Linear growth and decay

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.

Here, the coefficient of x is -7. For every increase of 1 in x, the value of y decreases by 7, so it represents linear decay. In linear growth, the coefficient of x is positive. Exam tip: In y = a + bx, the sign of b indicates growth or decay.

Given \(I(t)=3t+12\), substitute \(t=6\): \(I(6)=3\times6+12=18+12=30\). Therefore, 30 is the correct option. The value 24 may result from an incorrect addition after multiplication. Exam tip: substitute the given value for the variable, then follow the order of operations.

Substitute \(t=7\) in the given expression: \(O(7)=64-6\times 7=64-42=22\). Therefore, 22 is correct. The value 28 would result from using \(6\times 6\), which does not match the given value \(t=7\). Exam tip: Substitute the variable value first, then follow the correct order of multiplication and subtraction.

\(X(3)=5+4\times3=17\) and \(X(0)=5+4\times0=5\). Therefore, the difference is \(17-5=12\). Option \(17\) is the value of \(X(3)\), not the difference. Exam tip: For a linear expression, substitute both time values separately and then subtract the results.

Given \(N(t)=80-2t\), we get \(N(0)=80\) and \(N(10)=80-2\times10=60\). Therefore, \(N(0)-N(10)=80-60=20\). Option 60 is the value of \(N(10)\), not the required difference. Exam tip: First evaluate the function at both given times, then subtract the values carefully.

In the rule y=2x+7, the coefficient of x is 2. It means that y increases by 2 when x increases by 1. The number 7 is the constant term and gives the initial value of y when x=0, so option A describes the constant term, not the coefficient of x. Exam tip: In a linear rule y=mx+c, m is the rate of change (slope) and c is the initial value.

In the rule y=60-5x, 60 is the initial value. The coefficient of x is -5, so when x increases by 1, the value of y decreases by 5. Therefore, 5 represents the amount of decrease, not the initial value. Exam tip: In a linear rule y=a+bx, b gives the rate of change; a negative b indicates decay.

Substituting (t=0) gives (P(0)=a+3(0)=a). Since (P(0)=12) is given, (a=12). The number 3 is the rate of increase, not the initial value. Exam tip: In a linear expression (a+bt), putting (t=0) directly gives the initial value (a).

Given (R(t)=b-4t), substituting (t=0) gives (R(0)=b-4(0)=b). Since (R(0)=25), we get (b=25). The value 21 would be obtained at (t=1), not at the initial time. Exam tip: To find the constant term in a linear expression, often substitute 0 for the independent variable.

In \(y=10+kt\), \(k\) is the rate of change, or slope, of \(y\) with respect to \(t\). For linear growth, \(y\) must increase as \(t\) increases, so \(k>0\) is necessary. If \(k<0\), the relation shows linear decay, while \(k=0\) makes \(y\) constant. Exam tip: In \(y=a+mt\), the sign of \(m\) indicates growth or decay.