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Problem

Evaluate each expression. $$ \frac{2 \cdot 2^{5}-…

01:52

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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88 Problem 89 Problem 90 Problem 91 Problem 92 Problem 93 Problem 94 Problem 95 Problem 96 Problem 97 Problem 98 Problem 99 Problem 100 Problem 101 Problem 102 Problem 103 Problem 104 Problem 105 Problem 106 Problem 107 Problem 108 Problem 109 Problem 110 Problem 111 Problem 112 Problem 113 Problem 114 Problem 115 Problem 116 Problem 117 Problem 118 Problem 119 Problem 120 Problem 121 Problem 122 Problem 123 Problem 124 Problem 125 Problem 126 Problem 127 Problem 128 Problem 129 Problem 130 Problem 131 Problem 132

Problem 68 Medium Difficulty

Evaluate each expression.
$$
\frac{-3-(-1)}{-2+(-2)}
$$

Answer

$\frac{1}{2}$

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Video Transcript

So we're being asked to evaluate to give an expression. If you know this, we have a fracture. So our first goal is to get one number, a top and the number on the bottom by evaluating them both separately. And then we'll reduce after, if possible. Well, in the new more. Either we have negative three minus negative one. Well, in this case, we could add the opposite, so we're gonna add the opposite. So, really, this just means we need to do negative three plus one, which is negative, too. Now, in the denominator, we have negative two plus negative, too, Which is negative for. And now that we have evaluated both the numerator and denominator, now we need to try and simplify. Well, the first thing I know this is I have a negative getting divided by a negative. So those will become a positive. And then I have the fraction to force. Well, I can reduce this by dividing both the numerator and denominator by two. Well, two divided by two. It's one and four. Divided by two is too. So now we've found the

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Alan S. Tussy, R. David Gustafson

Elementary and Intermediate Algebra

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