Y=mx+b
6x+2y=-88
2y=-88-6x
y=-44-3x
y=-3x-44
m is the slope
b is the y-intercept
Answer:
i thinks it -11°F but not sure
(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.
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(b) The following functions are positive in quadrant III:
tangent, cotangent
The following functions are negative in quadrant III
cosine, sine, secant, cosecant
A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.
ok so here we go:
1) 
or 
2 )
so the answer is (0, -4) or -4
3) y=
4) y=[tex]\frac{3}{2}x+1[tex] so the answer is (0, 1) or 1
Dr.hong prescribed the lest and Dr.evans prescribed the most. that is the answer but this is how i got the answer ok so you see that hong gave 0.019 that is not at all the same as 0.02 so you would add a zero to Evans so it would be 0.020 so it is the same length of hongs so now do you know how i got that if not then message me
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