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\[ \frac{49^2-35^2}{14}= \]

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by M7MBA » Sat May 05, 2018 11:51 pm
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$$\frac{49^2-35^2}{14}=$$ A. 74
B. 76
C. 78
D. 79
E. 84

The OA is the option E.

What is the best way to solve this PS question? May anyone helps me? Please.
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by Vincen » Sun May 06, 2018 8:59 am
Hello M7MBA.

To solve this PS question, it is easier using the following formula: $$a^2-b^2=\left(a+b\right)\left(a-b\right).$$ Now, since the numerator is $$49^2-35^2$$ then we can rewrite the given expression as follows: $$\frac{49^2-35^2}{14}=\frac{\left(49+35\right)\left(49-35\right)}{14}=\frac{84\cdot14}{14}=84.$$ Hence, the correct answer is the option E .

I hope it helps.
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by Scott@TargetTestPrep » Tue May 08, 2018 3:15 pm
M7MBA wrote:$$\frac{49^2-35^2}{14}=$$ A. 74
B. 76
C. 78
D. 79
E. 84
Simplifying the expression we have:

(49 - 35)(49 + 35)/14

(14)(84)/14 = 84

Answer: E

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