Presentation is loading. Please wait.

Presentation is loading. Please wait.

EXAMPLE 5 Solve a real-world problem SOLUTION SCIENCE As part of an environmental study, you need to estimate the number of trees in a 150 acre area. You.

Published byGriselda Chapman Modified over 8 years ago

Similar presentations

Presentation on theme: "EXAMPLE 5 Solve a real-world problem SOLUTION SCIENCE As part of an environmental study, you need to estimate the number of trees in a 150 acre area. You."— Presentation transcript:

1 EXAMPLE 5 Solve a real-world problem SOLUTION SCIENCE As part of an environmental study, you need to estimate the number of trees in a 150 acre area. You count 270 trees in a 2 acre area and you notice that the trees seem to be evenly distributed. Estimate the total number of trees. Write and solve a proportion involving two ratios that compare the number of trees with the area of the land.

2 EXAMPLE 5 Solve a real-world problem Simplify. Cross Products Property There are about 20,250 trees in the 150 acre area. = 270 2 n 150 = 270 150 2 n n = 20,250 number of trees area in acres Write proportion.

3 EXAMPLE 6 Find a geometric mean Simplify. Definition of geometric mean Substitute 24 for a and 48 for b. Factor. Find the geometric mean of 24 and 48. SOLUTION = x abab = 24 48 = 24 24 2 = 224 The geometric mean of 24 and 48 is 24 33.9. 2

4 GUIDED PRACTICE for Examples 5 and 6 SOLUTION Write and solve a proportion involving two ratios that compare the number of trees with the area of the land. = 390 3 n 150 number of trees area in acres Write proportion. Cross Products Property = 390 150 3 n n = 19,500 Simplify. WHAT IF ? In Example 5, suppose you count 390 trees in a 3 acre area of the 150 acre area. Make a new estimate of the total number of trees. 8.

5 GUIDED PRACTICE for Examples 5 and 6 ANSWER There are about 19,500 in the 150 acre area

6 GUIDED PRACTICE for Examples 5 and 6 Find the geometric mean of the two numbers. Simplify. Definition of geometric mean Substitute 12 for a and 27 for b. Factor. SOLUTION = 18 The geometric mean of 12 and 27 is 18 = x ab 9. 12 and 27 = 12 27 = 12 3 9

7 GUIDED PRACTICE for Examples 5 and 6 Find the geometric mean of the two numbers. Simplify. Definition of geometric mean Substitute 18 for a and 54 for b. Factor. SOLUTION = 18 54= x ab 10. 18 and 54 =318 = 18 18 3 The geometric mean of 18 and 54 is 18 3

8 GUIDED PRACTICE for Examples 5 and 6 Find the geometric mean of the two numbers. Simplify. Definition of geometric mean Substitute 16 for a and 18 for b. Factor. SOLUTION = x ab 11. 16 and 18 =212 = 16 18 = 16 9 2 The geometric mean of 16 and 18 is 12 2

Download ppt "EXAMPLE 5 Solve a real-world problem SOLUTION SCIENCE As part of an environmental study, you need to estimate the number of trees in a 150 acre area. You."

Similar presentations

Identify the number of solutions of an equation

Identify the number of solutions of an equation
Write an equation given the slope and a point

Write an equation given the slope and a point
EXAMPLE 1 Use properties of proportions SOLUTION NP ST MN RS = Because 4 x = 8 10, then In the diagram, NP ST MN RS = Write four true proportions. By the.

EXAMPLE 1 Use properties of proportions SOLUTION NP ST MN RS = Because 4 x = 8 10, then In the diagram, NP ST MN RS = Write four true proportions. By the.
Example 1 Solve ax b c for x. Then use the solution to solve Solve a literal equation + = += 2x 5 11. SOLUTION STEP 1 Solve ax b c for x. += ax b c Write.

Example 1 Solve ax b c for x. Then use the solution to solve Solve a literal equation + = += 2x SOLUTION STEP 1 Solve ax b c for x. += ax b c Write.
Write an equation given the slope and a point EXAMPLE 2 Write an equation of the line that passes through (5, 4) and has a slope of –3. Because you know.

