Answer:
...
Step-by-step explanation:
Im going to need a question before i can properly answer this.
for 6
in a
given
slope(m)=1
point=(2,-4)=(x1,y1)
we know
equation of straight line
y-y1=m(x-x1)
y+4=1(x-2)
y+4=x-2
therefore x-y-6=0 is the required equation
substituting with ax+by+c=0 we get
a=1
b= -1
c= -6
Wdym? Can you be more specific?
Answer: sin u = -5/13 and cos v = -15/17
Step-by-step explanation:
The nice thing about trig, a little information goes a long way. That’s because there is a lot of geometry and structure in the subject. If I have sin u = opp/hyp, then I know opp is the opposite side from u, and the hypotenuse is hyp, and the adjacent side must fit the Pythagorean equation opp^2 + adj^2 = hyp^2.
So for u: (-5)^2 + adj^2 = 13^2, so with what you gave us (Quad 3),
==> adj of u = -12 therefore cos u = -12/13
Same argument for v: adj = -15,
opp^2 + (-15)^2 = 17^2 ==> opp = -8 therefore sin v = -8/17
The cosine rule for cos (u + v) = (cos u)(cos v) - (sin u)(sin v) and now we substitute: cos (u + v) = (-12/13)(-15/17) - (-5/13)(-8/17)
I am too lazy to do the remaining arithmetic, but I think we have created a way to approach all of the similar problems.
