MHB How many integers in between \sqrt{19} and \sqrt{90}

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The integers between $\sqrt{19}$ and $\sqrt{90}$ are determined by identifying the smallest integer greater than $\sqrt{19}$ and the largest integer less than $\sqrt{90}$. Since $\sqrt{19}$ is approximately 4.36, the first integer greater than it is 5. Meanwhile, $\sqrt{90}$ is approximately 9.49, making the largest integer less than it equal to 9. Therefore, the integers between these two values are 5, 6, 7, 8, and 9. In total, there are five integers between $\sqrt{19}$ and $\sqrt{90}$.
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how many integers are between \sqrt{19} and \sqrt{90}
 
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prasadini said:
how many integers are between $\sqrt{19}$ and $\sqrt{90}$

You can use math tags "$$$$" or "$$" to encase your typeset equations.

Have you had a go at the problem?
 
Do you understand what this question is asking? 19 lies between 16 and 25 so the first integer larger than [math]\sqrt{19}[/math] is what? 90 lies between 81 and 100 so the last integer less than [math]\sqrt{90}[/math] is what?

How many integers lie between [math]\sqrt{19}[/math] and [math]\sqrt{90}[/math]?
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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