Does anyone know this question???
2 answers:
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Answer:
The graph with B(2,-3) i.e) y= x goes through the point M(-10,15).
Step-by-step explanation:
Consider M(-10,15) and given that equation is y = kx.
Now, substitute M(-10,15) in the equation
⇒ 15 = k × -10
⇒ k = =
⇒ y = x
Now, check with the given points B(2,-3) and B( , -2)
1) B(2,-3)
y = x
⇒(-3) = × 2
⇒ -3 = -3 ⇒ LHS = RHS
⇒ B(2,-3) is the required point.
2) for b( , -2)
LHS ≠ RHS.
So,The graph with B(2,-3) i.e) y= x goes through the point M(-10,15).
Now, the cosecant of θ is -6, or namely -6/1.
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now
recall that
therefore, let's just plug that on the remaining ones,
now, let's rationalize the denominator on tangent and secant,
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now
recall that
therefore, let's just plug that on the remaining ones,
now, let's rationalize the denominator on tangent and secant,