17, and 19).

Checking Divisibility: We can check if T is divisible by each of these prime numbers. If it's not, we need to increase T until it becomes divisible by all of them.

Finding the Smallest T:

Start with a small number like T = 22 (which is divisible by 2 and 11). Check if it's divisible by 3, 5, 7, 13, 17, and 19. If not, increase T by 1 and repeat the checks.

The smallest T that satisfies all the divisibility conditions is T = 88.

Therefore, the smallest number of coins Cindy could have is 88​.

Let's analyze each condition to see what it tells us about the minimum number of marbles in the box.

Red Marbles: Lamant needs at least 20 marbles to guarantee at least one red marble. This means there could be a box with only 19 non-red marbles (blue and green combined).

Green Marbles: Similarly, at least 24 marbles are needed to guarantee at least one green marble. This implies there could be a box with only 23 non-green marbles (red and blue combined).

All Three Colors: Finally, 27 marbles are needed to ensure all three colors are present. This means there could be a box with only 26 marbles that don't have all three colors (a combination of red and blue only, or blue and green only, or red and green only).

Combining the Information:

Looking at these conditions, we see that the number of non-red marbles (blue and green combined) must be greater than 19 (from condition 1) but less than or equal to 23 (from condition 2).

Similarly, the number of non-green marbles (red and blue combined) must be greater than 23 (from condition 2) but less than or equal to 26 (from condition 3).

The only way to satisfy both constraints is if there are exactly 23 non-red marbles and exactly 23 non-green marbles. This means the box must have a total of:

Red marbles = Total - Non-red marbles = Total - 23

Green marbles = Total - Non-green marbles = Total - 23

Finding the Total:

Since the number of red and green marbles must be the same (both equal to Total - 23), the total number of marbles in the box (including red, blue, and green) must be a multiple of 2. The smallest multiple of 2 that satisfies both conditions is:

Total = Red marbles + Green marbles + Blue marbles

Total = (Total - 23) + (Total - 23) + Blue marbles

2 * Total = 46 + Blue marbles

Since the total needs to be a whole number, the number of blue marbles must be even (because multiplying by 2 gives an even number). The smallest even number that satisfies all the conditions is:

Blue marbles = 2

Therefore, the total number of marbles in the box is:

Total = (46 + Blue marbles) / 2

Total = (46 + 2) / 2

Total = 48 / 2

Lamant's box contains a total of 24​ marbles.

There are two ways to tackle this problem:

Method 1: Counting Valid Triangles

Total Points: We have 3 choices for x (1, 2, or 3) and 8 choices for y (3, 4, 5, ..., 10), resulting in a total of 3 * 8 = 24 points.

Choosing 3 Points: We need to choose 3 out of these 24 points to form a triangle.

The number of ways to do this can be calculated using combinations: nCr = n! / (r! * (n-r)!) where n is the total number of elements (24) and r is the number of elements to choose (3). 24C3 = 24! / (3! * (24 - 3)!) = 2280

Not All Triangles are Valid: However, not all combinations of 3 points will form valid triangles.\

The triangle inequality states that the sum of any two sides of a triangle must be greater than the third side. We need to check how many of the 2280 combinations violate this rule.

Loop through each combination of 3 points.

Calculate the distances between all three pairs of points using the distance formula (square root of the sum of squared differences in x and y coordinates).

If any of the distances violate the triangle inequality (i.e., the sum of two distances is less than the third distance), discard that combination.

Valid Triangles: After checking all combinations, count the remaining ones that satisfy the triangle inequality. This will be the number of valid triangles.

Method 2: Faster Approach (Counting Invalid Triangles)

Total Combinations: Same as method 1, there are 2280 total ways to choose 3 points from 24.

Invalid Triangles: We can count the number of triangles that violate the triangle inequality and subtract them from the total to get the valid ones. There are two main cases for invalid triangles:

All points on a straight line: If all three chosen points fall on the same line (e.g., (1, 3), (2, 3), (3, 3)), they cannot form a triangle.

Degenerate Triangle: If two points have the same x-coordinate or the same y-coordinate, and the third point does not lie above the line connecting them, they cannot form a valid triangle. (Imagine a straight line segment connecting the two points with the same coordinate. The third point must be above this line segment for a triangle to form).

Count the number of ways to choose 3 points that fall on the same horizontal or vertical line (3 choices for x-coordinate * 8 choices for y-coordinate = 24)

Count the number of ways to choose 2 points with the same x-coordinate (3 choices for x-coordinate * 7 choices for y-coordinate for the first point * 6 choices for y-coordinate for the second point, excluding the chosen y-coordinate = 126).

Do the same for points with the same y-coordinate.

Valid Triangles: The total number of valid triangles is then: Total Combinations - Invalid Triangles 2280 - (24 + 126 + 126) = 2004

Both methods will give you the answer: there are 2004​ triangles that can be formed using 3 of the points.

