Linear growth and decay

First, substitute \(3t\) for \(t\): \(K(3t)=120-5(3t)=120-15t\). Therefore, \(K(t)-K(3t)=(120-5t)-(120-15t)=120-5t-120+15t=10t\). Hence, the correct answer is \(10t\). The expression \(15t\) is only the value of \(5(3t)\), not the complete difference. Exam tip: use brackets while subtracting expressions so that the signs are handled correctly.

Given \(Y(x)=a+6x\), we get \(Y(3)=a+18\) and \(Y(7)=a+42\). Hence, \(Y(3)+Y(7)=2a+60=124\). Therefore, \(2a=64\), so \(a=32\). If 30 were used, the sum would be \(120\), not 124. Exam tip: Substitute each given \(x\)-value into the function separately before adding.

Given \(Z(x)=b-7x\), we get \(Z(2)=b-14\) and \(Z(6)=b-42\). Hence, \((b-14)+(b-42)=128\), so \(2b-56=128\). Therefore, \(2b=184\) and \(b=92\). If 96 were used, the sum would be 136, so it is not correct. Exam tip: Substitute each given \(x\)-value into the function separately before adding the results.

For the two polynomials to be equal, \(18+5x=7x-4\). Subtracting \(5x\) from both sides gives \(18=2x-4\), and adding 4 gives \(22=2x\). Hence, \(x=11\). Substituting \(x=10\) does not give equal values, so it is a close but incorrect option. Exam tip: Equate the expressions first, then collect the variable terms on one side.

Put p(x)=q(x): 150-8x=54+4x. Subtracting 54 from both sides gives 96-8x=4x, so 96=12x. Hence, x=8. If x=9 is substituted, the two expressions do not have equal values. Exam tip: First equate the two polynomials, then collect all x-terms on one side.

In the linear equation \(y=25+(k-4)x\), the coefficient of \(x\), or slope, is \(k-4\). For linear growth, the slope must be positive. Thus, \(k-4>0\), which gives \(k>4\). When \(k=4\), the slope is zero and \(y=25\) is a constant line, not growth. Exam tip: check whether the coefficient of \(x\) is positive for growth and negative for decay.

In the linear form \(y=a+mx\), the value of \(y\) decreases as \(x\) increases when the slope \(m\) is negative; this represents linear decay. Here, the slope is \(-(k-5)=5-k\). For decay, \(5-k<0\), which gives \(k>5\). If \(k<5\), the slope is positive and the relation shows growth, while for \(k=5\), the slope is zero and the value remains constant. Exam tip: To identify growth or decay, check the sign of the coefficient of \(x\).

In the linear function \(R(x)=65+mx\), the value at \(x=0\) is found directly by substituting zero. The term containing \(m\) disappears because \(m\times0=0\). Thus the constant term is the value where the graph meets the vertical axis. The extra condition involving two function values can be used to find the slope, but it is not needed to find \(R(0)\).

Indeed, \(R(0)=65+m(0)=65\), so option C follows immediately. The given difference also checks consistently: \(R(9)-R(4)=(65+9m)-(65+4m)=5m=55\), giving \(m=11\). This confirms the function but does not change its initial value. A value such as 55 or 11 describes the difference or slope, not \(R(0)\).

\(S(3)=s-18\) and \(S(10)=s-60\). Therefore, \(S(3)-S(10)=(s-18)-(s-60)=42\). The parameter \(s\) cancels out, so the given condition is true for every real \(s\). Values such as \(s=42\) or \(s=6\) are merely particular cases, not required values. Exam tip: In a difference of function values, check whether a common constant parameter cancels.

Find the difference: \(B(t)-A(t)=(40+7t)-(28+3t)=12+4t\). The coefficient of \(t\) is 4, so the difference increases by 4 for every increase of 1 in \(t\). The value 12 is only the initial difference, not the rate of increase. Exam tip: When subtracting linear expressions, change the signs of all terms in the expression being subtracted.

Find the difference: \(C(t)-D(t)=(170-4t)-(110-9t)=60+5t\). Since the coefficient of \(t\) is \(5\), the difference increases by \(5\) for every one-unit increase in \(t\). The number \(60\) is the initial difference, not the rate of change. Exam tip: When subtracting an expression, change the signs of all terms in the expression being subtracted.

In \(y=100-5x\), the coefficient \(-5\) means that \(y\) decreases by 5 units for every 1-unit increase in \(x\). Relations containing \(x^2\) or \(1/x\) are not linear. Exam tip: linear decay has a negative coefficient of \(x\).

