From Mathematica:<span>DSolve<span>[<span><span>y′</span>[x]=<span><span>(<span>y[x<span>]2</span>−<span>x2</span></span>)</span><span>2xy[x]</span></span>,y[x],x</span>]</span></span><span>{<span><span>{<span>y[x]→−<span><span>−<span>x2</span>+xC[1]</span><span>−−−−−−−−−−</span>√</span></span>}</span>,<span>{<span>y[x]→<span><span>−<span>x2</span>+xC[1]</span><span>−−−−−−−−−−</span>√</span></span>}</span></span>}</span>Take the second solution and square each side:<span>y[x<span>]2</span>=−<span>x2</span>+xC[1]</span>Move the -x^2 from the RHS to the LHS.<span>y[x<span>]2</span>+<span>x2</span>=xC[1]</span>If the same procedure is applied to my answer then the following is the result:<span>y[x<span>]2</span>+<span>x2</span>= 2cx+2C[1]</span><span>where C[1] is zero. I am not sure that their expression is totally correct.</span>
Answer:
b
Step-by-step explanation:
F(x) = 3x + 8
f(1) = 3 (1) + 8
f (1) = 3 + 8
f(1)= 11
Answer:
(5sqrt(2), 45 deg)
(-5sqrt(2), 225 deg)
Step-by-step explanation:
(x,y)=(5,5)
So theta=arctan(5/5)=arctan(1)=45 degrees
Now r! r=sqrt(x^2+y^2)=sqrt(5^2+5^2)=sqrt(50)=sqrt(25)sqrt(2)=5sqrt(2)
So one polar point is (5sqrt(2) , 45 degrees)
Now if we do 180+45=225 degrees... this puts us in the 3rd quadrant... to get back to quadrant 1 we just take the opposite of our r
so another point is (-5sqrt(2) , 225 degrees)
