Free math worksheets

- In function notation, the value inside the parentheses identifies the input. - Read each requested value from the column with the matching term number. - Think about how sequence positions are counted: first term, second term, third term, and so on.

1. The input is the term number \(n\), and the output is the term value \(q(n)\). 2. From the table, \(q(2)=8\) and \(q(4)=14\). 3. The input values shown are \(1,2,3,4\). A sequence uses whole-number term positions, so \(2.5\) is not a term number and is not in the domain.

a) Input: term number \(n\); output: term value \(q(n)\) b) \(q(2)=8\) and \(q(4)=14\) c) The shown inputs are \(1,2,3,4\). The value \(2.5\) is not a valid term number.

- Each table column represents one input-output pairing. - Keep the input first when you write an ordered pair. - After writing the pairs, collect the first coordinates and second coordinates separately.

1. Pair each term number with the value in the same table column: \((1,12)\), \((2,7)\), \((3,15)\), \((4,9)\), and \((5,11)\). 2. The domain is the set of inputs \(\{1,2,3,4,5\}\). 3. The range is the set of outputs \(\{7,9,11,12,15\}\).

Ordered pairs: \(\{(1,12),(2,7),(3,15),(4,9),(5,11)\}\). Domain: \(\{1,2,3,4,5\}\). Range: \(\{7,9,11,12,15\}\).

- Read a plotted sequence point as \((\text{term number},\text{term value})\). - Compare how the output changes each time the term number increases by \(1\). - Decide whether a term number can be halfway between two consecutive positive integers.

1. The plotted points show \(s(2)=5\) and \(s(5)=14\). 2. Each time \(n\) increases by \(1\), the output increases by \(3\). Since \(s(1)=2\), a matching rule is \(s(n)=3n-1\). 3. The inputs are term numbers, so the domain consists of positive integers. Inputs between consecutive integers are not sequence terms, so the graph uses isolated points rather than a connected line.

a) \(s(2)=5\) and \(s(5)=14\) b) \(s(n)=3n-1\) c) The sequence has positive-integer term numbers as inputs, so values between consecutive term numbers are not in the domain.

- Start at the middle plotted term and compare points the same horizontal distance to its left and right. - Record both the term numbers and the corresponding output values when you compare a pair. - A symmetry description should account for all displayed terms, not only one matching pair.

1. The least output is \(v(4)=2\). 2. Terms equally far from \(4\) have equal outputs: \(v(3)=v(5)=4\), \(v(2)=v(6)=6\), and \(v(1)=v(7)=8\). 3. Therefore the sequence is symmetric about term number \(4\): moving the same number of term positions left or right from \(4\) gives the same output.

The least output is \(2\) at \(n=4\). Equal-output pairs are \(n=3\) and \(5\), \(n=2\) and \(6\), and \(n=1\) and \(7\). The sequence is symmetric about term number \(4\).

- Translate the sequence into input-output pairs before judging the statements. - Check the function condition from the perspective of each input, not from how often an output repeats. - For the reverse relation, imagine swapping the coordinates of every pair.

1. The sequence pairs term numbers with values as \((1,4)\), \((2,1)\), \((3,4)\), \((4,9)\), \((5,4)\), and \((6,16)\). 2. Statement I is true because each term number is paired with exactly one output. 3. Statement II is false. Reversing the ordered pairs makes input \(4\) pair with outputs \(1\), \(3\), and \(5\), so the reverse relation gives one input more than one output. 4. Statement III is true because the inputs \(1\), \(3\), and \(5\) all produce output \(4\).

Statements I and III are true; statement II is false. The sequence is a function, \(4\) has preimages \(1,3,5\), and reversing the relation makes input \(4\) have three outputs.

- Compute the output change over more than one pair of consecutive term numbers. - Compare those one-step rates rather than judging from whether the graph would rise or fall. - When deciding whether a rate is constant, include the possibility that the rate could be zero.

1. Consecutive outputs are all \(7\), so every one-step output change is \(7-7=0\). 2. The input changes by \(1\) from one term number to the next, so the rate of change is consistently \(\frac{0}{1}=0\). 3. A constant rate of change of \(0\) is still a constant rate of change. Therefore the sequence is linear on its discrete domain, and the student's claim is incorrect.

The student is incorrect. The sequence has constant rate of change \(0\), so it is linear even though every output is the same.

- Separate the values for which an expression can be calculated from the inputs that the situation actually permits. - Use the description of the theater to determine which row numbers belong to the function. - Check an input at one end of the allowed set to provide a valid function value.

1. The stated domain is \(\{1,2,3,4,5,6\}\) because the theater has exactly six rows. 2. Although the algebraic expression \(12+2n\) can be evaluated at \(n=10\), the input \(10\) is outside the contextual domain, so \(S(10)\) does not describe this theater. 3. For example, \(S(1)=14\) and \(S(6)=24\); either is a valid endpoint value.

The student's conclusion is invalid because the contextual domain is \(\{1,2,3,4,5,6\}\), so row \(10\) is not an allowed input. A valid endpoint value is \(S(1)=14\) or \(S(6)=24\).

- Identify the length of one complete repeating block. - Compare the requested term number with earlier term numbers that occupy the same position in the cycle. - For part b), advance by one full cycle after locating the first occurrence of the target value.

1. The cycle has length \(3\). Term numbers \(2,5,8,11,\ldots\) occupy the second position in the cycle, whose value is \(5\). Therefore \(c(8)=5\). 2. Among the first \(12\) terms, the second position in each three-term block occurs at \(n=2,5,8,11\). Thus \(c(n)=5\) for those four inputs. 3. Any term number can be located by its position within a block of three. Terms in the same position of successive blocks have the same output.

a) \(c(8)=5\) b) \(n=2,5,8,11\) c) The cycle length is \(3\), so the position of a term within a three-term block determines its output.

- “First” means the smallest allowed term number that satisfies the condition. - Read \(P\) directly from its table and evaluate \(Q\) only as far as needed. - Compare the qualifying term numbers, not merely the qualifying output values.

1. From the table, \(P(4)=17\), while the earlier values \(6,8,13\) are below \(16\). Thus \(P\) first reaches the threshold at \(n=4\). 2. For \(Q\), the outputs on the finite domain are \(4,7,10,13,16\). Thus \(Q\) first reaches the threshold at \(n=5\). 3. Since \(4<5\), sequence \(P\) reaches the threshold first.

Sequence \(P\) reaches the threshold first. \(P\) first qualifies at \(n=4\); \(Q\) first qualifies at \(n=5\).

- Inspect the first coordinates before deciding what must change. - A valid repair must keep all four required input values in the domain. - Two functions can share most ordered pairs and still be different functions.

1. The original relation assigns two outputs to input \(2\), so one of the two pairs with first coordinate \(2\) must be removed. 2. Deleting \((2,9)\) gives second term \(7\). Deleting \((2,7)\) gives second term \(9\). 3. In each repaired relation, every input in \(\{1,2,3,4\}\) has exactly one output, so each relation is a function. They are different sequences because they assign different outputs to input \(2\).

Delete either \((2,9)\) or \((2,7)\). The resulting second term is respectively \(7\) or \(9\). Both repairs are functions because every allowed input has exactly one output, but the two functions differ at input \(2\).

- Pair each listed value with the position number used by that student. - Two functions are equal only when their domains and all corresponding input-output assignments agree. - Compare an input that appears in both domains after writing the ordered pairs.

1. Riley's ordered pairs are \(\{(1,5),(2,8),(3,11),(4,14)\}\). 2. Morgan's ordered pairs are \(\{(0,5),(1,8),(2,11),(3,14)\}\). 3. The functions are not the same. Their domains differ, and at the shared input \(1\), Riley's function gives \(5\) while Morgan's gives \(8\).

Riley: \(\{(1,5),(2,8),(3,11),(4,14)\}\). Morgan: \(\{(0,5),(1,8),(2,11),(3,14)\}\). They are different functions because the domains differ and they assign different outputs to input \(1\).

- Read the output at each plotted term number before comparing consecutive terms. - A one-step change is later output minus earlier output, so keep the sign of the result. - When ranking the changes, remember that a positive number is greater than a negative number and that ties are possible.

1. Read the graph values: \(w(1)=4\), \(w(2)=10\), \(w(3)=7\), \(w(4)=15\), and \(w(5)=12\). 2. The one-step changes are \(10-4=6\), \(7-10=-3\), \(15-7=8\), and \(12-15=-3\). 3. Ranked from greatest to least, the changes are \(8,6,-3,-3\). 4. The greatest increase is from term \(3\) to term \(4\). The greatest decrease is a tie: term \(2\) to term \(3\) and term \(4\) to term \(5\), each with change \(-3\).

Changes ranked greatest to least: \(8,6,-3,-3\). Greatest increase: term \(3\) to term \(4\). Greatest decrease: term \(2\) to \(3\) and term \(4\) to \(5\), tied at \(-3\).

- Work only with the six allowed inputs in the stated domain. - Keep the fraction form long enough to decide whether each value simplifies to a whole number. - The range is formed from output values, not from the term numbers.

1. Evaluate the rule on the stated domain: \(k(1)=12\), \(k(2)=6\), \(k(3)=4\), \(k(4)=3\), \(k(5)=2.4\), and \(k(6)=2\). 2. The range is \(\{12,6,4,3,2.4,2\}\). 3. Whole-number outputs occur at \(n=1,2,3,4,6\). The only noninteger output occurs at \(n=5\), where \(k(5)=2.4\).

Range: \(\{12,6,4,3,2.4,2\}\). Whole-number outputs occur for \(n=1,2,3,4,6\); the noninteger output occurs for \(n=5\).

- Calculate the first few terms and look for a repeating pattern. - Track how each new factor changes the preceding term. - Compare terms in even-numbered and odd-numbered positions.

1. The first four terms are \(-2, 1, -2, 1\). 2. Each pair of factors has product \((-2) \cdot (-0.5) = 1\), so every even-numbered term is \(1\), and every odd-numbered term is \(-2\). 3. Term 6 is positive and equals \(1\). 4. Term 7 is negative and equals \(-2\). 5. Since \(100\) is even, Term 100 is \(1\), so its absolute value is \(1\).

a) Positive; \(1\) b) Negative; \(-2\) c) \(1\), because every even-numbered term equals \(1\).

- Find the common difference between consecutive terms. - To test whether a number appears, set the term formula equal to that number and solve for \(n\). - Interpret the \(+1\) in \(3n+1\) in relation to multiples of \(3\).

1. The sequence starts at \(4\) and increases by \(3\), so \(a_n=4+3(n-1)=3n+1\). 2. To test \(82\), solve \(3n+1=82\). Then \(3n=81\), so \(n=27\). Since \(27\) is a positive whole number, \(82\) is the \(27\)th term. 3. The formula \(3n+1\) is one more than a multiple of \(3\), so every term has remainder \(1\) when divided by \(3\).

a) \(a_n=3n+1\) b) Yes. \(82\) is the \(27\)th term. c) The remainder is \(1\).

- Count how many equal increases occur between the first term and term \(n\). - An explicit rule should give the output directly from the term number without listing all earlier terms. - For the reverse question, set the explicit rule equal to the given output and solve for the term number.

1. From the first term to term \(n\), there are \(n-1\) increases of \(4\). Therefore, \(p(n)=7+4(n-1)=4n+3\). 2. \(p(25)=4(25)+3=103\). 3. Solve \(4n+3=83\). Then \(4n=80\), so \(n=20\).

a) \(p(n)=4n+3\) b) \(p(25)=103\) c) \(n=20\)

- Build the sequence in order because each new term depends on values that came before it. - For the reverse question, compare the target output with the term values you generated. - To decide whether the rule defines a function, focus on whether one allowed input can produce more than one output.

1. Using the two preceding terms each time gives \(r(3)=2+5=7\), \(r(4)=5+7=12\), \(r(5)=7+12=19\), and \(r(6)=12+19=31\). 2. Yes. The value \(12\) occurs when \(n=4\), so \(r(4)=12\). 3. The two starting values are fixed, and the rule determines each later term uniquely from earlier terms. Therefore every allowed term number has exactly one output.

a) \(r(3)=7\), \(r(4)=12\), \(r(5)=19\), \(r(6)=31\) b) Yes; the input is \(n=4\). c) The starting values and recursive rule determine exactly one output for every allowed term number.

