Concept explainers
Use the four-step procedure for solving variation problems given on page 445 to solve Exercises 21–36.
31. Body-mass index, or BMI, takes both weight and height into account when assessing whether an individual is underweight or overweight. BMI varies directly as one’s weight, in pounds, and inversely as the square of one’s height, in inches. In adults, normal values for the BMI are between 20 and 25, inclusive. Values below 20 indicate that an individual is underweight and values above 30 indicate that an individual is obese. A person who weighs 180 pounds and is 5 feet, or 60 inches, tall has a BMI of 35.15. What is the BMI, to the nearest tenth, for a 170-pound person who is 5 feet 10 inches tall? Is this person overweight?
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Chapter 3 Solutions
COLLEGE ALGEBRA ESSENTIALS-MYMATHLAB
- 1. If f(x² + 1) = x + 5x² + 3, what is f(x² - 1)?arrow_forward2. What is the total length of the shortest path that goes from (0,4) to a point on the x-axis, then to a point on the line y = 6, then to (18.4)?arrow_forwardموضوع الدرس Prove that Determine the following groups Homz(QZ) Hom = (Q13,Z) Homz(Q), Hom/z/nZ, Qt for neN- (2) Every factor group of adivisible group is divisble. • If R is a Skew ficald (aring with identity and each non Zero element is invertible then every R-module is free.arrow_forward
- Please help me with these questions. I am having a hard time understanding what to do. Thank youarrow_forwardAnswersarrow_forward************* ********************************* Q.1) Classify the following statements as a true or false statements: a. If M is a module, then every proper submodule of M is contained in a maximal submodule of M. b. The sum of a finite family of small submodules of a module M is small in M. c. Zz is directly indecomposable. d. An epimorphism a: M→ N is called solit iff Ker(a) is a direct summand in M. e. The Z-module has two composition series. Z 6Z f. Zz does not have a composition series. g. Any finitely generated module is a free module. h. If O→A MW→ 0 is short exact sequence then f is epimorphism. i. If f is a homomorphism then f-1 is also a homomorphism. Maximal C≤A if and only if is simple. Sup Q.4) Give an example and explain your claim in each case: Monomorphism not split. b) A finite free module. c) Semisimple module. d) A small submodule A of a module N and a homomorphism op: MN, but (A) is not small in M.arrow_forward
- I need diagram with solutionsarrow_forwardT. Determine the least common denominator and the domain for the 2x-3 10 problem: + x²+6x+8 x²+x-12 3 2x 2. Add: + Simplify and 5x+10 x²-2x-8 state the domain. 7 3. Add/Subtract: x+2 1 + x+6 2x+2 4 Simplify and state the domain. x+1 4 4. Subtract: - Simplify 3x-3 x²-3x+2 and state the domain. 1 15 3x-5 5. Add/Subtract: + 2 2x-14 x²-7x Simplify and state the domain.arrow_forwardQ.1) Classify the following statements as a true or false statements: Q a. A simple ring R is simple as a right R-module. b. Every ideal of ZZ is small ideal. very den to is lovaginz c. A nontrivial direct summand of a module cannot be large or small submodule. d. The sum of a finite family of small submodules of a module M is small in M. e. The direct product of a finite family of projective modules is projective f. The sum of a finite family of large submodules of a module M is large in M. g. Zz contains no minimal submodules. h. Qz has no minimal and no maximal submodules. i. Every divisible Z-module is injective. j. Every projective module is a free module. a homomorp cements Q.4) Give an example and explain your claim in each case: a) A module M which has a largest proper submodule, is directly indecomposable. b) A free subset of a module. c) A finite free module. d) A module contains no a direct summand. e) A short split exact sequence of modules.arrow_forward
- College Algebra (MindTap Course List)AlgebraISBN:9781305652231Author:R. David Gustafson, Jeff HughesPublisher:Cengage Learning