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\((-1)^{-3} \cdot 1^3 + (-3)^{-5} \cdot 3^5 + (-5)^{-7} \cdot 5^7\\ =-\frac{1}{1^3}\cdot1^3-\frac{1}{3^5}\cdot 3^5-\frac{1}{5^7}\cdot 5^7\\ =-1-1-1\\ \color{blue}=-3\)

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I actually found another solution on the website I got the problem from. Before I noticed that if two points are chosen that are at most a distance of 1 unit apart from each other, the angle that they make with the center point of the circle is at most 60 degrees (equilateral triangle). When I looked at the solution, I realized that that idea was actually a part of the solution. Let's say A is on the right side of the circle. That means point B could be chosen either above or below point A. This creates an arc of 120 degrees, and since there are 360 degrees in a circle, the correct solution is 120/360 = 1/3. Thank you all for giving your ideas!

A = 1 and B = -1, so A/B = -1.

The area of triangle XYD is at most 24 if and only if the height of triangle XYD is at most 2. The height of triangle XYD is at most 2 if and only if point D is within the shaded region below:

[asy] unitsize(1 cm);

pair X, Y, Z, D;

X = (0,0); Y = (12,0); Z = (6,6sqrt(3)); D = (6,3sqrt(3));

draw(X--Y--Z--X); draw(rightanglemark(X,Y,D,10)); draw(rightanglemark(Y,D,Z,10)); draw(X--D); draw(D--Y);

label("X", X, SW); label("Y", Y, NE); label("Z", Z, NW); label("D", D, S); [/asy]

The shaded region is a right triangle with legs of length 6 and 3, so its area is 1/2(6)(3)=9. Therefore, the probability that the area of triangle XYD is at most 24 is 9/12 = 3/4​​.

The sum is 9^(1/3)/4.

Let $s$ be the side length of equilateral triangle $XYZ$, and let $s'$ be the side length of equilateral triangle $X'Y'Z'$. Since the dilation centered at $G$ with scale factor $-3/4$ sends $X$, $Y$, and $Z$ to $X'$, $Y'$, and $Z'$, respectively, we have $GX' = -3/4 \cdot GX$, $GY' = -3/4 \cdot GY$, and $GZ' = -3/4 \cdot GZ$. Since $G$ is the center of equilateral triangle $XYZ$, we have $GX = GY = GZ = s/\sqrt{3}$.

Let $P$ be the intersection of lines $XZ$ and $Y'Z'$. Since $XZ$ and $Y'Z'$ are parallel, we have $\triangle GPZ' \sim \triangle GPX$, and it follows that $GPZ' = (3/4) GPX$. Similarly, we can show that $GPY' = (3/4) GPX$. Therefore, we have

A=[GPY′XZ′]=[GPY′]+[GPX]+[PZ′Y′X]=3/4[GPX]+[PZ′Y′X]

To find $[GPX]$, note that $\triangle GPX$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle with hypotenuse $GP = s/\sqrt{3}$. Therefore, we have $[GPX] = (1/2) \cdot (s/\sqrt{3}) \cdot (s/2) = s^2/(4\sqrt{3})$.

To find $[PZ'Y'X]$, note that $PZ' = 3/4 \cdot GX = -3s/(4\sqrt{3})$, and $Y'X = XZ = s$. Therefore, we have $[PZ'Y'X] = (1/2) \cdot (-3s/(4\sqrt{3})) \cdot s = -3s^2/(8\sqrt{3})$.

Substituting these values, we get

A=​ 3/4*s^2/(4*sqrt(3)) - 3s^2/(8*sqrt(3)) = s^2/(8*sqrt(3))

Therefore, we have $A/[XYZ] = A/(s^2\sqrt{3}/4) = 4/(8\sqrt{3}) = sqrt(3)/6.

Let $E$ be the vertex of the cube that lies on face $ABC$, and let $F$ be the opposite vertex of the cube. Let $s$ be the side length of the cube.

First, note that $\angle AOB = \angle AOC = \angle BOC = 90^\circ$, which implies that $ABC$ is a right triangle with right angle at $O$. Without loss of generality, assume that $AB < AC < BC$. Then, we have $AE = AF = s$, $BE = b - s$, $CE = c - s$, and $OE = a - s$. By the Pythagorean theorem, we have

\begin{align*} AE^2 + OE^2 &= OA^2 \ s^2 + (a-s)^2 &= a^2 \ 2as - s^2 &= a^2 - s^2 \ s^2 &= \frac{a^2}{2} - as \end{align*}

Similarly, we have

\begin{align*} BE^2 + OE^2 &= OB^2 \ (b-s)^2 + (a-s)^2 &= b^2 \ 2bs - 2as + a^2 + b^2 - 2as &= b^2 - s^2 \ s^2 &= 2as - 2bs + a^2 + b^2 - b^2 \ s^2 &= a^2 - 2as + b^2 \ s^2 &= \frac{b^2}{2} - as \end{align*}

and

\begin{align*} CE^2 + OE^2 &= OC^2 \ (c-s)^2 + (a-s)^2 &= c^2 \ 2cs - 2as + a^2 + c^2 - 2as &= c^2 - s^2 \ s^2 &= 2as - 2cs + a^2 + c^2 - c^2 \ s^2 &= a^2 - 2as + c^2 \ s^2 &= \frac{c^2}{2} - as \end{align*}

Adding the three equations, we get

3s^2 = a^2/2 + b^2/2 + c^2/2 - 3as

Solving for s, we obtain

s = (a + b + c) - sqrt(2)/2*sqrt(a^2 + b^2 + c^2 - ab - ac - bc).

Therrefore,

s = abc/(ab + ac + bc),

as desired.

4x-3+5=2x+7-8x      what is the solution for x in the equation?  

                                                                  4x – 3 + 5  =  2x + 7 – 8x    

This looks pretty straightforward. 

1st, put like terms on the same side.         4x – 2x + 8x  =  7 + 3 – 5   

Remember, when you move something    

to the other side of the equals mark,   

you have to change its polarity.  

Combine the like terms.                             10x  =  5  

Divide both sides by 10.                               x  =  1 / 2    (read as "x equals one-half")    

(five-tenths reduces to one-half)   

Check answer.                                   (4)(1/2) – 3 + 5  =  (2)(1/2) + 7 – (8)(1/2)  

                                                                    2 – 3 + 5  =  1 + 7 – 4   

                                                                                4  =  4       

NEVER trust an answer that doesn't show the work, like answer #1.   

This site is full of trolls and a** h***s ... that was a rhyme.  

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