lakofbrwn

its pretty hard to work with just these terms, but we can work around it :)

ok, so if we put the whole factor equal to 0 it'll be simple

and;

Y(x+2)^2=0

you can take the whole thing apart

so (x+2)^2=0 and Y=0

isolate for x 

x=-2 or +2

and for a Y root the x should be equal to 0 so;

replacing X with 0 in the initial formula 

Y=0

thats all i can work with :)

hope this helps

using pythagorean theorem 

SOH -->sinx=opposite/hypothenus 

CAH -->cosx=adjacent/hypthenus

TOA -->tanx=opposite/adjacent 

who's Arty?

45 x 2= 90

45 + 2 = 47

when to isolate such variable you need to work backwards from the outside and undo the equation

(2x^1/4)^2--->square root it

2x^1/4---> divide by 2

x^1/4---> put the whole equation to the 4

but x would be equal to 0 if you dont have anything on the other side 

undefined, divding anything by infinity is undifined or it go's towards zero if we look at it as a limits problem

correct me if im wrong, always convert it into meters/seconds when doing physics, just divde it by 3.6

g=ground velocity

a=airplane velocity

w= wind velocity

the general equation would be V(g)=V(a)+V(w)

since they're on angles you would have to split them up into 2 components x and y so;

x; V(g)= 138.8cos(180-45)+33.3cos(180+25)

y; V(g)= 138.8sin(180-45)+33.3sin(180+25)

solve them when you get both components you will have to find the vector between them so;

|V(g)|=sqr(t(x^2+y^2))

and you should get your vector 

hope it helped :)

its quite complicated to do it in your head but use the calculator, using these terms;

9.2*10^10-8.4*10^11

you are likely to get a negative answer

Do you know how to derive? its how you find the tangent(slope) of complex functions. If no, you can find 2 points on the line and divde (y2-y1)/(x2-x1)

I myself use derivatives, its simple, fast and useful, which you will figue later in your school year or lifetime.

can you be more clear, the question is a bit vague