Linear growth and decay

Given (f(x)=12+5x), we get (f(x+2)=12+5(x+2)=12+5x+10). Hence, (f(x+2)-f(x)=(12+5x+10)-(12+5x)=10). Option 5 is only the increase per one unit of x, whereas x increases by 2 here. Exam tip: For a linear function, multiply the change in x by the coefficient of x to find the change in the function.

In \(h(t)=24-3t\), the coefficient of \(t\) is \(-3\), so the quantity falls by 3 units for every 1-unit increase in time. \(24+3t\) shows growth. Exam tip: a negative linear coefficient indicates decay.

In the linear form \(y=mx+c\), \(m\) gives the change in \(y\) when \(x\) increases by 1. Since \(y\) must increase by 4 each time, \(m=4\). Also, \(y(0)=9\) gives \(c=9\). Therefore, the correct equation is \(y=9+4x\). In option C, the rate is \(-4\), so \(y\) decreases instead. Exam tip: To check \(y(0)\), substitute \(x=0\) in the equation.

For a linear relation, \(y=mx+c\). A decrease of 3 in \(y\) when \(x\) increases by 1 gives slope \(m=-3\). Also, \(y(0)=50\) means the constant term is \(c=50\). Hence, \(y=50-3x\) is correct. Option A has the initial value 50, but it increases by 3 rather than decreasing. Exam tip: substitute \(x=0\) to check \(y(0)\).

In \(y=12-4x\), the coefficient of x is \(-4\), so y decreases by 4 for every increase of 1 in x. Relations involving \(x^2\) or \(1/x\) are not linear. Exam tip: a negative x-coefficient indicates linear decay.

Substituting x=2 gives y=90-5(2)=80. Substituting x=7 gives y=90-5(7)=55. Therefore, the difference between the two y-values is 80-55=25. Here, 5 is the decrease in y for each increase of 1 in x; over an increase of 5 units in x, the total decrease is 5×5=25. Exam tip: For a linear expression, substitute both x-values and subtract the resulting values.

Substitute d=4: D(4)=2+1.5×4=2+6=8 GB. Therefore, 8 GB is correct. 7.5 GB accounts only for 1.5×4 and misses the initial 2 GB. Exam tip: after substitution in a linear expression, multiply first and then add.

Substitute h=6 in the model: T(6)=35-1.5(6)=35-9=26. Therefore, the temperature is 26. Option 29 can result from an incorrect calculation of 1.5×6. Exam tip: In a linear model, substitute the input value first, multiply, and then add or subtract.

Substitute 4 for x in the given expression: y=5+2.5(4)=5+10=15. Therefore, the correct answer is 15. The value 12.5 is not obtained when x=4; it results from an incorrect multiplication. In exams, multiply first and then add the constant term.

Substitute x=8 into the given equation: y=42-2.5×8. Since 2.5×8=20, y=42-20=22. Therefore, 22 is the correct option. A result such as 24 may arise from an error in multiplication or subtraction. Exam tip: First multiply the decimal coefficient by x, then subtract from the constant term.

When (k=0), we get (y=18+0x=18). Thus, y does not change as x changes, so the function is constant and there is no growth. Decay would require a negative slope, i.e., (k<0). Exam tip: In (y=a+kx), use the sign of k to identify growth, decay, or a constant value.

Putting \(k=0\) in \(y=70-kx\) gives \(y=70\). Its slope is \(0\), so \(y\) neither increases nor decreases as \(x\) changes. Linear decay requires a negative slope, whereas this graph is horizontal. Exam tip: in \(y=a-kx\), if \(k>0\), the slope \(-k\) is negative.

For 50 members, form the equation 32+2w=50. Thus, 2w=18 and w=9. Therefore, the number of members will be 50 after 9 weeks. At 8 weeks, the number is 32+2(8)=48, so it is not correct. Exam tip: Equate the given expression to the target value before solving for the variable.

