Linear growth and decay

Substituting t=5 gives A(5)=10+3(5)=25 and B(5)=25+5=30. Since 30 is greater than 25, B(t) has the greater value. The “both values are equal” option is incorrect because the values differ by 5. Exam tip: To compare linear models at a given t, substitute that value into both expressions before comparing.

Substituting \(t=5\), we get \(A(5)=10+3(5)=25\) and \(B(5)=25+5=30\). Since \(30>25\), \(B(5)>A(5)\) is correct. \(A(5)=B(5)\) is incorrect because the values differ by 5. Exam tip: Substitute the given value of \(t\) into each expression separately before comparing them.

Substituting \(t=5\), we get \(C(5)=80-4(5)=60\) and \(D(5)=70-2(5)=60\). Hence, both models give the same value. Options A and B are incorrect because neither value is larger than the other. Exam tip: Substitute the given value of \(t\) into each expression before comparing them.

Given \(T(t)=30+2t\) and \(T(t)=46\), set \(30+2t=46\). This gives \(2t=16\), so \(t=8\). Therefore, \(t=8\) is correct. For example, substituting \(t=6\) gives \(T(t)=42\), not 46. Exam tip: equate the function to the required target value first, then solve the linear equation for \(t\).

Given \(U(t)=64-4t\), set \(U(t)=28\): \(64-4t=28\). Thus, \(4t=36\), so \(t=9\). For example, at \(t=8\), \(U(8)=32\), not 28. Exam tip: substitute your final value back into the original expression to verify it.

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.

From (200-12d=80), (12d=120) and (d=10). In decay set the remaining amount equal to the target.

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.