The given equation is: 
To find the line perpendicular to it, we interchange coefficients and switch the signs of one coefficient.
The equation to a line perpendicular to it is:
$ 2y-x=c$
where, $c$ is some constant we have determine using the condition given.
It passes through $(2,-1)$
Put the point in our equation:
$2(-1)-(2)=c$
$c=-2-2$
$c=-4$
The final equation is:
$\boxed{ 2y-x=-4}$
Answer:
a^2+4a-5
Step-by-step explanation:
This is the expanded form
<span>Use the formula: r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ] where k = 0,1,2,3,4
</span><span>First 5th root:
k = 0
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]
2*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]
2*[ cos( (280+0)/5 ) + i*sin( (280+0)/5 ) ]
2*[ cos( 280/5 ) + i*sin( 280/5 ) ]
2*[ cos( 56 ) + i*sin( 56 ) ]
-------------------------------------------------------------------
Second 5th root:
k = 1
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]
2*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]
2*[ cos( (280+360)/5 ) + i*sin( (280+360)/5 ) ]
2*[ cos( 640/5 ) + i*sin( 640/5 ) ]
2*[ cos( 128 ) + i*sin( 128 ) ]
-------------------------------------------------------------------
Third 5th root:
k = 2
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]
2*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]
2*[ cos( (280+720)/5 ) + i*sin( (280+720)/5 ) ]
2*[ cos( 1000/5 ) + i*sin( 1000/5 ) ]
2*[ cos( 200 ) + i*sin( 200 ) ]
-------------------------------------------------------------------
Fourth 5th root:
k = 3
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]
2*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]
2*[ cos( (280+1080)/5 ) + i*sin( (280+1080)/5 ) ]
2*[ cos( 1360/5 ) + i*sin( 1360/5 ) ]
2*[ cos( 272 ) + i*sin( 272 ) ]
-------------------------------------------------------------------
Fifth 5th root:
k = 4
r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]
(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]
(32)^(1/5)*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]
2*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]
2*[ cos( (280+1440)/5 ) + i*sin( (280+1440)/5 ) ]
2*[ cos( 1720/5 ) + i*sin( 1720/5 ) ]
2*[ cos( 344 ) + i*sin( 344 ) ]</span>
Answer: 22, 8, 27
Step-by-step explanation: the sum of the 2 shorter sides must be greater than longest side
The answer is 35.08. she will have 35.08 yards of cloth left.
