• Submitted By: clane
  • Date Submitted: 02/23/2014 1:06 PM
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Real World Radical Formulas

MAT222 Week 3 Assignment

February 10, 2014

Real World Radical Formulas
I will be solving a radical formula for this week’s assignment. For my first part of this radical formula it is stated in # 103 on page 605(Dugopolski, 2012) that the capsize screening value C should be less than 2 is a boat is to be considered safe for ocean sailing. The formula given is C=4d-1/3b where d is the displacement in pounds and b is the beam width in feet. The exponent of -1/3 means that the cube root of d will be taken and then the reciprocal of that number will be used in the multiplication.
a.) For the first example, I must find the capsize screening value for the Tartan 4100, which has a displacement of 23,245 pounds and a beam of 13.5 feet.
C=4(23,245)-13(13.5) I put in the values given. According to order of operations,
the exponents are solved first ( solved using a calculator).
C=4(.035)(13.5) Now I have just two multiplications.
C=.14(13.5) Multiply
C=1.89 The capsize screening value is less than 2.
b.) For this part I have to solve for d using the same formula in part a.
C=4d-1/3b Original equation
C4b=4d-1/3b4b Divide both sides by 4b
(C4b)-3=(d-13)-3 Raise all parts of both sides to the 3 power.
C-34b-3=d The 4b-3 on the left side becomes its reciprocal to the third.
64b3C3 Final answer. The formula has now been solved for d
c.) Now I have to solve for what displacement is the Tartan 4100 safe for ocean sailing?
d=64b3C3 My formula for this problem
d=64(13.5)3(1.89)3 I substituted my variables from parts a and be into the equation to solve
for d. I will solve the exponents first.
d= 64(2460.38)6.75 Now multiply 64 * 2460.38
d=157464.326.75 Divide