Читать книгу Algebra I All-in-One For Dummies - Mary Jane Sterling, Mary Sterling Jane - Страница 71

1 5. First do the subtraction in the parentheses. Then multiply the result by 3. Finally, add 14.

2 77. Work inside the brackets first. Perform the two operations in the parentheses. Then perform the multiplications on the two results. Add the products. Then multiply that result by 4. Finally, subtract 11.

3 180. Multiplying by the 5 outside the braces will come last. First, do the subtraction in the parentheses. Then add inside the brackets. Multiply that result by 8, and then subtract 4. Finally, multiply by 5.

4 Your last step will be to divide the result in the numerator by the result in the denominator. In the numerator, first do the subtraction in the parentheses, multiply that result by 3, and then subtract that product from 19. Finally, find the square root of the difference. In the denominator, do the subtraction in the parentheses, multiply the result by 4, subtract that product from 8, and then multiply the difference by 6. Finally, subtract 1 from the product. Then you can do the division indicated by the fraction line.

5 –56. Working inside the absolute value bars, first do the subtraction in the parentheses, multiply the difference by 3, and then subtract the product from 6. Find the absolute value of the result before multiplying by 5. Then subtract the product from 4.

6 8. Work on the numerator and denominator separately — leaving the division for the last step. In the numerator, do the subtraction in the first parentheses and do the multiplication in the last term. Multiply the remainder by 5. Then subtract the second term from the first. In the denominator, do the subtraction under the first radical. Then look at the radical-within-the-radical and do the addition. Evaluate the two radical values you’ve formed. Next, find the difference under the remaining radical and evaluate it. Perform the subtraction in the denominator — and, finally, divide.

If you’re ready to test your skills a bit more, take the following chapter quiz that incorporates all the chapter topics.

7 . Multiply each term in the parentheses by 4.

8 . Multiply each term in the parentheses by –3.

9 . This problem involves operations with fractions. You will see more on fractions in Chapter 4.

10 –5.

11 .

12 .

13 47. Regroup so that the first two numbers are together. Their sum is 0!

14 5. Regroup so that the second two numbers are together.

15 70. Regroup so that the first two numbers are together. The fraction reduces to make the multiplication easier.

16 110. Regroup so that the second two numbers are together. The multiplication is simple!

17 5. Switch the order of the second and third numbers.

18 470. Switch the order of the second and third numbers.

19 78. Switch the order of the second and third numbers.

20 8. Switch the order of the second and third numbers.

21 560. There’s a lot of switching that goes on here to make the problem easier. Your goal is to be able to multiply fractions times numbers that will eliminate their denominators. In this particular problem, I’m bringing the up next to the 16 and the up next to the 18.

Then I group, reduce, multiply, and find the product of the three “nice” numbers!

22 –11 and . The sum of 11 and –11 is 0; the product of 11 and is 1.

23 and –3. The sum of and is 0; the product of and –3 is 1.

24 and . The sum of and is 0; the product of and is 1.

25 1 and –1. The sum of –1 and 1 is 0; the product of –1 and –1 is 1. The number is its own multiplicative inverse.

26 and or . The sum of and 4.5 is 0. The product of 4.5 and is 1. You usually don’t want to write decimals in fractions, so you can change by writing it with a fractional denominator and simplifying:

You’ll see more on working with fractions in Chapter 4.

27 and . The additive inverse is just the positive version of the same number. To write the multiplicative inverse, you change the mixed number to an improper fraction and just flip it:

There’s a full coverage of fractions in Chapter 4.

28 Use 8. The additive inverse of –8 is 8. If you add 8 to the expression, you have . Use the associative property to group the –8 and 8 together: . The sum of a number and its additive inverse is 0, so the expression becomes . Because 0 is the additive identity, .

29 Use –6. The additive inverse of 6 is –6. If you add –6 to the expression, you have . Use the associative property to group the 6 and –6 together: . The sum of a number and its additive inverse is 0, so the expression becomes 0 – 3x. Because 0 is the additive identity, .

30 Use . The multiplicative inverse of –7 is . If you multiply the expression by , you have . Use the commutative property to rearrange the factors and the associative property to group the –7 and together: . The product of a number and its multiplicative inverse is 1, so the expression becomes 1x. Because 1 is the multiplicative identity, .

31 Use 4. The expression can be written as . The multiplicative inverse of is 4. If you multiply the expression by 4, you have

An added bonus here is that the constant also becomes an integer.

32 6.

33 6. By definition, the value of 0! is 1.

34 639,200.

35 147. The 9, which will be dropped with the 5 following it, is greater than 5. Round the 6 in the ones place to a 7.

36 146. The greatest integer smaller than 146.95 is 146.

If you’re ready to test your skills a bit more, take the following chapter quiz that incorporates all the chapter topics.