parmen

We can solve this problem by considering two cases:

Case 1: One family has siblings in each row

There are 3 families (let's call them A, B, and C). In this case, we need to place one child from each family in each row. There are 3 choices for the first seat in the first row (child from any family).

Once the first seat is filled, there are 2 choices left for the second seat (child from one of the remaining families). The third seat is automatically filled by the remaining child.

Similarly, for the second row, there are again 3 choices for the first seat, and so on. Therefore, the number of arrangements for this case is:

Arrangements (Case 1) = 3 * 2 * 3 * 2 = 36

Case 2: Two families have siblings in one row

There are again 3 choices for which family will have siblings in the same row (A, B, or C). Without loss of generality, let's say family A has siblings in the same row.

There are now 2 choices for which child of family A sits in the first seat of the first row. The other child of family A must sit in the second seat.

The remaining child from family B can sit in the third seat. Now, for the second row, there are 2 choices for the first seat (child from either family B or C). Following the same logic as before, the number of arrangements for this case is:

Arrangements (Case 2) = 3 * 2 * 1 * 2 * 2 = 24

Total Arrangements

To get the total number of arrangements, we add the number of arrangements from both cases:

Total Arrangements = Arrangements (Case 1) + Arrangements (Case 2)

Total Arrangements = 36 + 24

Total Arrangements = 60​

There are 5!=120 ways to order the numbers 1 through 5. However, this counts an arrangement as valid if it is just a rotation of another valid arrangement.

For instance, the arrangement 1<2>3<4>5 is the same arrangement as 5<1>2<3>4 if we just rotate the boxes cyclically.

There are 5 ways to rotate the boxes, so we have overcounted exactly that many times. Therefore, the number of valid arrangements is 120 ÷ 5 = 24​.

For question 2:

Absolutely, we can find the positive values of n for which both n and 3n are four-digit base-9 integers.

Understanding the Problem:

Base-9 Integers: Numbers are represented in base-9 using digits 0 to 8, just like base-10 uses 0 to 9.

Four-digit Integers: We're interested in n such that both n and 3n have four digits in base-9.

Conditions for Four-digit Numbers:

For a base-b number to be four-digit, it must satisfy:

b^3 ≤ n < b^4

In this case, the base is b = 9. Therefore:

9^3 ≤ n < 9^4

729 ≤ n < 6561

Constraint for 3n:

We also need 3n to be a four-digit base-9 integer. This translates to:

9^3 ≤ 3n < 9^4

243 ≤ 3n < 6561

Dividing the second inequality by 3, we get:

81 ≤ n < 2187

Combining Conditions:

To satisfy both conditions, n must lie between 729 and 2187 (inclusive):

729 ≤ n ≤ 2187

Counting Possible Values of n:

Within this range, n must be a multiple of 3 (since 3n is a multiple of 3 and needs to be a four-digit integer).

The first multiple of 3 within the range is 732, and the last multiple of 3 is 2187.

To count the number of multiples of 3 in this range, we can divide the last multiple by the first multiple and subtract 1 (since we only consider integers from 732 to 2187, inclusive):

(2187 / 732) - 1 = 3

Answer:

Therefore, there are 3 positive values of n (732, 735, and 738) for which both n and 3n are four-digit base-9 integers.

For question 1:

Here's how to find the number of positive five-digit integers containing the digit grouping "24" at least once:

Total Possible Numbers:

There are 9 choices (excluding 0) for each digit in a five-digit number (1, 2, 3, 4, 5, 6, 7, 8, 9).

So, the total number of possible five-digit positive integers is 9 * 9 * 9 * 9 * 9 = 59,049.

Unwanted Scenarios (Numbers without "24"):

No "24" group: There are 8 choices for the digit in the place where "24" could be (all digits except 2 and 4). For the remaining 4 digits, there are 9 choices each. So, there are 8 * 9 * 9 * 9 = 5832 cases where "24" doesn't appear in any of the five digits.

"24" only appears once, but not next to each other:

Choose the digit that will be between "2" and "4": There are 7 choices (all digits except 0, 2, and 4).

Choose the positions for "2" and "4" (not next to each other): There are 4 choices for the first position (remaining slots after fixing one digit). There are 3 choices for the second position (remaining slots after fixing the first and the middle digit). So, there are 4 * 3 = 12 ways to arrange "2" and "4" with another digit in between.

