Linear growth and decay

The governing concept is a linear decay model. In I(d) = 200 − 12d, the initial amount is 200 when d = 0, and the constant slope −12 means that 12 items disappear each day. To find the day on which 80 items remain, set the model equal to 80: 200 − 12d = 80. Subtracting 200 gives −12d = −120, and dividing by −12 gives d = 10. Substitution confirms the result: I(10) = 200 − 12(10) = 80. Therefore option C is correct. For comparison, d = 8 gives 104 items, d = 9 gives 92 items, and d = 12 gives 56 items, so none of those alternatives reaches the target.

A linear rule has the form y = a + bx, where b is the slope or constant rate of change. For y to show growth as x increases, b must be positive; for the rule to be non-constant, b must also be nonzero. Option C has y = 18 + 2x, so its slope is 2. Every increase of 1 in x raises y by 2, which demonstrates positive linear growth. Option A is constant because its slope is 0. Option B simplifies to y = 18 and is also constant. Option D has slope −2, so it represents linear decrease rather than growth. Hence option C is the unique correct answer, and the intercept 18 only gives the starting value; the sign of the coefficient of x determines growth or decline.

In H(d)=15+3d, the coefficient of d is +3. For every one-unit increase in d, the plant’s height increases by 3 cm, so it represents linear growth. Linear decay would have a negative coefficient of d, such as 15-3d. Exam tip: in y=a+bx, b>0 indicates growth and b<0 indicates decay.

In W(t)=90-5t, the coefficient of t is -5. This means the amount of water decreases by 5 litres for every unit increase in time. Since the rate of decrease is constant, it is linear decay. A constant quantity has no t-term, while exponential decay has the variable in the exponent. Exam tip: In y=a+bt, b<0 indicates linear decay.

The initial value is the value when the independent variable is zero. Putting \(n=0\), we get \(A(0)=22+4(0)=22\), so 22 is correct. Here, 4 is the rate of increase per unit of \(n\), while 26 is the value at \(n=1\). Exam tip: In a linear rule \(a+bn\), the initial value is the constant term \(a\).

In a linear rule, the coefficient of t gives the rate of change per unit. In B(t)=130-8t, the coefficient of t is -8, so B decreases by 8 whenever t increases by 1. The value 8 is only the magnitude of the decrease; the rate itself is -8. Exam tip: in a linear expression a+bt, the rate of change is always the coefficient b of t.

A linear polynomial has degree 1 and is of the form \(ax+b\), where \(a\ne0\). In \(5x-3\), the highest power of x is 1. \(x²+2x+1\) is quadratic, while 7 is constant. Exam tip: identify the degree by checking the highest exponent.

Given \(R(h)=84-6h\). Substituting \(h=7\), we get \(R(7)=84-6\times7=84-42=42\). Therefore, 42 is correct. The value 48 would result from an incorrect calculation instead of applying the multiplication and subtraction properly. Exam tip: substitute the given value first, then follow the order of operations.

In \(y=18+6x\), the coefficient of \(x\) is \(+6\). Therefore, whenever \(x\) increases by 1, \(y\) increases by 6, giving equal growth of 6 per step. In \(y=18-6x\), there is a decrease of 6 per step, while \(y=x^2+6\) does not have a constant rate of growth. Exam tip: In a linear rule \(y=a+bx\), \(b\) gives the change per step.

In \(y=60-4x\), the coefficient of \(x\) is \(-4\). Therefore, when \(x\) increases by 1, \(y\) decreases by 4 at each step. In \(y=60+4x\), the value increases by 4, while \(y=60\) has no change. Exam tip: In a linear rule \(y=a+bx\), the coefficient \(b\) gives the change per step.

The expression is \(C(d)=17+2d\). The coefficient of \(d\) is \(2\), so when \(d\) increases by \(1\), \(C(d)\) increases by \(2\times1=2\). The constant term \(17\) does not affect the rate of increase. Exam tip: in a linear expression \(a+bd\), the change per unit increase in \(d\) is the coefficient \(b\).

