Now cos⁻¹(0.7) is about 45.6°, that's on the first quadrant.
keep in mind that the inverse cosine function has a range of [0, 180°], so any angles it will spit out, will be on either the I quadrant where cosine is positive or the II quadrant, where cosine is negative.
however, 45.6° has a twin, she's at the IV quadrant, where cosine is also positive, and that'd be 360° - 45.6°, or 314.4°.
now, those are the first two, but we have been only working on the [0, 360°] range.... but we can simply go around the circle many times over up to 720° or 72000000000° if we so wish, so let's go just one more time around the circle to find the other fellows.
360° + 45.6° is a full circle and 45.6° more, that will give us the other angle, also in the first quadrant, but after a full cycle, at 405.6°.
then to find her twin on the IV quadrant, we simply keep on going, and that'd be at 360° + 360° - 45.6°, 674.4°.
and you can keep on going around the circle, but only four are needed this time only.
Answer:
ok did you really just say brain fart I'm dying I cant stop laughing
In point slope form, the equation is y-7=(-10/3)(x+9). In slope-intercept form, it is y=(-10/3)x-23.
First find the slope of the line. The formula for slope is
m=(y₂-y₁)/(x₂-x₁)
Using our points, we have
m=(-3-7)/(-6--9) = -10/3
Plug this into point slope form:
y-y₁=m(x-x₁)
y-7=(-10/3)(x--9)
y-7=(-10/3)(x+9)
Using the distributive property:
y-7=(-10/3)*x+(-10/3)*9
y-7=(-10/3)x-90/3
y-7=(-10/3)x-30
Add 7 to both sides:
y=(-10/3)x-23
