Answer:
y = x + 4
Step-by-step explanation:
y = mx + b
y = 1x + b
5 = 1 + b
b = 4
y = x + 4
Answer:
40.20cm approx
Step-by-step explanation:
Step one
given data
We are told that the dimension of a computer screen
diagonal= 52 cm
height= 33cm
Length = ?
Required
The length of the base of the screen
Step two
Let us use Pythagoras theorem to find the length of the base
Hyp^2= Opp^2+ Adj^2
52^2= 33^2+ L^2
2704= 1089+L^2
L^2= 2704-1089
L^2= 1615
L=√1615
L= 40.20cm approx
<u>The length is 40.20cm approx.</u>
If the diameter of the smaller circle is 3.5cm, that means the radius is 1.75cm.
The area of the smaller circle is π × 1.75², which is ≈ 9.62112750162.
The diameter of the larger circle is 12.5 cm, so the radius would be half that, which is 6.25cm. The area of the larger circle would be π × 6.25², which is ≈ 122.718463031.
So now that we know the area of the larger circle and the smaller circle, we can find the area of the shaded region by subtracting the area of the smaller circle from the area of the larger circle, which is basically just 122.718463031 - 9.62112750162, which is = 113.097336. The closest answer here is 113.04, hence the answer is 113.04cm².
A) The signs of the first derivative (g') tell you the graph increases as you go left from x=4 and as you go right from x=-2. Since g(4) < g(-2), one absolute extreme is (4, g(4)) = (4, 1).
The sign of the first derivative changes at x=0, at which point the slope is undefined (the curve is vertical). The curve approaches +∞ at x=0 both from the left and from the right, so the other absolute extreme is (0, +∞).
b) The second derivative (g'') changes sign at x=2, so there is a point of inflection there.
c) There is a vertical asymptote at x=0 and a flat spot at x=2. The curve goes through the points (-2, 5) and (4, 1), is increasing to the left of x=0 and non-increasing to the right of x=0. The curve is concave upward on [-2, 0) and (0, 2) and concave downward on (2, 4]. A possible graph is shown, along with the first and second derivatives.
First one. PNO =22 MNO =40