Linear growth and decay

Given \(D(d)=4+1.5d\). Substituting \(d=8\), we get \(D(8)=4+1.5\times8=4+12=16\) GB. Therefore, 16 is correct. The value 14 does not include the fixed initial 4 GB and is not the total. Exam tip: Substitute the given value in every term of a linear expression, then multiply before adding.

Substituting h=10 gives T(10)=44-2.5(10)=44-25=19. Therefore, 19 is the correct option. Getting 21 would be incorrect because 2.5 multiplied by 10 is 25, not 23. In exams, multiply the rate by the input first and then subtract it from the initial value.

Substitute x=6 in the given equation: y=9+3.5(6)=9+21=30. Therefore, the correct answer is 30. Option 27 would result from using 9+3.5×5, so it is not correct for x=6. Exam tip: After substitution, perform multiplication before addition or subtraction.

Substitute x=10: y=75-4.5(10)=75-45=30. Therefore, 30 is the correct answer. The value 35 can result from incorrectly taking 4.5×10 as 40. In exams, take care of the decimal place when multiplying by 10.

Here, k is the coefficient of x, or the slope of the line. Substituting k=0 gives y=32, so y does not change when x changes. Therefore, it is neither growth nor decay; it is a constant function. For decay, the slope must be negative, i.e., k<0. Exam tip: In y=a+kx, the sign of k tells whether the quantity grows, decays, or remains constant.

Substituting k=0 gives y=140-0x=140. Thus, y does not change when x changes, so the relation is neither growth nor decay; it is a constant function. For linear decay in y=140-kx, k must be positive so that the slope −k is negative. Exam tip: substitute the given value of k first and then check the slope.

Set the number of members equal to 100: \(60+4w=100\). Thus, \(4w=40\), so \(w=10\). If \(w=9\), the membership would be \(60+4(9)=96\), not 100. Exam tip: In a linear expression, substitute the target value first and then isolate the variable term.

For 120 items to remain, set \(I(d)=120\). Thus, \(300-20d=120\), so \(20d=180\) and \(d=9\). Therefore, the correct option is \(d=9\). At \(d=8\), 140 items would remain, so it is a close but incorrect distractor. Exam tip: In a linear expression problem, equate the expression to the required target value and solve.

Given \(f(x)=6x+11\), we get \(f(4)=6\times4+11=35\) and \(f(8)=6\times8+11=59\). Therefore, \(f(4)+f(8)=35+59=94\). The value 96 can result from incorrectly handling or omitting the constant term 11. Exam tip: evaluate each function value separately before adding them.

Given \(g(x)=90-4x\), \(g(5)=90-4(5)=70\) and \(g(12)=90-4(12)=42\). Therefore, \(g(5)+g(12)=70+42=112\), so option C is correct. An answer such as 106 may result from an error in subtraction or multiplication. Exam tip: substitute each input into the function separately before adding the outputs.

For the two expressions to be equal, set \(24+7x=48+4x\). Subtracting \(4x\) from both sides gives \(24+3x=48\), so \(3x=24\) and \(x=8\). Therefore, \(x=8\) is correct. For example, at \(x=6\), the two values are not equal. Exam tip: equate the expressions first, then collect like terms on the same side.

For the two values to be equal, \(180-9x=120-3x\). Rearranging gives \(60=6x\), so \(x=10\). Hence, the correct option is \(x=10\). For example, at \(x=12\), \(C(x)=72\) and \(D(x)=84\), so they are not equal. Exam tip: equate the two expressions first, then collect variable terms on one side.

Given y=36+8x and y=100, substitute 100 for y: 36+8x=100. Thus, 8x=64 and x=8. Checking, x=8 gives y=36+8(8)=36+64=100. For x=7, y=92, so it is not correct. Exam tip: Substitute the target value of y first, then isolate the term containing x.

Given \(y=165-12x\) and \(y=45\), we get \(165-12x=45\). Subtracting 165 from both sides gives \(-12x=-120\), so \(x=10\). If \(x=9\), then \(y=57\), not 45. Exam tip: Track signs carefully when transposing or simplifying negative terms.

The increase in (x) is (14-4=10). The coefficient of (x) is (5), so the change in (y) is (5×10=50). Checking directly, (y=80) at (x=4) and (y=130) at (x=14); hence (y) increases by (50). The constant term (60) affects the starting value, not the change. Exam tip: for a linear equation (y=mx+c), use (Δy=mΔx).

Here, the increase in x is 13−3=10. The coefficient of x in the equation is −11, so the change in y is −11×10=−110. Therefore, y decreases by 110. A decrease of 99 is incorrect because the change in x is 10, not 9. Exam tip: In a linear equation y=a+bx, the total change is b×Δx.

Given \(F(t)=26+rt\) and \(F(4)=58\), we get \(26+4r=58\). Thus, \(4r=32\) and \(r=8\). Now \(F(9)=26+8\times9=26+72=98\). Therefore, 98 is the correct option. A value such as 90 can result from an error while finding \(r\) or calculating \(9r\). Exam tip: first substitute the given function value to find the unknown coefficient, then evaluate the required value.

Given \(G(3)=99\), we get \(q-7(3)=99\), so \(q-21=99\) and hence \(q=120\). Now, \(G(10)=120-7(10)=120-70=50\). Therefore, 50 is correct. The value 57 can result from using an incorrect decrease instead of accounting for the full change from \(t=3\) to \(t=10\). Exam tip: first find the unknown constant from the given function value, then substitute the required input.

The change in (x) is (7-2=5) and the growth rate is (10). Therefore the total increase is (10\times5=50).

In y = 220 - 13x, the coefficient -13 represents a decrease of 13 units in y for every one-unit increase in x. The change in x from 4 to 9 is 9 - 4 = 5 units. Thus the total decrease is 13 × 5 = 65 units. Checking by substitution gives y(4) = 168 and y(9) = 103, whose difference is 168 - 103 = 65. Therefore option B is correct. Options A, C, and D arise from multiplying the rate by an incorrect number of x-intervals.