Linear growth and decay

In \(y=80-3t\), the coefficient \(-3\) means the quantity falls by 3 in every time unit, so it is linear decay. \(0.9^t\) represents exponential decay. Exam tip: identify linear relations in the form \(a+bt\).

Here, W(t)=75-5t is a linear expression because the power of t is 1. The coefficient of t is -5, so the amount of water decreases by 5 litres for every 1-unit increase in t. Therefore, it represents linear decay. A constant amount would have no t-term, while in exponential change the variable appears in the exponent. Exam tip: in the linear form at+b, a negative value of a indicates decay.

The initial value is the value obtained when the independent variable is 0. Substituting \(n=0\), we get \(P(0)=40+6(0)=40\), so 40 is correct. The number 6 is the rate of increase (slope), not the initial value. Exam tip: In a rule of the form \(a+bn\), the constant term \(a\) is the initial value.

In a linear rule, the coefficient of t gives the rate of change per unit. In S(t)=120-8t, the coefficient of t is -8, so S decreases by 8 when t increases by 1. Here, 120 is the initial value, not the rate; 8 gives only the magnitude of the decrease, so the negative sign is essential. Exam tip: In the form y=a+bt, the rate of change is b.

Given \(A(t)=9+7t\). Substituting \(t=3\), we get \(A(3)=9+7\times3=9+21=30\). Hence, the correct answer is 30. The value 21 is only \(7\times3\); the constant term 9 must also be added. Exam tip: To evaluate a function, replace the variable with the given value and simplify step by step.

Given \(B(h)=96-6h\). Substituting \(h=4\), we get \(B(4)=96-6(4)=96-24=72\). Therefore, 72 is correct. The value 24 is only \(6\times4\); it still needs to be subtracted from 96. Exam tip: To evaluate a function, substitute the given value for the variable and then follow the order of operations.

In \(y=14+9x\), the coefficient of \(x\) is \(+9\). Therefore, whenever \(x\) increases by 1, \(y\) increases by 9 each time. In \(y=30-9x\), the value decreases by 9, while in \(y=x^2+9\), the increase is not constant. Exam tip: In a linear rule \(y=a+bx\), the coefficient \(b\) gives the change per step.

In \(y=50-7x\), the coefficient of \(x\) is \(-7\). Therefore, when \(x\) increases by 1, \(y\) decreases by 7. In \(y=50+7x\), the value increases by 7, not decreases. Exam tip: In a linear rule \(y=a+bx\), a negative value of \(b\) indicates decrease.

In C(d)=25+5d, the coefficient of d is 5. When d increases by 1, the term 5d increases by 5, so C(d) also increases by 5. The constant term 25 does not change when d changes. Exam tip: In a linear expression a+bd, b gives the change in the value for a 1-unit change in the variable.

The governing idea is the constant rate of change of a linear function. In D(t) = 88 − 4t, the coefficient of t is −4, so increasing t by one unit changes D by −4. This can be shown directly: D(t+1) = 88 − 4(t+1) = 88 − 4t − 4 = D(t) − 4. Therefore the new value is 4 units smaller than the old value, and the amount of decrease is the positive number 4. Option C is correct. Option B, −4, is the signed change, not the amount by which the function decreases. Option A is the constant term and option D is not a general change; it could only be a value for a particular input, not the requested one-unit decrease.

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.