6 th grade test prep

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Algebra

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6 th Grade Test Prep. Algebra. 5.A.2 Translate simple expressions into algebraic expressions. Eighteen less than twice a number. 2t - 18. A number squared, plus nineteen. n² + 19. The product of eleven and a number, divided by sixty. 11 × k ÷ 60. - PowerPoint PPT Presentation

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Page 1: 6 th  Grade Test Prep

Algebra

Page 2: 6 th  Grade Test Prep

Eighteen less than twice a number

A number squared, plus nineteen

2t - 18

n² + 19

The product of eleven and a number, divided by sixty 11 × k ÷ 60

One hundred more than the quotient of a number and two hundredths n ÷ 0.02 + 100

5.A.25.A.2

Translate simple expressions into algebraic Translate simple expressions into algebraic expressionsexpressions

Page 3: 6 th  Grade Test Prep

Joanna receives $8 per hour when she works at the coffee shop. Last week she earned $256. Write an equation that can be used to find the number of hours Joanna worked last week.

Write an equation to represent the following. The sum of 7 and a number, divided by 15, is equal to 33.

Let h = Hours Joanna worked8h = 256

Let n = the number(7 + n) ÷ 15 = 33

Page 4: 6 th  Grade Test Prep

Evaluate 8a – 2b when a = 7 and b = 4

Evaluate x² + y³ when x = 3 and y = 2

How do you evaluate 6c + d² when c = 9 and d = 5

5.A.3

Page 5: 6 th  Grade Test Prep

Step 1: Use the inverse operation (division) to ‘move’ the 5.5t ÷ 5 = 25 ÷ 5

Step 2: Simplifyt = 5

Step 3: Check your work by substituting and solving. t = 55t = 255(5) = 2525 = 25

Solve for t: 5t = 25

5.A.45.A.4 5.A.55.A.5

One step equationsOne step equations

Page 6: 6 th  Grade Test Prep

Step 1: Start by identifying like terms. Use the inverse operation (subtraction) to ‘move’ the 5.

5 + (-5) + 4t = 25 + (-5)4t = 20

Step 2: To isolate the variable, use the inverse operation (division), to ‘move’ the 4.

4t ÷ 4 = 20 ÷ 4 t = 5

Step 3: Check your work by substituting and solving. t = 55 + 4(5) = 255 + 20 = 25

Solve for t: 5 + 4t = 25

Page 7: 6 th  Grade Test Prep

Step 1: Start with the term that does NOT include the variable. Use the inverse operation (subtraction) to ‘move’ the 19.

n/8 – 19 (+ 19) = 7 + 19n/8 = 26

Step 2: To isolate the variable, use the inverse operation (multiplication), to ‘move’ the 8.

n/8 (* 8) = 26 (* 8)n = 208

Step 3: Check your work by substituting and solving. n = 208208 / 8 – 19 = 7 √

Solve for n: - 19 = 78

n

Page 8: 6 th  Grade Test Prep

A car uses 9 gallons of gasoline for a 162-mile drive. How many gallons of gasoline will the same car use in a 216-mile drive?

Step 1: Write the ratio given in the problem.The ratio is “162 miles uses 9 gallons.”As a fraction, the ratio miles/ gallons is 162/9

Step 2: Represent the unknown by a variable.g = gallons of gasoline for a 216-mile drive

Step 3: Write the ratio and fraction for the unknown variable. The ratio is “216 miles uses ‘g’ gallons.”As a fraction, the ratio miles/ gallons is 216 / g

Step 4: Write the ratios as proportions. Cross-multiply to solve.

162g = 1944 g = 12The car will use 12 gallons of gasoline for a 216-mile drive .

Page 9: 6 th  Grade Test Prep

Julia’s sister drove 224 miles in 4 hours. If Julia’s sister drove at a constant speed, how fast did she drive? Use the formula d = r t.

Terry just got a $1,200 three-year loan. The loan has a 7% interest rate. How much interest will Terry pay on the loan? Use the formula I = P r t.

P = $1,200r = 7%t = 3 years I = ?I = P r t

d = 224 milest = 4 hoursr = ?d = r t

224 = r 4224 ÷ 4 = r 56 miles per hour = r

I = (1200)(.07)(3)I = 252Terry will $252 as interest over the 3 years.

Page 10: 6 th  Grade Test Prep

Which temperature is warmer 34°C or 95°F? Use the formula °C = 5/9 * (°F – 32)

°C = 34°F = 95°C = 5/9 * (°F – 32)

°C = 5/9 * (95 – 32)°C = 5/9 * (63)°C = 35°

Thus 95°F = 35°C, which is warmer than 34°C.