In \(y=18-3x\), the coefficient of x is the constant \(-3\), so y falls at the same rate as x increases. A relation containing \(x^2\) is not linear. Exam tip: for linear decay, look for x to power 1 with a negative coefficient.

Given \(f(x)=7x+18\), we get \(f(2x)=7(2x)+18=14x+18\). Hence, \(f(2x)-f(x)=(14x+18)-(7x+18)=7x\). Setting \(7x=42\) gives \(x=6\). Therefore, option C is correct. For \(x=5\), the difference would be \(35\), not \(42\). Exam tip: while finding \(f(2x)\), substitute the entire \(2x\) in place of \(x\).

Given \(g(x)=144-8x\), we get \(g(2x)=144-8(2x)=144-16x\). Therefore, \(g(x)-g(2x)=(144-8x)-(144-16x)=8x\). Setting \(8x=64\) gives \(x=8\), so option C is correct. For \(x=7\), the difference is \(56\), not 64. Exam tip: while finding \(g(2x)\), substitute the entire expression \(2x\) for \(x\).

This is a linear function. Here, \(P(3)=42+6\times3=60\) and \(P(11)=42+6\times11=108\), so \(P(3)+P(11)=168\). Also, \(P(7)=42+6\times7=84\); hence \(2P(7)=2\times84=168\). Therefore, the correct answer is \(P(7)\). \(P(6)\) is a close distractor, but \(2P(6)=156\), not 168. Exam tip: for a linear function, the function value at the midpoint of two inputs equals the average of their function values.

Since this is a linear function, the average of the function values at two inputs equals the function value at their midpoint. The midpoint of 2 and 12 is 7. Checking directly, \(Q(2)=150\) and \(Q(12)=100\), so \(Q(2)+Q(12)=250\). Also, \(Q(7)=125\), hence \(2Q(7)=250\). Therefore, \(Q(7)\) is correct. \(Q(6)\) is close, but \(2Q(6)=260\). Exam tip: for a linear function, first find the average of the two input values in questions of this form.

The change in x is 12 - 5 = 7. In the linear expression y = 90 + ax, the change in y is aΔx. Hence, 7a = 63, so a = 9. The constant term 90 does not affect the change in y. Exam tip: First find Δx, then use Δy = aΔx in such questions.

The change in \(x\) is \(12-4=8\). In \(y=220-bx\), every increase of 1 in \(x\) decreases \(y\) by \(b\). Hence, the total decrease is \(8b=96\), giving \(b=12\). If \(b=11\), the decrease would be \(8\times11=88\), not 96. Exam tip: for a linear relation, check the rate using \(\frac{\text{total change}}{\text{change in input}}\).

“\(A(t)\) is 5 more than \(B(t)\)” means \(A(t)=B(t)+5\). Thus, \(20+4t=35+3t+5\), which gives \(t=20\). At \(t=18\), the difference is only \(3\), so it is close but not correct. Exam tip: In “more than” questions, first write the difference as \(A(t)-B(t)\).

“12 less than (D(t))” means (C(t)=D(t)-12). Therefore, (180-7t=60+5t-12). Simplifying gives (180-7t=48+5t), so (132=12t) and hence (t=11). Therefore, option C is correct. At (t=10), the difference between the two functions is not 12. Exam tip: In “less than” questions, subtract the stated difference from the reference expression first.

The sum is ((a+10)+(a+20)+(a+40)=3a+70). From (3a+70=126), (a=\frac{56}{3}), so none of the listed options is correct.

Given \(g(t)=b-4t\), we get \(g(1)=b-4\), \(g(5)=b-20\), and \(g(9)=b-36\). Their sum is \((b-4)+(b-20)+(b-36)=3b-60\). Hence, \(3b-60=156\), so \(3b=216\) and \(b=72\). Therefore, option D is correct. If \(b=68\), the sum would be \(144\), not 156. Exam tip: Substitute each input value separately before adding the function values.

In \(Q=80-5t\), the coefficient of \(t\) is the constant \(-5\), so the quantity falls by 5 units per time interval. \(80+5t\) shows growth, not decay. Exam tip: linear decay has a constant negative slope.

In \(y=-3x+12\), the coefficient of \(x\) is \(-3\), so \(y\) falls by 3 units for every increase of 1 in \(x\). Substituting \(x=0\) gives 12, which is positive. Option C has a negative initial value. Exam tip: decay requires a negative slope.