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You are welcome :)

Of course I haven't proven anything.

It is just an assumption that the pattern will continue.

There are four possible outcomes when flipping a fair coin two times:

HH (heads on both flips)

HT (heads on the first flip, tails on the second flip)

TH (tails on the first flip, heads on the second flip)

TT (tails on both flips)

Let X be the length of the longest run in a sequence of two coin flips. We can compute the probability of each possible value of X:

X = 1 if the two coin flips are different (HT or TH), which occurs with probability 1/2 + 1/2 = 1.

X = 2 if the two coin flips are the same (HH or TT), which occurs with probability 1/4 + 1/4 = 1/2.

The expected value of X is the sum of the possible values of X multiplied by their respective probabilities:

E(X) = 1*(1/2) + 2*(1/2) = 1.5

Therefore, the expected length of the longest run in a sequence of two coin flips is 1.5.

To solve this problem, we need to apply the function C three times to n, and then check if the result is equal to 16. That is, we need to find all n such that C(C(C(n))) = 16.

Let's start with an arbitrary positive integer n and apply the function C to it. If n is even, then C(n) = n/2. If n is odd, then C(n) = 3n + 1. Let's consider these two cases separately.

Case 1: n is even

If n is even, then C(n) = n/2. Applying the function C again, we get:

C(C(n)) = C(n/2) = (n/2)/2 = n/4

Applying the function C a third time, we get:

C(C(C(n))) = C(n/4) = (n/4)/2 = n/8

Therefore, in this case, we need to solve the equation n/8 = 16, which gives n = 128.

Case 2: n is odd

If n is odd, then C(n) = 3n + 1. Applying the function C again, we get:

C(C(n)) = C(3n + 1) = (3n + 1)/2, since 3n + 1 is even

Applying the function C a third time, we get:

C(C(C(n))) = C((3n + 1)/2) = (3n + 1)/4

Therefore, in this case, we need to solve the equation (3n + 1)/4 = 16, which gives n = 63.

So the two solutions to the equation C(C(C(n))) = 16 are n = 128 and n = 63.

Lets pretend that the 3 flavors are A, B, and C

how many ways can 2 children pick A? 6c2 = 15

How many ways can the leftover 2 of 4 children pick B? 4c2 = 6

How many ways can the last 2 pick B? 2c2 = 1

multiply that together you get 90

1-9 is 9 digits

10-99 is 90 numbers with 2 digits each, 180 digits total.

180+9 is 189 digits

There are 189 digits

There are 189 digits, 9 of which are 0: 10, 20, 30, 40, 50, 60, 70, 80, 90

9/189 when simplified is 1/21

1/21 probability of choosing a 0

This is an arithmetic sequence, to find the sum, take the smallest number and add it on to the largest number, then multiply that by the number of numbers there are and divide it by 2. 

So take what I just said and the expression will look like this 

((1+99)99)/2

=(100x99)/2

=9900/2

=4450 is the answer

A) 189 digits

B) 1/21 choosing a 0

C)4450 sum of all digits

since 67 pounds of seeds can plant a 9 acre field the equation kinda looks likes this

67 pounds of seed =  9 acres

so now if we are looking for 1 pound, divided the pounds of seed and acres by 67

1 pound of seed = 9/67 acres

so the fraction is 9/67 acres per pound of seed

incorrect

Let's label the points in the diagram as follows:

The vertices of the parallelogram are A, B, C, and D, with AB parallel to CD and AD parallel to BC.

The points where the parallel lines intersect the sides of the parallelogram are E, F, G, and H, as shown in the diagram.

The side length of the equilateral triangle is s.

Since the three regions have the same area, the area of each region is 1/3 of the total area of the parallelogram. Let's denote the area of the parallelogram as A.

The area of the equilateral triangle is (s^2√3)/4, and since it is one of the regions, we know that:

(s^2√3)/4 = A/3

Simplifying this equation, we get:

s^2 = 4A/(3√3)

Now, let's look at the parallelogram as a whole. Since AB is parallel to CD, we know that triangle AEF is similar to triangle BGF, and since AD is parallel to BC, we know that triangle ADE is similar to triangle BCF. This means that:

AE/EB = BF/FG AD/DE = BC/CF

Let's use these ratios to express AE, BF, AD, and BC in terms of s:

AE = s/3 (since the perimeter of the equilateral triangle is 1) EB = 2s/3 BF = 2s/3 FG = s/3 AD = (2s + 2EB)/3 = (2s + 4s/3)/3 = 10s/9 DE = (2s + 2AE)/3 = (2s + 2s/3)/3 = 8s/9 BC = (2s + 2FG)/3 = (2s + s/3)/3 = 7s/9 CF = (2s + 2BF)/3 = (2s + 4s/3)/3 = 10s/9

Now we can find the perimeter of the parallelogram:

Perimeter = 2(AB + AD) = 2(EB + BC) = 2(2s/3 + 7s/9) = 4s/3 + 14s/9 = (22s/9)

Substituting s^2 = 4A/(3√3) from the earlier equation, we get:

Perimeter = (22/9) sqrt(3)

So the perimeter of the original parallelogram is 22/9*sqrt(3).

