Use the graph of y =g(x) to graph the given function.
1 answer:
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Answer:
Step-by-step explanation:
Important information:
Circumference of a circle=
Method:
To work out the diameter, we would need to divide the circumference of 34.54 by pi, which in this case is 3.14, this gives us 11cm. This is because we are simply rearanging the formula of the circumference of a circle to make the diameter the subject.
Step By Step:
1) Rearange the formula to make the diameter the subject.
2) Divide 34.54 by 3.14().
Answer:
- (a) see attached
- (b) width: 2x; height: y = 9-x²
- (c) A=2x(9-x²) . . . 0 ≤ x ≤ 3
- (d) dA/dx = -6x² +18; x=±√3
- (e) 12√3 units²
Step-by-step explanation:
(a) The attachment shows the graph of the parabola in blue. It also shows an inscribed rectangle in black.
(b) The upper right point of the rectangle is shown in the attachment as (x, y). The dimension y is the height of the rectangle. The x-dimension is half the width of the rectangle, which is symmetrical about the y-axis. Hence the width is 2x.
(c) As with any rectangle, the area is the product of length and width:
... A = (2x)(9 -x²) . . . . . the attachment shows a graph of this
... A = -2x³ +18x . . . . . expanded form suitable for differentiation
A suitable domain for A is where both x and A are non-negative: 0 ≤ x ≤ 3.
(d) The derivative of A with respect to x is ...
... A' = -6x² +18
This is defined everywhere, so the critical values will be where A' = 0.
... 0 = -6x² +18
... 3 = x² . . . . . . . divide by -6, add 3
... √3 = x . . . . . . . -√3 is also a solution, but is not in the domain of A
(e) The rectangle will have its largest area where x=√3. That area is ...
... A = 2x(9 -x²) = 2√3(9 -(√3)²) = 2√3(6)
... A = 12√3 . . . . square units . . . . ≈ 20.785 units²