Given f(x)=3x+5, f(4)=3(4)+5=17 and f(6)=3(6)+5=23. Therefore, f(4)+f(6)=17+23=40. The value 35 can result from incorrectly handling the constant term 5 while evaluating one of the function values. Exam tip: Substitute each input separately into the function before adding the results.

Given \(g(x)=60-4x\), \(g(3)=60-4\times3=48\) and \(g(8)=60-4\times8=28\). Therefore, \(g(3)+g(8)=48+28=76\). Option 80 can result from an error in multiplication or subtraction. Exam tip: substitute each input into the function separately before adding the results.

For the two expressions to be equal, set \(10+4x=20+2x\). Subtracting \(2x\) from both sides gives \(10+2x=20\), so \(2x=10\) and \(x=5\). Checking: \(A(5)=30\) and \(B(5)=30\). Option \(x=4\) is a close distractor, but it gives 26 and 28, which are not equal. Exam tip: To find where two linear expressions are equal, equate them first and solve for \(x\).

For the two expressions to be equal, set \(C(x)=D(x)\): \(100-5x=80-x\). This gives \(20=4x\), so \(x=5\). Therefore, option B is correct. Substituting \(x=4\) does not give equal values, so that close distractor is incorrect. Exam tip: To find where two linear expressions are equal, equate them first and then solve for \(x\).

Given y=15+5x and y=55, substitute 55 for y: 15+5x=55. Thus, 5x=40 and x=8. Therefore, 8 is the correct option. For example, if x=7, then y=15+35=50, not 55. Exam tip: In a linear expression, substitute the given value of y and isolate the term containing x first.

Given (y=96-8x) and (y=40), we get (96-8x=40). Subtracting 96 from both sides gives (-8x=-56), so (x=7). If x=6, then y=48, not 40. Exam tip: When moving or simplifying negative terms, track the signs on both sides carefully.

Here, the increase in x is 11 - 1 = 10. In y = 2x + 30, the coefficient of x is 2, so the change in y is 2 × 10 = 20. Therefore, y increases by 20. The constant term 30 changes the starting value of y, not the amount of increase. Exam tip: In a linear equation y = mx + c, multiply the change in x by m to find the change in y.

As x changes from 3 to 13, the change in x is 10. In y = 150 - 6x, the coefficient of x is -6, so y decreases by 6 for every increase of 1 in x. Therefore, the total change in y is -6 × 10 = -60; hence, y decreases by 60. Checking directly, y(3) = 132 and y(13) = 72. The value 78 is not the final value of y, and y does not increase. Exam tip: In a linear equation, total change = coefficient of x × change in x.

In linear decay, the power of \(x\) is 1 and its coefficient is negative. In \(y=45-3x\), the coefficient of \(x\) is \(-3\), so \(y\) decreases by 3 for every one-unit increase in \(x\). \(y=45+3x\) shows linear growth, while \(y=45-3x^2\) is quadratic rather than linear. Exam tip: for \(y=a+bx\), check that \(b<0\) to identify linear decay.

Given F(t)=18+rt and F(2)=30, we get 18+2r=30. Thus, 2r=12 and r=6. Now, F(5)=18+6×5=48. Therefore, 48 is the correct answer. The value 42 would result from using t=4, so it is not correct here. Exam tip: first find r from the given function value, then substitute the required value of t.

Given \(H(t)=24+5t\). Substituting \(t=4\), \(H(4)=24+5(4)=24+20=44\) cm. Therefore, 44 cm is correct. The value 49 cm would result from an error in multiplication or addition. Exam tip: To evaluate a linear expression, substitute the given value for the variable and calculate step by step.

For a linear polynomial \(ax+b\), the sign of coefficient \(a\) determines its direction of change. In \(-3x+7\), the coefficient is \(-3\), so its value decreases as \(x\) increases. A positive coefficient, as in \(5x+2\), indicates growth. Exam tip: check the coefficient of \(x\) first.