proyaop

\(x(x+6)>x^2-x+30\) 

\(x^2+6x>x^2-x+30\)

\(6x>-x+30\)

\(7x>30\)

\(x>{30\over{7}}\)

In interval notation, your answer is (30/7, inf)

Let the midpoint of AB be M and the midpoint of BC be N, let the center of the smaller circle be O, and the point tangent to circle O with arc BC be point P:

Common fallacies:

NOM is a right triangle with angle NOM = 90 degrees.

P, O, and N are collinear.

These assumptions should not be made, especially when our diagram is not to scale- also be clear in your statements of your variables so that you also don't get lost in your work... One can quickly use the Law of Cosines to see that r cannot be 2... good attempt still Boseo!! 

Hey boseo--- there are a couple of inaccuracies here... If O is the center of the (smaller) circle, and D by your construction is the midpoint of BC, BOD is not necessarily a right triangle and thus you cannot operate pythagorean theorem. Additionally, DO is clearly not r, but r + 3.

Additionally, COD is most definitely not isosceles, and there appears to be a random pythagorean again... 

OH! Thanks for catching the error CPhill!!! I thought a and b were the roots!!! Welp kelhaku, if you want to know Vieta's it's above, but if you want the right answer read CPhill's response--- 

\(y\) is a function of \(x\), so \(y = f(x) \):

\(f(-3)=9a-3b+c=8\)                               Equation 1

\(f(-1)=a-b+c=5\)                                   Equation 2

\(f(2)=4a+2b+c=11\)                                Equation 3

This is a system of 3 equations and 3 variables, so it is solvable (and there is no redundancy).

Let's get solving!

First we want to change it into a system of 2 variables and 2 equations by manipulating to get rid of the variable c via elimination.

Equation 1 - Equation 2: \(9a-3b+c-(a-b+c)=8-5=8a-2b=3\)                 Equation 1.2

Equation 3 - Equation 2: \(4a +2b + c - (a-b+c)=11-5=3a+3b=6\); \(a+b=2\)       Equation 3.2

Then, we can eliminate b to find the value of a: Equation 1.2 + 2 * (Equation 3.2):

\(8a-2b+2(a+b)=3+2*2=10a=7\); \(a = {7\over{10}}\)                                 Value of \(a\)

Substitute this back into our equation \(a + b = 2\):

\({7\over{10}}+b=2={20\over{10}}\); \(b={20\over{10}}-{7\over{10}}={13\over{10}}\)                                    Value of \(b\)

Substitute both variables' values into the equation \(a - b + c = 5\) (it doesnt matter which equation you choose; take the easier one).

\({7\over{10}}-{13\over{10}}+c=5={50\over{10}}\); \(c={50\over{10}}+{13\over{10}}-{7\over{10}}={56\over10}\)                       Value of \(c\)

\(a+b+c={7\over10}+{13\over10}+{56\over{10}}={76\over{10}}=7.6\)

\(\frac{x+4}{x+5} = \frac{x-3}{2x + 8} - 7\)

\((x+4)(2x+8)=(x-3)(x+5)-7(2x+8)(x+5)\)

\(2x^2+16x+32=x^2+2x-15-7(2x^2+18x+40)\)

\(2x^2+16x+32=x^2+2x-15-14x^2-126x-280=-13x^2-124x-295\)

\(15x^2+140x+327=0\)

\(x = {-140\pm\sqrt{-20}\over{30}}={-70\pm\sqrt{5}i\over{15}}\)

Combine like terms: \(19x^2-6x+31=0\)

By vieta's formulas, the sum of the roots of a quadratic ax^2 + bx + c = -b/a, and their product = c/a.

Here, our roots, A and B, have a product of the term c/a = 31/19, thus A * B = 31/19 (capitalized such that it isn't confused for the coefficients of the quadratic.)

Combine like terms: \(x^2+23x+k\)

Because this is a quadratic, and the problem states it has a double root, we know the expression can be factored to \((x-r)^2\).

When expanded, it is expanded to \(x^2-2xr+r^2\). Setting these expressions equal to each other, we get -2xr = 23x, r = -23/2.

We also get that r^2 = k, so k = (-23/2)^2 = 529/4.

If there are 3 skaters, including 3 Americans, in how many ways can the medals be awarded to three of the 3 skaters if exactly one of the Americans wins a medal...

As you can see, there are only three competitors of which all are American, so assuming none get disqualified and complete it, we will have 3 winners, each American, so the question conradicts itself saying that only exactly one American can win a medal...

0 ways.

Law of Cosines Method: Trigonometry Rules!!!

Let the center of the semicircle be P, and the center of the small circle be Q. Additionally, let the midpoint of AB = M, and the midpoint of BC = N. I'm about to spit out constructions, so try to follow along:

We can construct a triangle MQP, with side lengths MQ = 2 + r, QP = 5 - r, and PM = 5 - 2 = 3.

Additionally, triangle NQP, with side lengths NQ = 3 + r, QP = 5 - r, and PN = 5 - 3 = 2.

Using the Law of Cosines (which I expect that you should know already) on NQP:

\((3 + r)^2 = (5 - r)^2+2^2-2(5-r)(2)\cos({QPN})=r^2+6r+9=r^2-10r+29-(20-4r)\cos({QPN})\)

\({{16r-20}\over{4r-20}}=\cos({QPN})={4r-5\over{r-5}}\)

Using the Law of Cosines on MQP: (and substitution)

\((2+r)^2=3^2+(5-r)^2-2(3)(5-r)\cos({QPM})=r^2+4r+4=r^2-10r+34+{(30-6r)(4r-5)\over{(r-5)}}\)

\(14r-30={-(6r-30)(4r-5)\over{(r-5)}}=-6(4r-5)=-24r+30\)

Note that cos(QPM) = -cos(QPN)

Solving this equation, you get 38r = 60, and r = 30/19.