Step-by-step explanation:
Starting with the first one, we know that 3²=9 and 4²=16, so √10 is between 3 and 4. Similarly, √27 is a little over 5. √3 is close to 2, √2 is close to 1, and √9=3, so for simplicity, we can write this as ![\frac{3.something*5.something}{1.something*1.something*3}](https://tex.z-dn.net/?f=%5Cfrac%7B3.something%2A5.something%7D%7B1.something%2A1.something%2A3%7D)
The 3.something and the 3 kind of cross out, and we're left with 5.something over 2 1.somethings, which can be closer to 2 or 1, which is somewhat unclear -- therefore, our answer can be D or E, and we'll wait on this one
For the second one, we can cross out the √6s to get
-- π is a little greater than 3, so this is E, making the first one D
For the third one, we can cross out the √3 to get (3-2)/2 = 1/2, or A as it is between 0 and 1
For the last one, note the multiplication -- cross out the √3 to get (3*2√2)/6, and we get 6√2/6=√2, or C
Answer:
quantity of 16 plus 18 i over 145
Step-by-step explanation:
Answer:
y= 9x + 8
Step-by-step explanation: