Respuesta :
1a.
[tex]0.42 = \frac{0.42}{1} = \frac{42}{100} = \frac{21}{50} [/tex]
1b. Manipulate the numbers using algebra.
[tex]100x = 42.2222. \\ 10x = 4.2222...[/tex]
Then, subtract and solve for x.
[tex]90x = 38 \\ x = \frac{38}{90} \\ x = \frac{19}{45} [/tex]
1c. Same thing, manipulate the numbers.
[tex]100x = 42.4242... \\ x = 0.4242...[/tex]
Again, subtract and solve for x.
[tex]99x = 42 \\ x = \frac{42}{99} \\ x = \frac{14}{33} [/tex]
2a. Jacob is pulling the 3 out of the root operation without getting the root of it.
[tex] \sqrt{75} = \sqrt{3} \times \sqrt{25 } = 5 \sqrt{3} [/tex]
2b. Without using a calculator, the only way that is feasible is by trial and error, using perfect squares. For example: you know
[tex] \sqrt{3} [/tex]
is somewhere between 1 and 2, as these are the closest perfect square roots to it. Try a value, say 1.5.
[tex] {1.5}^{2} = 2.25[/tex]
Not bad, seems a bit low. Lets try 1.6.
[tex] {1.6}^{2} = 2.56[/tex]
Seems like we can go bigger. 1.7 this time.
[tex] {1.7}^{2} = 2.89[/tex]
Alright, we are getting close to 3. 1.8.
[tex] {1.8}^{2} = 3.24[/tex]
Oops, too big. But no problem, because we now know that the value we are looking for is between 1.7 and 1.8, meaning the first decimal place must be 7. Let's keep going.
[tex] {1.75}^{2} = 3.0625 \\ {1.74}^{2} = 3.0276 \\ {1.73}^{2} = 2.9929[/tex]
Alright, the value is between 1.73 and 1.74.
[tex] {1.735}^{2} = 3.01... \\ {1.734}^{2} = 3.006... \\ {1.733}^{2} = 3.003... \\ {1.732}^{2} = 2.999...[/tex]
Bingo, your approximation is 1.732.
3. Convert each number to a tenth-rounded decimal.
[tex]\pi = 3.1[/tex]
[tex] \sqrt{3} = 1.7[/tex]
Find the square root of 5 first, as then you only have to double it for the remaining number.
[tex] {2.5}^{2} = 6.25 \\ {2.3}^{2} = 5.29 \\ {2.2}^{2} = 4.84 \\ {2.25}^{2} = 5.0625 \\ \sqrt{5} = 2.2[/tex]
The reason I take the 2.5 is to see whether to round up or down.
Then the last one is double the previous.
[tex]2 \sqrt{5} = 4.4[/tex]
3b.
[tex] \sqrt{3} < \sqrt{5} < \pi< 2 \sqrt{5} [/tex]
Hope this all helps.