Answer: 9 units, so option A is correct. In D(t)=72−9t, the coefficient of t is −9. This means that every increase of 1 in t changes D by −9, or makes D nine units smaller. Show it algebraically: D(t+1)=72−9(t+1)=72−9t−9=D(t)−9. Thus the signed change is −9, while the amount of decrease is the positive number 9. Option B, 72, is the constant term and is not the decrease. Option C, −9, describes the signed change, but the wording “by how much will it decrease” asks for the magnitude of the decrease. Option D, 63, is not a fixed decrease; it may be the value for a particular input, but no input was specified. Memory cue: a negative coefficient means decrease, and its absolute value is the decrease per one unit.

Given \(Q(w)=30+8w\). Substituting \(w=5\), \(Q(5)=30+8\times5=30+40=70\). Therefore, the student will have solved 70 questions. Option 40 represents only \(8\times5\) and misses the initial 30 questions. Exam tip: after substituting a value in a linear expression, multiply first and then add or subtract.

Given \(S(d)=64-3d\). Substituting \(d=8\), \(S(8)=64-3\times8=64-24=40\) kg. Therefore, 40 kg is correct. The value 56 kg would result from subtracting 8 from 64, which does not follow the given expression. Exam tip: Write the substitution with brackets and perform multiplication before subtraction.

Substitute 6 for x: M(6)=25+5(6)=25+30=55. Therefore, 55 is the correct option. The value 30 is only 5×6; the constant term 25 must also be added. In exams, substitute the given value first, then multiply before adding.

Given \(N(x)=58-2x\). Substituting \(x=9\), we get \(N(9)=58-2(9)=58-18=40\). Hence, the correct answer is \(40\). The close distractor \(42\) would result from subtracting 16, but \(2\times9=18\). Exam tip: while evaluating a function, substitute the given value first and complete multiplication before subtraction.

A linear rule has the form \(y=a+bx\), where \(a\) is the initial value and \(b\) is the growth rate. In \(y=33+4x\), the constant term is \(33\) and the coefficient of \(x\) is \(+4\), so it is the correct rule. Although \(y=33-4x\) starts at 33, its rate is \(-4\), which represents decay rather than growth. Exam tip: put \(x=0\) to identify the initial value, and check the coefficient of \(x\) for the growth or decay rate.

A linear rule has the form \(y=a+bx\), where \(a\) is the initial value. A decay rate of 12 means that the value decreases by 12 for every increase of 1 in \(x\), so the coefficient of \(x\) must be \(-12\). Therefore, the rule is \(y=96-12x\). The rule \(y=96+12x\) represents growth at a rate of 12, not decay. Exam tip: put \(x=0\) to check the initial value, and look for a negative slope for decay.

Substitute t=0: F(0)=41+6(0)=41. Therefore, the correct answer is 41. The number 6 is the coefficient of t and represents the rate of increase, not the initial value. Exam tip: for a linear expression a+bt, its value at t=0 is always a.

To find the initial value, substitute \(t=0\): \(G(0)=115-5(0)=115\). Therefore, the correct answer is 115. The value 110 is obtained when \(t=1\), so it is a close but incorrect option. Exam tip: for \(G(0)\), the constant term gives the initial value.

Given L(t)=13+4t, L(7)=13+4(7)=41 and L(3)=13+4(3)=25. Therefore, L(7)-L(3)=41-25=16. Option 28 may result from confusing the value at 7 with the required difference, but the question asks for the difference of two function values. Exam tip: For a linear expression a+bt, L(x)-L(y)=b(x-y).

Given T(t)=140-10t, we get T(2)=140-10(2)=120 and T(6)=140-10(6)=80. Therefore, T(2)-T(6)=120-80=40. The value 60 can result from an incorrect subtraction or an error while evaluating the 10t term. Exam tip: Substitute each value of t separately first, then subtract the two function values.

Here, the coefficient of x is 0, so y remains 27 even when x changes. Therefore, it represents a constant value. For linear growth or decay, the coefficient of x must be positive or negative respectively, not zero. Exam tip: In y=a+bx, the value is constant when b=0.

In a linear expression \(Z(t)=a+bt\), the coefficient \(b\) of \(t\) represents the growth rate. Here, the coefficient of \(t\) is \(9\), so the growth rate is 9. The number \(16\) is the initial value, not the growth rate. Exam tip: In a linear function, identify the number multiplying the variable.

The coefficient of t is -18, so V(t) decreases by 18 units for every unit increase in time. The magnitude of the decay rate ignores the negative sign, so it is 18. Option -18 is the signed rate of change, not its magnitude. Exam tip: In a linear expression a+bt, the rate is b and its magnitude is |b|.