Part A:
Function A:
Slope = (7-3)/(5-3) = 4/2 = 2
Equation:
y - 3 = 2(x - 3)
y - 3 = 2x - 6
y = 2x - 3
Funcion B:
(0, 3 ) and (-5, 0)
Slope = (3 - 0)/(0 + 5) = 3/5
y-intercept (0,3) so b = 3
Equation:
y = 3/5 x + 3
Function C:
y = 3x + 1
Part B:
Rate of change is the change in y over the change in x (rise/run). It's also the slope
Function A: rate of change = 2
Function B: rate of change = 3/5 (smallest)
Function C: rate of change = 3 (largest)
Order linear functions based on rate of change from least to greatest.
Function B: y = 3/5 x + 3
Function A: y = 2x - 3
Function C: y = 3x + 1
Answer:
Yes, because
is equivalent to 
Step-by-step explanation:
and 

Hope this helps!
Well, should be likee thiss correct meh if im wrong or what ever, but
the answer would be, last one (x + 8)(x - 8), well 1, is because the 64 is a negative so if you would just take that 64 and bring it down then look at this onece the work is donee then when u add the 2 numbers that x would be x² or x2 n that - 64 is that + 8 (-8) witch would equal out so if you really look at it the CORRECT ANSWER WOULD BE B. (x - 8)(x -8)
i realli understand if youu don't then comment in the sec beloww↓
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,
Step-by-step explanation:
<u>y</u> - <u>1</u> = <u>-</u><u>3</u>
x - 0 2
2y - 2 = -3x
2y = -3x + 2