The equation that is represented by the graph below is y = e^x - 4.
<h3>How to illustrate the graph?</h3>
From the graph, it can be seen that the only exponential function n to hat intercept the function at equal to 3 is y = e^x - 4.
Let's equate y to 0.
y = e^x - 4.
e^x - 4 = 0
e^x = 4
In(e^x) = In(4)
x = 1.386
The x intercept is located at (1.386, 0).
In conclusion, the correct option is D.
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Answer:
Answer: It can be written as 1.47x10-³
Step-by-step explanation:
0.00147 = 1.47
1.47 = 1.47 x 10 -^3
Answer:
<u>Mass</u>

<u>Center of mass</u>
<em>Coordinate x</em>

<em>Coordinate y</em>

<em>Coordinate z</em>

Step-by-step explanation:
Let W be the wire. We can consider W=(x(t),y(t),z(t)) as a path given by the parametric functions
x(t) = t
y(t) = 4 cos(t)
z(t) = 4 sin(t)
for 0 ≤ t ≤ 2π
If D(x,y,z) is the density of W at a given point (x,y,z), the mass m would be the curve integral along the path W

The density D(x,y,z) is given by

on the other hand

and we have

The center of mass is the point 
where

We have

so