I guess you are talking about:

\(T_f = \{f^{2021}(s): s \in S\}\), find the remainder when \(\displaystyle \sum_{f\in F} |T_f|\) is divided by 2017, am I correct?

The key to this problem is finding the least common multiple (LCM) of 360, 450, and 540. This represents the smallest number of days that will take for all three planets to complete one full rotation around the sun relative to their starting positions.

Here's why:

If all planets complete their rotations in multiples of their individual periods (e.g., 360 days for X, 900 days for Y, etc.), they might not be lined up

again.

The LCM ensures that after that specific number of days, X will have completed exactly m rotations (for some integer m), Y will have completed exactly n rotations (for some integer n), and Z will have completed exactly o rotations (for some integer o).

To find the LCM, we can use various methods. In this case, since the numbers are relatively close, you might be able to find it by inspection.

However, a more general approach is to use the Euclidean Algorithm:

Find the greatest common divisor (GCD) of the largest two numbers (here, 450 and 540).

Divide the LCM of the initial three numbers (which doesn't exist yet) by the GCD you just found. This gives you the LCM of the first two numbers.

Now, find the GCD of the LCM you obtained in step 2 and the remaining number (here, 360).

The final LCM is the product of the GCDs you found in steps 1 and 3.

Following these steps, you'll find the GCD of 450 and 540 to be 90. The LCM of 450 and 540 is then (450 * 540) / 90 = 300.

Finally, the GCD of 300 and 360 is 60. Therefore, the LCM of 360, 450, and 540 is (300 * 60) = 18000.

However, 18000 is not the minimum positive number of days. Why? Because all three planets might be lined up again at a smaller common multiple. We need to find the least common multiple.

Looking closer, we see that 540 (Z's rotation period) is divisible by both 360 (X's period) and 450 (Y's period).

This means that every time Z completes a full rotation, X will have completed some integer number of rotations and Y will have completed some other integer number of rotations. In other words, they will all be lined up again after every 540 days.

Therefore, the minimum positive number of days before they are on the same line again is 540​ days.

Simplifying and completing the square, we have

\(t^2 -9t - 36 + 13t - 60\\ = t^2 + 4t - 96\\ = t^2 + 4t + 4 - 100\\ = (t + 2)^2 - 100\)

Note that \((t + 2)^2 \geq 0\) for any real t, and equality holds when \(t + 2 = 0\), i.e., \(t = -2\).

Hence, the smallest value is produced when t = -2.

I assume you meant largest.

Simplifying the expression and completing the square, we have

\(2s^2 - 8s + 19 - 7s^2 - 8s + 20\\ = -5s^2 - 16s + 39\\ = -5(s^2 + \frac{16}5s) + 39\\ = -5(s + \frac85)^2 + 5(\frac 85)^2 + 39\\ =-5(s + \frac85)^2 + \frac{259}5\)

Note that \(-5(s + \frac85)^2 \leq 0\) for any real s. Then the largest possible value is \(\dfrac{259}5\).

There is no smallest value. The expression can be arbitrarily small.

I assume you mean minimum.

Simplifying and completing the square,

\(-t^2 +8t-4+5t^2-4t+18\\ =4t^2 +4t + 14\\ =4t^2 +4t+1 +13\\ =(2t+1)^2 + 13\\ \)

Note that \((2t + 1)^2 \geq 0\) for all real t. Then the minimum is 13. 

The expression can be arbitrarily large so there is no maximum.

Understanding the Slopes:
 

Slope of line segment OP = 1: This means for every 1 unit you move up on the y-axis, you also move 1 unit to the right on the x-axis.

Since O is the origin (0,0), point P will be at (x, x) for some positive value x.

Slope of line segment PQ = 3: This means for every 3 units you move up on the y-axis, you move 1 unit to the right on the x-axis.

Finding Coordinates of P:

As mentioned earlier, P lies at (x, x).

Connecting P and Q:

We know the slope of PQ is 3. We need to find the y-coordinate of point Q relative to P (i.e., how much we move up from P). Let the y-coordinate of Q be y.

Relating Coordinates using Slope:

The difference in y-coordinates (y - x) is 3 times the difference in x-coordinates (which is just x in this case).

Setting up the Equation:

y - x = 3x

Combining Like Terms:

y = 4x

Finding the Slope of PQ:

Slope of PQ = (Change in y) / (Change in x) = (y_Q - y_P) / (x_Q - x_P)

Since both P and Q lie on the positive x-axis (first quadrant), x_Q and x_P are both positive values. Therefore, we can divide both sides of the equation y = 4x by x (assuming x is not zero) to get:

y/x = 4

This gives us the slope of line segment PQ, which is 4.