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3 years ago

14

Which formula is used to find circular mil area?

1 answer:

krok68 [10]3 years ago

7 0

The term circular mil is common in expressing the cross sectional area of a wire. Electrical wires have very minute diameters that are often measured a thousandth of an inch. Hence, when you find the area of the circular wire, for convenience, the unit used is circular mil which is equivalent to one-thousandth of an inch.

For example, a wire has a diameter of 6×10⁻⁵ inches. To find its area:

A = πr² = π( 6×10⁻⁵ /2)² = 2.83×10⁻⁹ in²

Since 1 mil = 1/1000 (inch) or 0.001 inch

A = (2.83×10⁻⁹ in²)* (1 mil/ 0.001 inch)²
A = 0.00283 circular mils

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"What is the probability that the next two employees that join the office staff will take the bus"

castortr0y [4]

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Car 18

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3 years ago

What is the largest possible integral value in the domain of the real-valued function

kotegsom [21]

Answer:

Max Value: x = 400

General Formulas and Concepts:

<u>Algebra I</u>

Domain is the set of x-values that can be inputted into function f(x)

<u>Calculus</u>

Antiderivatives
Integral Property: $\int {cf(x)} \, dx = c\int {f(x)} \, dx$
Integration Method: U-Substitution
[Integration] Reverse Power Rule: $\int {x^n} \, dx = \frac{x^{n+1}}{n+1} + C$

Step-by-step explanation:

<u>Step 1: Define</u>

$f(x) = \frac{1}{\sqrt{800-2x} }$

<u>Step 2: Identify Variables</u>

<em>Using U-Substitution, we set variables in order to integrate.</em>

$u = 800-2x\\du = -2dx$

<u>Step 3: Integrate</u>

Define: $\int {f(x)} \, dx$
Substitute: $\int {\frac{1}{\sqrt{800-2x} } } \, dx$
[Integral] Int Property: $-\frac{1}{2} \int {\frac{-2}{\sqrt{800-2x} } } \, dx$
[Integral] U-Sub: $-\frac{1}{2} \int {\frac{1}{\sqrt{u} } } \, du$
[Integral] Rewrite: $-\frac{1}{2} \int {u^{-\frac{1}{2} }} \, du$
[Integral - Evaluate] Reverse Power Rule: $-\frac{1}{2}(2\sqrt{u}) + C$
Simplify: $-\sqrt{u} + C$
Back-Substitute: $-\sqrt{800-2x} + C$
Factor: $-\sqrt{-2(x - 400)} + C$

<u>Step 4: Identify Domain</u>

We know from a real number line that we cannot have imaginary numbers. Therefore, we cannot have any negatives under the square root.

Our domain for our integrated function would then have to be (-∞, 400]. Anything past 400 would give us an imaginary number.

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