- Compare the horizontal coordinates that are actually included in each representation. - The algebraic rule tells how to obtain an output once an input is allowed; it does not by itself specify the domain. - For part b, use the rule with the allowed input \(4\).

1. The continuous function accepts real inputs, and \(2(2.5)-1=4\), so \((2.5,4)\) lies on the line. The sequence accepts only the listed whole-number term numbers, so \(2.5\) is not in its domain. 2. \(s(4)=2\cdot4-1=7\). 3. The displayed sequence has domain \(\{1,2,3,4,5\}\). The two functions use the same output rule but have different domains, so they are different functions and have different graphs.

a) \((2.5,4)\) belongs to the continuous line but not to the sequence because \(2.5\) is not a sequence input. b) \(s(4)=7\) c) Domain: \(\{1,2,3,4,5\}\). The same formula with different domains defines different functions.

- Count the number of changes that have occurred before term \(n\), rather than counting the term number itself as a change. - A correct explicit rule must reproduce the given first term. - Check a later term to make sure the repeated change is represented consistently.

1. The first term is \(30\), and \(n-1\) decreases have occurred by term \(n\). Therefore \(d(n)=30-4(n-1)\), equivalently \(d(n)=34-4n\). 2. At \(n=1\), \(d(1)=30-4(0)=30\), as required. 3. At \(n=5\), \(d(5)=30-4(4)=14\), which agrees with four successive decreases from \(30\). 4. The incorrect expression \(-4n+30\) gives \(26\) at \(n=1\), because it applies one decrease too early.

\(d(n)=30-4(n-1)\), or \(d(n)=34-4n\). It gives \(d(1)=30\) and \(d(5)=14\). The rule \(-4n+30\) is shifted one term because it subtracts \(4\) already at \(n=1\).

- Start with the stated domain and ask which panel displays points only at those allowed inputs. - Look between two consecutive whole-number term numbers in the connected panel. - A graph represents every point that is drawn, not only the marked endpoints.

1. Panel a) is the more accurate graph because it shows only the four isolated input-output pairs in the stated domain. 2. Panel b) appears to allow every real input between \(1\) and \(4\), such as \(1.5\), \(2.3\), and \(3.8\). 3. The line segments create additional points between the sequence terms. Those additional points assign outputs to inputs that are not in the sequence's domain, so the connected drawing represents a different, continuous piecewise-linear function.

a) Panel a) b) It appears to allow all real inputs between \(1\) and \(4\). c) Connecting the dots adds input-output pairs that are not part of the discrete sequence, so it changes the represented function.

- Track the changes between consecutive terms, not only the term values themselves. - The next change continues the pattern in the list of amounts being added. - For linearity, ask whether equal one-step changes in the input produce equal changes in the output.

1. After \(12\), add \(5\) to get \(17\), then add \(6\) to get \(23\). The sixth and seventh terms are \(17\) and \(23\). 2. Start with \(2\). To move from term \(n\) to term \(n+1\), add \(n\). This produces additions \(1,2,3,4,\ldots\). 3. The consecutive differences are \(1,2,3,4,5,6,\ldots\), which are not constant. Therefore the sequence is nonlinear.

a) Sixth term: \(17\); seventh term: \(23\) b) Start at \(2\); after term \(n\), add \(n\) to obtain the next term. c) No. Its consecutive differences change rather than remaining constant.

- Generate only as many successive bounce heights as you need to locate the threshold. - Distinguish between multiplying by the same factor and subtracting the same amount. - Test the coach's statement against more than one transition between bounces.

1. The bounce heights are \(96,48,24,12,6,\ldots\). The fourth bounce is still above \(10\) inches, while the fifth is \(6\) inches, so bounce \(5\) is the first below \(10\) inches. 2. The loss from bounce \(1\) to bounce \(2\) is \(96-48=48\) inches, but the loss from bounce \(2\) to bounce \(3\) is \(48-24=24\) inches. The losses are not a constant \(48\) inches. “Half as high” describes a constant multiplicative factor, not a constant additive loss.

a) Bounce \(5\) b) The claim is false. The first two losses are \(48\) inches and \(24\) inches, so the additive loss changes even though each new height is half the preceding height.

