Answer:
?≤y≤?
Step-by-step explanation:
the answer is
x=8y/3(1-8y)
y=3x/8(1+3x)
hope this helped l used cymath to solve this problem.
good day : )
1. The direction of the vector
relative to the positive x-axis satisfies
![\tan\theta=\dfrac35\implies\theta\approx30^\circ](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cdfrac35%5Cimplies%5Ctheta%5Capprox30%5E%5Ccirc)
2. A vector with magnitude
and direction
has component form
![\mathbf v=\langle\|\mathbf v\|\cos\theta,\|\mathbf v\|\sin\theta\rangle](https://tex.z-dn.net/?f=%5Cmathbf%20v%3D%5Clangle%5C%7C%5Cmathbf%20v%5C%7C%5Ccos%5Ctheta%2C%5C%7C%5Cmathbf%20v%5C%7C%5Csin%5Ctheta%5Crangle)
We have
, but "angle of 60 degrees with the negative x-axis" is a bit ambiguous. I would take it to mean "60 degrees counterclockwise relative to the negative x-axis", so that the direction is
. Then
![\mathbf v=\langle5\cos240^\circ,4\sin240^\circ\rangle\approx\langle-2.5,-4.3\rangle](https://tex.z-dn.net/?f=%5Cmathbf%20v%3D%5Clangle5%5Ccos240%5E%5Ccirc%2C4%5Csin240%5E%5Ccirc%5Crangle%5Capprox%5Clangle-2.5%2C-4.3%5Crangle)
But this doesn't match any of the options, so more likely it means the angle is 60 degrees clockwise relative to the negative x-axis, in which case
and we'd get
![\mathbf v=\langle5\cos120^\circ,4\sin120^\circ\rangle\approx\langle-2.5,4.3\rangle](https://tex.z-dn.net/?f=%5Cmathbf%20v%3D%5Clangle5%5Ccos120%5E%5Ccirc%2C4%5Csin120%5E%5Ccirc%5Crangle%5Capprox%5Clangle-2.5%2C4.3%5Crangle)
3. Same as question 2, but now we're using
notation. The vector has magnitude
and its direction is
. So the vector is
![\mathbf v=14\cos(-30^\circ)\,\mathbf i+14\sin(-30^\circ)\,\mathbf j=12.1\,\mathbf i-7\,\mathbf j](https://tex.z-dn.net/?f=%5Cmathbf%20v%3D14%5Ccos%28-30%5E%5Ccirc%29%5C%2C%5Cmathbf%20i%2B14%5Csin%28-30%5E%5Ccirc%29%5C%2C%5Cmathbf%20j%3D12.1%5C%2C%5Cmathbf%20i-7%5C%2C%5Cmathbf%20j)
4. The velocity of the plane relative to the air,
is 175 mph at 40 degrees above the positive x-axis. The velocity of the air relative to the ground,
, is 35 mph at 60 degrees above the positive x-axis. We want to know the velocity of the plane relative to the ground,
.
We use the relationship
![\mathbf v_{P/G}=\mathbf v_{P/A}+\mathbf v_{A/G}](https://tex.z-dn.net/?f=%5Cmathbf%20v_%7BP%2FG%7D%3D%5Cmathbf%20v_%7BP%2FA%7D%2B%5Cmathbf%20v_%7BA%2FG%7D)
Translating the known vectors into component form, we have
![\mathbf v_{P/G}=\langle145\cos40^\circ,145\sin40^\circ\rangle+\langle35\cos60^\circ,35\sin60^\circ\rangle\approx\langle152,143\rangle](https://tex.z-dn.net/?f=%5Cmathbf%20v_%7BP%2FG%7D%3D%5Clangle145%5Ccos40%5E%5Ccirc%2C145%5Csin40%5E%5Ccirc%5Crangle%2B%5Clangle35%5Ccos60%5E%5Ccirc%2C35%5Csin60%5E%5Ccirc%5Crangle%5Capprox%5Clangle152%2C143%5Crangle)
This vector has magnitude and direction
![\|\mathbf v_{P/G}\|\approx\sqrt{152^2+143^2}\approx208\,\mathrm{mph}](https://tex.z-dn.net/?f=%5C%7C%5Cmathbf%20v_%7BP%2FG%7D%5C%7C%5Capprox%5Csqrt%7B152%5E2%2B143%5E2%7D%5Capprox208%5C%2C%5Cmathrm%7Bmph%7D)
![\tan\theta\approx\dfrac{143}{152}\implies\theta\approx43^\circ](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%5Capprox%5Cdfrac%7B143%7D%7B152%7D%5Cimplies%5Ctheta%5Capprox43%5E%5Ccirc)
(or 43 degrees NE)
The language of oh yea doesn't exist :)