Let's draw a diagram to visualize the problem:

We can see that the line segment connecting the centers of the two circles, OP, is perpendicular to the line l at H, the point of tangency. We can use this fact to find the length of OP using the Pythagorean theorem.

Let x be the length of OP. Then we have:

OH = 3 (the radius of circle O) PH = 5 (the radius of circle P)

By the Pythagorean theorem, we have:

x^2 = OH^2 + PH^2 = 3^2 + 5^2 = 9 + 25 = 34

Therefore, only possible lengths of OP is sqrt(34).

Thanks to Guest for giving this problem a shot, but Melody's solution was correct! Thanks so much, I probably would've never noticed that pattern if it weren't for your solution

Let's first draw a diagram of the regular tetrahedron ABCD and the point M, the midpoint of CD.

Since ABCD is a regular tetrahedron, we know that all of its edges are congruent and all of its faces are equilateral triangles. Let's call the length of an edge of ABCD "a".

]We can use trigonometry to find the tangent of the angle AMB. To do so, we need to find the length of AM and BM.

Since M is the midpoint of CD, we know that DM = MC = a/2.

We can use the Pythagorean theorem to find the length of AM:

AM^2 = AB^2 - BM^2

Since ABCD is a regular tetrahedron, we know that triangle ABC is an equilateral triangle with side length a. Therefore, we can use the Pythagorean theorem to find the length of AB:

AB^2 = AM^2 + BM^2 AB^2 = 4AM^2 (because AB = AC = BC = BD = CD = a)

Substituting the second equation into the first, we get:

AM^2 = 4AM^2 - BM^2 BM^2 = 3AM^2 BM = AM * sqrt(3)

Now that we know the lengths of AM and BM, we can find the tangent of angle AMB:

tan(AMB) = BM / AM tan(AMB) = (AM * sqrt(3)) / AM tan(AMB) = sqrt(3)

Therefore, the tangent of angle AMB is sqrt(3).

The perimeter of OSTQ is 34.

The perimeter is 68.

The area of rectangle OSTQ is 60.

Sorry, should be meters per hour.

The route may be hilly, up then down then up then level then down again, and so on.

Suppose that the total distance uphill is U, the the total distance on level ground is L and the total downhill is D.

The time taken for the outward trip T1 would be equal to U/100 + L/120 + D/150.

On the return journey, uphill becomes downhill and downhill becomes uphill, level stays the same.

The time taken for the return trip T2 will then be U/150 + L/120 + D/100.

The total time for the round trip will be T1 + T2 =  U/60 + L/60 + D/60 = ( U + L +D)/60

The total distance TD = 2(U + L + D), so the average speed will be TD/(T1 + T2) = 120 km/ hr.

I get 4

(a,b,c,d) = (8,3,-5,3)

The number of possible arrangements is 2,468,220.

Triangle ABC is a right triangle because it has it's hypotenuse on a diameter of a circle. (point C is backwards on my graph, but that doesnt affect the result because its a reflection of the original image)

Finding AB with pythagorean theorem, we get \(AB = \sqrt{4^2 + 8^2} = 4\sqrt{5}\).

We also Get ACM=MCB = 45 degrees 

Because interior angles of a circle are congruent, we get AMB = BAM = 45. Now we see that ABM is a right isoceles triangle. Also, CAMB is a cyclic quadrilateral. We can calculate AM and BM too. \(AM = BM = \frac{4\sqrt{5}}{\sqrt{2}} = 2\sqrt{10}\)

Because ACBM is a cyclic, Ptolemy's theorem can be applied. 

Set the length of CM as X

then \(4*2\sqrt{10}+8*2\sqrt{10}=x*4\sqrt{5}\) and solving the equation we get x = \(x = 6\sqrt{2}\)

Using the alterating pattern of i^n, the sum works out to i.

To calculate Carissa's average speed during the entire round trip, we need to first determine the total distance she walked and the time it took her to complete the trip.

Let's assume that the distance between the café and boba shop is x kilometers. When walking uphill, Carissa's speed is 100 meters per hour, so it would take her x/0.1 = 10x hours to cover this distance. When walking across flat ground, her speed is 120 meters per hour, so she would cover the same distance in x/0.12 = 10/1.2 x = 25/3 x hours. When walking downhill, her speed is 150 meters per hour, so she would cover the same distance in x/0.15 = 20/3 x hours.

Thus, the total time it took Carissa to walk from the café to the boba shop was 10x + 25/3 x + 20/3 x = 100/3 x hours. When she walked back from the boba shop to the café, she covered the same distance, so the total time for the round trip was 200/3 x hours.

The total distance of the round trip was 2x kilometers. Therefore, Carissa's average speed during the entire round trip was:

average speed = total distance / total time = 2x / (200/3 x) = 6/5 x/hour = 1.2x kilometers per hour.

So Carissa's average speed during the entire round trip was 1.2x kilometers per hour.

The number of ways is 520,080.