MHB Approximation for π and sqrt{2}

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The discussion evaluates two mathematical statements using approximations for π and sqrt{2}. For the first statement, substituting π with 3.1 shows that 2 is less than or equal to (π + 1)/2, confirming it as TRUE. The second statement involves comparing sqrt{7} to 2, where it's established that since 7 is greater than 4, sqrt{7} is indeed greater than 2, making this statement TRUE as well. Participants confirm the correct approach to simplifying and applying the approximations. Both statements are validated as TRUE based on the provided approximations.
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Say whether each statement is TRUE OR FALSE. Do not use a calculator or tables; use instead the approximations sqrt{2} is about 1.4 and π is about 3.1.

1. 2 < or = (π + 1)/2

2. sqrt{7} - 2 > or = 0

For question 1, I replace π with 3.1, and then simplify, right?

How do I apply the approximation given for sqrt{2} to the sqrt{7} to answer question 2?
 
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RTCNTC said:
Say whether each statement is TRUE OR FALSE. Do not use a calculator or tables; use instead the approximations sqrt{2} is about 1.4 and π is about 3.1.

1. 2 < or = (π + 1)/2

2. sqrt{7} - 2 > or = 0

For question 1, I replace π with 3.1, and then simplify, right?
Yes.
How do I apply the approximation given for sqrt{2} to the sqrt{7} to answer question 2?
7 is larger than 4 so sqrt{7} is larger than 2.
 
RTCNTC said:
Say whether each statement is TRUE OR FALSE. Do not use a calculator or tables; use instead the approximations sqrt{2} is about 1.4 and π is about 3.1.

1. 2 < or = (π + 1)/2

2. sqrt{7} - 2 > or = 0

For question 1, I replace π with 3.1, and then simplify, right?

How do I apply the approximation given for sqrt{2} to the sqrt{7} to answer question 2?

For 2 all you need to remember is that since $\displaystyle \begin{align*} 4 < 7 \end{align*}$ that means $\displaystyle \begin{align*} \sqrt{4} < \sqrt{7} \end{align*}$...
 
Question 1

2 < or = (π + 1)/2

Let pi be about 3.1

2 < or = (3.1 + 1)/2

2 < or = 4.1/2

2 < or = 2.05

TRUE
 
Question 2

sqrt{7} - 2 > or = 0

Like you said, sqrt{4} < sqrt{7} because 4 < 7.

Then the answer is TRUE.
 
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