Linear growth and decay

Given \(D(d)=3+2.5d\). Substituting \(d=6\), \(D(6)=3+2.5\times6=3+15=18\) GB. Hence, 18 GB is correct. The value 15 GB represents only \(2.5\times6\); the initial 3 GB must also be added. Exam tip: after substitution in a linear expression, multiply before adding.

Substitute h=8 in the model: \(T(8)=38-2.5\times 8=38-20=18\). Therefore, the temperature is 18. Option 20 may result from multiplying 2.5 incorrectly. Exam tip: In a linear model, perform multiplication before addition or subtraction.

Substituting x=10 gives y=7+1.5(10)=7+15=22. Therefore, the correct answer is 22. The value 18 may result from incorrectly adding or multiplying 7 and 1.5; here, 1.5 must first be multiplied by 10. Exam tip: after substitution, perform multiplication before addition or subtraction.

Given \(y=64-3.5x\). Substituting \(x=12\), we get \(y=64-3.5\times12=64-42=22\). Therefore, 22 is correct. The value 24 would result only if the amount subtracted were 40, but \(3.5\times12=42\). Exam tip: Multiply the decimal rate by the given value first, then subtract.

When k=0, y=25+0x=25. The value of y does not change as x changes, so this is a constant function, not growth. Linear growth requires k>0, whereas k<0 represents linear decay. Exam tip: In y=a+kx, k is the slope or rate of change; its sign indicates growth or decay.

Substituting \(k=0\) gives \(y=85-0x=85\). Thus, \(y\) does not decrease as \(x\) changes, so the relation is not linear decay. It is a constant function with slope \(0\). Option B describes the graph, but since the question asks what is correct about decay, A is the most appropriate answer. Exam tip: In \(y=a-kx\), decay occurs only when \(k>0\), because the slope \(-k\) is then negative.

Set the number of members equal to 72: \(45+3w=72\). Thus, \(3w=27\), so \(w=9\). If \(w=8\), the number of members would be \(45+24=69\), not 72. Exam tip: equate the function to the target value first, then solve for the variable.

From (250-15d=100), (15d=150) and (d=10). In decay set the remaining amount equal to the target.

Given \(f(x)=4x+9\), \(f(5)=4\times5+9=29\) and \(f(7)=4\times7+9=37\). Therefore, \(f(5)+f(7)=29+37=66\). A value such as 72 can result from incorrect substitution while evaluating the function. Exam tip: evaluate each function value separately before adding them.

Given \(g(x)=75-5x\), \(g(4)=75-5(4)=55\) and \(g(9)=75-5(9)=30\). Therefore, \(g(4)+g(9)=55+30=85\). Getting 95 would result from an error in substitution or arithmetic. Exam tip: evaluate the function separately at each input before adding the results.

For the two expressions to be equal, set \(18+5x=30+3x\). Subtracting \(3x\) and 18 from both sides gives \(2x=12\), so \(x=6\). Checking: \(A(6)=48\) and \(B(6)=48\). The nearby option \(x=5\) does not make the two values equal. Exam tip: To find when two linear expressions are equal, equate them first and solve for \(x\).

For the two values to be equal, set \(140-8x=100-3x\). This gives \(40=5x\), so \(x=8\). On checking, \(C(8)=140-64=76\) and \(D(8)=100-24=76\). At \(x=6\), their values are different, so it is not correct. Exam tip: to find where two linear expressions are equal, equate them first and solve for \(x\).

Given \(y=21+7x\) and \(y=77\), we get \(21+7x=77\). Thus, \(7x=56\), so \(x=8\). Therefore, 8 is the correct option. For instance, if \(x=7\), then \(y=70\), not 77. In exams, substitute the given value of \(y\) into the equation and solve for \(x\).

Put y=55 in the equation: 132-11x=55. Therefore, 11x=132-55=77, so x=7. Hence, 7 is the correct option. If x=6, the value of y is 66, not 55. Exam tip: Substitute the target value of y first, then isolate the x-term.

In \(y=18-2x\), the coefficient of x is \(-2\), so y decreases by a constant 2 for every increase of 1 in x. B shows growth, while C is not linear. Exam tip: check the sign of the x-coefficient.

The change in x is 14 - 4 = 10. The coefficient of x is -9, so for every increase of 1 in x, y decreases by 9. Thus, the total change in y is -9 × 10 = -90, meaning y decreases by 90. The value 10 is the change in x, not in y. Exam tip: In a linear equation y = a + bx, multiply b by Δx to find the total change in y.

Given \(F(t)=20+rt\) and \(F(3)=41\), we get \(20+3r=41\). Thus, \(3r=21\) and \(r=7\). Now \(F(7)=20+7\times7=69\). The value 62 would result from using an incorrect rate or value of \(t\). Exam tip: first find \(r\) from the given function value, then substitute the required \(t\).

Given \(G(t)=q-5t\). Substituting \(t=4\), we get \(G(4)=q-20=60\), so \(q=80\). Now, substituting \(t=9\), \(G(9)=80-5(9)=80-45=35\). Hence, 35 is the correct answer. The nearby option 40 does not result from the correct subtraction. Exam tip: first find the unknown constant from the given function value, then substitute the required input.

The change in (x) is (6-1=5) and the growth rate is (8). Therefore the total increase is (8\times5=40).

The equation y = 160 - 12x is linear, and the coefficient of x is -12. This means y decreases by 12 units whenever x increases by 1 unit. From x = 2 to x = 7, the increase in x is 7 - 2 = 5. Therefore, the total decrease is 12 × 5 = 60 units. The starting value and final value also confirm this: y(2) = 136 and y(7) = 76, so 136 - 76 = 60. Hence option B is correct; the other values result from using an incorrect interval or rate.