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2 years ago

What is the circumference of a circle with a diameter of 9 inches.

Mathematics

1 answer:

Dennis_Churaev [7]2 years ago

8 0

Answer:

28.26 inches

Step-by-step explanation:

Given

diameter (d) = 9 inches

circumference (c)

= π d

= 3.14 * 9

= 28.26 inches

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3 3/4 divided by 5 5/8

Musya8 [376]

Answer:

*Me Dividing 3 3/4 divided by 5 5/8* there you go.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

1. Find the slope of the line that goes through the points (2, 5) and (6, 7)

ella [17]

Answer:

$\displaystyle m = \frac{1}{2}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)
Slope Formula: $\displaystyle m = \frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Identify

Point (2, 5)

Point (6, 7)

Step 2: Find slope m

Substitute in points [Slope Formula]: $\displaystyle m = \frac{7-5}{6-2}$
[Fraction] Subtract: $\displaystyle m = \frac{2}{4}$
[Fraction] Simplify: $\displaystyle m = \frac{1}{2}$

7 0

2 years ago

Please help me solve this

Ira Lisetskai [31]

I’m sorry I do not know the other person was rude

6 0

2 years ago

An accident at an oil drilling platform is causing a circular oil slick. The slick is 0.07 foot thick, and when the radius of th

ololo11 [35]

Answer: $43.54 ft^3/min$

Step-by-step explanation:

Given

Thickness of oil slick=0.07 foot

radius of slick=110 ft

$\frac{\mathrm{d} r}{\mathrm{d} t}=0.9 ft/s$

Let V be the volume of oil slick

so, $V=0.07\pi \cdot r^2$

rate of oil flowing is

$\frac{\mathrm{d} V}{\mathrm{d} t}=0.07\times 2\pi \cdot r\frac{\mathrm{d} r}{\mathrm{d} t}$

$\frac{\mathrm{d} V}{\mathrm{d} t}=0.07\times 2\pi \times 110\times 0.9=43.54 ft^3/min$

7 0

3 years ago

3. At noon Lan is in his blue mini-van 300 km north of Makenna in her Bugatti. Lan heads south

LiRa [457]

Answer:

The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h

Step-by-step explanation:

At noon the location of Lan = 300 km north of Makenna

Lan's direction = South

Lan's speed = 60 km/h

Makenna's direction and speed = West at 75 km/h

The distance Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km

The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km

The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km

Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.

By Pythagoras' theorem, we have;

s² = x² + y²

The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61

ds²/dt = dx²/dt + dy²/dt

2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt

2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900

ds/dt = 900/(2×30·√61) ≈ 1.92

The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h

5 0

2 years ago