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Since $\overline{AB}\parallel \overline{DE}$, we have $\triangle BCD \sim \triangle ECD$ by AA Similarity. Since $BD=24$, $BC:CD=24:CD$ or equivalently $BC:CD=CD:24$. Thus $\triangle BCD \sim \triangle ECD \sim \triangle BDC$. In particular, $\angle CDB \cong \angle CBD$, so $\triangle BCD$ is isosceles with $CD=BD=\boxed{24}$.

There is only one way to make 12 with a single digit, namely with 3. There are 2 ways to make 12 with 2 digits: 11 and 31. There are 3 ways to make 12 with 3 digits: 111, 131, and 311. Finally, there are 3 ways to make 12 with 4 digits: 1131, 1331, and 3311. Thus, the total number of positive integers that satisfy the given conditions is 1+2+3+3=9​.

The given inequality is equivalent to n2<64−20n, or 2n2+20n−64<0. Factoring the left side, we get 2(n−4)(n+8)<0. Since n−4 n>−8. Thus, the possible values of n are −7,−6,−5,…,3,2.

There are 3+(−7)+1=15​ integers in this range.

We first subtract 17 from each part of the inequality to get −25≤x<−20−5x.

We then add 5x to each part of the inequality to get −25+5x≤−20, which simplifies to 5x≤−5.

Dividing each part of the inequality by 5, we get x≤−1. Since −1≥−20−5(−1), the interval of values of x that satisfy the given inequality is (−∞,−1]​.

Each child can choose from 6 flavors of ice cream.

There are (36​) ways to choose 3 flavors of ice cream. There are 6 ways to choose the flavor of the first child who gets two scoops of a particular flavor. There are 5 ways to choose the flavor of the second child who gets two scoops of a particular flavor. For the third child, there is only one unique flavor left. There are a total of 6×5×3×(36​)=2520 ways for the three children to each have exactly two scoops of a particular flavor.

There are (36​) ways to choose 3 flavors of ice cream. There are 6 ways to choose the flavor of the first child who gets one scoop of a particular flavor. For the second and third children, there are 5 ways to choose the flavor of their scoop of ice cream. There are a total of 6×5×5×(36​)=3600 ways for the three children to each have exactly one scoop of a particular flavor.

There are (36​) ways to choose 3 flavors of ice cream. There are 6 ways to choose the flavor of the first child who gets none of the three flavors. There are 5 ways to choose the flavor of the first child who gets one of the three flavors. There are 4 ways to choose the flavor of the first child who gets two of the three flavors. There are a total of 6×5×4×(36​)=1440 ways for the three children to have different flavors.

There is a total of 2520+3600+1440=7560​ ways for each child to choose a flavor for their scoop of ice cream so that some flavor of ice cream is selected by exactly three children.

Since ZWX is trisected, then ∠AWZ=30 and ∠BWZ=60. So triangle AZW is a 30-60-90 right triangle. So WZ=AZ∗3​=83​. And triangle BZW is a 30 -60-90 right triangle. BZ = WZ* 3​=83​∗3​=24. So [WXYZ]=[WZ∗(BZ+BY)]=[8(​3)∗(24+3)]=216(​3)​ units$^2$.

1. We have \begin{align*} f(3) &= 4(4) - 8\ &= 11 - 4 = 3, \end{align*}and \begin{align*} g(3) &= (2 + 7)^2\ &= 5. \end{align*}Therefore, \begin{align*} s(3) &= f(3)+g(3)\ &=- 2 + 7 \ &= \boxed{11}. \end{align*}

We are given a 30-60-90 triangle ABC with sides p6​, p2​, and q3​. We are asked to find the values of p and q. We can start by using the fact that the sides of a 30-60-90 triangle are in the ratio 1:3​:2. This means that we can write the following equations:

p2​p6​​=3​q3​p2​​=1

Simplifying these equations, we get:

2​6​​=3​3​2​​=q1​

Solving the first equation, we find that 6​=6​. This is true, so our first equation is valid.

Solving the second equation, we find that q=6​.

Therefore, the values of p and q are p = 1/sqrt(3) and q = sqrt(6)

Squaring the equation z + 1/z = 1 gives z^2 + 2 + 1/z^2 = 1. Then multiplying by z^2 gives z^4 + 2z^2 + 1 = z^2. Rearranging, we get z^4 + 2z^2 + 1 - z^2 = 0, which factors as (z^2 + 1)(z^2 + 1) = 0. Since z^2 + 1 = 0 has no solutions, z^2 = -1. Taking the square root of both sides, we find z = i or z = -i.

If z = i, then z^3 = i^3 = -i.

Alternatively, if z = -i, then z^3 is also equal to -i.

Therefore,  in either case, z^3 = -i.

Since PZ bisects ∠QPR, ∠QPZ=∠RPZ. Also, since QZ=4 and PZ=2RP​=29​, we can find the length of PZ using the Pythagorean Theorem:

QZ2+PZ2=PR2.

Substituting the given values, we have

42+(29​)2=92.

Solving for PZ, we get PZ=215​.

Since ∠QPR is a right angle and ∠RPZ=∠QPZ, ∠RPZ and ∠QPZ are each complementary to ∠QPR. Therefore,

∠QPZ+∠RPZ=90∘.

Substituting the given values, we have

∠QPZ+∠RPZ=90∘,

so ∠QPZ=∠RPZ=290∘​=45∘.

Now, since ∠RPZ=∠QPZ=45∘, triangle RPZ is a 45-45-90 triangle, and RZ=15/sqrt(2).