- Ask what operation was performed last when moving forward from one term to the next. - Undo one operation at a time rather than trying to guess an earlier term. - Check a recovered term by applying the forward rule to it.

1. To reverse “double, then subtract \(1\),” first add \(1\), then divide by \(2\). From \(u(5)=17\): \(u(4)=\frac{17+1}{2}=9\), \(u(3)=\frac{9+1}{2}=5\), \(u(2)=\frac{5+1}{2}=3\), and \(u(1)=\frac{3+1}{2}=2\). 2. Forward from \(u(5)=17\), \(u(6)=2\cdot17-1=33\). 3. The forward rule performs multiplication before subtraction. Undoing it must reverse both the operations and their order: undo the subtraction by adding \(1\), then undo the multiplication by dividing by \(2\).

a) \(u(4)=9\), \(u(3)=5\), \(u(2)=3\), \(u(1)=2\) b) \(u(6)=33\) c) Reverse the operations in reverse order: add \(1\), then divide by \(2\).

- Decide which rule applies before substituting each term number. - Organize the outputs in term-number order after evaluating the two cases. - For the function question, ask whether any allowed input can fall into both cases at once.

1. For odd inputs, \(p(1)=6\), \(p(3)=8\), and \(p(5)=10\). For even inputs, \(p(2)=4\), \(p(4)=8\), and \(p(6)=12\). The outputs in order are \(6,4,8,8,10,12\). 2. \(p(3)=8\) and \(p(4)=8\), so consecutive term numbers \(3\) and \(4\) have the same output. 3. Every allowed input is either odd or even, but not both. Therefore exactly one of the two formulas applies to each input, giving exactly one output.

a) \(6,4,8,8,10,12\) b) \(n=3\) and \(n=4\) c) Odd and even inputs form separate cases, so exactly one rule applies to each allowed input.

- Compare the change in Sequence B's output between neighboring plotted term numbers. - After you find a rule for B, solve the equality equation and then check whether the resulting input belongs to the stated sequence domain. - To find the first term where A is larger, compare the neighboring integer inputs around the crossover. - The graph of a sequence contains only the inputs that are actual term numbers.

1. The outputs of Sequence B decrease by \(1\) each time \(n\) increases by \(1\), and \(B(1)=14\). Thus \(B(n)=15-n\) for \(n=1,2,\ldots,8\). 2. Solve \(2n+1=15-n\). Then \(3n=14\), so \(n=\frac{14}{3}\). This is not an integer term number, so the two sequence functions never have the same value at an allowed input. 3. At \(n=4\), \(A(4)=9\) and \(B(4)=11\), so A is still smaller. At \(n=5\), \(A(5)=11\) and \(B(5)=10\), so \(n=5\) is the first term number for which \(A(n)>B(n)\). 4. If the isolated points were connected, the continuous lines would appear to intersect at \(n=\frac{14}{3}\). But \(\frac{14}{3}\) is not in the sequence domain, so that apparent intersection is not an equality of sequence terms.

a) \(B(n)=15-n\) for \(n=1,2,\ldots,8\) b) The real-number equation gives \(n=\frac{14}{3}\), so there is no allowed term number at which the sequences are equal. c) \(n=5\) d) A connected graph would create an apparent intersection at a noninteger input that is not in the sequence domain.

- A proposed rule must agree with every displayed input-output pair, not only the first ones checked. - Evaluate each candidate at later plotted term numbers and compare the outputs with the graph. - Think about what extra assumption makes two points sufficient to determine a function rule.

1. Kai's rule gives \(T(3)=10\), but the graph shows \(T(3)=12\), so his rule fails. Mira's rule gives \(2,6,12,20,30\) for \(n=1,2,3,4,5\), matching all five plotted terms. Therefore \(T(n)=n(n+1)\) matches the displayed sequence. 2. \(T(8)=8 \cdot 9=72\). 3. Two input-output pairs do not determine a unique rule among all possible sequence functions. They determine a unique line only when the function is already known to be linear. Since no linearity condition was given, matching the first two terms was not enough evidence.

a) \(T(n)=n(n+1)\) matches all five plotted terms; \(T(n)=4n-2\) does not. b) \(T(8)=72\) c) Two terms can fit more than one possible sequence rule; two points determine a unique rule only within a restricted family such as linear functions.