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Questions in mathematics

[Free] If $f(x)$ and its inverse function, $f^{-1}(x)$, are both plotted on the same coordinate plane, where is their point of intersection? A. $(0,6)$ B. $(1,4)$ C. $(2,2)$ D. $(3,0)

[Free] Solve and check $\frac{c-4}{c-2}=\frac{c-2}{c+2}-\frac{1}{2-c}$ The solution is $c=$ $\square$ How many extraneous solutions are there? $\square$

[Free] What is the value of $\left(\frac{3}{4}-\frac{1}{2}\right)+5.25$ expressed as a fraction?

[Free] What is the range of the function $f(x)=-2\left(6^x\right)+3$? A. $(-\infty,-2]$ B. $(-\infty, 3)$ C. $[-2, \infty)$ D. $[3, \infty)$

[Free] Line segment $PQ$ is a directed line segment beginning at $P(6, -5)$ and ending at $Q(-2, 4)$. Find point $R$ on the line segment $PQ$ that partitions it into the segments $PR$ and $RQ$ in the ratio $3:2$. A. $\left(\frac{6}{5}, \frac{2}{5}\right)$ B. $\left(\frac{14}{5},-\frac{7}{5}\right)$ C. $\left(-\frac{6}{6}, \frac{2}{5}\right)$ D. $\left(\frac{14}{6}, \frac{7}{5}\right)$

[Free] [tex]\frac{3}{17}+6[/tex]

[Free] Which sequence of transformations could be used to map quadrilateral RSTU onto R"S"T"U"? A. [tex]R_{0,90^{\circ}} \circ T_{(-3,-1)}[/tex] B. [tex]T_{(-3,-1)} \circ R_{0,90^{\circ}}[/tex] C. [tex]R_{0,270^{\circ}} \circ T_{(3,1)}[/tex] D. [tex]T_{(3,1)} \circ R_{0,270^{\circ}}[/tex]

[Free] A triangle has side lengths measuring [tex]$2 x+2 ft$[/tex], [tex]$x+3 ft$[/tex], and [tex]$n ft$[/tex]. Which expression represents the possible values of [tex]$n$[/tex], in feet? Express your answer in simplest terms. [tex]$x-1\ \textless \ n\ \textless \ 3 x+5$[/tex]

[Free] What is the difference? $\frac{2 x+5}{x^2-3 x}-\frac{3 x+5}{x^3-9 x}-\frac{x+1}{x^2-9}$ $\frac{(x+5)(x+2)}{x^3-9 x}$ $\frac{(x+5)(x+4)}{x^3-9 x}$ $\frac{-2 x+11}{x^3-12 x-9}$ $\frac{3(x+2)}{x^2-3 x}$

[Free] The one-to-one functions [tex]g[/tex] and [tex]h[/tex] are defined as follows. [tex] \begin{array}{l} g=\{(-8,7),(-7,-2),(4,-8),(6,1)\} \\ h(x)=4 x-13 \end{array} [/tex] Find the following. [tex] \begin{array}{l} g^{-1}(-8)=4 \\ h^{-1}(x)=\frac{x+13}{4} \\ \left(h \circ h^{-1}\right)(-5)=\square \\ \hline \end{array} [/tex]