Write an equation given the slope and a point EXAMPLE 2 Write an equation of the line that passes through (5, 4) and has a slope of –3. Because you know.
EXAMPLE 1 Identify arithmetic sequences

EXAMPLE 1 Identify arithmetic sequences
EXAMPLE 4 Solve proportions SOLUTION a. 5 10 x 16 = Multiply. Divide each side by 10. a. 5 10 x 16 = = 10 x5 16 = 10 x80 = x8 Write original proportion.

EXAMPLE 4 Solve proportions SOLUTION a x 16 = Multiply. Divide each side by 10. a x 16 = = 10 x5 16 = 10 x80 = x8 Write original proportion.
EXAMPLE 4 Solve proportions SOLUTION a. 5 10 x 16 = Multiply. Divide each side by 10. a. 5 10 x 16 = = 10 x5 16 = 10 x80 = x8 Write original proportion.

EXAMPLE 4 Solve proportions SOLUTION a x 16 = Multiply. Divide each side by 10. a x 16 = = 10 x5 16 = 10 x80 = x8 Write original proportion.
Solve an equation with variables on both sides

Solve an equation with variables on both sides
Use the cross products property EXAMPLE 1 Write original proportion. 8 15 = x 6 Solve the proportion =. 8 x 6 15 Cross products property Simplify. 120.

Use the cross products property EXAMPLE 1 Write original proportion = x 6 Solve the proportion =. 8 x 6 15 Cross products property Simplify. 120.
EXAMPLE 2 Using the Cross Products Property 6.8 15.4 = 40.8 m Write original proportion. Cross products property Multiply. 6.8m 6.8 = 628.32 6.8 Divide.

EXAMPLE 2 Using the Cross Products Property = 40.8 m Write original proportion. Cross products property Multiply. 6.8m 6.8 = Divide.
Solve an equation using subtraction EXAMPLE 1 Solve x + 7 = 4. x + 7 = 4x + 7 = 4 Write original equation. x + 7 – 7 = 4 – 7 Use subtraction property of.

Solve an equation using subtraction EXAMPLE 1 Solve x + 7 = 4. x + 7 = 4x + 7 = 4 Write original equation. x + 7 – 7 = 4 – 7 Use subtraction property of.
Standardized Test Practice EXAMPLE 4 ANSWER The correct answer is B. DCBA Simplify the expression 4(n + 9) – 3(2 + n). 4(n + 9) – 3(2 + n) = Distributive.

Standardized Test Practice EXAMPLE 4 ANSWER The correct answer is B. DCBA Simplify the expression 4(n + 9) – 3(2 + n). 4(n + 9) – 3(2 + n) = Distributive.
Standardized Test Practice

Standardized Test Practice
EXAMPLE 1 Collecting Like Terms x + 2 = 3x x + 2 –x = 3x – x 2 = 2x 1 = x Original equation Subtract x from each side. Divide both sides by 2. 2 2 2x2x.

EXAMPLE 1 Collecting Like Terms x + 2 = 3x x + 2 –x = 3x – x 2 = 2x 1 = x Original equation Subtract x from each side. Divide both sides by x2x.
Standardized Test Practice

Standardized Test Practice
EXAMPLE 2 Write a rule for the nth term Write a rule for the nth term of the sequence. Then find a 7. a. 4, 20, 100, 500,... b. 152, –76, 38, –19,... SOLUTION.

EXAMPLE 2 Write a rule for the nth term Write a rule for the nth term of the sequence. Then find a 7. a. 4, 20, 100, 500,... b. 152, –76, 38, –19,... SOLUTION.
EXAMPLE 4 Find perimeters of similar figures Swimming

EXAMPLE 4 Find perimeters of similar figures Swimming
EXAMPLE 2 Write a rule for the nth term a. 4, 9, 14, 19,... b. 60, 52, 44, 36,... SOLUTION The sequence is arithmetic with first term a 1 = 4 and common.

EXAMPLE 2 Write a rule for the nth term a. 4, 9, 14, 19,... b. 60, 52, 44, 36,... SOLUTION The sequence is arithmetic with first term a 1 = 4 and common.
6.1 – Ratios, Proportions, and the Geometric Mean Geometry Ms. Rinaldi.

6.1 – Ratios, Proportions, and the Geometric Mean Geometry Ms. Rinaldi.

Similar presentations

Ads by Google