The Japanese 5 yen coin has a hole in the middle.

Mathematics

1 answer:

alexandr402 [8]2 years ago

5 0

Answer:

Subtract the two diameters and multiply by 0.5

Step-by-step explanation:

Calculating the Radius of the hole gives distance to the inner edges WHILE the Radius of the coin gives us the distance to the outer edges.

By subtracting the output we can then get the difference between both distances.

If hole diameter = 5mm and coin diameter = 22mm

Difference = [(22 - 5)mm] ÷ 2

17mm / 2 = 8.5mm

You might be interested in

Need help!! ASAP!! Geometry

kodGreya [7K]

Total 52 students.

52 - 17 = 35 students left
35 - 5 = 30 students left
30 - 2 = 28 students left

Each class has 11 students that play ONE sport, which means 22 total in 2 classes. You're on your own now! It's not that hard, I'm sure you'll figure it out.

7 0

3 years ago

Solve the Anti derivative.

Alex Ar [27]

Answer:

$\displaystyle \int {\frac{1}{9x^2+4}} \, dx = \frac{1}{6}arctan(\frac{3x}{2}) + C$

General Formulas and Concepts:

Algebra I

Factoring

Calculus

Antiderivatives - integrals/Integration

Integration Constant C

U-Substitution

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Trig Integration: $\displaystyle \int {\frac{du}{a^2 + u^2}} = \frac{1}{a}arctan(\frac{u}{a}) + C$

Step-by-step explanation:

Step 1: Define

$\displaystyle \int {\frac{1}{9x^2 + 4}} \, dx$

Step 2: Integrate Pt. 1

[Integral] Factor fraction denominator: $\displaystyle \int {\frac{1}{9(x^2 + \frac{4}{9})}} \, dx$
[Integral] Integration Property - Multiplied Constant: $\displaystyle \frac{1}{9} \int {\frac{1}{x^2 + \frac{4}{9}}} \, dx$

Step 3: Identify Variables

Set up u-substitution for the arctan trig integration.

$\displaystyle u = x \\ a = \frac{2}{3} \\ du = dx$

Step 4: Integrate Pt. 2

[Integral] Substitute u-du: $\displaystyle \frac{1}{9} \int {\frac{1}{u^2 + (\frac{2}{3})^2} \, du$
[Integral] Trig Integration: $\displaystyle \frac{1}{9}[\frac{1}{\frac{2}{3}}arctan(\frac{u}{\frac{2}{3}})] + C$
[Integral] Simplify: $\displaystyle \frac{1}{9}[\frac{3}{2}arctan(\frac{3u}{2})] + C$
[integral] Multiply: $\displaystyle \frac{1}{6}arctan(\frac{3u}{2}) + C$
[Integral] Back-Substitute: $\displaystyle \frac{1}{6}arctan(\frac{3x}{2}) + C$