This is probley easy...
2 answers:
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Answer:
- See attachment for table values
- y₁ = y₂ for x = 6
Step-by-step explanation:
In each case, put the x-value in the formula and do the arithmetic. If you're allowed, you can save some time and effort by realizing that the solution (x) will have to be an even number.
y₁ is an integer value for all integer values of x. y₂ is an integer value for even values of x only. y₁ and y₂ will both be integers (and possibly equal) only when x is even.
For example, for x = 6, we have
... y₁ = 3·6 - 8 = 18 -8 = 10
... y₂ = 0.5·6 +7 = 3 +7 = 10
That is, for x = 6, both columns of the table have the same number (10). That is, y₁ = y₂ for x = 6. The solution to the equation
... y₁ = y₂
is
... x = 6.
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,