(i) (b^(2)-25)/(2b+10)÷((b-5)^(2))/(5b^(2))

Question

(i)  (b^(2)-25)/(2b+10)÷((b-5)^(2))/(5b^(2))

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Recognize Division Rule: First, we need to recognize that dividing by a fraction is the same as multiplying by its reciprocal. So, we will rewrite the division as a multiplication by the reciprocal of the second fraction.Rewrite as Multiplication: Rewrite the expression as a multiplication by the reciprocal of the second fraction: ((b^2 - 25) / (2b + 10)) * (5b^2 / (b - 5)^2)Factor Numerator and Denominator: Next, we can factor the numerator of the first fraction and the denominator of the second fraction: ((b - 5)(b + 5) / (2b + 10)) * (5b^2 / (b - 5)(b - 5))Cancel Common Factors: Now, we can simplify the expression by canceling out common factors. The (b - 5) term in the numerator of the first fraction and one (b - 5) term in the denominator of the second fraction cancel out. Also, (2b + 10) in the denominator of the first fraction can be factored out as 2(b + 5), which cancels with the (b + 5) in the numerator.Simplify Expression: After canceling the common factors, the expression simplifies to: (1 / 2) * (5b^2 / (b - 5))Multiply Remaining Terms: Now, multiply the remaining terms: (5/2) * (b^2 / (b - 5))Final Simplified Form: Finally, we can leave the expression in its simplified form: (5b^2) / (2(b - 5))

(i) $\frac{b^{2}-25}{2 b+10} \div \frac{(b-5)^{2}}{5 b^{2}}$

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(i) b2−252b+10÷(b−5)25b2 \frac{b^{2}-25}{2 b+10} \div \frac{(b-5)^{2}}{5 b^{2}} 2b+10b2−25​÷5b2(b−5)2​

(i) $\frac{b^{2}-25}{2 b+10} \div \frac{(b-5)^{2}}{5 b^{2}}$