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Bingel [31]
1 year ago
13

a rectangular lawn has an area of a^3 - 125. use the difference of cubes to find out the dimensions of the rectangle.

Mathematics
1 answer:
ANEK [815]1 year ago
6 0

The area of a rectangle is the product of its dimensions

The dimensions of the rectangle are: \mathbf{Length = a -5} and \mathbf{Width = a^2 + 5a + 25}

The area is given as:

\mathbf{Area = a^3 - 125}

Express 125 as 5^3

\mathbf{Area = a^3 - 5^3}

Apply difference of cubes

\mathbf{Area = (a - 5)(a^2 + 5a + 5^2)}

\mathbf{Area = (a - 5)(a^2 + 5a + 25)}

The area of a rectangle is:

\mathbf{Area = Length \times Width}

So, by comparison:

\mathbf{Length = a -5}

\mathbf{Width = a^2 + 5a + 25}

Read more about areas at:

brainly.com/question/3518080

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Step-by-step explanation:

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onsider the graph of the line of best fit, y = 0.5x + 1, and the given data points. A graph shows the horizontal axis numbered n
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What is the ratio of the area of sector ABC to the area of sector DBE?
shusha [124]

We have to find the" ratio of the area of sector ABC to the area of sector DBE".

Now,

the general formula for the area of sector is

Area of sector= 1/2 r²θ

where r is the radius and θ is the central angle in radian.


180°= π rad

1° = π/180 rad


For sector ABC, area= 1/2 (2r)²(β°)

= 1/2 *4r²*(π/180 β)

= 2r²(π/180 β)

For sector DBE, area= 1/2 (r)²(3β°)

= 1/2 *r²*3(π/180 β)

= 3/2 r²(π/180 β)

Now ratio,

Area of sector ABC/Area of sector DBE =\frac{2r^{2}*\ \frac{\pi}{180} beta}{3/2 r^{2}*\ \frac{\pi}{180}beta}

= 4/3

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2 years ago
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A poll found that 5% of teenagers (ages 13 to 17) suffer from arachnophobia and are extremely afraid of spiders. At a summer cam
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Answer:

a) Calculate the probability that at least one of them suffers from arachnophobia.

x = number of students suffering from arachnophobia

= P(x ≥ 1)

= 1 - P(x = 0)

= 1 - [0.05⁰ x (1 - 0.05)¹¹⁻⁰ ]

= 1 - (0.95)¹¹

= 0.4311999 = 0.4312

b) Calculate the probability that exactly 2 of them suffer from arachnophobia? 0.08666

=  P(x = 2)

=  (¹¹₂) x (0.05)² x (0.95)⁹

where ¹¹₂ = 11! / (2!9!) = (11 x 10) / (2 x 1) = 55

= 55 x 0.0025 x 0.630249409 = 0.086659293 = 0.0867

c) Calculate the probability that at most 1 of them suffers from arachnophobia?

P(x ≤ 1)

= P(x = 0) + P(x = 1)    

= [(¹¹₀) x 0.05⁰ x 0.95¹¹] + [(¹¹₁) x 0.05¹ x 0.95¹⁰]

= (1 x 1 x 0.5688) + (11 x 0.05 x 0.598736939) = 0.5688 + 0.3293 = 0.8981

4 0
2 years ago
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