Linear growth and decay

Here, the coefficient of x is -7. For every increase of 1 in x, the value of y decreases by 7, so it represents linear decay. In linear growth, the coefficient of x is positive. Exam tip: In y = a + bx, the sign of b indicates growth or decay.

Given \(I(t)=3t+12\), substitute \(t=6\): \(I(6)=3\times6+12=18+12=30\). Therefore, 30 is the correct option. The value 24 may result from an incorrect addition after multiplication. Exam tip: substitute the given value for the variable, then follow the order of operations.

Substitute \(t=7\) in the given expression: \(O(7)=64-6\times 7=64-42=22\). Therefore, 22 is correct. The value 28 would result from using \(6\times 6\), which does not match the given value \(t=7\). Exam tip: Substitute the variable value first, then follow the correct order of multiplication and subtraction.

\(X(3)=5+4\times3=17\) and \(X(0)=5+4\times0=5\). Therefore, the difference is \(17-5=12\). Option \(17\) is the value of \(X(3)\), not the difference. Exam tip: For a linear expression, substitute both time values separately and then subtract the results.

Given \(N(t)=80-2t\), we get \(N(0)=80\) and \(N(10)=80-2\times10=60\). Therefore, \(N(0)-N(10)=80-60=20\). Option 60 is the value of \(N(10)\), not the required difference. Exam tip: First evaluate the function at both given times, then subtract the values carefully.

In the rule y=2x+7, the coefficient of x is 2. It means that y increases by 2 when x increases by 1. The number 7 is the constant term and gives the initial value of y when x=0, so option A describes the constant term, not the coefficient of x. Exam tip: In a linear rule y=mx+c, m is the rate of change (slope) and c is the initial value.

In the rule y=60-5x, 60 is the initial value. The coefficient of x is -5, so when x increases by 1, the value of y decreases by 5. Therefore, 5 represents the amount of decrease, not the initial value. Exam tip: In a linear rule y=a+bx, b gives the rate of change; a negative b indicates decay.

Substituting (t=0) gives (P(0)=a+3(0)=a). Since (P(0)=12) is given, (a=12). The number 3 is the rate of increase, not the initial value. Exam tip: In a linear expression (a+bt), putting (t=0) directly gives the initial value (a).

Given (R(t)=b-4t), substituting (t=0) gives (R(0)=b-4(0)=b). Since (R(0)=25), we get (b=25). The value 21 would be obtained at (t=1), not at the initial time. Exam tip: To find the constant term in a linear expression, often substitute 0 for the independent variable.

In \(y=10+kt\), \(k\) is the rate of change, or slope, of \(y\) with respect to \(t\). For linear growth, \(y\) must increase as \(t\) increases, so \(k>0\) is necessary. If \(k<0\), the relation shows linear decay, while \(k=0\) makes \(y\) constant. Exam tip: In \(y=a+mt\), the sign of \(m\) indicates growth or decay.

In \(y=80-3t\), the coefficient \(-3\) means the quantity falls by 3 in every time unit, so it is linear decay. \(0.9^t\) represents exponential decay. Exam tip: identify linear relations in the form \(a+bt\).

Here, W(t)=75-5t is a linear expression because the power of t is 1. The coefficient of t is -5, so the amount of water decreases by 5 litres for every 1-unit increase in t. Therefore, it represents linear decay. A constant amount would have no t-term, while in exponential change the variable appears in the exponent. Exam tip: in the linear form at+b, a negative value of a indicates decay.

The initial value is the value obtained when the independent variable is 0. Substituting \(n=0\), we get \(P(0)=40+6(0)=40\), so 40 is correct. The number 6 is the rate of increase (slope), not the initial value. Exam tip: In a rule of the form \(a+bn\), the constant term \(a\) is the initial value.

In a linear rule, the coefficient of t gives the rate of change per unit. In S(t)=120-8t, the coefficient of t is -8, so S decreases by 8 when t increases by 1. Here, 120 is the initial value, not the rate; 8 gives only the magnitude of the decrease, so the negative sign is essential. Exam tip: In the form y=a+bt, the rate of change is b.

Given \(A(t)=9+7t\). Substituting \(t=3\), we get \(A(3)=9+7\times3=9+21=30\). Hence, the correct answer is 30. The value 21 is only \(7\times3\); the constant term 9 must also be added. Exam tip: To evaluate a function, replace the variable with the given value and simplify step by step.

Given \(B(h)=96-6h\). Substituting \(h=4\), we get \(B(4)=96-6(4)=96-24=72\). Therefore, 72 is correct. The value 24 is only \(6\times4\); it still needs to be subtracted from 96. Exam tip: To evaluate a function, substitute the given value for the variable and then follow the order of operations.

In \(y=14+9x\), the coefficient of \(x\) is \(+9\). Therefore, whenever \(x\) increases by 1, \(y\) increases by 9 each time. In \(y=30-9x\), the value decreases by 9, while in \(y=x^2+9\), the increase is not constant. Exam tip: In a linear rule \(y=a+bx\), the coefficient \(b\) gives the change per step.

In \(y=50-7x\), the coefficient of \(x\) is \(-7\). Therefore, when \(x\) increases by 1, \(y\) decreases by 7. In \(y=50+7x\), the value increases by 7, not decreases. Exam tip: In a linear rule \(y=a+bx\), a negative value of \(b\) indicates decrease.

In C(d)=25+5d, the coefficient of d is 5. When d increases by 1, the term 5d increases by 5, so C(d) also increases by 5. The constant term 25 does not change when d changes. Exam tip: In a linear expression a+bd, b gives the change in the value for a 1-unit change in the variable.

The coefficient is (-4), so the value decreases by (4) each step. The amount of decrease is (4) in positive form.