Linear growth and decay

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 \(b\).

The coefficient is (-9), so the value decreases by (9) each step. The amount of decrease is (9) in positive form.

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.