A = -16 b = 0 c = 541
Use the quadratic formula which is
x = [-b +- sq root (b^2 -4*a*c)] / 2*a
x = [-0 +-sq root (0 -4*-16*541)] / 2 * -16
x = + - sq root (0 -4*-16*541) / -32
x = + - sq root (34,624) / -32
x = + - 186.075 / -32
x1 = 186.075 / -32 =
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-5.81484375
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x2 = -186.075 / -32 =
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5.81484375
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</span>
Answer: ![\frac{\sqrt[4]{10xy^3}}{2y}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B2y%7D)
where y is positive.
The 2y in the denominator is not inside the fourth root
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Work Shown:
![\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B5x%7D%7B8y%7D%7D%5C%5C%5C%5C%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B5x%2A2y%5E3%7D%7B8y%2A2y%5E3%7D%7D%5C%20%5C%20%5Ctext%7B....%20multiply%20top%20and%20bottom%20by%20%7D%202y%5E3%5C%5C%5C%5C%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B10xy%5E3%7D%7B16y%5E4%7D%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B16y%5E4%7D%7D%20%5C%20%5C%20%5Ctext%7B%20...%20break%20up%20the%20fourth%20root%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%7D%20%5C%20%5C%20%5Ctext%7B%20...%20rewrite%20%7D%2016y%5E4%20%5Ctext%7B%20as%20%7D%20%282y%29%5E4%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B2y%7D%20%5C%20%5C%20%5Ctext%7B...%20where%20y%20is%20positive%7D%5C%5C%5C%5C%5C%5C)
The idea is to get something of the form 
in the denominator. In this case, 
To be able to reach the 
, your teacher gave the hint to multiply top and bottom by 
For more examples, search out "rationalizing the denominator".
Keep in mind that ![\sqrt[4]{(2y)^4} = 2y](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%20%3D%202y)
only works if y isn't negative.
If y could be negative, then we'd have to say ![\sqrt[4]{(2y)^4} = |2y|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%20%3D%20%7C2y%7C)
. The absolute value bars ensure the result is never negative.
Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.
The answer to this question would be 3d-2=31 and d=11
Answer:
eeeeeeeeeeeeeeeeeeeeeeeeeee nose perdon xd
Answer:
50%
Step-by-step explanation:
Given that:
P(late) = P(L) = 50% = 0.5
P(Eat) = P(E) =?
P(LnE) = 40% = 0.4
P(LuE) = 60% = 0.6
P(LuE) = P(L) + P(E) - P(LnE)
0.6 = 0.5 + x - 0.4
0.6 = 0.1 + x
0.6 - 0.1 = x
0.5 = x
P(E) = 0.5
0.5 * 100% = 50%
