Answer:
24h+6
Step-by-step explanation:
Answer:
You didnt identify the volume so we can find the radius
Step-by-step explanation:
lol
The second one would be the answer. You want to plug in for this question. You can divide over the five leaving you with (x-2)^2 =25. Since you are squaring this answer, you want to get either 5 or negative five. Both -3 and 7 will do that.
answer: -5x+12=20
-12 -12
-5x/-5= 8/-5
x= -1.6
Check: rewrite the orignal problem
plug your variable
1) The function is
3(x + 2)³ - 32) The
end behaviour is the
limits when x approaches +/- infinity.3) Since the polynomial is of
odd degree you can predict that
the ends head off in opposite direction. The limits confirm that.
4) The limit when x approaches negative infinity is negative infinity, then
the left end of the function heads off downward (toward - ∞).
5) The limit when x approaches positive infinity is positivie infinity, then
the right end of the function heads off upward (toward + ∞).
6) To graph the function it is important to determine:
- x-intercepts
- y-intercepts
- critical points: local maxima, local minima, and inflection points.
7)
x-intercepts ⇒ y = 0⇒ <span>
3(x + 2)³ - 3 = 0 ⇒ (x + 2)³ - 1 = 0
</span>
<span>⇒ (x + 2)³ = -1 ⇒ x + 2 = 1 ⇒
x = - 1</span>
8)
y-intercepts ⇒ x = 0y = <span>3(x + 2)³ - 3 =
3(0 + 2)³ - 3 = 0 - 3×8 - 3 = 24 - 3 =
21</span><span>
</span><span>
</span><span>9)
Critical points ⇒ first derivative = 0</span><span>
</span><span>
</span><span>i) dy / dx = 9(x + 2)² = 0
</span><span>
</span><span>
</span><span>⇒ x + 2 = 0 ⇒
x = - 2</span><span>
</span><span>
</span><span>ii)
second derivative: to determine where x = - 2 is a local maximum, a local minimum, or an inflection point.
</span><span>
</span><span>
</span><span>
y'' = 18 (x + 2); x = - 2 ⇒ y'' = 0 ⇒ inflection point.</span><span>
</span><span>
</span><span>Then the function does not have local minimum nor maximum, but an
inflection point at x = -2.</span><span>
</span><span>
</span><span>Using all that information you can
graph the function, and I
attache the figure with the graph.
</span>