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If (G(t)=260-rt) and (G(3)=G(11)+104), what is (r)?

Answer and explanation

Correct answer: 13

Given \(G(t)=260-rt\), we have \(G(3)=260-3r\) and \(G(11)=260-11r\). Hence, \(G(3)-G(11)=8r\). The question states that this difference is \(104\), so \(8r=104\), giving \(r=13\). If \(r=16\), the difference would be \(8\times16=128\), so it is not correct. Exam tip: when subtracting values of a linear function at two times, the constant term \(260\) cancels out.

Related tags

Linear FunctionsRate Of ChangeAlgebraic EquationsPolynomial IntroductionLinear Growth And DecayIntroduction To PolynomialsMathematicsClass 9 Mcq

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