Linear growth and decay

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 \(b>0\).

In \(P=15+4t\), the power of \(t\) is 1 and 4 is the constant growth rate, so \(P\) rises by 4 per time unit. \(15t^2\) has a changing rate. Exam tip: identify linear forms as \(a+bt\).

Given H(t)=36+4t, substitute t=10: H(10)=36+4(10)=36+40=76. Therefore, option B is correct. The value 72 would result from incorrectly adding 36 and 36; here, 4t becomes 4×10=40. In exams, substitute the input value first and then simplify carefully.

Substitute 8 for t: \(K(8)=144-12\times 8=144-96=48\). Therefore, the correct answer is 48. The value 96 is only \(12\times 8\); it must still be subtracted from 144. Exam tip: To evaluate a function, substitute the given value for the variable and follow the order of operations.

The coefficient of \(x\) is \(-3\), so \(y\) decreases as \(x\) increases; this is linear decay. A positive slope would indicate growth. Exam tip: the coefficient of \(x\) is the slope.

In \(y=50-3x\), the coefficient of \(x\) is \(-3\), so \(y\) falls by 3 for each one-unit rise in \(x\). Relations involving \(x^2\) or \(1/x\) are not linear. Exam tip: identify decay by a constant negative slope.

The initial number is 5, so it is the constant term. Since the number rises by 6 at each step, the total increase after \(x\) steps is \(6x\). Hence, the rule is \(y=5+6x\). In \(y=6+5x\), the initial value and the rate of increase are interchanged. Exam tip: in a linear rule, the constant term is the starting value and the coefficient of \(x\) is the change per step.

Given \(y=42+3x\), substitute \(x=6\): \(y=42+3(6)=42+18=60\). Hence, 60 is correct. The value 48 results from doing \(42+6\) and incorrectly omitting multiplication by 3. Exam tip: after substitution in a linear expression, perform multiplication before addition or subtraction.

Given \(y=85-5x\), substitute \(x=9\): \(y=85-5(9)=85-45=40\). Therefore, the correct value is 40. The value 45 is only \(5\times9\); it must be subtracted from 85. Exam tip: When evaluating a linear expression, substitute the given value first, multiply, and then perform addition or subtraction.