Answer:
18
Step-by-step explanation:
1/6 due to 3 appitiezers and 3 desserts 1/6
162 m^2
you can make a triangle in the corner with legs length s/2 and the angles are 45 degrees. in a 45 45 90 triangle, the legs are h/sqrt(2) where h is the hypotenuse. this means that s/2=9/2 sqrt(2), s=9 sqrt(2)
and s^2=162
Answer:
40
Step-by-step explanation:
4x2=8
8x5=40
if we have a number like say hmm 4, and we say hmmm √4 is ±2, it simply means, that if we multiply that number twice by itself, we get what's inside the root, we get the 4, so (+2)(+2) = 4, and (-2)(-2) = 4, recall that <u>minus times minus = plus</u>.
so, any when we're referring to even roots like ![\bf \sqrt[2]{~~},\sqrt[4]{~~},\sqrt[6]{~~}....](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B2%5D%7B~~%7D%2C%5Csqrt%5B4%5D%7B~~%7D%2C%5Csqrt%5B6%5D%7B~~%7D....)
, the positive number, that can multiply itself an even amount of times, will produce a valid value, BUT the negative number that multiply itself an even amount of times, will also produce a valid value.
now, that's is not true for odd roots like ![\bf \sqrt[3]{~~},\sqrt[5]{~~},\sqrt[7]{~~}....](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B3%5D%7B~~%7D%2C%5Csqrt%5B5%5D%7B~~%7D%2C%5Csqrt%5B7%5D%7B~~%7D....)
, because the multiplication of the negative number will not produce a valid value, let's put two examples on that.
![\bf \sqrt[3]{27}\implies \sqrt[3]{3^3}\implies 3\qquad because\qquad (3)(3)(3)=27 \\\\\\ however\qquad (-3)(-3)(-3)\ne 27~\hspace{8em}(-3)(-3)(-3)=-27 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \sqrt[3]{-125}\implies \sqrt[3]{-5^3}\implies -5\qquad because\qquad (-5)(-5)(-5)=-125 \\\\\\ however\qquad (5)(5)(5)\ne -125~\hspace{10em}(5)(5)(5)=125](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B3%5D%7B27%7D%5Cimplies%20%5Csqrt%5B3%5D%7B3%5E3%7D%5Cimplies%203%5Cqquad%20because%5Cqquad%20%283%29%283%29%283%29%3D27%0A%5C%5C%5C%5C%5C%5C%0Ahowever%5Cqquad%20%28-3%29%28-3%29%28-3%29%5Cne%2027~%5Chspace%7B8em%7D%28-3%29%28-3%29%28-3%29%3D-27%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B-125%7D%5Cimplies%20%5Csqrt%5B3%5D%7B-5%5E3%7D%5Cimplies%20-5%5Cqquad%20because%5Cqquad%20%28-5%29%28-5%29%28-5%29%3D-125%0A%5C%5C%5C%5C%5C%5C%0Ahowever%5Cqquad%20%285%29%285%29%285%29%5Cne%20-125~%5Chspace%7B10em%7D%285%29%285%29%285%29%3D125)
so, when the root is an odd root, you will always get only one number that will produce the radicand.
