Answer:
6 1/2
Step-by-step explanation:
i don't know the rest but I know that that one might be 6 1/2
Sum/difference:
Let
![x = 5 + (-3\sqrt{8}) = 5-3\sqrt{8}](https://tex.z-dn.net/?f=%20x%20%3D%205%20%2B%20%28-3%5Csqrt%7B8%7D%29%20%3D%205-3%5Csqrt%7B8%7D%20)
This means that
![3\sqrt{8} = 5-x \iff \sqrt{8} = \dfrac{5-x}{3}](https://tex.z-dn.net/?f=%203%5Csqrt%7B8%7D%20%3D%205-x%20%5Ciff%20%5Csqrt%7B8%7D%20%3D%20%5Cdfrac%7B5-x%7D%7B3%7D%20)
Now, assume that
is rational. The sum/difference of two rational numbers is still rational (so 5-x is rational), and the division by 3 doesn't change this. So, you have that the square root of 8 equals a rational number, which is false. The mistake must have been supposing that
was rational, which proves that the sum/difference of the two given terms was irrational
Multiplication/division:
The logic is actually the same: if we multiply the two terms we get
![x = -15\sqrt{8}](https://tex.z-dn.net/?f=%20x%20%3D%20-15%5Csqrt%7B8%7D%20)
if again we assume x to be rational, we have
![\sqrt{8} = -\dfrac{x}{15}](https://tex.z-dn.net/?f=%20%5Csqrt%7B8%7D%20%3D%20-%5Cdfrac%7Bx%7D%7B15%7D%20)
But if x is rational, so is -x/15, and again we come to a contradiction: we have the square root of 8 on one side, which is irrational, and -x/15 on the other, which is rational. So, again, x must have been irrational. You can prove the same claim for the division in a totally similar fashion.
Answer:
Step-by-step explanation:
Square Root of 85
Answer:
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Step-by-step explanation:
thannnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnks forrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr thhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhe frrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrree poinnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnts
Answer:
Step-by-step explanation:
3x+2x+7=22
5x+7-7=22-7
5x=15
5x/5=15/5
X=3