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Answer: 12 

Step-by-step explanation: Because $\overline{PQ}$, $\overline{UV}$, and $\overline{SR}$ are all perpendicular to $\overline{QR}$, we have $\overline{PQ} \parallel \overline{UV} \parallel \overline{SR}$. Therefore, we have $\angle UPQ = \angle UTS$ and $\angle UQP = \angle UST$, which means that $\triangle UPQ \sim \triangle UTS$. So, we have $UQ/US = PQ/ST$.

Because $ST/SR = 2/3$ and $PQ = SR$, we have

\[\frac{UQ}{US} = \frac{PQ}{ST} = \frac{SR}{ST} = \frac{3}{2}.\]Since $UQ/US = 3/2$, we have $UQ/QS = 3/5$.

We have $\triangle UQV \sim \triangle SQR$ by AA Similarity, so $UV/SR = UQ/QS = 3/5$. Therefore, we have $UV = (3/5)SR = \boxed{12}$.

g^{-1}(-7) = 30.

The answer is m = 23.

I will unlock the original posts for you. I only answered one of them.

Please explain what you do not understand about my answer on the original post 

This has been answered in a nearby post.

\( |x| = -\dfrac 1 2 x + 4\\ |x| = \frac {-x+8}{2}\\ \text{square both sides}\\ x^2=\frac{x^2-16x+64}{4}\\ 4x^2=x^2-16x+64\\ 3x^2+16x-64=0\\ x=\frac{-16\pm\sqrt{256+768}}{6}\\ x=\frac{-16\pm32}{6}\\ x=\frac{-8\pm16}{3}\\ x=-8\quad or \quad x=\frac{8}{3}\\ \text{BUT these answers need to be checked for validity.} \)

Check x=-8

RHS= 4+4=8=LHS     works

check x=8/3

RHS= -8/6 +4 = -4/3 +4 = (-4+12)/3 = 8/3 = LHS   also works

LaTex

 |x| = -\dfrac 1 2 x + 4\\
 |x| = \frac {-x+8}{2}\\
\text{square both sides}\\
x^2=\frac{x^2-16x+64}{4}\\
4x^2=x^2-16x+64\\
3x^2+16x-64=0\\
x=\frac{-16\pm\sqrt{256+768}}{6}\\
x=\frac{-16\pm32}{6}\\
x=\frac{-8\pm16}{3}\\
x=-8\quad or \quad x=\frac{8}{3}\\
\text{BUT these answers need to be checked for validity.}

g^-1(7)   does not exist.

Thanks for the answer. 

First, we can find the length of AB. Since AOB is a right angle, we can use the Pythagorean Theorem to get:

AB^2 = OA^2 + OB^2 = 1^2 + 1^2 = 2 AB = sqrt(2)

Similarly, we can find the length of AC and BC:

AC^2 = OA^2 + OC^2 = 1^2 + 1^2 = 2 AC = sqrt(2)

BC^2 = OB^2 + OC^2 = 1^2 + 1^2 = 2 BC = sqrt(2)

Now, let's consider the cube inscribed in the tetrahedron. Let D be the vertex of the cube opposite from O, and let E, F, and G be the vertices of the cube on faces AOB, AOC, and BOC, respectively. Since AOB, AOC, and BOC are all right angles, we know that the line segments DE, DF, and DG are perpendicular to faces AOB, AOC, and BOC, respectively.

Let x be the side length of the cube. Then, we have:

DE^2 = (AB - x)^2 + (OB - x)^2 = (sqrt(2) - x)^2 + (1 - x)^2 DF^2 = (AC - x)^2 + (OA - x)^2 = (sqrt(2) - x)^2 + (1 - x)^2 DG^2 = (BC - x)^2 + (OC - x)^2 = (sqrt(2) - x)^2 + (1 - x)^2

Note that these expressions are all equal, since DE, DF, and DG are simply three different line segments perpendicular to three different faces of the cube. Thus, we have:

(sqrt(2) - x)^2 + (1 - x)^2 = DG^2 = (sqrt(2) - x)^2 + (1 - x)^2

Expanding and simplifying, we get:

2(sqrt(2) - x)^2 + 2(1 - x)^2 = 2 2(3 - 2sqrt(2)x + x^2) + 2(1 - 2x + x^2) = 2 6 - 4sqrt(2)x + 2x^2 + 2 - 4x + 2x^2 = 2 4x^2 - 4sqrt(2)x + 1 = 0

Solving for x using the quadratic formula, we get:

x = (2sqrt(2) ± sqrt(8 - 4))/8 x = (2sqrt(2) ± sqrt(4))/8 x = (2sqrt(2) ± 2)/8

Since x must be positive, we take the positive root:

Since g(x) is a piecewise function, we need to find the inverse of each piece separately.

The inverse of g(x)=5x2 is x=5g(x)​​.

The inverse of g(x)=21+x is x=g(x)−21.

The inverse of g(x)=2−x​ is x=4−4x​.

Therefore, g−1(7) is equal to \begin{align*} \sqrt{\frac{7}{5}} &= \sqrt{\frac{7}{5}} \ 21 + \sqrt{\frac{7}{5}} &= 21 + \sqrt{\frac{7}{5}} \ 4 - 4\sqrt{\frac{7}{5}} &= 4 - 4\sqrt{\frac{7}{5}}. \end{align*}The only solution in the domain of g^{−1}(7) is 4 - 4*sqrt(7/5).

Since the absolute value of any real number is non-negative, it follows that $|x| \geq 0$. Therefore, the equation $|x| = -\frac{1}{2}x + 4$ can only be satisfied if $-\frac{1}{2}x + 4 \geq 0$, which is equivalent to $x \leq 8$.

Now we consider two cases:

Case 1: $x \geq 0$. In this case, $|x| = x$, so the equation becomes $x = -\frac{1}{2}x + 4$. Solving for $x$ gives $x = \frac{8}{3}$.

Case 2: $x < 0$. In this case, $|x| = -x$, so the equation becomes $-x = -\frac{1}{2}x + 4$. Solving for $x$ gives $x = -8$. However, this value does not satisfy the original condition $x < 0$. Therefore, there are no solutions in this case.

Therefore, the only solution is $x = \boxed{\frac{8}{3}}$.  The total number of solutions is 1.

The answer is the interval (-1,1).

The smallest m that works is 5.

We see that g^−1(−7) is the value of x such that g(x)=−7. Since −7 lies in the interval (−3,10), we use the second piecewise definition of g(x), which gives −7=21+x. Solving for x, we find x=−28, so g^−1(−7) = -28​.

Since a1​+a3​+a5​=−12, we can write

$a_1 + a_3 + a_5 = 3 \cdot \frac{a_1 + a_3 + a_5}{3} = 3 \cdot \frac{a_1 + (a_1 + d) + (a_1 + 2d)}{3} = 3a_1 + d^2 = -12$.

Then d2=−36, so d=±6. If d=6, then a1​+a3​+a5​=0, which is a contradiction, so d=−6. Then

a_{10} = a_1 + 9d = a_1 - 54 = -12 - 54 = -66.

That's not correct

im pretty sure one value of n is 128

Since C(n) is always odd, C(C(n)) must be a multiple of 3. Therefore, n must be a multiple of 3. We can check that C(16)=16 and C(C(16))=16, so n=16​ is a solution. We can also check that C(C(C(48)))=16, so n=48​ is also a solution. These are the only solutions, since C(C(C(n))) is always a multiple of 9, and there are no positive integers that are a multiple of 9 and a multiple of 3 but not a multiple of 27.

So the only solutions are 16 and 48.

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