Linear growth and decay

The coefficient of t is 3, which is positive. Therefore, for every increase of 1 unit in t, the plant’s height increases by 3 cm, so this is linear growth. In linear decay, the coefficient of t would be negative, such as 20-3t. Exam tip: In y=a+bt, the sign of b indicates growth or decay.

In W(t)=50-4t, the coefficient of t is -4. This means that the amount of water decreases by 4 litres for every unit increase in time. Since the rate of change is constant and negative, it represents linear decay. For a constant amount, the coefficient of t would be 0. Exam tip: in y=a+bt, b<0 indicates linear decay.

The initial value is the value when the independent variable is 0. Here, putting d=0 gives C(0)=15+2(0)=15, so the correct answer is 15. The number 2 is the coefficient of d; it represents the rate of increase per unit, not the initial value. Exam tip: in the form y=a+bx, a is the initial value.

In the linear expression P(t)=80-5t, the coefficient of t is -5. Therefore, when t increases by 1 unit, P decreases by 5 units, so the rate of change is -5. The value 5 gives only the magnitude of the decrease; the negative sign shows that the value is decreasing. Exam tip: In a linear rule of the form ax+b, a is the rate of change.

Given \(S(t)=10+6t\). Substituting \(t=4\), \(S(4)=10+6(4)=10+24=34\). Therefore, 34 is correct. A value such as 30 can result from incorrectly calculating \(6\times4\). Exam tip: substitute the value of the variable first, then multiply before adding.

Given \(M(t)=90-7t\). Substituting \(t=5\), \(M(5)=90-7(5)=90-35=55\). Therefore, the correct answer is 55. The value 35 is only \(7\times5\); it must be subtracted from 90. Exam tip: To evaluate a function, substitute the given value for the variable and follow the order of operations carefully.

In \(B(n)=100+8n\), the coefficient of \(n\) is 8. It means that the number of books increases by 8 whenever \(n\) increases by 1. The constant term 100 gives the initial number of books, not the growth rate. Exam tip: in a linear expression \(a+bn\), \(b\) represents the rate of change per unit of \(n\).

To find the initial amount, put t=0: L(0)=60-3(0)=60 litres. Therefore, the correct answer is 60 litres. The value 57 litres is obtained at t=1, when 3 litres have already decreased. Exam tip: in a linear expression a+bt, the value at t=0 is the constant term a.

A linear rule has the form \(y=a+bx\). When \(b\) is positive, \(y\) increases at a constant rate as \(x\) increases. In \(y=12+5x\), the coefficient of \(x\) is \(5\), so it represents linear growth. \(y=30-2x\) is linear but represents decay because its coefficient is negative, while \(y=x^2+1\) is not linear. Exam tip: check the sign of the coefficient of \(x\) to identify linear growth.

A linear decay rule has the form \(y=a-bx\), where \(b>0\). In \(y=25-3x\), the coefficient of \(x\) is \(-3\), so \(y\) decreases by 3 for every increase of 1 in \(x\). \(y=18+4x\) represents linear growth because its coefficient is positive, while \(y=x^2-3\) is quadratic rather than linear. Exam tip: check whether the coefficient of \(x\) is negative to identify linear decay.

In A(t)=12+4t, the coefficient of t is 4. Therefore, when t increases by 1, A(t) increases by 4×1=4. The constant term 12 does not affect the rate of increase. Exam tip: In a linear expression a+bt, the change per 1 unit is b.

In a linear rule R(t)=a+bt, the coefficient b gives the change in R for each increase of 1 in t. Here b=-6, so R decreases by 6 units whenever t increases by 1. Directly, R(t+1)-R(t)=[40-6(t+1)]-[40-6t]=-6, confirming a decrease of 6. Thus option A is correct; 40 is the initial value, not the change.

Substitute \(w=3\) in the given expression: \(N(3)=30+5(3)=30+15=45\). Therefore, the correct answer is 45. The value 40 would be obtained for \(w=2\), so it is a close but incorrect option. Exam tip: To evaluate an expression, substitute the given value of the variable carefully.

The given function is \(B(h)=100-10h\). Substituting \(h=6\), \(B(6)=100-10(6)=100-60=40\). Therefore, 40% battery remains. The option 50% is incorrect because the decrease in 6 hours is \(10\times6=60\) percentage points, not 50. Exam tip: In linear-decay questions, substitute the given value of \(h\) first, then multiply and subtract.

Given \(P(n)=200+25n\), substitute \(n=2\): \(P(2)=200+25(2)=200+50=250\). Hence, 250 is correct. The value 225 would result from adding the increase only once, but at \(n=2\), the increase of 25 is counted twice. Exam tip: To evaluate a polynomial, substitute the given value, multiply first, and then add or subtract.

Given \(D(t)=120-15t\). Substituting \(t=4\), \(D(4)=120-15\times4=120-60=60\) km. Hence, 60 km of distance is left after 4 time units. The value 75 km would result from using \(t=3\), not \(t=4\). Exam tip: Substitute the given value of \(t\) in every term of the linear expression carefully.

Here, the coefficient of x is 0, so y remains 9 even when x changes. Therefore, this is a constant-value rule. In linear growth or decay, the coefficient of x is non-zero. Exam tip: In y=a+bx, the rule is constant when b=0.

In a linear expression, the growth rate is the coefficient of t. In (G(t)=5+9t), the coefficient of t is 9, so the growth rate is 9. The number 5 is the initial value, not the growth rate. Exam tip: in an expression of the form at+b, a is the growth rate.

The rate is (-8), and its magnitude is (8). In decay, the amount of decrease may be asked as a positive value.

The initial value is the constant term (25), and the rate is the coefficient (4). So (y=25+4x) is correct.