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Questions in Grade High-school

[Done] 7. $5 \underline{x}-4=21$ 8. $42=6+9 x$ 9. $8-2 x=10$ 10. $22=3 x-5$ 11. $\overrightarrow{5}=5 x-20$ 12. $5 x-3=28$

[Done] Substitute the given values into the formula, and [tex]A=\frac{1}{2} h(B+b)[/tex] ; A=56, B=4, b=3

[Done] 8) $45 x^2+120 x+80$

[Done] $3 x^2 - (9x - 8x^2) + 4x - 4

[Done] Blake is doing an experiment to find the relationship between string tension [tex]F_T[/tex], string linear density [tex]D[/tex], frequency of oscillation ([tex]f[/tex]) and wavelength ([tex]\lambda[/tex]) for a wave on a string. The student keeps the string tension and density constant. He sets the frequency of the string's oscillation and measures the wavelength. The known relationship is: [tex]\sqrt{\frac{F_T}{D}}=f \lambda[/tex] Rearrange the equation so the dependent variable is by itself on one side. What variables could he graph to get a linear relationship? x : y : Fill in the data table below with any additional data you need to linearize it.

[Done] Keyaree is able to swim [tex]$3 \frac{2}{5}$[/tex] laps every 4 minutes. Assuming that Reyaree swims at a constant rate, how many laps (or fractions of a lap) does she swim per minute?

[Done] Which fraction represents the decimal $0.8888 \ldots$ ? A. $\frac{80}{99}$ B. $\frac{1,111}{1,250}$ C. $\frac{8}{9}$ D. $\frac{9}{8}$

[Done] The distance it takes a truck to stop can be modeled by the function [tex]d(v)=\frac{2.15 v^2}{64.4 f}[/tex] [tex]d=[/tex] stopping distance in feet [tex]v=[/tex] initial velocity in miles per hour [tex]f=[/tex] a constant related to friction When the truck's initial velocity on dry pavement is 40 mph, its stopping distance is 138 ft. Determine the value of [tex]f[/tex], rounded to the nearest hundredth.

[Done] Conversion Practice To convert to a smaller unit, move decimal point to the right or multiply \begin{tabular}{|c|c|c|c|} \hline \begin{tabular}{c} Kilo- (k) \\ 1000 \\ units \end{tabular} & \begin{tabular}{c} Hecro- (h) \\ 100 \\ units \end{tabular} & \begin{tabular}{c} Deka- (D) \\ 10 \\ units \end{tabular} & \begin{tabular}{c} Basic \\ Unit \end{tabular} \\ \hline \begin{tabular}{c} To convert to a larger unit, move \\ decimal point to the left or divide \end{tabular} & \begin{tabular}{c} Gram (g) \\ Meter (m) \\ Liter (L) \end{tabular} \\ \hline \end{tabular} Try these conversions using the ladder method. [tex]$\begin{array}{l} 1000 mg=\frac{1.0}{160} g \\ 160 cm=\frac{169}{109 g}=\frac{1.109}{1.2467} Dm \\ 2467 cm=\frac{0.25}{4.5} cm \\ 45 hm= \end{array}$[/tex] [tex]$\begin{array}{l} 1 L=1,000 \\ 14 km={ }^{250 m} \\ 250 m \\ 1,456,980 mg={ }^{m} kg \\ 5 L=5,0000 kL \end{array}$[/tex] [tex]$\begin{array}{l} 35.465 g={ }^{mg} \\ 3.4 L=\frac{0.34}{1 Dm=} \frac{dL}{m} \\ 35.25 dL={ }_{1.050}^{cm} \\ 1000 km= \end{array}$[/tex]

[Done] If [tex]f(x)=16 x-30[/tex] and [tex]g(x)=14 x-6[/tex], for which value of [tex]x[/tex] does [tex](f-g)(x)=0[/tex]?