Question

question 1-1 simplify the expression (7^-2) + (7^-5). what is the result? 7^3 7^-7 7^-3 7^5

Explanation:

Step1: Recall exponent - addition rule

When adding terms with the same base \(a^m + a^n\) (where \(a\) is the base and \(m,n\) are exponents), we cannot directly add the exponents. However, if it is \(a^m\times a^n=a^{m + n}\). In this case, for \(7^{-2}+7^{-5}\), we first find a common - denominator. But if it is \(7^{-2}\times7^{-5}\), according to the rule \(a^m\times a^n=a^{m + n}\), where \(a = 7\), \(m=-2\) and \(n = - 5\).

Step2: Calculate the sum of exponents

\(m + n=-2+( - 5)=-2-5=-7\). So \(7^{-2}\times7^{-5}=7^{-7}\). But if it is addition \(7^{-2}+7^{-5}=\frac{1}{7^{2}}+\frac{1}{7^{5}}=\frac{7^{3}+1}{7^{5}}\). Since the problem seems to be a mis - type and should be multiplication, if we assume it is multiplication: \(7^{-2}\times7^{-5}=7^{-2+( - 5)} = 7^{-7}\). If we assume the problem is as written (addition), there is no correct option. Assuming it is a multiplication problem and the options are mis - written in terms of positive exponents, the correct result of \(7^{-2}\times7^{-5}\) is \(7^{-7}\). But if we consider the closest match in the given options based on the wrong operation assumption (addition is wrong here, multiplication is more likely), and using the correct multiplication rule, the result of \(7^{-2}\times7^{-5}\) is \(7^{-7}\). If we assume the options are just wrong in sign representation of exponents for the multiplication operation, the correct operation \(7^{-2}\times7^{-5}\) gives \(7^{-7}\). If we assume the problem is a multiplication problem and the options are written in a non - standard way for negative exponents, the closest conceptually is when we use the rule \(a^m\times a^n=a^{m + n}\).

Answer:

None of the given options are correct for the addition operation \(7^{-2}+7^{-5}\). If the operation was meant to be multiplication \(7^{-2}\times7^{-5}\), the correct result is \(7^{-7}\) and there is no correct option among the given ones.