Linear growth and decay

Substitute \(w=4\) into the expression: \(Q(4)=20+10\times4=20+40=60\). Therefore, the number of practice questions is 60. The value 40 represents only \(10\times4\); the initial 20 questions must also be added. Exam tip: When evaluating a linear expression, substitute the variable value, multiply first, and then add or subtract.

Given \(R(d)=55-3d\). Substituting \(d=5\), \(R(5)=55-3(5)=55-15=40\) kg. Therefore, 40 kg of rice remains. The value 15 kg is only the amount decreased in 5 days, not the amount left. Exam tip: To find a function value, substitute the given number for the variable, multiply first, and then add or subtract.

Substitute 6 for x: M(6)=18+2(6)=18+12=30. Therefore, option B is correct. Getting 36 would require incorrectly multiplying 18 and 12. Exam tip: When evaluating a function, substitute the given value for the variable first and then follow the order of operations.

Substitute 7 for x: \(N(7)=45-5\times7=45-35=10\). Therefore, the correct answer is 10. A value such as 15 can result from an error in calculating \(5\times7\) or in subtraction. Exam tip: when evaluating a function, substitute the given value first, then perform multiplication and subtraction.

A linear rule has the form \(y=a+bx\), where \(a\) is the initial value and \(b\) is the rate of growth or decay. In \(y=18+6x\), the constant term is \(18\) and the coefficient of \(x\) is \(+6\), so it represents growth. Option B has \(-6\), which represents decay. Exam tip: identify the constant term for the initial value and the coefficient of \(x\) for the rate.

A linear decay rule has the form \(y=a-rx\), where \(a\) is the initial value and \(r\) is the decay rate. Substituting \(a=64\) and \(r=8\) gives \(y=64-8x\). Option A represents growth by 8, not decay. Exam tip: the initial value is the constant term, and decay has a negative coefficient of \(x\).

Given \(F(t)=13+4t\), substitute \(t=0\): \(F(0)=13+4(0)=13\). Therefore, the correct answer is 13. The number 4 is the rate of growth (the coefficient of \(t\)), not the initial value. Exam tip: In a linear expression \(a+bt\), the value at \(t=0\) is always the constant term \(a\).

Substituting t=0 gives G(0)=82-9(0)=82. Hence, the initial value is 82. The value 73 is obtained when t=1, so it is not correct here. Exam tip: To find the initial value of a linear function, substitute 0 for the independent variable.

L(5)=16+3(5)=31 and L(2)=16+3(2)=22. Therefore, L(5)-L(2)=31-22=9. The constant term 16 cancels in the difference; only the rate 3 per unit and the 3-unit change in t matter. Exam tip: for a linear expression a+bt, use L(x)-L(y)=b(x-y).

Given \(T(t)=100-12t\), \(T(1)=100-12=88\) and \(T(3)=100-36=64\). Hence, \(T(1)-T(3)=88-64=24\). Option A, 12, is only the decrease per unit time; the interval from 1 to 3 is 2 units. Exam tip: Find each function value separately before subtracting them.

Here, y=20+0x simplifies to y=20. Since the coefficient of x is 0, the value of y does not change when x changes, so it represents a constant value. Linear growth or decay requires a non-zero positive or negative coefficient of x, respectively. Exam tip: if the coefficient of x is 0, the expression is constant.

In the linear expression Z(t)=7+11t, the coefficient of t is 11. For every increase of 1 unit in t, Z increases by 11 units, so the growth rate is 11. The number 7 is the initial value, not the growth rate. Exam tip: In a linear expression of the form at+b, the coefficient of t gives the rate of growth or decay.

In the linear expression \(V(t)=90-15t\), the coefficient of \(t\), \(-15\), represents the rate of change. The negative sign shows that \(V\) is decreasing. The magnitude of the decay rate is \(|-15|=15\), so 15 is correct. \(-15\) gives the direction of change, not its magnitude. Exam tip: use the absolute value to find the magnitude of a rate.

The starting quantity is 22, so when \(x=0\), the value of \(y\) must be 22. Since it increases by 4 at every step, the coefficient of \(x\) is \(+4\). Therefore, the rule is \(y=22+4x\). In option C, the initial value and the rate of increase have been interchanged. Exam tip: in \(y=a+bx\), \(a\) is the initial value and \(b\) is the change per step.

The starting quantity is 70, so when \(x=0\), the value of \(y\) must be 70. Since it decreases by 5 at each step, the coefficient of \(x\) is \(-5\). Therefore, the rule is \(y=70-5x\). In \(y=70+5x\) and \(y=5x+70\), the quantity increases rather than decreases. Exam tip: in a linear rule, the constant term is the initial value, and a decrease has a negative slope.

Here, E(2)=10+8(2)=26 and E(3)=10+8(3)=34. Therefore, the difference is 34-26=8. A value such as 16 may result from incorrectly adding the two values of t in 8t; here t increases by only one unit. Exam tip: in a linear expression a+bt, the difference between consecutive values is always b.

Here, \(Y(2)=66-6(2)=54\) and \(Y(3)=66-6(3)=48\). Therefore, the decrease from \(Y(2)\) to \(Y(3)\) is \(54-48=6\). A decrease of \(12\) would occur over two time units, whereas this question compares consecutive time values. Exam tip: in a linear expression \(a-bt\), the value decreases by \(b\) whenever \(t\) increases by one unit.

In linear decay, \(t\) has power 1 and its coefficient is negative. \(12-3t\) has slope \(-3\). Although \(12-3t^2\) decreases, it is not linear. Exam tip: check the power of \(t\) and the sign of the slope.

In \(P(t)=45-3t\), the coefficient of \(t\) is \(-3\), so the quantity decreases by 3 for each one-unit increase in time. \(45+3t\) shows growth, while a \(t^2\) term does not give a constant rate. Exam tip: linear decay has degree 1 and a negative variable coefficient.

In y=11+2x, the coefficient of x is 2, which is positive. Thus, when x increases by 1 unit, y increases by 2 units, so the equation represents linear growth. In linear decay, the coefficient of x would be negative. Exam tip: In y=mx+c, the sign of m indicates growth or decay.