Answer:
Mid point is: (2.8;4.9)
Step-by-step explanation:
To find midpoint of a line segment we can use the general equation:

Where the point of the line are: (x₁;y₁) and (x₂;y₂).
In the problem, x₁ = 2.6, y₁ = 5.1 and x₂ = 3 and y₂ = 4.7. Replacing in the equation:

<h3>Mid point is: (2.8;4.9)</h3>
Answer:
n = 18 cycles
Step-by-step explanation:
To know how many times does the current oscillates in the given time, you take into account that the number of oscillation can be calculated by using the following expression:
(1)
f: frequency of the oscillation of the current
t: time = 0.30s
The frequency is the number of cycles per second, that is, f = 60 cycles/s
You replace the values of f and t in the equation (1):

In 0.30s the current oscillates 18 times
Answer:
7(x-10)
Step-by-step explanation:
7×(x-10)
you ×7by whatever x-10 is
Answer:
40 miles
Step-by-step explanation:
In the attached diagram, Point A is the starting point and C is the end point. We want to determine the distance from A to C.
The path driven forms a right triangle in which AC is the hypotenuse.
We therefore use the<u> Pythagorean Theorem</u> to solve for the AC.
Pythagorean Theorem: 

The straight line distance from the starting point is 40 miles.
Answer:
We can find if a critical point is a local minimum or maximum by looking at the second derivatives.
Step-by-step explanation:
If you take the first derivative, you will find the slope at the given point, which if it is a minimum or a maximum will be 0.
Then we take the second derivative. If that number is a positive number, then we have a local minimum. If it is a negative number, then it is a local maximum.