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I solved this problem!  The answer is 100*sqrt(2).

I solved this problem!  The answer is 5/2.

I solved this problem!  The answer is 12/5.

I solved this problem!  The answer is 16*sqrt(15).

I solved this problem!  The answer is 3/8.

Im pretty sure the original question is:

How many ordered triples of positive integers \((a,b,c)\) have \(\text{GCF}(a, b, c) = 2020\) and \(\text{LCM}(a, b, c) = 2020^2\)

Because, 2020 has factors of 2, 2, 5, 101 and 4020 has factors of 2, 2, 5, 3, 67 and that wouldn't result in an integer.

but the answer to the "real question" would be to:

Note that the prime factorization of 2020 is \(2^2,5,101\) and that the prime factorization of \(2020^2\) is thus \(2^4 \cdot 5^2 \cdot 101^2\)

Thus the exponents of 2 in the prime factorizations of a,b,c must be 2, m, 4 in some order, where \(2 \le m \le 4\)That way, we can ensure that the GCD has two factors of 2 and the LCM has four factors of 2 If m=2 there are 3 distinct shufflings of (2, 4, 4.) If m=3 there are 6 distinct shufflings of (2,3,4) If m=4 there are 3 distinct shufflings of (2,4,4) So there are 3 allowed arrangements of powers of 2 among a,b ,and c .

The exponents of 5 for a,b,c must be some shuffling of (1, 2, 3) or (1, 2, 2) by the same logic, giving us 6 allowed arrangements of powers of 5. Likewise, there are 6 allowed arrangements for the powers of 101 by similar reasoning.

Thus the total number of ordered triples is the product of these three counts: \(12\times6\times6\) =   432  ordered triples.

We can say that he has completed 60 games, because it is 60% of 100. He so far won 40% of them, so they have won 24 games. To get 50% in total for the next 40 games, he must win 26 games. 26 / 40 * 100 = 26 / 2 * 5 = 13 * 5 = 65, so he must win 65% of the next 40% of his games.

There are two cases to consider:

Case 1: Two cards are from the same color and one card from a different color.

Choose one of the four colors for the two matching cards: There are 4 ways to do this.

Choose two out of the seven cards of that chosen color: There are ⁷C₂ ways to do this (combination, order doesn't matter).

Choose one card from each of the remaining three colors: There are 3 * 3 * 3 = 27 ways to do this (since there are 3 colors left and 3 cards of each color).

Case 2: Three cards are from different colors.

Choose one card from each of the four colors: There are 4 ways to do this.

Adding the number of ways for each case:

Total hands = (Ways in Case 1) + (Ways in Case 2)

Total hands = (4 * ⁷C₂ * 27) + 4

We can calculate ⁷C₂ using the combination formula: ⁷C₂ = ⁷! / (2! * (⁷ - 2)!) = 21

Total hands = (4 * 21 * 27) + 4

Total hands = 2520 + 4

Total hands = 2524​

The answer is 2524.

We can think about this problem in two steps:

Step 1: Arrange the mathematicians and scientists as groups.

Since the mathematicians must sit together and the scientists must sit together, we can treat them as single units for now. This leaves us with 3 units (mathematicians, scientists, journalists).

There are 3! ways to arrange these 3 units around the circular table.

Step 2: Arrange the people within each group.

The mathematicians can arrange themselves in 4! ways within their designated section.

The scientists can arrange themselves in 3! ways within their designated section.

However, we've overcounted the arrangements because rotating the entire table doesn't create a new seating arrangement.

Step 3: Correct for overcounting due to circular symmetry.

Since there are 9 people sitting around the table, there are 9 ways to rotate the table so it appears as a new arrangement.

Total Arrangements:

To get the final number of arrangements, we take the number of arrangements from steps 1 and 2 and divide by the number of times we've overcounted in step 3:

Arrangements = (Arrangements of groups) * (Arrangements within groups) / (Overcounting)

Arrangements = (3!) * (4! * 3!) / (9)

Arrangements = 6 * 24 * 6 / 9

Arrangements = 864 / 9

Arrangements = 96​

The answer is 96.