Choose the remaining 2 digits: There are 9 choices each.

Total unwanted cases with "24" not next to each other: 7 * 12 * 9 * 9 = 5832.

Total Numbers with "24" at least once:

We want to find the number with "24" at least once. So, subtract the unwanted cases from the total:

Total numbers - Unwanted cases = Numbers with "24"

59,049 - (5832 + 5832) = 47,385

Therefore, there are 47,385 positive five-digit integers containing the digit grouping "24" at least once.

We can solve this problem using the properties of isosceles triangles and the Pythagorean theorem.

Isosceles Triangle Properties:

Since AB = AC in an isosceles triangle ABC, angles B and C are congruent.

Dividing the Base:

The height from A to BC is the perpendicular bisector of the base (divides it into two segments of equal length) in an isosceles triangle.

Segment Lengths:

Given that BC = 8 cm and BD = 3 cm, we know DC = BC - BD = 8 cm - 3 cm = 5 cm.

Finding the Height:

We can use the Pythagorean theorem on right triangle ABD (since the height is perpendicular to the base).

Pythagorean Theorem:

a^2 + b^2 = c^2 (where a and b are the lengths of the legs and c is the length of the hypotenuse)

In this case:

a (length of leg BD) = 3 cm

c (length of hypotenuse AB) = unknown (represents the height of the triangle)

b (length of leg AD) = unknown

Solving for Height (c):

b^2 = c^2 - a^2 (rearranging the equation)

b^2 = c^2 - 3^2 (substitute known values)

Since AD is half of the base (property of isosceles triangle):

AD = 1/2 * BC = 1/2 * 8 cm = 4 cm

Substitute the value of AD (b) in the equation:

4^2 = c^2 - 3^2

Simplify:

16 = c^2 - 9

Add 9 to both sides:

25 = c^2

Take the square root of both sides (remember to consider both positive and negative since squaring either results in 25):

c = ± 5

Positive Height:

Since the height represents a distance, we take the positive value:

c (height of the triangle) = 5 cm

Therefore:

Length of segment DC = 5 cm

Height of the triangle = 5 cm

The answer is

\[\binom{38}{5} + \binom{38}{32} = \binom{40}{5}.\]

Finding the Slope (m):

The slope (m) of the line passing through points (-4, 2) and (1, -1) can be found using the following formula:

m = (y2 - y1) / (x2 - x1)

Here, (x1, y1) = (-4, 2) and (x2, y2) = (1, -1)

m = (-1 - 2) / (1 - (-4)) = -3 / 5

Finding the y-intercept (b):

We can use the point-slope form of the equation to find the y-intercept (b). Here, we can use either of the given points. Let's use (-4, 2).

y = mx + b

2 = (-3/5) * (-4) + b

2 = 12/5 + b

b = 2 - 12/5 = -2/5

Writing the Equation in Ax + By = C form:

Now, we can express the equation in the desired form:

A(x) + B(y) = C

Substitute the slope (m) and y-intercept (b):

A(x) + (-3/5)(y) = -2/5

Finding A, B, and C with GCD = 1 and A positive:

To ensure the greatest common divisor (GCD) of A, B, and C is 1 and A is positive, we can multiply the entire equation by the least common multiple (LCM) of the denominators of m and b (which is 5 in this case).

5A(x) - 3y = -2

Since A is positive, we can simply set A = 5:

5(x) - 3y = -2

Therefore, the equation of the line in the desired form is:

5x - 3y = -2

Let's say the distance between the café and the boba shop is d, and Carissa walks the same distance in reverse on her way back.

Let's break the trip down into three parts: walking uphill, walking across flat ground, and walking downhill.

When walking uphill, Carissa walks at a speed of 100 meters per hour, so it takes her d/100 hours to complete this part of the trip.

When walking across flat ground, Carissa walks at a speed of 120 meters per hour, so it takes her d/120 hours to complete this part of the trip.

When walking downhill, Carissa walks at a speed of 150 meters per hour, so it takes her d/150 hours to complete this part of the trip.

So, the total time for the round trip is:

d/100 + d/120 + d/150 + d/100 + d/120 + d/150 = (11/600)d

To find the average speed, we need to divide the total distance (2d) by the total time, which is (11/600)d:

average speed = 2d / ((11/600)d) = 1200 / 11 ≈ 109.1 meters per hour

Therefore, Carissa's average speed during the entire round trip is approximately 109.1 meters per hour.

