Chrystal L Jones
MAT222: Intermediate Algebra (ACR1442F)
Instructor: James Condor
November 3, 2014
Changing the Real World with Radical Formulas
During the course of this class, we have ventured into solving proportions and two-variable inequalities. Now we get to solve and evaluate “real world radical formulas”, to be perfectly honest I don’t when I will ever have to use any of these algebraic aspects but I trying my hardest to comply with the class and do my best to complete all the given tasks.
There are several “real” world applications that use radical formulas. Radical formulas are very useful in making it easier to arrive at a correct answer for complicated problems. Radical formulas help by giving a more creative way to get to the correct answer with fewer steps. This week’s assignment was on page 605 and was #103 from the textbook (Dugopolski, 2012). The problem has three parts which I will solve and show my work step by step.
a.) For the first part of this problem, we are asked to find the capsize screening value when the displacement is 23,245 lbs. and a beam of 13.5 ft. The C variable represents the capsize screening value, d is the displacement, and b is the beam length. Begin with the given equation. Bring the d variable into the denominator to make the exponent positive. Plug in the beam length (b) and the displacement (d). Convert the denominator to radical notation. This is easier to work with than the exponential form. This gives us me the cube root of the displacement in the denominator. Solve the equation for C.
This value for C is the capsize screening value for the sailboat when the displacement is 23,245 lbs. and the beam is 13.5 ft. The problem states that in order for a sailboat to be considered safe for ocean sailing, the capsize screening value must be less than 2. The sailboat can be considered safe for ocean sailing when the displacement is 23,245 lbs. and the beam is 13.5 ft.
b.) In the second part of this problem, we are asked to solve the given formula for d. Begin with the given equation. First, I put the d variable in the denominator so that I am only working with positive exponents. Next, I multiplied both sides of the equation by . This leaves me with the C and d variables on the left side of the equation. In order to get the d variable alone on the left side of the equation, I divide both sides by C. Now, I have only the d variable on the left side of the equation. Because the variable carries an exponent of 1/3, I can convert it into radical notation. This gives me the cube root of d. In order to solve the equation for d, I cubed both sides of the equation. This leaves the d variable with no radical on the left side of the equation. The equation can now easily be solved…