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At first glance, it seems like Heron's formula is needed, but notice that \(\frac{15\cdot20}{2} = 150\). That means 15 and 20 are the length of the legs of a right triangle.

The hypotenuse of that triangle is \(\sqrt{15^2+20^2}=25\), using the Pythagorean theorem.

The perimeter of ABC is therefore equal to \(15+20+25=\boxed{60}\)

First, to find out what k is, just use the remainder theorem: if x=2 is a root, then when you evaluate the polynomial at 2 you should get zero. Plugging 2 in for x:

\(2\cdot2^3 - 9\cdot2^2+13\cdot2+k = 0\\6+k=0\\k=-6\)

Knowing that, just use polynomial division to divide this into a quadratic:

\(\hspace{2.35cm}2x^2 - 5x + 3 \\x-2|\overline{2x^3-9x^2+13x-6}\\ \hspace{1.15cm}2x^3 - 4x^2\\ \hspace{2cm} \overline{-5x^2+13x}\\ \hspace{1.8cm} -5x^2 + 10x\\ \hspace{3.75cm} \overline{3x-6}\\ \hspace{3.75cm}3x-6 \\ \hspace{4.75cm} \overline{0}\)

Then, factor the quadratic:

\(2x^2-5x+3 = (2x-3)(x-1)\)

Therefore, the two other roots are \(\boxed{\frac{3}{2}, 1}\)

First, simplify the terms in the numerator of the fraction.

\(\frac{5}{\sqrt{80}} = \frac{5\sqrt{80}}{80} = \frac{5\cdot4\sqrt{5}}{80}=\frac{\sqrt{5}}{4}\)

\(\frac{\sqrt{845}}{9} = \frac{13\sqrt{5}}{9}\)

\(\sqrt{45} = 3\sqrt{5}\)

The numerator then becomes

\(\frac{\sqrt{5}}{4} + \frac{13\sqrt{5}}{9} + 3\sqrt{5}\)

Since the denominator contains \(\sqrt{5}\), we can just divide that from the numerator:

\(\frac{1}{4} + \frac{13}{9} + 3 = \frac{169}{36}\)

Now, just take the square root of that to get the final answer:

\(\sqrt{\frac{169}{36}} = \frac{\sqrt{169}}{\sqrt{36}} = \boxed{\frac{13}{6}}\)

The number of 10-digit binary numbers that end with 2 ones is \(2^7=128\), since the first digit must be 1, the last 2 digits must be 1, and the middle 7 digits can be either 1 or 0.

The number of 10 digit binary numbers that start with 2 ones is \(2^8 = 256\), since the first 2 digits must be 1 and the other 8 digits can be either 1 or 0.

If you add the 2 numbers together, you get \(128+256 = 384\), but that is not the answer because the number of 10-digit binary numbers satisfying both conditions is overcounted and needs to be subtracted.

The number of 10-digit binary numbers that satisfy both conditions is \(2^6 = 64\), since the first 2 and the last 2 digits must be equal to 1 and the middle 6 digits can be equal to either 1 or 0. 

Therefore, the answer is \(384-64=\boxed{320}\)

Use Simon's favorite factoring trick; it's going to make it so much easier:

\((a+2)(b+2)=-12\\ (a+2)(c+2)=30\\ (b+2)(c+2)=-10\)

Multiply all the equations to get:

\(((a+2)(b+2)(c+2))^2=3600\\ (a+2)(b+2)(c+2)= \pm 60\)

In the first case where the right-hand side of the equation above is positive, to solve for a, divide \((a+2)(b+2)(c+2)=60\) by \((b+2)(c+2)=-10\) to get \(a+2=-6 \rightarrow a=-8\). Do that the same with b and c to get \(b = 0\) and \(c=-7\).

In the second case where the right-hand side of the equation is negative, do the exact same thing mentioned above to get \(a = 4, b = -4, c = 3\).

Therefore, the solutions to this system of the equation are:

\(\boxed{a = -8, b = 0, c = -7 \text{ or } a = 4, b = -4, c = 3} \)

I'm really sorry that I got it wrong. Could you tell me the answer so I can see where I made a mistake?

Edit: It seems like you mistyped the problem according to this:

https://web2.0calc.com/questions/help_12091

The solution would have been:

\(x(\sqrt{2}+1)+y(\sqrt{2}+1)=\frac{1}{2}(\sqrt{2}+1)\\ x+y=\frac{1}{2}\)

Since the sum of the circumference of the circle is 2 times the radius times pi, just multiply 2pi to get the final answer:

\(\frac{1}{2} \cdot 2 \cdot \pi = \boxed{\pi}\)

ElectricPavlov nice solution!

Solution for 2:

Since the perimeter of the square is \(2\sqrt{2}\), the side length of one side of the square is \(\frac{\sqrt{2}}{2}\). The diagonal of that square is then equal to \(\frac{\sqrt{2}}{2} \cdot \sqrt{2} = 1\)

Let x be the radius of one circle, and let y be the radius of the other circle. 

Notice that the diagonal is equal to \(x\sqrt{2} + x + y\sqrt{2} + y\). Therefore, 

\(x\sqrt{2} + x + y\sqrt{2} + y = 1\\ x(\sqrt{2}+1)+ y(\sqrt{2}+1)=1\\ x+y=\frac{1}{\sqrt{2}+1} = \sqrt{2}-1\)

Since the sum of the circumference of the circle is 2 times the radius times pi, just multiply 2pi to get the final answer:

\(\boxed{2\sqrt{2}\pi-2\pi}\)

The numerator does not matter in this case; only if the denominator is zero for some real number, the domain will not be set for all real numbers.

To prevent the denominator from having real roots, the discriminant must be negative, which means \(2^2-4\cdot-5\cdot k = 4+20k\) is negative.

That will only happen if \(\boxed{k<-\frac{1}{5}}\).

A regular decagon is rotationally symmetrical by 360/10 = 36 degrees.

To take I to E, that rotation needs to be done 4 times, so alpha is equal to \(4(36)=\boxed{144}\)