\left[x _{2}\right] = \left[ \frac{-1+i \,\sqrt{3}+2\,by+\left( -2\,i \right) \,\sqrt{3}\,by}{2^{\frac{2}{3}}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{24}+\left( \frac{-1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{\sqrt[3]{2}}\right][x2]=⎣⎢⎢⎢⎢⎡2323√(432by+√(−6912+41472by+103680by2+55296by3))−1+i√3+2by+(−2i)√3by+3√224−3√(432by+√(−6912+41472by+103680by2+55296by3))+(24−1i)√33√(432by+√(−6912+41472by+103680by2+55296by3))⎦⎥⎥⎥⎥⎤
totally answer.
Not sure what you’re asking ? but
move the terms
-5x+20-a
divide both sides
x=-4+1/5a
solution:
x=-4+1/5a
Answer:
The distance of the two runners from the flagpole is 0.0882 km
Step-by-step explanation:
We are going to use the minus sign when the runner is on the west and the plus sign when the runner is on the east.
Then, Taking into account the runner A is 6 km west and is running with a constant velocity of 3 Km/h east, the distance of the runner A from the flagpole is given by the following equation:
Xa = -6 Km + (3 Km/h)* t
Where Xa is the position of the runner A from the flagpole and t is the time in hours.
At the same way the distance of the runner B, Xb, from the flagpole is given by the following equation:
Xb = 7.4 Km - (3.8 Km/h)*t
Then, the two runners cross their path when Xa is equal to Xb, so if we solve this equation for t, we get:
Xa = Xb
-6 + (3*t) = 7.4 - (3.8*t)
(3.8*t) + (3*t) = 7.4 + 6
(6.8*t) = 13.4
t = 13.4/6.8
t = 1.9706
Therefore, at time t equal to 1.9706 hours, both runners cross their path. The distance of the two runners from the flagpole can be calculated replacing the value of t in equation for Xa or in equation for Xb as:
Xa = -6 Km + (3 Km/h)* t
Xa = -6 Km + (3 Km/h)*(1.9706 h)
Xa = -0.0882 Km
That means that both runners are 0.0882 Km west of a flagpole.
5n=20 is the equation 5 times n equals 20
The fraction for that is 1/2
