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3 years ago

The circumference of a circle is 15 meters. Find the area. Use 3.14 for π.

Mathematics

2 answers:

PilotLPTM [1.2K]3 years ago

7 0

Answer:

2.4 m AND 18.4 m

Step-by-step explanation:

2.4 and 18.4

SVETLANKA909090 [29]3 years ago

3 0

Answer:

Given that:

circumference of circle = 15 meters

π = 3.14

As we know that:

circumference of circle = 2 πr

By putting values:

15 = 2 (3.14) r

By simplifying:

15 = 6.28 r

Dividing both sides by 6.28 we get:

r = 2.39 meters

Now for finding area:

Area = π r^2

Putting values:

Area = (3.14) (2.39)^2

By simplifying we get:

Area = 17 .935 meters^2

So by rounding off radius to nearest tenth:

radius = r = 2.4 meters

Rounding off area to nearest tenth:

area = a = 17.9 meters^2

i hope it will help you!

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An engineer is trying to design a runway for an airport. He knows that airplanes have the ability to accelerate at 3.3 m/s ^ 2 a

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Answer:

The proposed 1600 meters runway length is not long enough

Step-by-step explanation:

The given parameters are;

The amount by which a plane can accelerate = 3.3 m/s²

The time it takes for a plane to lift off the ground = 35 seconds

The length of the runway on the blueprint = 1,600 meters

Therefore, from the equation of motion for distance traveled, s, acceleration, a, and time taken, t, we have;

s = u·t + 1/2·a·t²

Where;

u = The initial velocity

t = 35 seconds

a = 3.3 m/s²

Given that all planes will start motion from rest, we have;

u = 0 m/s

Therefore;

s = 0 × 35 + 1/2 × 3.3 × 35² = 2021.25 meters

Which indicates that the length of the runway a plane would need for take off should be at least 2021.25 meters long, therefore, the blueprint's 1600 meters runway length is not long enough.

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3 years ago

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brainly being slow today so i did my work on this attachment

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│Hope this helped _____________________│

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3 years ago

Read 2 more answers

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8090 [49]

Yes they are equivalent

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3 years ago

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

6 0

3 years ago

Question 3 (1 point)

solong [7]

Answer:

let's do it

4 * 8 - 4 *7 =?

32-28=4

7 0

3 years ago