Respuesta :
If Moussa thinks of a number whose root is a multiple of 3, then there are many possible numbers, like 9, 36, 81, etc.
The numbers between 100 and 200 that have an integer square root are (121, 144, 169 and 196).
The number that Jiale thinks of is 64.
There are three separate, unrelated questions.
Firstly, Moussa thinks of a number whose root is a multiple of 3. Let us assume that the number is "y" and the concerned multiple of 3 be "3x" where (x = 1, 2, 3, 4, 5...)
Hence, as per the question statement,
[tex]\sqrt{y} =3x \\or, y = (\sqrt{y})^{2} =(3x)^{2} \\or,y=9x^{2}...(i)[/tex]
Using (x = 1, 2, 3, 4, 5...) in equation (i), we can have our possible set of answers, i.e., (9, 36, 81...)
Second, we are required to determine all the numbers between 100 and 200 that have an integer square root. We know that, [(10)² = 100]. Therefore, we will have to calculate the squares of numbers above 10 till the squared value reaches or crosses 200, i.e.,
[tex]11^{2}=121, (100 < 121 < 200)\\12^{2}=144, (100 < 144 < 200)\\13^{2}=169, (100 < 169 < 200)\\196^{2}=196, (100 < 196 < 200)\\15^{2}=225, (200 < 225)\\[/tex]
Hence, the numbers between 100 and 200 that have an integer square root are (121, 144, 169 and 196).
Third, Jiale thinks of a number whose square root of cube root is 2. Let the number be "z". As per question statement,
[tex]\sqrt{\sqrt[3]{x}}=2\\ or,(x^{\frac{1}{3}})^\frac{1}{2} =2\\or,(x)^(\frac{1}{3}*\frac{1}{2})=2\\or, (x)^\frac{1}{6}=2\\or,x=(2)^6\\or,x=64[/tex]
And to show that Jiale would have got the same number (2) by calculating the the square root first and then the cube root,
[tex]\sqrt{64} =8,\\\sqrt[3]{8}=2[/tex].
- Multiple: A multiple of "x" will be a number which will be exactly divided by x without any remainder.
- Integer: Any number on the number line which is not a fraction is an Integer.
To learn more about Multiples & Integers, click on the link below.
https://brainly.com/question/11009848
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