Subscribe in seconds and receive Math Homework Answers - Recent questions and answers's news feed updates in your inbox, on your phone or even read them from your own news page here on follow.it.
You can select the updates using tags or topics and you can add as many websites to your feed as you like.
And the service is entirely free!
Follow Math Homework Answers - Recent questions and answers: Math Homework Answers
Is this your feed? Claim it!
The integral wrt y is best tackled by using the power series for sine and substituting y3 for the argument:
sin(y3)=y3-(y3)3/3!+(y3)5/5!-...=n=1Σn=∞(-1)n-1y6n-3/(2n-1)!
On integration we get n=1Σn=∞(-1)n-1y6n-2/[(6n-2)(2n-1)!] to which we apply the limits.
Let y=csc(x/2) then 4y2-2y-2=2(2y2-y-1)=2(2y+1)(y-1).
Now replace y: 2(2csc(x/2)+1)(csc(x/2)-1). This is the simplest form of the given expression.
[The zeroes of this expression are found from y=-½ and y=1, that is:
csc(x/2)=-½⇒sin(x/2)=-1...
x12=10.85. 12 is an even exponent so x6=±√10.85=±3.2939338 approx.
x3=±√3.2939338=±1.8149198 approx, and ±i√3.2939338 where i=√-1 (imaginary) if we are considering complex as well as real numbers.
Algebra 1 curriculum may not include complex numbers, so x3=±√3.29393...
2x2+3xy-4y2=2(22)+3(2)(-4)-4(-4)2=2×4-24-4×16=8-24-64=-80.
I think you mean the y-intercept. To get it plug in x=0: 2y=12, so the y-intercept is 6 (=12/2).