Free Math Worksheets

- Use nearby multiplication facts to locate the boundary value. - Check the digits immediately below and above that boundary. - Pay close attention to whether each inequality uses \(<\) or \(>\).

1. For a), \(40 \times 6 = 240\) and \(40 \times 7 = 280\). Therefore, the solutions are the digits \(0\) through \(6\). 2. For b), \(700 \times 5 = 3500\) and \(700 \times 6 = 4200\). Therefore, the solutions are \(6, 7, 8\), and \(9\). 3. For c), \(90 \times 8 = 720\) and \(90 \times 9 = 810\). Therefore, the solutions are the digits \(0\) through \(8\).

a) \(0, 1, 2, 3, 4, 5, 6\) b) \(6, 7, 8, 9\) c) \(0, 1, 2, 3, 4, 5, 6, 7, 8\)

- Compare each product with \(400\) using related basic multiplication facts. - Estimate the boundary value by dividing \(400\) by \(6\). - Once one multiple of \(10\) works, decide whether larger multiples also work. - More than one value may work.

1. Test the nearby multiple of \(10\): \(6 \times 60 = 360\), which is not greater than \(400\). 2. The next value works because \(6 \times 70 = 420 > 400\). 3. Larger choices also work: \(6 \times 80 = 480\) and \(6 \times 90 = 540\). 4. Therefore, the values are \(70\), \(80\), and \(90\).

\(x = 70\), \(80\), or \(90\).

- Identify the integers to the right of \(-7\). - Pay attention to which endpoint is included. - Test each integer against both parts of the inequality.

1. The condition \(z>-7\) excludes \(-7\), so the least possible integer is \(-6\). 2. The condition \(z\le -2\) includes \(-2\). 3. Therefore, the integers are \(-6,-5,-4,-3,-2\).

\(-6,-5,-4,-3,-2\)

- Picture the interval between \(-9\) and \(-1\) on a number line. - Test each listed number against both inequalities. - The endpoints are not included.

1. A qualifying number must be greater than \(-9\) and less than \(-1\). 2. Testing the listed values shows that \(-8\), \(-5\), and \(-2\) meet both conditions.

\(-8,-5,-2\)

- Rewrite the outside fractions with denominator \(12\). - With equal positive denominators, compare the numerators. - The strict inequality means the endpoints are not included.

1. Rewrite both boundary fractions with denominator \(12\): \(-\frac{3}{4} = -\frac{9}{12}\) and \(-\frac{1}{3} = -\frac{4}{12}\). 2. The inequality becomes \(-\frac{9}{12} < \frac{k}{12} < -\frac{4}{12}\). 3. Since the denominators are equal and positive, compare numerators: \(-9 < k < -4\). 4. The integers in this interval are \(-8, -7, -6, -5\).

\(k \in \{-8, -7, -6, -5\}\)

- Solve each inequality separately. - List the digit solutions for the first inequality. - List the digit solutions for the second inequality. - Keep only the digits that appear in both lists.

1. For \(30d > 100\), \(30 \times 3 = 90\) and \(30 \times 4 = 120\), so the digit solutions are \(4, 5, 6, 7, 8\), and \(9\). 2. For \(80d < 500\), \(80 \times 6 = 480\) and \(80 \times 7 = 560\), so the digit solutions are \(0, 1, 2, 3, 4, 5\), and \(6\). 3. The digits in both solution sets are \(4, 5\), and \(6\).

\(4, 5, 6\)

- Analyze the two conditions separately. - Express the integer immediately after \(x\) as \(x+1\). - Express the integer immediately before \(x\) as \(x-1\). - Find the integers common to both solution sets.

1. The integer after \(x\) is \(x+1\). The condition \(x+1>-3\) gives \(x>-4\), so an integer solution must satisfy \(x\geq-3\). 2. The integer before \(x\) is \(x-1\). The condition \(x-1<-2\) gives \(x<-1\), so an integer solution must satisfy \(x\leq-2\). 3. The integers satisfying both conditions are \(-3\) and \(-2\).

\(x=-3\) or \(x=-2\)