Linear growth and decay

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|.

The starting quantity is 12, so the constant term must be 12. Since the quantity increases by 5 at each step, the coefficient of x is +5. Therefore, the rule is \(y=12+5x\). In \(y=12-5x\), the quantity decreases, so it does not represent growth. Exam tip: In a linear rule, the starting value is the constant term and the change per step is the coefficient of x.

A linear rule has the form \(y=a+mx\), where \(a\) is the initial value and \(m\) is the change per step. Here, the initial value is 66 and the quantity decreases by 3 each step, so \(m=-3\). Therefore, the rule is \(y=66-3x\). The rule \(y=66+3x\) represents an increase of 3 per step, so it is not correct. Exam tip: the phrase “decreases by” indicates a negative slope.

Given E(t)=21+11t, E(4)=21+11×4=65 and E(5)=21+11×5=76. Therefore, the difference is 76−65=11. The constant term 21 does not affect the change between consecutive values; it is determined by the coefficient 11. Exam tip: in a linear expression a+bt, the difference between consecutive values is always b.

Y(3)=96-12(3)=60 and Y(4)=96-12(4)=48. Therefore, the decrease from 60 to 48 is 60-48=12. The term -12t means that Y decreases by 12 whenever t increases by 1; 24 would be the decrease over two steps. Exam tip: for consecutive values in a linear expression, use the magnitude of the coefficient of t.

In \(12-3t\), the coefficient of \(t\) is \(-3\), so the value falls by 3 for each unit increase in time. \(3t+12\) represents growth. Exam tip: for linear decay, check for a negative coefficient of \(t\).

In \(P(t)=15-6t\), the coefficient of \(t\) is \(-6\), so the quantity falls by 6 per time unit. Option A shows growth; in exams, a negative slope indicates linear decay.

The equation y=19+7x is linear, and the coefficient of x is +7. Thus, when x increases by 1 unit, y increases by 7 units, so it represents linear growth. In linear decay, the coefficient of x would be negative. Exam tip: In y=a+bx, the sign of b indicates growth or decay.

In y = 52 - 4x, the coefficient of x is -4. When x increases by 1, y decreases by 4, so the relation represents linear decay. It is not a constant function because y changes when x changes. Exam tip: In y = a + bx, a positive b indicates linear growth and a negative b indicates linear decay.

Given U(t)=9t+14, substitute t=7: U(7)=9×7+14=63+14=77. Hence, 77 is the correct option. The value 63 is only 9×7 and does not include the constant term 14. Exam tip: To evaluate a function, substitute the given input first and then follow the order of operations.

Given \(J(t)=100-6t\). Substituting \(t=11\), \(J(11)=100-6(11)=100-66=34\). Therefore, the correct answer is 34. Getting 44 would result from an error in multiplication or subtraction. Exam tip: substitute the given value first, multiply, and then perform addition or subtraction.

Here, \(X(6)=18+8\times6=66\) and \(X(0)=18\). Therefore, the difference is \(66-18=48\). Option 66 is the value of \(X(6)\), not the difference. Exam tip: when finding the difference between two values of a linear expression \(a+bt\), the constant term \(a\) cancels out.

Here, K(0)=160-8(0)=160 and K(12)=160-8(12)=64. Therefore, K(0)-K(12)=160-64=96. Option 64 is the value of K(12), not the difference. Exam tip: in linear decay, 8×12=96 directly gives the total decrease.

In \(y=20-2x\), the coefficient of x is \(-2\), so y falls by 2 for every increase of 1 in x. This is linear decay at a constant rate. Exam tip: a linear rule has x to the power 1.

In

y=125-5x

, the term

-5x

means that when

x

increases by 1, the value of

y

decreases by 5. Therefore, 5 is the amount of decrease per step. The number 125 is the initial value; a positive coefficient would indicate growth. Exam tip: In a linear rule

y=a+bx

, a negative

b

indicates decrease, and its magnitude gives the decrease per step.

Substitute t=0: P(0)=a+7(0)=a. Since P(0)=31, a=31. The number 7 is the rate of increase, not the initial value. Exam tip: in a linear expression a+bt, the value at t=0 is always a.

Given \(R(t)=b-10t\). Substituting \(t=0\) gives \(R(0)=b-10(0)=b\). Since \(R(0)=75\), \(b=75\). Option 65 could be the value of \(R(1)\), not the initial value. Exam tip: in a linear expression, substituting \(t=0\) gives the constant term, or initial value.

In \(y=29+kt\), \(k\) is the rate of change, or slope, of \(y\) with respect to \(t\). For linear growth, \(y\) must increase as \(t\) increases, so \(k\) must be positive. If \(k<0\), the relation shows linear decay, while \(k=0\) gives a constant value of \(y\). Exam tip: in \(y=a+bt\), growth requires \(b>0\).