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1. circle each value that is a perfect square. 50 81 289 360 4 100 75 2…

Question

  1. circle each value that is a perfect square.

50 81 289 360 4 100 75 224
directions: find each square root.

  1. $sqrt{36}$
  2. $-sqrt{225}$
  3. $-sqrt{64}$
  4. $sqrt{324}$
  5. $sqrt{121}$
  6. $-sqrt{169}$
  7. $sqrt{\frac{16}{9}}$
  8. $sqrt{\frac{81}{400}}$
  9. $-sqrt{\frac{1}{100}}$

directions: identify the two consecutive integers in which each square root lies between.

  1. $sqrt{95}$
  2. $sqrt{320}$
  3. $-sqrt{17}$
  4. $-sqrt{156}$
  5. $sqrt{48}$
  6. $-sqrt{249}$

Explanation:

Step1: Recall perfect - square definition

A perfect square is a number that can be written as the square of an integer.
$81 = 9^2$, $289=17^2$, $4 = 2^2$, $100 = 10^2$. So the perfect - squares among the given numbers are 81, 289, 4, 100.

Step2: Calculate square roots

For $\sqrt{36}$, since $6\times6 = 36$, $\sqrt{36}=6$.
For $-\sqrt{225}$, since $15\times15 = 225$, $-\sqrt{225}=- 15$.
For $-\sqrt{64}$, since $8\times8 = 64$, $-\sqrt{64}=-8$.
For $\sqrt{324}$, since $18\times18 = 324$, $\sqrt{324}=18$.
For $\sqrt{121}$, since $11\times11 = 121$, $\sqrt{121}=11$.
For $-\sqrt{169}$, since $13\times13 = 169$, $-\sqrt{169}=-13$.
For $\sqrt{\frac{16}{9}}$, since $\frac{4}{3}\times\frac{4}{3}=\frac{16}{9}$, $\sqrt{\frac{16}{9}}=\frac{4}{3}$.
For $\sqrt{\frac{81}{400}}$, since $\frac{9}{20}\times\frac{9}{20}=\frac{81}{400}$, $\sqrt{\frac{81}{400}}=\frac{9}{20}$.
For $-\sqrt{\frac{1}{100}}$, since $\frac{1}{10}\times\frac{1}{10}=\frac{1}{100}$, $-\sqrt{\frac{1}{100}}=-\frac{1}{10}$.

Step3: Find consecutive integers for square - roots

For $\sqrt{95}$, since $9^2 = 81$ and $10^2 = 100$, $\sqrt{95}$ lies between 9 and 10.
For $\sqrt{320}$, since $17^2=289$ and $18^2 = 324$, $\sqrt{320}$ lies between 17 and 18.
For $-\sqrt{17}$, since $4^2 = 16$ and $5^2 = 25$, $-\sqrt{17}$ lies between - 5 and - 4.
For $-\sqrt{156}$, since $12^2 = 144$ and $13^2 = 169$, $-\sqrt{156}$ lies between - 13 and - 12.
For $\sqrt{48}$, since $6^2 = 36$ and $7^2 = 49$, $\sqrt{48}$ lies between 6 and 7.
For $-\sqrt{249}$, since $15^2 = 225$ and $16^2 = 256$, $-\sqrt{249}$ lies between - 16 and - 15.

Answer:

  1. Perfect - squares: 81, 289, 4, 100
  2. 6
  3. - 15
  4. - 8
  5. 18
  6. 11
  7. - 13
  8. $\frac{4}{3}$
  9. $\frac{9}{20}$
  10. $-\frac{1}{10}$
  11. 9 and 10
  12. 17 and 18
  13. - 5 and - 4
  14. - 13 and - 12
  15. 6 and 7
  16. - 16 and - 15