The line defined by the equation 2y+3=-(2/3)(x-3) is tangent to the graph of g(x) at x=-3. What is the value of the limit as x a
1 answer:
<span>the limit as x approaches -3 of [g(x)-g(-3)]over(x+3) is the same as the derivative, or slope, of g(x) at the point x=-3, or g'(-3).
Since you are given the equation of the tangent line, the answer is just the slope of that line.
</span><span>2y+3=-(2/3)(x-3)
</span><span>6y+9=-2(x-3)
6y+9=-2x+6
6y=-2x-3
y= (-2x-3)/6
slope is -2/6 = </span>
Since you are given the equation of the tangent line, the answer is just the slope of that line.
</span><span>2y+3=-(2/3)(x-3)
</span><span>6y+9=-2(x-3)
6y+9=-2x+6
6y=-2x-3
y= (-2x-3)/6
slope is -2/6 = </span>
You might be interested in
Answer:
C) 2,6,24,120,720
Step-by-step explanation:
Here the n-th term An =
A1 is the first term
A1 = (1 + 2)!/(1 + 2)
= 3!/3
A1 = 2
A2 is the second term
A2 = (2 +2)!/(2 +2)
= 4!/4
A2 = (1*2*3*4) /4
A2 = 6
A3 is the third term
A3 = (3 + 2)!/(3 +2)
A3 = 5!/5
A3 = 24
A4 is the fourth term
A4 = (4 + 2)!/(4 + 2)
A4 = 6!/6
A4 = 120
A5 is the fifth term
A5 = (5 + 2)!/(5 +2)
A5 = 7!/7
A5 = 720
Answer: C) 2,6,24,120,720
Thank you.