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Identify parts of expressions

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5223336
State the coefficient in each expression. a) \(-x\) b) \(1.2y\) c) \(\frac{z}{10}\) d) \(-\frac{4}{5}a\) e) \(b\)

Hints

- The coefficient is the number multiplying the variable. - A minus sign in front of a variable represents an implied factor. - Rewrite division by a number as multiplication by its reciprocal. - Multiplying a variable by \(1\) does not change its value.

Solution

1. A coefficient is the numerical factor multiplying a variable. 2. In \(-x\), the implied numerical factor is \(-1\), so the coefficient is \(-1\). 3. In \(1.2y\), the coefficient is \(1.2\). 4. Rewrite \(\frac{z}{10}\) as \(\frac{1}{10}z\), so the coefficient is \(\frac{1}{10}\), or \(0.1\). 5. In \(-\frac{4}{5}a\), the coefficient is \(-\frac{4}{5}\), or \(-0.8\). 6. The expression \(b\) means \(1\times b\), so the coefficient is \(1\).

Answer

a) \(-1\) b) \(1.2\) c) \(\frac{1}{10}\), or \(0.1\) d) \(-\frac{4}{5}\), or \(-0.8\) e) \(1\)
5227496
Rewrite each expression as a sum of signed terms. Keep the sign with each term. 1) \(7x-4y\) 2) \(-a^2+5a-9\) 3) \(3u^2-uv+2v^2\)

Hints

- Rewrite subtraction as addition of a negative term. - Keep each minus sign attached to the term that follows it. - Identify every term before rewriting the expression.

Solution

1. The signed terms in \(7x-4y\) are \(7x\) and \(-4y\). Written as a sum, the expression is \((7x)+(-4y)\). 2. The signed terms in \(-a^2+5a-9\) are \(-a^2\), \(5a\), and \(-9\). Written as a sum, the expression is \((-a^2)+(5a)+(-9)\). 3. The signed terms in \(3u^2-uv+2v^2\) are \(3u^2\), \(-uv\), and \(2v^2\). Written as a sum, the expression is \((3u^2)+(-uv)+(2v^2)\).

Answer

1) \((7x)+(-4y)\) 2) \((-a^2)+(5a)+(-9)\) 3) \((3u^2)+(-uv)+(2v^2)\)
5223976
For each expression, identify the operation performed last and use that operation to name the expression as a sum, difference, product, quotient, or power. 1) \(x(y+5)\) 2) \(a^2-10\) 3) \((p-q)\div4\) 4) \(7+3n\) 5) \((m+1)^2\)

Hints

- Use the order of operations to decide which operation is performed last. - Imagine substituting numbers for the variables and completing the calculation. - The last operation determines the name of the entire expression.

Solution

1. In \(x(y+5)\), the addition inside parentheses is performed first and multiplication is performed last. The expression is a product. 2. In \(a^2-10\), exponentiation is performed first and subtraction is performed last. The expression is a difference. 3. In \((p-q)\div4\), subtraction inside parentheses is performed first and division is performed last. The expression is a quotient. 4. In \(7+3n\), multiplication is performed before addition, so addition is performed last. The expression is a sum. 5. In \((m+1)^2\), addition inside parentheses is performed first and exponentiation is performed last. The expression is a power.

Answer

1) Multiplication; product 2) Subtraction; difference 3) Division; quotient 4) Addition; sum 5) Exponentiation; power
5227506
Consider the expression \(-5x^2+x-0.8\). a) Rewrite the expression as a sum of signed terms. b) State the numerical factor of each term. For the constant term, use the constant itself as its numerical factor. What is the coefficient of the \(x\)-term?

Hints

- A variable with no visible numerical factor has an implied coefficient. - Include each term’s sign as part of its numerical factor. - Rewrite subtraction as addition of a negative term.

Solution

1. Rewrite each subtraction as addition of a negative term: \((-5x^2)+x+(-0.8)\). 2. The numerical factor of \(-5x^2\) is \(-5\). 3. The term \(x\) means \(1\times x\), so its coefficient is \(1\). 4. The numerical factor of the constant term \(-0.8\) is \(-0.8\).

Answer

a) \((-5x^2)+x+(-0.8)\) b) The numerical factors are \(-5\), \(1\), and \(-0.8\). The coefficient of the \(x\)-term is \(1\).
5128046
Consider the expression \(4(x-2.5)\). a) Describe the expression in words. b) Evaluate it when \(x=5.5\). c) What value of \(x\) makes the expression equal to \(0\)? Explain.

Hints

- Identify the operation inside the parentheses before the multiplication. - Substitute the given value for part b). - A product is \(0\) when at least one factor is \(0\).

Solution

1. The expression means four times the difference between \(x\) and \(2.5\). 2. For \(x=5.5\), \(4\times(5.5-2.5)=4\times 3=12\). 3. Since \(4\ne 0\), the product equals \(0\) only when \(x-2.5=0\). 4. Therefore, \(x=2.5\).

Answer

a) Four times the difference between \(x\) and \(2.5\) b) \(12\) c) \(x=2.5\), because the factor \(x-2.5\) must equal \(0\).

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