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Questions in mathematics

📝 Answered - The set of $\qquad$ numbers is $\{0,1,2,3,4,5, \ldots\}$.

📝 Answered - Question 2: Calculate the maximum height (MH) of a cricket ball if its height is given by the function [tex]$h(x)=2 x^3-12 x^2+1$[/tex] using the second derivative test. a) [tex]$MH =4$[/tex] b) [tex]$MH =3$[/tex] c) [tex]$MH =2$[/tex] d) [tex]$MH -1$[/tex]

📝 Answered - 18. [tex]x^4-5 x^3 \leq x^2-5 x[/tex]

📝 Answered - Point P is on line segment [tex]$O Q$[/tex]. Given [tex]$O P=6, O Q=4 x-3$[/tex], and [tex]$P Q=3 x$[/tex], determine the numerical length of [tex]$\overline{O Q}$[/tex].

📝 Answered - \begin{tabular}{|l|l|} \hline $x$ & $f(x)$ \\ \hline -2 & 4 \\ \hline 0 & 5 \\ \hline 2 & $a$ \\ \hline 3 & 7 \\ \hline \end{tabular} $f$ is an even function. $a=?

📝 Answered - Find the sum: $\frac{x-2}{x^2+1}+\frac{x+3}{x^2+1}$ A. $\frac{2 x+1}{2 x^2+2}$ B. $\frac{1}{x^2+1}$ C. $\frac{2 x+1}{x^2+1}$ D. $\frac{2}{x}$

📝 Answered - The base of a solid right pyramid is a regular hexagon with a radius of $2 x$ units and an apothem of $x \sqrt{3}$ units. Which expression represents the area of the base of the pyramid? A. $x^2 \sqrt{3}$ units ${ }^2$ B. $3 x^2 \sqrt{3}$ units ${ }^2$ C. $4 x^2 \sqrt{3}$ units ${ }^2$ D. $6 x^2 \sqrt{3}$ units $^2$

📝 Answered - A scatterplot is used to display data where [tex]$x$[/tex] is the amount of time, in minutes, one member can tolerate the heat in a sauna, and [tex]$y$[/tex] is the temperature, in degrees Fahrenheit, of the sauna. Which interpretation describes a line of best fit of [tex]$y=-1.5 x+173$[/tex] for the data? A. The member can tolerate a temperature of [tex]$173^{\circ}$[/tex] Fahrenheit for 0 minutes. B. The amount of time the member can tolerate the heat in a sauna is 173 minutes. C. The time increased 1.5 minutes for every degree Fahrenheit the temperature increased. D. The time decreased 1.5 minutes for every degree Fahrenheit the temperature decreased.

📝 Answered - Which equation correctly uses the law of cosines to solve for $y$? $9^2=y^2+19^2-2(y)(19) \cos \left(41^{\circ}\right)$ $y^2=9^2+19^2-2(y)(19) \cos \left(41^{\circ}\right)$ $9^2=y^2+19^2-2(9)(19) \cos \left(41^{\circ}\right)$ $y^2=9^2+19^2-2(9)(19) \cos \left(41^{\circ}\right)$ Law of cosines: $a^2=b^2+c^2-2 b c \cos (A)$

📝 Answered - Water flows through a pipe at a rate of 1900 cubic feet per day. Express this rate of flow in quarts per second. Round your answer to the nearest hundredth. *Note: you must use these exact conversion factors to get this question right. \begin{tabular}{|l|l|} \hline Volume & Time \\ \hline 1 cup (cup) $=8$ fluid ounces (floz) & 1 minute $(\min )=60$ seconds $( sec )$ \\ \hline 1 pint (pt) $=2$ cups (cups) & 1 hour $( hr )=60$ minutes $( min )$ \\ \hline 1 quart ( qt ) $=2$ pints ( pt ) & 1 day (day) $=24$ hours (hr) \\ \hline 1 gallon (gal) = 4 quarts ( qt ) & 1 week (week) = 7 days (days) \\ \hline 1 cubic foot $\left( ft ^3\right)=7.481$ gallons (gal) & 1 month (month) $=30$ days (days) \\ \hline 1 liter (L) = 1000 milliliters (mL) & 1 year (year) $=365$ days (days) \\ \hline 1 cubic meter $\left(m^3\right)=1000$ liters (L) & \\ \hline 1 gallon (gal) $=3.785$ liters (L) & \\ \hline 1 fluid ounce (fl oz) $=29.574$ milliliters (mL) & \\ \end{tabular}