Linear growth and decay

The equation y=48-6x is linear, and the coefficient of x is -6. This means that when x increases by 1, y decreases by 6, so it represents linear decay. It is not a constant function because y changes with x. Exam tip: In a linear equation y=a+bx, a positive b indicates growth and a negative b indicates decay.

Given \(U(t)=5t+15\). Substituting \(t=7\), \(U(7)=5\times7+15=35+15=50\). Therefore, 50 is correct. Getting 45 is incorrect because adding 15 to \(5\times7=35\) gives 50. Exam tip: To find a function value, substitute the given number for the variable, then follow the order of operations.

Given \(J(t)=81-7t\). Substituting \(t=8\), we get \(J(8)=81-7(8)=81-56=25\). Therefore, 25 is the correct option. The value 32 could result from incorrectly calculating \(7\times8\) as 49. Exam tip: substitute the value first, then perform multiplication before subtraction.

Here, \(X(4)=8+6\times4=32\) and \(X(0)=8\). Therefore, the difference is \(32-8=24\). Option 32 is the value of \(X(4)\), not the difference. Exam tip: In a linear expression \(a+bt\), multiply the change in \(t\) by the coefficient \(b\) to find the change in the value quickly.

Here, P(0)=90-3(0)=90 and P(12)=90-3(12)=54. Therefore, P(0)-P(12)=90-54=36. Option 30 is incorrect because multiplying the decrease rate 3 by 12 gives 36. Exam tip: Substitute both values of t separately in a linear expression, then find their difference.

In the rule (y=3x+10), the coefficient of x is 3. It means that y increases by 3 units for every 1-unit increase in x. The number 10 is the constant term and gives the initial value when x=0, so it is not the rate of increase. Exam tip: in the linear form y=mx+c, m is the rate of change and c is the initial value.

In
(y=85-10x), the coefficient is
(-10). This means that when
(x) increases by 1, the value of
(y) decreases by 10. Hence, 10 represents the decrease per step. The initial value is 85 because
(y=85) when
(x=0). Exam tip: In a linear rule
(y=a+bx),
(b) gives the rate of change; a negative
(b) indicates a decrease.

Given \(P(t)=a+5t\), substitute \(t=0\): \(P(0)=a+5(0)=a\). Since \(P(0)=18\), we get \(a=18\). The number 5 is the rate of increase, not the initial value. Exam tip: in a linear expression \(a+bt\), putting \(t=0\) gives the initial value \(a\).

In \(P(t)=60-4t\), the coefficient of \(t\) is \(-4\), so the quantity decreases by 4 for every one-unit increase in time. This is linear decay. \(60(0.9)^t\) shows exponential decay. Exam tip: in a linear expression, the variable has power 1.

In \(y=15+kt\), \(k\) is the constant rate of change (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 \(y=a+bt\), the sign of \(b\) tells whether there is growth or decay.

In (y=52-kt), the coefficient of (t) is (-k). Since (k>0), (-k) is negative. As (t) increases, (kt) increases, so a larger value is subtracted from 52; therefore, (y) decreases. (y) would remain constant only if (k=0). Exam tip: in a linear expression, a positive coefficient indicates increase, while a negative coefficient indicates decrease.

Given H(t)=32+2t, substitute 10 for t: H(10)=32+2(10)=32+20=52. Therefore, 52 is the correct answer. The value 42 would result from adding 10 to 32 and incorrectly missing the multiplication 2×10. Exam tip: To evaluate a function, replace the variable with the given value and then perform the operations in order.

Given \(K(t)=110-11t\). Substituting \(t=5\), \(K(5)=110-11(5)=110-55=55\). Therefore, the correct answer is 55. The value 66 would result only if the amount subtracted were 44, but here \(11\times 5=55\). Exam tip: When evaluating a function, substitute the given value for the variable using brackets.

The coefficient of \(x\) is the slope, and here it is \(-3\). For \(\Delta x=1\), \(\Delta y=-3\times1=-3\), so \(y\) decreases by 3 units. The 7 is only the initial value. Exam tip: a negative slope indicates decay.

In \(y=30-2x\), the coefficient of x is the constant negative number \(-2\), so y decreases by 2 for every 1-unit rise in x. Expressions with \(x^2\) or \(1/x\) are not linear. Exam tip: identify the form \(y=a+bx\).

At the start, that is, when \(x=0\), the number is \(2\). Since \(3\) is added at every step, the total increase after \(x\) steps is \(3x\). Therefore, the rule is \(y=2+3x\). The expression \(y=2-3x\) represents decay, while \(y=3x-2\) has an initial value of \(-2\). Exam tip: in a linear rule, the constant term is the starting value and the coefficient of \(x\) is the change per step.

Given y=28+5x, substitute x=4: y=28+5(4)=28+20=48. Therefore, 48 is correct. Option 52 would result from incorrectly treating 5×4 as 24, whereas it equals 20. Exam tip: Substitute the given value for the variable and perform multiplication before addition.

In \(y=12-5x\), the coefficient \(-5\) means that \(y\) decreases by 5 units whenever \(x\) increases by 1. \(y=5x+12\) shows constant growth instead. Exam tip: a negative coefficient of \(x\) identifies linear decay.

A linear polynomial has the form \(a+bx\). Here \(b=-3\), so the value decreases by 3 for every increase of 1 in x. \(12-3x^2\) also decreases but is not linear. In exams, check that the power of x is 1 and its coefficient is negative.

In \(y=12-3x\), the constant coefficient of \(x\) is \(-3\), so \(y\) falls by 3 for every 1-unit rise in \(x\). In \(12-3x^2\), the rate of decrease is not constant. Exam tip: a negative constant slope indicates linear decay.