hairyberry

A coefficient of x can only be formed, by only a constant term times, a linear term.

We see that this can occur in two ways:

1*(-23x) = -23x

x(14) = 14x

Summing these up, we get -23x+14x=-9x.

The Coefficient of x is -9.

IGNORE THE OTHER ANSWER!!!

Guys, Sorry, I realized i made a mistake in one of the steps, the answer is not corrrect .

In my solution, i substitued x-10, instead of 10-x for AF. Instead \({AB}^{2}=49+{(10-x)}^{2}-2*7*(10-x)*\cos({60}^{\circ})\). So Simplifying, \({AB}^2={x}^2-13x+79\).

Ssing pythagorean theorem we form a equation, our previous responses, \(37+{x}^2-3x+9={x}^{2}-13x+79\). Solving, we get \(x=\frac{33}{10}\). Ignore the other answer!!! this answer is correct. The answer is 33/10.

Looking at this problem, we don't see we have that many options, so we use our backup plan for any geometry problem: trigonometry. 

Set AE as x. The side length of triangle DEF is 10, so BF = 7, and AF = 10-x.

We want to write some equation, with x, so we look at the triangle in the middle, BCA, and hope to use the pythagorean theorem.

Now in each of these triangles, BDC, AEC, BFA, we have 2 sides and an angle, and we want to find the side opposite to the angle.

Luckily, for us we have the Law of Cosines!

\({BC}^{2}=9+49-2*3*7*\cos({60}^{\circ})\). Simplifying, we get, \({BC}^{2}=37\). 

Similarly, \({AC}^2=9+{x}^{2}-2*3*x*\cos({60}^{\circ})\). Simplifying, we, get \({AC}^{2}=9+{x}^{2}-3x\).

Also, \({AB}^2=49+{(x-10)}^{2}-2*7*(x-10)*cos({60}^{\circ})\). Simplifying, we get, \({AB}^{2}={x}^{2}-27x+219\).

Using pythagorean theorem, we have \(37+{x}^2-3x+9={x}^{2}-27x+219\). Solving and cancelling, we get \(x=\frac{173}{24}\). Our answer is 173/24.

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Simplifying the equation, we get \({x}^{2}+19x+5=0\). We know, that \(a+b=-19, ab=5\), by Vieta's formula.

Lets see if we can use this to our advantage here. A quadratic with roots \(\frac{1}{a+1}, \frac{1}{b+1}\) will have a sum of roots of \(\frac{1}{a+1} + \frac{1}{b+1} = \frac{b+1+a+1}{(a+1)(b+1)}=\frac{a+b+2}{ab+a+b+1}\). Plugging in the values we got from Vieta's, we get \(\frac{-19+2}{5-19+1}=\frac{-17}{-13}=\frac{17}{13}\). 

A quadratic with roots \(\frac{1}{a+1},\frac{1}{b+1}\), will have a product of roots of \((\frac{1}{a+1})(\frac{1}{b+1})=\frac{1}{ab+a+b+1}\). Again, plugging in the values from Vieta's, we get \(\frac{1}{5-19+1}=-\frac{1}{13}\). 

If you're looking for a monic quadratic, a possible answer would be \({x}^{2}-\frac{17}{13}x-\frac{1}{13}\). An equalivent, non monic quadratic is \(13{x}^{2}-17x-1\). Note, there are infinitely many solutions because of different possibilites of leading coefficients.

The volume of the resulting shape, is the area of the cube - 8*the area of each pyramid. (8 corners).

The volume of the cube, is 6*6*6=216.

We see that each pyramid is made out of 3 right isoceles triangles, and one equaliteral. We see that each right isoceles triangle has a leg length of 3, because 2 leg lengths make 6.

To find the volume of each pyramid, we use the formula, \(Volume = \frac{1}{3}base*height\), but we should choose carefully what we want our base and height to be. If we let the equilateral triangle be our base, and the distance from that triangle to the corner be our height, that is some really hard calculations. Instead, we see that each pyramid corner that is cut off is made from 3 right isosceles sides, with one equilateral triangle triangle. If we let an right isoceles triangle be our base, we get much easier calculations.

