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To solve this, let's first consider the case where there are no restrictions on the seating arrangements. In this case, there are 6! ways to arrange the 6 siblings in the 6 chairs.

Next, let's consider the number of arrangements where siblings are seated next to each other. For each pair of siblings, there are 2 ways to arrange them (either sibling can sit in the front or back chair). Since there are 3 pairs of siblings, there are 23=8 ways to arrange the siblings in pairs.

However, we have double-counted the arrangements where two pairs of siblings are seated next to each other. For each arrangement of two pairs of siblings, there are 2 ways to arrange the third pair. Therefore, we have overcounted the number of arrangements with two pairs of siblings by 8⋅2=16.

Finally, let's consider the number of arrangements where all three pairs of siblings are seated next to each other. There are 2 ways to arrange the three pairs of siblings (either all three pairs can be in the front row or all three pairs can be in the back row). Once the pairs are arranged, there are 2 ways to arrange each pair within their respective row. Therefore, there are 2⋅23=16 arrangements where all three pairs of siblings are seated next to each other.

Therefore, the total number of arrangements where siblings cannot be seated next to each other is 6!−8⋅2−16=552​.

Okay. Let's consider the siblings as A1 and A2, B1 and B2, C1 and C2. SInce they are sitting in two rows we will fill the first row in first. So A1 is in the first, B1 is in the second, and C1 is in the third. Now we move on to the second row which is A2, B2, and C2. 

However, there is another possible arrangement for the first row. That is A2 for the first, B1 for the second, and C1 for the third. Then the second row is A1, B2, and C2. 

These are the only two possible arrangments that will satisy the conditions given. Thus, there are two ways to seat three siblings so that siblings don't sit next to each other. 

sorry, that's wrong. The problem is asking for any way to make it 12, and i dont think that your answers match that. It asks how many possibilities are there that could work.

Let $x = AC$. Since $\overline{BC}\parallel\overline{DE}$, we have that $\angle BAD$ and $\angle BCD$ are corresponding angles, and thus congruent. Therefore, $\triangle ABD \sim \triangle DCB$ by AA similarity.

From the similarity, we have \[
\frac{AB}{BD}=\frac{BD}{CB}\quad\Rightarrow\quad \frac{x}{20}=\frac{20}{x+20}.
\] Cross-multiplying gives $x^2+20x = 20^2$ so $x^2+20x-400=0$. Factoring the quadratic gives $(x-20)(x+40) = 0$, so $x= \boxed{20}$. 

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​.

Your question is incomplete.

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]​.

Hint:

Expanded, you an get 

\(x^2+6x>49-5x\) which simplfies into

\(x^2+11x-49>0\)

Which can be factored - You can then draw a graph on a numberline where the 2 roots of the quadratic are points on a parabola, and find the ranges on the numberline for which the Y value of the parabola is positive.

Sorry, I don't know how to draw the graph on web2.0calc.

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*}

Since we defined  g(x)=f(x/2), a real number x is in the domain of g if and only if x/2  is in the domain of f . Therefore, the domain of g consists of all x such that

-8<=x<=8

The solutions to this inequality are given by -16<=x<=16.

so , the domain of g is interval of width 16-(-16)=32.

If $t$ is a real number, what is the maximum possible value of the expression $-t^2 + 8t -4 +5t^2 - 4t + 18$?  

When you combine like terms, you get 4t² plus some other   

stuff, but the square is all we're interested in for this problem. 

The 4t² indicates that this curve is a parabola that opens  .  

upward.  Since it opens upward, there is no maximum value,   

_.

Hint 1: Use the quadratic formula, most notably, the discrimant.

Hint 2: Part a) is a single number, the rest are inequalities.

Here's the answer for part (a):

If you put the 2 equations together, you will get \(x^2-9x+c=8x^2-13x\), from which you will get \(7x^2-4x-c=0\). In order for their to be only 1 solution, the discriminant must be equal to \(0\). The discrimant of any quadratic is \(b^2-4ac\) (It is the part in the square root of the quadratic formula, which is the formula that pops up whenever you press the LaTeX button on your screen.)

\(b^2-4ac = 0\), so

\(16+28c=0\), from which you can solve

\(c=-\frac{16}{28}=-\frac{4}{7}\)

Since AB=AC, △ABC is isosceles, so ∠CAB=∠CBA, and the two medians CD and BE intersect at the incenter I. Also, since CD is a median, D is the midpoint of BC, so BD=14/2=7.

Since △ABC is isosceles, CD is also an altitude of △ABC. Since CD is an altitude, ∠ICD=90∘. Since ∠ICD=90∘, rectangle ICBD is formed, so CI=BD=7. Since IC=7, CDBD​=21​.

Now, since CDBD​=21​, △ICD is similar to △ABC. Since △ICD is similar to △ABC,

ABCI​=BCCD​.

Substituting the given values, we have

257​=14CD​.

Solving for CD, we get CD=2514⋅7​=2598​.

Since CD=2AC​=2598​, AC=2⋅CD=2598​⋅2=25196​.

Since ∠CAB=∠CBA, △ABC is 45-45-90 triangle. Since △ABC is a 45-45-90 triangle, BC=2​⋅AC=2​⋅25196​=5249​​=5142​​.

Since BCCD​=21​,

5142​​2598​​=21​.

Simplifying both sides, we have

272​​=1425​.

Multiplying both sides by 2142​​, we get

72​=7175​.

Squaring both sides, we have

49⋅2=4930625​.

Solving for 2, we get 2=49625​.

Then $BI = 7*sqrt(2)/5$.

Sorry I meant \(z^3+1=0\), and \(z^2-z=-1\).

 Sorry if I confused anyone.