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\(2\log{(a -2b)} = \log{a} + \log{b}\\ \log((a-2b)^2)=\log(ab)\\ (a-2b)^2=ab\\ a^2-4ab+4b^2=ab\\ a^2-5ab+4b^2=0\\ \)

You can factor this polynomial like this:

\((a-4b)(a-b)=0\)

So the two solutions for a/b are:

\(a-4b=0\\a=4b\\ \boxed{\frac{a}{b}=4}\)

and

\(a-b=0\\a=b\\\boxed{\frac{a}{b}=1}\)

if \(f(f(x)) = x\) for all x, then f(x) is an involution. The properties of an involution include the fact that it is reflexive across the line y=x, which means that y=ax+b must be a reflected version of y=9-2x through the line y=x. To reflect that line, just find the inverse of that line by changing y for x and solving for y:

\(x=9-2y\\2y=9-x\\y=4.5-0.5x\)

a = -0.5, b = 4.5, so a+b \(\boxed{4}\)

Since we wish to get rid of the irrational numbers in the polynomial, we set the other 2 roots to be 1+sqrt(2) and 2-sqrt(7), since they are the conjugates of those roots, and when an irrational number multiplies the conjugate of that number, it becomes rational. So the monic polynomial is:

\((x-(1-\sqrt{2}))(x-(1+\sqrt{2}))(x-(2+\sqrt{7}))(x-(2-\sqrt{7}))=\\ (x^2-x(1+\sqrt{2})-x(1-\sqrt{2})+(1-\sqrt{2})(1+\sqrt{2}))(x^2-x(2+\sqrt{7})-x(2-\sqrt{7})+(2+\sqrt{7})(2-\sqrt{7}))=\\ (x^2-2x-1)(x^2-2x-3)=\\ \boxed{x^4-6x^3+4x^2+10x+3}\)

Notice that AC is equal to the sum of the radius and EC. We know that AB is the radius, and AB is 5 ft, so AC must be equal to 5 ft + 8 ft = 13 ft.

Since BC is tangent to circle A, AB is perpendicular to BC, so ABC is a right triangle. 

Let BC equal x. To find BC, use the Pythagorean theorem:

\(5^2+x^2=13^2\\25+x^2=169\\x^2=144\\x=\boxed{12}\)

First question: 

The child will only get 10 cents if he selects 2 nickels. The probability of that happening is \(\frac{{6 \choose 2}}{16 \choose 2}=\frac{15}{120} = \boxed{\frac{1}{8}}\)

Second question:

This is basically using the same logic as the previous question. 

X could be equal to 2, 6, 10, 11, 15, or 20, and their probabilities are \(\frac{{5 \choose 2}}{{16 \choose 2}}, \frac{5\cdot6}{{16 \choose 2}}, \frac{{6 \choose 2}}{{16 \choose 2}}, \frac{5\cdot5}{{16 \choose 2}} , \frac{5\cdot6}{{16 \choose 2}}, \) and \(\frac{{5 \choose 2}}{{16 \choose 2}}\), respectively. Multiply the value of X to the corresponding probability to get:

\(2\cdot\frac{{5 \choose 2}}{{16 \choose 2}}+ 6\cdot\frac{5\cdot6}{{16 \choose 2}}+ 10\cdot\frac{{6 \choose 2}}{{16 \choose 2}}+ 11\cdot\frac{5\cdot5}{{16 \choose 2}} + 15\cdot\frac{5\cdot6}{{16 \choose 2}} + 20\cdot\frac{{5 \choose 2}}{{16 \choose 2}}\\= \frac{1}{12}\cdot(2+20)+\frac{1}{8}\cdot10+\frac{1}{4}\cdot(6+15)+\frac{5}{24}\cdot11\\=\boxed{\frac{85}{8}}\)

Let the constant term of the 3rd degree polynomial equal a. Since the constants in the 3rd degree polynomial multiply to equal the constant in the 6th-degree polynomial, \(a^2=81\). Since none of the signs in the 6th-degree polynomial are negative, the polynomials of Jamie and Billy must both have all negative signs or all positive signs, but since it specified monic polynomials, then it must be that all of them are positive, so the answer is just \(\boxed{9}\)

Notice that only 2^n rows (where n is a positive integer) can obtain this property, where the row only contains even numbers except for the beginning and the end. (For example, only row 2^1 = 2, row 2^2=row 4, row 2^3=row 8, and so on). There are 4 numbers that can be expressed as 2^n (where n is a positive integer) under 20 (which are 2, 4, 8, and 16), so the answer is \(\boxed{4}\).

My intuition for the fact that only 2^n rows (where n is a positive integer) can obtain this property uses the fact that the numbers in the pascal triangle can be expressed as \(\frac{n!}{p!(n-p)!}\), where n is the nth row and p is the pth number in that row. All odd numbers obviously don't work, because if p = 1, then the result is just going to be equal to n, and since n is odd, the pascal triangle contains odd numbers. Even numbers that cannot be expressed as 2^n (where n is an integer) also cannot have only even numbers, because it must also have odd factors, and if it has an odd factor, a certain value of p will cancel out all the even factors and leave the number odd, so only 2^n would work.

Also a very similar question was answered a long time ago: https://web2.0calc.com/questions/another-pascal-s-triangle-question

How about when the exponent is equal to zero (when x is equal to \(\pm 5\))?

The expression \(-n^2+22n-21\) will only be positive if \(1\leq x\leq21\). (If it's negative then the left-hand side is undefined), which means that both sides of the expression must be positive, so we can square both sides without worrying about changing signs.

\(49(-n^2+22n-21)\leq n^2+78n+1521\\-49n^2+1078n-1029\leq n^2+78n+1521\\ 0\leq50n^2-1000n+2550\\0\leq n^2-20n+51\\0\leq (x+3)(x+17)\)

The zeroes of the quadratic of the left-hand side are -3 and -17, which means that the answer is \(\boxed{15}\)

If \(f(x) = y\), for 2 numbers x and y, then \(f^{-1}(y)=x\).

Now that we know that, we can find the value of \(f^{-1}(4)\), which is just 3.

Also, \(f(5) =1\) (pretty straightforward).

Can you take it from there?