Linear growth and decay

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.

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.

This situation is represented by a linear decay function because the inventory decreases by the same amount, 15 items, for each increase of one day. To determine when the inventory is 100, substitute the target into the model: 250 − 15d = 100. Subtracting 100 from 250 gives 15d = 150. Dividing both sides by 15 yields d = 10. A substitution check gives I(10) = 250 − 15(10) = 250 − 150 = 100, confirming the answer. Hence option C is correct. The alternatives give different amounts: I(8) = 130, I(9) = 115, and I(15) = 25. Therefore none of them satisfies the stated target of 100 items.

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 expression \\(y=34+8x\\) is a linear rule. Its coefficient of \\(x\\), namely 8, tells us that \\(y\\) increases by 8 whenever \\(x\\) increases by 1. We need the total change, not the value of \\(y\\) at just one endpoint. From \\(x=1\\) to \\(x=6\\), the change in \\(x\\) is \\(6-1=5\\).

Multiplying the constant rate by this change gives the increase: \\(8\times5=40\\). Directly, \\(y(1)=34+8=42\\) and \\(y(6)=34+48=82\\), so the difference is \\(82-42=40\\) as well. The constant term 34 cancels when finding a difference. Hence option B, 40, is correct; 48 would incorrectly use the endpoint value of the variable part without accounting for the starting point.

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.

Given \(H(t)=30+6t\). Substituting \(t=5\), \(H(5)=30+6(5)=30+30=60\) cm. Therefore, 60 cm is correct. 66 cm would result from using \(t=6\), but the given value is \(t=5\). Exam tip: after substitution, perform multiplication before addition.

Given \(W(t)=180-12t\). Substituting \(t=9\), \(W(9)=180-12(9)=180-108=72\) litres. Therefore, 72 litres is correct. The nearby distractor 84 litres does not result from substituting 9 correctly in the given expression. Exam tip: in a linear decay expression, substitute the time value first, multiply, and then subtract the decrease from the initial amount.