Chapter 9: Problem 38
Simplify the expression. $$\frac{\sqrt{9}}{\sqrt{49}}$$
Chapter 9: Problem 38
Simplify the expression. $$\frac{\sqrt{9}}{\sqrt{49}}$$
All the tools & learning materials you need for study success - in one app.
Get started for freeCHANGING C-VALUES In Exercises 21-23, find values of \(c\) so that the equation will have two solutions, one solution, and no real solution. Then sketch the graph of the equation for each value of \(c\) that you chose. $$2 x^{2}+3 x+c=0$$
You see a firefighter aim a fire hose from 4 feet above the ground at a window that is 26 feet above the ground. The equation \(h=-0.01 d^{2}+1.06 d+4\) models the path of the water when \(h\) equals height in feet. Estimate, to the nearest whole number, the possible horizontal distances \(d\) (in feet) between the firefighter and the building.
Solve the equation or write no solution. Write the solutions as integers if possible. Otherwise write them as radical expressions. $$6 x^{2}-54=0$$
Use the following information. Scientists simulate a gravity-free environment called microgravity in free- fall situations. A similar microgravity environment can be felt on free-fall rides at amusement parks or when stepping off a high diving platform. The distance \(d\) (in meters) that an object that is dropped falls in \(t\) seconds can be modeled by the equation \(d=\frac{1}{2} g\left(t^{2}\right),\) where \(g\) is the acceleration due to gravity (9.8 meters per second per second). How are these formulas similar? \(d=\frac{1}{2} g\left(t^{2}\right)\) when \(d\) is distance, \(g\) is gravity, and \(t\) is time \(h=-16 t^{2}+s\) when \(h\) is height, \(s\) is initial height, and \(t\) is time
INTERPRETING THE DISCRIMINANT Consider the equation \(\frac{1}{2} x^{2}+\frac{2}{3} x-3=0\) How many solutions does the equation have?
What do you think about this solution?
We value your feedback to improve our textbook solutions.