There are (312​)=220 ways to choose the three white balls on your ticket. For each of these choices, there is one way to choose the red SuperBall. There is also a way to choose the three white balls and the red SuperBall in the same order that they were drawn, so there are also (312​) winning ticket combinations in which your ticket matches at least two of the white balls drawn.

Finally, there are 8 winning ticket combinations in which your ticket matches the red SuperBall, but does not match any of the white balls drawn. Therefore, there are a total of 220+220+8=448 winning ticket combinations.

Since there are a total of (312​)⋅8=840 ways to choose three white balls and one red ball, the probability that your ticket is a winning ticket is 448/840 = 56/105​​.

Let s=2x3+x−3, then x3+s3=3, or 3s3⋅x3−3s=0. Factoring out x3, we have x3⋅(3s2−1)=0.

Since s=2x3+x−3=0, we have x3=0, or x=0​.

By the quadratic formula, the roots of x2−27​x+413​ are [ \frac{7\pm\sqrt{\left(\frac{7}{2}\right)^2-4\cdot1\cdot\frac{13}{4}}}{2}=\frac{7\pm\sqrt{1}}{2}=\frac{7}{2}\pm\frac{\sqrt{1}}{2} ]

Therefore, the values of x at which the graphs of f and g intersect are x=2, x=27+1​​ and x=27−1​​.

We can plug these x values back into f(x) to get that f(2)=21, f(27+1​​)=825+1​​ and f(27−1​​)=825−1​​, so the points of intersection are [ \left(2,21\right),\left(\frac{7+\sqrt1}{2},\frac{25+\sqrt{1}}{8}\right),\left(\frac{7-\sqrt1}{2},\frac{25-\sqrt{1}}{8}\right). ]

Since the x-intercept of the line is (10,0), we know that the equation of the line is of the form y=mx+n for some value of m and n. Since the y-intercept of the line is (0,−2), we have that −2=m⋅0+n. Hence, n=−2. Therefore, the equation of the line is y=mx−2.

We know that the point (a,a2) lies on the line, so a2=ma−2. We also know that the point (b,b2) lies on the line, so b2=mb−2. Subtracting these two equations, we get b2−a2=m(b−a). Since a 0, so m=b−ab2−a2​.

Since the line passes through the point (10,0), we have that 0=m(10)−2. Substituting the expression for m into this equation, we get 0=b−ab2−a2​(10)−2. This simplifies to b2−a2=5a−5b. We also have the equation a2=ma−2. Substituting m=b−ab2−a2​, we get a2=b−ab2−a2​(a)−2. This simplifies to (b−a)a2−(b2−a2)a+2(b−a)=0. This factors as (a−2)(b−a)a2=0.

Since a

Substituting a=0 into the equation b2−a2=5a−5b, we get b2=5b. This factors as b2−5b=0, which factors as b(b−5)=0. Therefore, b=0 or b=5. Since a

Therefore, (a,b)=(0,5)​.

The distance from the center of the circle to the line is equal to the radius of the circle. To find the radius, we need to find the perpendicular distance from the origin to the line.

The equation of the line in slope-intercept form is y=21​x+215​. The slope of the line is 21​, and its negative reciprocal is −2. Therefore, the equation of the perpendicular line that passes through the origin is y=−2x. Setting the two equations equal to each other, we get −2x=21​x+215​. Solving for x, we get that x=−5.

Substituting this value of x back into the equation of the line, we get that y=5, so the origin is (0,5). Therefore, the radius of the circle is 5, so the area of the circle is π⋅52=25π​ square units.

Case 1: The number has an even number of digits.

Since the sum of the digits is 12, this is impossible.

Case 2: The number has an odd number of digits, and the units digit is 1.

In this case, the first digit must be 3, and the digits in between must add to 9. There are (25​)=10 ways to choose the two numbers that add to 9, and 22=4 ways to order them. Thus the number of possibilities in this case is 10⋅4=40.

Case 3: The number has an odd number of digits, and the units digit is 3.

In this case, the first digit must be 1, and the digits in between must add to 3. There are (13​)=3 ways to choose the number that adds to 3, and 2 ways to place it. Thus the number of possibilities in this case is 3⋅2=6.

The total number of possibilities is 40+6=46​.