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2 years ago

14. Jeremy plants flowers in a circular bed,

Mathematics

1 answer:

GrogVix [38]2 years ago

6 0

The area of the circular shaped patio is: 172.7 ft²

<h3>What is the Area of a Circle?</h3>

Area of a circle = πr² [r is given as the radius of the circle].

The area of the patio = area of the larger circle - area of the smaller circle.

Radius of the larger circle = 8 cm

Radius of the smaller circle = 3 cm

π = 3.14

Area of smaller circle = πr² = (3.14)(3²)

Area of smaller circle = 28.26 cm²

Area of bigger circle = πr² = (3.14)(8²)

Area of bigger circle = 200.96 cm²

The area of the patio = 200.96 - 28.26

The area of the patio = 172.7 ft²

Learn more about area of rectangle on:

brainly.com/question/25292087

#SPJ1

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dimulka [17.4K]

F(x + 1) = (5/2)*f(x)

f(x + 1) = 2.5*f(x), f(1) = 3.2

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3 years ago

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3 years ago

What is the 79th term of this arithmetic sequence? 3, 9, 15, 21,

hodyreva [135]

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3 years ago

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Westkost [7]

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3 years ago

A 10-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Alex777 [14]

Answer:

The area is changing at 15.75 square feet per second.

Step-by-step explanation:

The triangle between the wall, the ground, and the ladder has the following dimensions:

H: is the length of the ladder (hypotenuse) = 10 ft

B: is the distance between the wall and the ladder (base) = 6 ft

L: the length of the wall (height of the triangle) =?

dB/dt = is the variation of the base of the triangle = 9 ft/s

First, we need to find the other side of the triangle:

$H^{2} = B^{2} + L^{2}$

$L = \sqrt{H^{2} - B^{2}} = \sqrt{(10)^{2} - B^{2}} = \sqrt{100 - B^{2}}$

Now, the area (A) of the triangle is:

$A = \frac{BL}{2}$

Hence, the rate of change of the area is given by:

$\frac{dA}{dt} = \frac{1}{2}[L*\frac{dB}{dt} + B\frac{dL}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} + B\frac{d(\sqrt{100 - B^{2}})}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} - \frac{B^{2}}{(\sqrt{100 - B^{2}})}*\frac{dB}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - 6^{2}}*9 - \frac{6^{2}}{\sqrt{100 - 6^{2}}}*9]$

$\frac{dA}{dt} = 15.75 ft^{2}/s$

Therefore, the area is changing at 15.75 square feet per second.

I hope it helps you!

5 0

3 years ago