Since I is the incenter, then ∠BIM=∠CIN=90∘. Also, since MN is parallel to BC, then ∠AMN=∠ABC and ∠ANM=∠ACB.

By the Law of Cosines on triangle ABC, [\cos C = \frac{AB^2 + AC^2 - BC^2}{2AB \cdot AC} = \frac{5^2 + 5^2 - 8^2}{2 \cdot 5 \cdot 5} = -\frac{3}{25}.]Then sinC=1−cos2C​=2546​​.

pen_spark

Let x=BM and y=CN. Then by the Law of Sines on triangle ABM, [\frac{BM}{\sin C} = \frac{AB}{\sin \angle BAM} = \frac{AB}{\sin 90^\circ} = AB,]so BM=5⋅sinC=546​​.

Similarly, by the Law of Sines on triangle ACN, [\frac{CN}{\sin C} = \frac{AC}{\sin \angle CAN} = \frac{AC}{\sin 90^\circ} = AC,]so CN=5⋅sinC=546​​.

Then triangle AMN is a right triangle with legs of length 546​​, so its area is [\frac{1}{2} \cdot \frac{4 \sqrt{6}}{5} \cdot \frac{4 \sqrt{6}}{5} = \boxed{\frac{32}{5}}.]

We can solve the inequality by distributing the terms, combining terms on each side, and isolating x. Here's how:

Distribute the terms on the left side: -4(x + 4) = -4x - 16

Combine terms on both sides: -4x - 16 > x + 7 + 8x + 21

Combine like terms: -4x - 16 > 9x + 28

Isolate x by getting all the x terms on one side: -13x - 16 > 28

Add 16 to both sides to isolate x further: -13x > 44

Divide both sides by -13 to isolate x (remembering to flip the inequality sign since we're dividing by a negative number): x < - \dfrac{44}{13}

Simplifying the answer: x < - \dfrac{33}{9}

Therefore, the solution to the inequality is the set of all real numbers less than - \dfrac{33}{9}. We can express this mathematically as: x ∈ {real numbers | x < - \dfrac{33}{9}}.

Let x be the maximum weight each of my neighbor's small trucks can carry. Then my truck can carry at most x+500 pounds.

The total amount her four trucks can carry is 4x pounds. Since my truck's maximum load is at most 41​ of the total her trucks can carry, we know x+500≤41​⋅4x=x.

Simplifying the inequality on the right gives x+500≤x, which means 500≤0. Since this is not possible, there must be no value of x that satisfies both of our original inequalities. Therefore, there is no upper bound on the amount my truck can carry, so the greatest load it can carry is infinite​ pounds. However this makes no sense in the real world, so we can determine that there is an error in the original problem statement.

Simplifying the right side of the inequality gives: 2x - 5 <= -8x + 30

Combining terms with x on the left side: 10x - 5 <= 30

Adding 5 to both sides: 10x <= 35

Dividing both sides by 10 (and flipping the inequality since we're dividing by a positive number): x <= 3.5

Therefore, the solution is the set of all real numbers less than or equal to 3.5. We can express this mathematically as: x ∈ {real numbers | x ≤ 3.5}.

By the distance formula,

\begin{align*} PA^2 &= (x - 4)^2 + (y + 4)^2, \ PB^2 &= (x - 3)^2 + (y - 5)^2, \ PC^2 &= (x + 1)^2 + (y - 2)^2. \end{align*}

Hence, the given equation becomes \begin{align*} (x - 4)^2 + (y + 4)^2 + (x - 3)^2 + (y - 5)^2 + (x + 1)^2 + (y - 2)^2 &= 3((x - Q)^2 + (y - Q)^2) + k \ \Rightarrow \quad 3x^2 + 3y^2 - 14x - 14y + 12 &= 3x^2 - 6Qx + 3Q^2 + 3y^2 - 6Qy + 3Q^2 + k \ \Rightarrow \quad -14x - 14y &= -6Qx - 6Qy + k - 12 \ \Rightarrow \quad 7x + 7y &= 3Q(x + y) + \frac{k - 12}{7}. \end{align*}

Since this equation holds for all points P, the coefficients of x and y must be zero.

Hence, Q=−7/3 and k−12=0, so k = 12​.

b equals to 1.769 and -1.769