SA 6 s 2 SA 6 4 2 SA 616 SA 96 The surface area of the cube is 96 cm 2 b The

# Sa 6 s 2 sa 6 4 2 sa 616 sa 96 the surface area of

This preview shows page 24 - 28 out of 32 pages.

SA = 6 s 2 SA = 6( 4 ) 2 SA = 6(16) SA = 96 The surface area of the cube is 96 cm 2 . b) The diagram models the Pythagorean relationship. The relationship between the areas of the squares on the sides of a right triangle is represented by the formula c 2 = a 2 + b 2 , where a and b are the legs of the triangle and c is the hypotenuse. c 2 = a 2 + b 2 c 2 = 5 2 + 12 2 c 2 = 25 + 144 c 2 = 169 The area of the square attached to the hypotenuse is 169 cm 2 . 3.4 Using Exponents to Solve Problems MHR 115
Use a formula to solve each problem. a) A right triangle has two shorter sides that measure 8 cm and 15 cm. What is the area of a square attached to the hypotenuse of the right triangle? b) What is the surface area of a cube with an edge length of 3 m? Show You Know c) Method 1: Calculate in Stages The formula for the area of a square is A = s 2 , where s is the side length of the square. A = s 2 A = 20 2 A = 400 The area of the square is 400 cm 2 . The formula for the area of a circle is A = π r 2 , where r is the radius of the circle. The diameter of the circle is 20 cm. Therefore, the radius is 10 cm. A = π r 2 A = π ( 10 ) 2 A = π (100) A 314 … The area of the circle is approximately 314 cm 2 . Calculate the area of the shaded region. You can subtract the area of the circle from the area of the square. 400 - 314 = 86 The area of the shaded region is about 86 cm 2 . Method 2: Evaluate One Expression Calculate the area of the shaded region. You can subtract the area of the circle from the area of the square. A = s 2 - π r 2 A = 20 2 - π ( 10 ) 2 A = 400 - π (100) A 400 - 314 A 86 The area of the shaded region is about 86 cm 2 . 15 cm 8 cm 3 m 116 MHR Chapter 3
Example 2: Develop a Formula to Solve a Problem A dish holds 100 bacteria. It is known that the bacteria double in number every hour. How many bacteria will be present after each number of hours? a) 1 b) 5 c) n Solution a) After 1 h, the bacteria population doubles. 100 × 2 = 200 After 1 h, there will be 200 bacteria. b) In a period of 5 h, the bacteria population doubles five times. 100 × 2 × 2 × 2 × 2 × 2 = 100(2 5 ) = 100(32) = 3200 After 5 h, there will be 3200 bacteria. c) After n hours, the bacteria population doubles n times. Number of bacteria = 100(2 n ) After n hours, there will be 100(2 n ) bacteria. A type of bacterium is known to triple every hour. There are 50 bacteria to start with. How many will there be after each number of hours? a) 3 b) 5 c) t Show You Know Use a Variable Strategies Key Ideas Powers are found in many formulas. When repeated multiplication is present in a formula, it is represented as a power. The use of powers keeps the formula as short as possible. Many patterns that involve repeated multiplication can be modelled with expressions that contain powers. s s s Volume of a cube = s 3 3.4 Using Exponents to Solve Problems MHR 117
Check Your Understanding Communicate the Ideas 1. The surface area, SA , of a sphere can be calculated using the formula SA = 4 × π × r × r , where r is the radius. Rewrite the formula using powers and no multiplication signs. Identify the coefficient, variable, and exponent in your formula.

#### You've reached the end of your free preview.

Want to read all 32 pages?

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern