Linear growth and decay

Given \(A(t)=a+7t\). Substituting \(t=4\) gives \(A(4)=a+7(4)=a+28\). Since \(A(4)=50\), we get \(a+28=50\), so \(a=22\). Option 28 is the increase over 4 time units, not the initial value \(a\). Exam tip: Substitute the given value of \(t\) first, then solve the resulting linear equation.

Given R(6)=42, substitute t=6 in R(t)=b-8t: b-8(6)=42. Thus, b-48=42, so b=42+48=90. Option 84 is incorrect because 48 has not been added back correctly to 42. Exam tip: Substitute the given value of t into a linear expression and then isolate the unknown constant.

In \(N(t)=120-8t\), the coefficient of \(t\) is \(-8\), so the quantity decreases by 8 for every unit of time. Option A shows growth; for linear decay, check for a negative variable coefficient.

In \(L(h)=36-3h\), the coefficient of \(h\) is the rate of change. Its negative sign shows decay, so the candle loses 3 cm each hour. The value 36 is the initial length. Exam tip: always check the sign of the slope.

For linear growth the coefficient must be positive and the initial value must be (16). (y=16+6x) satisfies both.

For a rule to show linear decay, it must be linear and have a negative coefficient of x. Its initial value is found by substituting x=0. In option A, y=72-9x, y(0)=72 and the slope is -9, so the value falls by 9 for every unit increase in x. Option B increases, option C starts at 9, and option D is quadratic.

Given \(M(3)=39\), moving from \(t=3\) to \(t=8\) increases the input by 5. Since \(M(t)\) grows at a rate of 6 per unit, the total increase is \(5\times6=30\). Therefore, \(M(8)=39+30=69\). Option 72 would require an increase of 33, which does not match the given rate. Exam tip: for a linear function, change in output = rate × change in input.

Given \(N(t)=n-4t\). Substituting \(t=5\) gives \(N(5)=n-20=35\), so \(n=55\). Now, substituting \(t=12\), \(N(12)=55-4(12)=55-48=7\). Therefore, the correct answer is 7. The value 11 can result from incorrectly handling the decrease of \(4\times 7=28\) from \(t=5\) to \(t=12\). Exam tip: first find the unknown constant \(n\) using the given value, then substitute the required value of \(t\).

Given \(y=30+kx\), substitute \(x=4\) and \(y=54\): \(54=30+4k\). Thus, \(4k=24\), so \(k=6\). Option \(24\) is the value of \(4k\), not of \(k\). Exam tip: To find an unknown coefficient in a linear expression, substitute the given values of \(x\) and \(y\) first.

At \(x=5\), we are given \(y=110\). Substituting gives \(110=150-5k\). Hence, \(5k=40\), so \(k=8\). Note that \(40\) is the value of \(5k\), not of \(k\). Exam tip: Substitute the given values of \(x\) and \(y\) first, then isolate the unknown term.

Substituting t=6 gives A(6)=15+4×6=39 and B(6)=35+2×6=47. Since 47 is greater than 39, the value of model B is greater. Although model A has a higher growth rate, its lower initial value means that it has not overtaken B at t=6. Exam tip: substitute the same value of t into both expressions and compare the resulting numbers directly.

Substituting \(t=5\), we get \(C(5)=100-6\times5=70\) and \(D(5)=85-3\times5=70\). Therefore, the two model values are equal, so \(C(5)=D(5)\) is correct. They are not zero; their common value is 70. Exam tip: evaluate both expressions at the given value of \(t\) before comparing them.

Given \(T(t)=42+3t\), set the target value equal to 72: \(42+3t=72\). Thus, \(3t=30\), so \(t=10\). Therefore, \(t=10\) is correct. The closest distractor, \(t=9\), gives \(T(9)=42+27=69\), not 72. Exam tip: Substitute the required output value into the function and solve the resulting linear equation.

Given \(U(t)=120-10t\). For \(U(t)=50\), set \(120-10t=50\). Thus, \(-10t=-70\), so \(t=7\). If \(t=6\), then \(U(t)=60\), not 50. Exam tip: isolate the variable term by performing the same operation on both sides of a linear equation.

Given f(x)=18+4x, we get f(x+3)=18+4(x+3)=18+4x+12. Therefore, f(x+3)-f(x)=(18+4x+12)-(18+4x)=12. Option 4 is only the coefficient of x; for an increase of 3 in x, the change is 4×3=12. Exam tip: For a linear function ax+b, increasing x by h changes the function by ah.

In \(120-5x\), the constant term 120 gives the initial value, while the coefficient of x is \(-5\), so the quantity falls by 5 per unit of x. \(120+5x\) represents growth. Exam tip: decay has a negative x-coefficient.

In the linear form \(y=b+mx\), \(b=y(0)\), while \(m\) gives the change in \(y\) when \(x\) increases by 1. Here, \(y(0)=14\), so the constant term is 14; and \(y\) increases by 6 each time, so the coefficient is 6. Therefore, \(y=14+6x\) is correct. In \(y=6+14x\), the growth rate is 14, while \(y=14-6x\) represents a decrease of 6. Exam tip: substitute \(x=0\) to check the initial value, then check the coefficient of \(x\).

In a linear expression, the constant term gives \(y(0)\), so it must be \(80\). Since \(y\) must decrease by 5 when \(x\) increases by 1, the coefficient of \(x\) must be \(-5\). Hence, \(y=80-5x\) is correct. Option A has the correct initial value, but it makes \(y\) increase by 5 instead. Exam tip: check the constant term for \(y(0)\) and the coefficient of \(x\) for the rate of change.

The relation is linear: \(y=55+3x\). The difference between the two \(x\)-values is \(9-2=7\). Since \(y\) increases by 3 for every increase of 1 in \(x\), the difference in \(y\) is \(3\times7=21\). Option 18 would result from incorrectly taking the difference in \(x\) as 6. Exam tip: for a linear expression \(a+bx\), the difference between two \(y\)-values is \(b\) times the difference between the corresponding \(x\)-values.

At \(x=3\), \(y=130-8(3)=106\). At \(x=10\), \(y=130-8(10)=50\). Therefore, the difference between the two \(y\)-values is \(106-50=56\). Choosing 48 would result from using an incorrect change in \(x\) instead of the actual difference, \(10-3=7\). Exam tip: for a linear expression, calculate both values of \(y\) and subtract them to find the difference reliably.