If \(p=4-\sqrt{99}\), in which of the following intervals does \(p\) lie on the number line?
Answer and explanation
Correct answer: Between −6 and −5
Since \(9^2=81<99<100=10^2\), we have \(9<\sqrt{99}<10\). Therefore, \(-10< -\sqrt{99}< -9\), and adding 4 gives \(-6<p<-5\). Hence, \(p\) lies between −6 and −5. Exam tip: when multiplying an inequality by \(-1\), reverse its direction; this prevents choosing the nearby interval −5 to −4.
Frequently asked questions
What is the correct answer to this question?
Between −6 and −5
Why is this the correct answer?
Since \(9^2=81<99<100=10^2\), we have \(9<\sqrt{99}<10\). Therefore, \(-10< -\sqrt{99}< -9\), and adding 4 gives \(-6<p<-5\). Hence, \(p\) lies between −6 and −5. Exam tip: when multiplying an inequality by \(-1\), reverse its direction; this prevents choosing the nearby interval −5 to −4.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Representing real numbers on the number line.
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