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.
The equation which can be used to find the number of quarters and nickels is; n + q = 84, 0.05n + 0.25q = 17.80.
<h3>Quarters and nickels</h3>
- Total coins = 84
- Total value = $17.80
let
- Number of quarters = q
- Number of nickels = n
n + q = 84
0.05n + 0.25q = 17.80
From equation (1)
n = 84 - q
Substitute into (2)
0.05n + 0.25q = 17.80
0.05(84 - q) + 0.25q = 17.80
4.2 - 0.05q + 0.25q = 17.80
- 0.05q + 0.25q = 17.80 - 4.2
0.20q = 13.60
q = 13.60 / 0.20
q = 68
Substitute q = 68 into
n + q = 84
n + 68 = 84
n = 84 - 68
n = 16
Therefore, the number of quarters and nickels are 68 and 16 respectively.
Learn more about quarters and nickels:
brainly.com/question/17127685
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16.75 square yards because area of a triangle is half the base times the height. 40.2÷4.8 is 8.375. Since that is half the base, multiply by two. 8.375×2 is 16.75. Hope that helps! :D
Answer:
x = (8√3)/3
Step-by-step explanation:
Reference angle = 30°
Hypotenuse = x
Side length adjacent to reference angle = 4
Apply CAH, thus:
Cos 30 = Adj/Hyp
Cos 30 = 4/x
x × Cos 30 = 4
x = 4/Cos 30
x = 4/(√3/2) (cos 30 = √3/2)
x = 4 × 2/√3
x = 8/√3
Rationalize
x = (8 × √3)/(√3 × √3)
x = (8√3)/3
Answer:
Im not sure If your suppose to calculate the whole area but for FGH Its 9
Step-by-step explanation:
