What are the measures of EBG and EBC? Classify each angle as acute,right, abtuse or straight

The solution's average current density is 1 ampere per square decimeter.

We need to convert this value to ampere per square meter.

Since 1 decimeter equals meters,
squaring gives us:

1 square decimeter equals square meters.

Hence, the current density now is .

Therefore, the current density of the electrolyte solution is 100 amperes per square meter.

The count of people serves as the independent variable. The picnic's starting cost (initial fee) is $40. When more people attend, the picnic's overall expense rises. 

Answer:

(1, 3)

Step-by-step explanation:

Provided:

A circle containing these specified points-

(4, 2), (-2, 2), (-1, 4), (1, 1), (1, 3), (2, -3).

A function resembles a mechanism that produces an output for each unique input. Each input must yield a distinct output.

The function's input is characterized as the domain, and the output corresponds to its range.

The domain is indicated by the 'x' coordinate while the range is represented by the 'y' coordinate.

This means the domain serves as the independent variable, and the range is contingent upon the domain’s values.

In this case, a relation or set of points constitutes a function only if every 'x' in the ordered pairs is unique.

Within the described relation, the pairs (1, 1) and (1, 3) share the same input. Thus, removing either one will convert the relation into a function.

Therefore, choosing (1, 3) is the right answer.

.... pertains to the first choice

Answer:

E(t) = [ 4cos(t), 2 sin(t) ]

Step-by-step explanation:

The equation for the ellipse can be represented as:

 (x² ÷ a² ) + (y² ÷ b² ) = 1    Here, a and b symbolize the semi-major and semi-minor axes.

In this specific instance, we have    x²/16 + y²/ 4 = 1

This expression indicates that it follows the form of (x/a)² + (y/b)² =1

Specifically in our case,    x²/16  + y²/4 = 1, which resembles                 sin²α + cos²α = 1.

By setting x = 4 cos(t), we can proceed to calculate y.

Utilizing the equation x²/16  + y²/4 = 1, we find that:

x²/16  + y²/4 = 1   ⇒    (x²  +  4y²) ÷ 16  = 1

Solving for y yields: (x²  +  4y²)  = 16    ⇒  y²  = ( 16 - x²) ÷4

Substituting x = 4 cos(t) gives us y²  =  (16 - 16cos²(t)) ÷ 4, leading to y = √4 (1 - cos²(t)).

Consequently, we have y = 2 sin(t).

Thus, the vector parameterization of the ellipse can be stated as:

E(t) = [ 4cos(t), 2 sin(t) ]