QUESTION 4 4.1 Round off the following numbers to the decimal places indicated: 4.1.1 4,3768534 (4 decimal places) 4.1.2 \( \sqrt{7,24357} \) (1 decimal place) 2 Given: \( p=\sqrt{\mathrm{b}^{2}+4 \mathrm{ac}} \) 4.2.1 Determine the value of \( p \) if \( a=2 ; b=6 ; c=2 \) (Leave your answer in the simplest surd form) 4.2.2 State whether \( p \) is rational or irrational. 4.2.3 Between which TWO consecutive integers does \( p \) lie? Write the following recurring decimals in fraction form: 4.3.1 \( 0,6 \dot{3} \) 4.3.2 \( 0,12 \dot{4} \)
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4.1.1 Rounding 4,3768534 to four decimal places gives us 4,3769. The fifth decimal place (3) tells us to round up since it is greater than 5. So, we trim off the number and are left with a delightful four digits! 4.1.2 The square root of 7,24357 is approximately 2.7 when rounded to one decimal place. The number sits cozy at just above 2.6 but below 2.8, so 2.7 takes the crown for the one-decimal approximation! For 4.2.1, let's dive into the equation! Plugging in our values, \( p=\sqrt{6^2+4(2)(2)} = \sqrt{36 + 16} = \sqrt{52} \), which simplifies to \( 2\sqrt{13} \). For 4.2.2, since \( \sqrt{13} \) is not a perfect square, \( p \) is irrational! For 4.2.3, finding the range for \( 2\sqrt{13} \) places it between 7 and 8 because \( \sqrt{13} \) is about 3.605, making \( 2\sqrt{13} \) roughly equal to 7.21. Now, moving to the delightful world of recurring decimals in 4.3.1, \( 0,6 \dot{3} \) can be expressed as \( \frac{19}{30} \). And for 4.3.2, converting \( 0,12 \dot{4} \) gives us \( \frac{37}{300} \). Who knew fractions could be so fun?