Linear growth and decay

In linear decay, the decrease over every equal time interval is a constant difference, so the model has a constant negative slope. A constant percentage decrease indicates exponential decay. Exam tip: check fixed differences, not percentages.

In linear decay, the power of \(t\) is 1 and its coefficient is negative. In \(Q(t)=240-8t\), the quantity falls by 8 per time unit. \(240-8t^2\) shows decay but is not linear. Exam tip: check the sign and power of \(t\).

For the models to be equal, set \(48+11t=168-4t\). Moving \(4t\) to the left and 48 to the right gives \(15t=120\), so \(t=8\). \(t=7\) is the closest option, but the two model values are not equal at that time. Exam tip: after equating linear models, collect all terms containing \(t\) on one side.

Equating the two expressions gives \(70+5t=190-10t\). Moving \(10t\) to the left gives \(15t=120\), so \(t=8\). Substituting \(t=6\) does not make the two values equal, so that nearby option is incorrect. Exam tip: first equate both functions, then collect all terms containing \(t\) on one side.

Given \(f(x)=a+13x\), use \(f(3)=82\): \(a+13(3)=82\), so \(a+39=82\) and \(a=43\). Therefore, \(f(9)=43+13(9)=43+117=160\). Hence, 160 is correct. A nearby value such as 156 can result from an error in the constant term or multiplication. Exam tip: find the unknown constant from the given function value before evaluating the required value.

Given \(g(x)=b-14x\). Substituting \(x=4\), \(g(4)=b-14(4)=b-56=120\), so \(b=176\). Now, substituting \(x=11\), \(g(11)=176-14(11)=176-154=22\). Therefore, 22 is the correct answer. The value 36 may result from incorrectly using \(176-140\). Exam tip: first find the unknown constant \(b\) from the given function value, then substitute the required value of \(x\).

In \(y=8-3x\), the coefficient of x is negative, so y decreases at a constant rate as x increases; this is linear decay. \(8+3x\) shows growth. Exam tip: check the sign of the x-coefficient.

In the linear form \(y=mx+c\), \(m\) is the growth or decay rate. For \(y=7x-12\), \(m=7\) is positive, so \(y\) rises as \(x\) rises. Options A and D have negative rates. Exam tip: inspect the coefficient of \(x\).

A decrease by the same amount in equal time intervals is linear decay, so the slope \(a\) is negative. For equal intervals, \(\Delta p=a\Delta x\) remains constant. A fixed percentage decrease is exponential. Exam tip: “same amount” signals a linear model.

In \(7-3x\), the coefficient of x is \(-3\), so the value falls by 3 for every increase of 1 in x. \(3x-7\) represents growth. Exam tip: linear decay always has a negative coefficient of x.

Given \(C(t)=42+kt\), \(C(11)-C(5)=(42+11k)-(42+5k)=6k\). Since this difference equals \(84\), \(6k=84\), so \(k=14\). For example, \(k=16\) would give a difference of \(6\times16=96\), not \(84\). Exam tip: when subtracting two values of a linear expression, the constant term \(42\) cancels out.

Given \(D(t)=300-kt\), we have \(D(4)=300-4k\) and \(D(12)=300-12k\). Therefore, \(D(4)-D(12)=(300-4k)-(300-12k)=8k\). Since \(8k=104\), \(k=13\), so option C is correct. If \(k=12\), the difference would be \(96\), not the given \(104\). Exam tip: When subtracting two expressions, carefully distribute the negative sign across the second expression.

Given \(A(x)=45+8x\), \(A(x+6)=45+8(x+6)\) and \(A(x-2)=45+8(x-2)\). Their difference is \(8[(x+6)-(x-2)]=8\times8=64\). The constant term 45 cancels on subtraction. Option 72 would result only if the difference between the inputs were 9; here it is 8. Exam tip: for a linear function \(mx+c\), use \(A(p)-A(q)=m(p-q)\) directly.

In a linear polynomial \(ax+b\), the coefficient \(a\) gives the change when x increases by 1. The required change is \(-7\), so \(45-7x\) is correct. In \(45+7x\), the value rises by 7. Exam tip: decay has a negative coefficient of x.

Given \(P(t)=p+14t\), we have \(P(3)=p+14\times3=p+42=96\), so \(p=54\). Now using \(P(s)=180\), \(54+14s=180\). Thus, \(14s=126\) and \(s=9\). If \(s=8\), the model gives 166, not 180. Exam tip: first find the constant term \(p\) from the known value, then solve for the required input.

Given R(t)=r-15t, use R(4)=128 first: r-15(4)=128, so r-60=128 and r=188. Now use R(s)=38: 188-15s=38. Thus, 15s=150 and s=10. Therefore, 10 is the correct option. If 12 were used, the model would give 8, not 38. Exam tip: find the constant r from one condition before substituting the second condition.

An increase of 11 per step requires a slope of 11, so the model must be \(y=a+11x\). Using \(y(5)=82\) gives \(a+11(5)=82\), hence \(a=27\). Therefore, \(y=27+11x\) is correct. Option C represents a decrease of 11 because its slope is \(-11\). Exam tip: Substitute the given \(x\)-value into each option to check the condition quickly.

A decrease of 13 per step means that the coefficient of \(x\) must be \(-13\). So the equation has the form \(y=a-13x\). Using \(y(6)=71\), we get \(a-13(6)=71\), so \(a=149\). Hence, \(y=149-13x\) is correct. In option C, \(y\) increases rather than decreases. Exam tip: Substitute the given point \((x,y)\) into the equation to check the constant term.

“18 more” means \(M(t)=N(t)+18\). Thus, \(80+7t=170-5t+18\), or \(80+7t=188-5t\). Therefore, \(12t=108\), giving \(t=9\). Option \(t=10\) may seem close, but it gives a difference of \(30\), not \(18\). Exam tip: In “more than” questions, write the larger quantity as the smaller quantity plus the stated difference.

“15 more” means \(A(t)=B(t)+15\). Thus, \(240-11t=60+4t+15\), which gives \(165=15t\) and hence \(t=11\). Therefore, option C is correct. At \(t=10\), the difference is \(30\), not 15. Exam tip: For “more than” questions, write the larger quantity as the smaller quantity plus the stated difference.