Linear growth and decay

A linear rule has the form \(y=a+mx\), where \(a\) is the initial value and \(m\) is the rate of change. A decay rate of 6 means that the rate is \(-6\). Therefore, the rule with initial value 90 is \(y=90-6x\). In \(y=90+6x\), the value increases rather than decays. Exam tip: check the constant term for the initial value and a negative coefficient of \(x\) for decay.

Given (T(d)=18+2d), substitute (d=0): (T(0)=18+2(0)=18). Hence, the correct answer is 18. The value 20 would be obtained for d=1, since (T(1)=18+2=20). Exam tip: To find the initial value of a function, set the independent variable equal to 0.

Given \(V(t)=45-5t\). Substituting \(t=0\), we get \(V(0)=45-5(0)=45\). Therefore, 45 is correct. The value 40 is obtained for \(t=1\), so it is a close but incorrect option. Exam tip: in a linear expression, the constant term gives the initial value at \(t=0\).

In \(y=7+3x\), the coefficient of \(x\) is \(3\). Therefore, when \(x\) increases by 1, \(y\) increases by 3 each time. In \(y=7-3x\), the value decreases by 3, while \(y=7+x\) grows by only 1 per step. Exam tip: in a linear equation \(y=a+bx\), \(b\) gives the change per step.

In \(y=10-4x\), the coefficient of \(x\) is \(-4\). Therefore, when \(x\) increases by 1, \(y\) decreases by 4. In \(y=4x-10\), the coefficient is \(+4\), so it represents an increase of 4, not a decrease. Exam tip: In a linear expression \(y=a+bx\), the coefficient \(b\) gives the change per step.

In linear growth, a quantity increases by a fixed difference over equal time intervals. Here it rises by 3 cm each week. Doubling indicates exponential growth. Exam tip: look for a constant difference, not a constant ratio.

First find the two values: \(K(2)=70-3\times2=64\) and \(K(5)=70-3\times5=55\). Therefore, \(K(2)-K(5)=64-55=9\), so option C is correct. The distractor 15 may result from taking only \(3\times5\) instead of evaluating and subtracting the function values. Exam tip: evaluate each function value separately before finding their difference.

In linear growth, the value increases at a constant positive rate as the independent variable increases. Therefore, the line has a positive slope and rises from left to right. A line that falls from left to right represents linear decay. Exam tip: a positive slope means the line rises as you move to the right.

In linear decay, a quantity decreases by an equal amount over equal intervals. Hence, its graph is a straight line with a negative slope, moving downward from left to right. A line parallel to the x-axis represents a constant quantity, not decay. Exam tip: identify linear decrease by its negative slope.

Given (Q(x)=6+2x), substitute (x=1): \(Q(1)=6+2(1)=6+2=8\). Therefore, the correct answer is 8. The value 10 would result if the term were 4x instead of 2x. Exam tip: To evaluate a polynomial, replace the variable with the given number throughout the expression.

Substitute 4 for x: J(4)=25-2(4)=25-8=17. Therefore, 17 is the correct option. The value 19 would result from an incorrect calculation of 2×4. Exam tip: When evaluating a function, put the given value in brackets and perform multiplication first.

At the start, that is, when \(d=0\), the plant’s height is \(10\) cm. Since it increases by \(2\) cm per day, its height after \(d\) days is \(h(d)=10+2d\). Option B represents a decrease in height, so it does not model growth. Exam tip: in a linear rule, the constant term is the initial value and the coefficient of \(d\) is the change per day.

The box starts with 40 candies, so the constant term must be 40. Since 5 candies are removed each day, the daily change is \(-5\). Therefore, after \(d\) days, the number of candies is \(c(d)=40-5d\). The rule \(c(d)=40+5d\) represents an increase, not a decrease. Exam tip: in a linear rule, the constant term is the initial value and a decrease has a negative slope.

Given \(E(t)=11+3t\), we get \(E(2)=11+3(2)=17\) and \(E(3)=11+3(3)=20\). Therefore, the difference is \(20-17=3\). The value 6 would arise if \(t\) increased by 2, but here it increases only from 2 to 3. Exam tip: in a linear expression \(a+bt\), the difference between consecutive values is always \(b\).

Here, Z(1)=100-9(1)=91 and Z(2)=100-9(2)=82. Therefore, the decrease from Z(1) to Z(2) is 91-82=9. A decrease of 18 would occur over two time steps, whereas the question asks about only one step, from t=1 to t=2. Exam tip: In a linear expression, the change between consecutive values of t equals the coefficient of t in magnitude.

In \(y=a+bx\), \(b\) is the rate of change (slope) of \(y\) with respect to \(x\). In linear growth, \(y\) increases as \(x\) increases, so \(b>0\). A negative \(b\) represents linear decay, while \(b=0\) gives a constant value of \(y\). Exam tip: growth has a positive slope; decay has a negative slope.

In (y=a-bx), the coefficient of (x) is (-b). Since (b>0), the slope is negative. Therefore, as (x) increases, (bx) increases, so a larger amount is subtracted from (a) and (y) decreases. (y) can be 0 only for a particular value of (x), not always. Exam tip: In a linear equation, a negative coefficient of (x) means that (y) decreases as (x) increases.

In linear decay, the quantity falls by the same amount in equal time intervals, so its slope is negative. A fixed percentage decrease is exponential decay, not linear. Exam tip: in \(Y(t)=mt+c\), \(m<0\) indicates decay.

This represents linear decay because increasing \(t\) by 1 changes \(N\) by the fixed amount \(-b\). Thus, the quantity decreases by \(b\) per time unit. Option D has a changing rate, so it is not linear. Exam tip: a negative coefficient of \(t\) indicates decay.

In y=14+6x, the coefficient of x is 6, which is positive. Thus, for every increase of 1 in x, y increases by 6, 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 tells whether there is growth or decay.