lpieleanu

By the British Flag Theorem, if \(X\) is a point in the interior of a rectangle \(ABCD\), then $$AX^2+CX^2=BX^2+DX^2.$$

It follows that $$9+25=16+DX^2 \implies DX^2=18 \implies DX = \boxed{3\sqrt{2}}.$$

Since triangle \(ABC\) is equilateral, \(PX\) is also a median, bisecting side \(QR\). Thus, \(XR=\frac{9}{2}.\) Note that triangle \(YXR\) is a 30-60-90 triangle with right angle at \(Y\) and 30 degree angle at \(\angle{YXR}\). Thus, \(YR = \frac{\frac{9}{2}}{2} = \frac{9}{4}.\) Since the side opposite the \(60^\circ\) angle in a 30-60-90 triangle is \(\sqrt{3}\) times the side opposite the \(30^\circ\) angle, we have \(XY = \frac{9}{4} \cdot \sqrt{3} = \boxed{\frac{9\sqrt{3}}{4}}.\)

Every season is 3 months, so 9 months ago was 3 seasons ago. Since now is fall, one season ago was summer, two seasons ago was spring, and therefore, three seasons ago was winter. 

Actually the answer is C. A describes a family of similar triangles, B can be a parallelogram or a square, and D can be two different triangles.

Actually, the answer is \( (\frac{\frac{4}{5}}{2})^2 = (\frac{2}{5})^2 = \frac{4}{25}.\)

I'm assuming \(C\) is the right angle, \(h\) is \(CH\), and \(x\) and \(y\) are the lengths of \(AH\) and \(HB\)? There's a lot of information missing from the problem...

I claim that the system of equations has no solutions. Assume, for the sake of contradiction, that the equation does have solutions. Subtracting \(q\) from both sides in the second equation yields \(-2r = -2 \rightarrow r=1.\) Subtracting $p$ from both sides in the last equation yields \(r=9,\) contradiction, so the system does not have a solution \((p,q,r).\)

Note that since \(BY\) is both an angle bisector and a median, triangle \(ABC\) is isosceles with \(AB=BC\). Therefore, \(BC = \boxed{16}\). A simple application of the angle bisector theorem (using \(CZ\) as the angle bisector) yields \(\frac{BZ}{AZ} = \frac{BC}{AC} = \frac{1}{2},\) so \(BZ = \frac{1}{3} \cdot 16 = \boxed{ \frac{16}{3}}.\)

Let \(x\) be the total number of people polled. We have the equation \(\frac{7}{12}x = 280\) which yields the solution \(x=480\) so there are \(480\) people in total. Since \(\frac{7}{12}\) of all people are bored by cable news, \(\frac{5}{12}\) are not, so the answer to the problem is \(\frac{5}{12} \cdot 480 = \boxed{200}.\)

Note that there is no way to get the same amount from two different combinations of coins. Thus, it suffices to find the nmber of ways to choose some nonzero number of coins from the coins Steve has where coins of the same value are considered identical.

We can have zero or one quarter, so there are two ways to choose how many quarters.

We can have zero, one, or two nickels, so there are three ways to choose how many nickels.

We can have any where from zero to three pennies, so there are four ways to choose the pennies.

However, note that we included the case when we choose no coins at all, which is not allowed, so we subtract \(1\) from the product of \(2,3,\)and \(4\), to get \(2 \cdot 3 \cdot 4 - 1 = \boxed{23}.\)

1. By the area formula for a triangle, the area is \(\frac{6 \cdot 8}{2} = 24\)

2. Since the sum of the angles in any triangle is \(180^\circ\), we have \(\angle{BAC} = 180^\circ - \angle{ABC} - \angle{ACB} = 180^\circ - 135^\circ - 30^\circ = 15^\circ.\)