Algebra Questions from Dec 18,2024

Browse the Algebra Q&A Archive for Dec 18,2024, featuring a collection of homework questions and answers from this day. Find detailed solutions to enhance your understanding.

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An object is thrown upward at a speed of 140 feet per second by a machine from a height of 12 feet off the ground. The height \( h \) of the object after \( t \) seconds can be found using the equation \( h=-16 t^{2}+140 t+12 \) When will the height be 217 feet? When will the object reach the ground? Question Help: Video An object is thrown upward at a speed of 68 feet per second by a machine from a height of 10 feet off the ground. The height \( h \) of the object after \( t \) seconds can be found using the equation \( h=-16 t^{2}+68 t+10 \) When will the height be 54 feet? When will the object reach the ground? One side of a rectangle is 10 in longer than two times another side. The area of the rectangle is \( 408 \mathrm{in}^{2} \). Find the length of the shorter side. \( \square \) Solve the following equation for the indicated variables. \( b=\sqrt{\frac{16 a T}{d}} \) for \( T \) \( T=\square \) Question Help: \( \square \) Message instructor Submit Ouestion Regroup the like terms. \( \begin{aligned} \frac{1}{4} x+1+\frac{3}{4} x-\frac{2}{3}-\frac{1}{2} x & =\frac{1}{4} x+1+\frac{3}{4} x+\left(-\frac{2}{3}\right)+\left(-\frac{1}{2} x\right) \\ & =\left(\frac{1}{4} x+\frac{1}{2} x+-\frac{1}{2} x\right)+\left(\frac{1}{3}+-\frac{2}{3}\right)\end{aligned} \) Solve the equation for the indicated variable. \( V=\frac{r B}{e^{2}} \) for \( e \) The two answers are \( \square \) Question Help: \( \square \) Message instructor Submit Question 4) Write an expression that is equi 4egroup the like terms. \( \begin{aligned} \frac{1}{4} x+1+\frac{3}{4} x-\frac{2}{3}-\frac{1}{2} x & =\frac{1}{4} \\ & =\end{aligned} \) b) \( \sqrt[6]{x+16}=x-4 \) Solve the formula for the specified variable. \( S=24 \pi r^{2} \) for \( r \). The two answers are \( \square \) and Question Help: \( \square \) Message instructor b) The \( \mathrm{m}<2=6 \mathrm{x} \) and the \( \mathrm{m}<7=2 \mathrm{x}+40 \). Find \( \mathrm{m}<7 \)
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