The area of the base is. \(\frac{3*3}{2}=\frac{9}{2}\), and using the volume formula, each pyramid has an area of \(\frac{1}{3}*\frac{9}{2}*3=\frac{9}{2}\). So the area of the resulting polyhedron is \(216-\frac{9}{2}*8=180\).

What do we know about tan(x) and sec(x)? We know \(\tan(x)=\frac{\sin(x)}{\cos(x)}, sec(x)=\frac{1}{cos(x)} \). Now use these, and replace them in the equation. 

We get \(\frac{sin(x)}{\cos(x)}+\frac{1}{cos(x)}=0\). Multiplying by cos(x), with the restriction, cos(x) is not equal to 0, gives us \(\sin(x)+1=0\), or \(sin(x)=-1\). So, now we use the formula, \({\sin}^{2}(x)+{\cos}^{2}(x)=1\), we get \(cos^2(x)=0, cos(x)=0\). Notice there is only one solution, because +/- 0 is always 0.

But we see from ealier that cos(x) is not 0, so we have no solution.\(\)

We see that triangles EAD~EBF by AA similarity.

But we also know that they have the same area, so they must be congruent.

So, AE = BE = 1/2AB = 1/2DC.

The area of triangle FDC is \(4*\frac{{DC}^{2}}{{EB}^{2}}\), or 4 times the square of the ratio of the bases. And because DC = 2EB, the area of FDC is 4*4=16.

The area of ABCD is 16-4+4 = 16.

Median is the sum of two bases/2.

One base is 12, the other base is 12+16=28.

So the median is \(\frac{28+12}{2}=20\).

Scroll down to see answer, very long.

I'll give this my best attempt 

We see that without the condition, the number of total possibilities of arranging the indistinguishable fruits is \(\frac{8!}{4!4!}=70\).

Now add this condition in, and we need to find the total number of cases that don't satisfy this condition, or otherwise that there is at least one instance of an apple that is surrounded by apples.

Lets split that into 3 sub-conditions 

Condition 1) An apple is first and the second fruit is also an apple

Condition 2) An apple is last and the second to last fruit is also an apple

Condition 3) An apple is not first or last, but the two fruits next to it are apples. (3 apples in a row)

Now we have lots of cases to consider.

Satisfying condition 1, we have \(\frac{6!}{4!2!}=15\) cases for condition 1.

Satisfying condition 2, we also have \(\frac{6!}{4!2!}=15\) cases for condition 2.

Condition 3 is more difficult, because 3 apples in a row, also includes 4 apples in a row, but we overcount, the cases for 4 apples in a row.

First we count the cases for 3 apples which, we treat 3 apples a group, or a seperate variable, so we order the group, one apple, and 4 pears getting \(\frac{6!}{4!1!1!}=30\), but we overcount many cases with 4 apples in a row. Each case with 4 apples in a row can be formed 2 ways, for example consider the sequence PAAAAPPP where P is pear and A is apple. This sequence can be formed by P + AAA +APPP or also PA + AAA+PPP with our counting, so we count each 4 in a row case twice, so we need to subtract once the cases with 4 in a row.

Again, to count the number of 4 in a row cases, we treat the 4 as a group, a seperate item, so there is a total of \(\frac{5!}{4!1!} = 5\) 4 in a row cases. Subtracting this, we get \(30-5=25\) total cases for condition 3.

Now we count the overlaps. We see that there is 1 case of condition 1 and condition 2, AAOOOOAA.

We see there is 5 cases of condition 1 and condition 3. AAA + an A in any of the 5 remaining spaces, yields 5 cases.

We see there is also 5 cases of condition 1 and condition 3. an A in any of the first 5 spaces + AAA, yields an additional 5 cases.

We see there is not possible a case of condition 1 and condition 2 and condition 3, because after AAOOOOAA we can't satisfy condition 3.

So now we have. \(15+15+25-5-5-1=44\)cases of apples being surrounded by apples. 

This is the exact opposite of the case we want to find, so we do total cases - cases of apples being surrounded by apples for a total of \(70-44=26\) cases. Our final answer is 26 arrangements.