Answer:
6 5/12
Step-by-step explanation:
Given the expression :
4 2/3 + 1 3/4
The sum of the numbers :
4 2/3 + 1 3/4
14/3 + 7/4
L.C.M of 3 and 4 = 12
(56 + 21) / 12
77 / 12
= 6 5/12
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>
A) Perimeter I’m 90% sure
I say that becuase if she is lining the OUTER EDGE of her driveway then she would need to know the outside og her driveway which would be perimeter.
Is there some question supposed to be attached?
Answer:

Step-by-step explanation:
Given
---- the perimeter of fencing
Required
The maximum area
Let


So, we have:

This gives:

Divide by 2

Make L the subject

The area (A) of the fence is:

Substitute 

Open bracket

Differentiate with respect to W

Set to 0

Solve for 2W

Solve for W

Recall that:




So, the maximum area is:


